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� {y0} ∪ {x0} ∗ Y ) and S(X ∧ Y )/S({x0} ∧ {y0}) are homeomorphic, and deduce that X ∗ Y ≃ S(X ∧ Y ). 25. homotopy equivalent to Y If X is a CW complex with components Xα, show that the suspension SX is In the case that X is a ﬁnite α SXα for some graph Y. graph, show that SX is homotopy equivalent to a wedge sum of circles and 2 spheres. W 26. Use Corollary 0.20 to show that if (X, A) has the homotopy extension property, then X × I deformation retracts to X × {0} ∪ A× I. Deduce from this that Proposition 0.18 holds more generally for any pair (X1, A) satisfying the homotopy extension property. 27. Given a pair (X, A) and a homotopy equivalence f : A→B, show that the natural map X→B ⊔f X is a homotopy equivalence if (X, A) satisﬁes the homotopy extension [Hint: Consider X ∪ Mf and use the preceding problem.] An interesting property. case is when f is a quotient map, hence the map X→B ⊔f X is the quotient map identifying each set f −1(b) to a point. When B is a point this gives another proof of Proposition 0.17. 28. Show that if (X1, A) satisﬁes the homotopy extension property, then so does every pair (X0 ⊔f X1, X0) obtained by attaching X1 to a space X0 via a map f : A→X0. 29. In case the CW complex X is obtained from a subcomplex A by attaching a single cell en, describe exactly what the extension of a homotopy ft : A→Y to X given by the proof of Proposition 0.16 looks like. That is, for a point x ∈ en, describe the path ft(x) for the extended ft. Algebraic topology can be roughly deﬁned as the study of techniques for forming algebraic images of topological spaces. Most often these algebraic images are groups, but more elaborate structures such as
rings, modules, and algebras also arise. The mechanisms that create these images — the ‘lanterns’ of algebraic topology, one might say — are known formally as functors and have the characteristic feature that they form images not only of spaces but also of maps. Thus, continuous maps between spaces are projected onto homomorphisms between their algebraic images, so topo- logically related spaces have algebraically related images. With suitably constructed lanterns one might hope to be able to form images with enough detail to reconstruct accurately the shapes of all spaces, or at least of large and interesting classes of spaces. This is one of the main goals of algebraic topology, and to a surprising extent this goal is achieved. Of course, the lanterns necessary to do this are somewhat complicated pieces of machinery. But this machinery also has a certain intrinsic beauty. This ﬁrst chapter introduces one of the simplest and most important functors of algebraic topology, the fundamental group, which creates an algebraic image of a space from the loops in the space, the paths in the space starting and ending at the same point. The Idea of the Fundamental Group To get a feeling for what the fundamental group is about, let us look at a few preliminary examples before giving the formal deﬁnitions. 22 Chapter 1 The Fundamental Group Consider two linked circles A and B in R3, as shown in the ﬁgure. Our experience with actual links and chains tells us that since the two circles are linked, it is impossi- ble to separate B from A by any continuous motion of B, such as pushing, pulling, or twisting. We could even take B to be made of rubber or stretchable string and allow completely general continu- ous deformations of B, staying in the complement of A at all times, and it would still be impossible to pull B oﬀ A. At least that is what intuition suggests, and the fundamental group will give a way of making this intuition mathematically rigorous. Instead of having B link with A just once, we could make it link with A two or more times, as in the ﬁgures to the right. As a further variation, by assigning an orientation to B we can speak of B linking A a positive or a negative number of times, say positive when B comes forward through A and negative for the reverse direction. Thus for each nonzero integer n we have an oriented circle Bn
linking A n times, where by ‘circle’ we mean a curve homeomorphic to a circle. To complete the scheme, we could let B0 be a circle not linked to A at all. Now, integers not only measure quantity, but they form a group under addition. Can the group operation be mimicked geometrically with some sort of addition op- eration on the oriented circles B linking A? An oriented circle B can be thought of as a path traversed in time, starting and ending at the same point x0, which we can choose to be any point on the circle. Such a path starting and ending at the same point is called a loop. Two diﬀerent loops B and B′ both starting and ending at the same point x0 can be ‘added’ to form a new loop B + B′ that travels ﬁrst around B, then around B′. For example, if B1 and B′ 1 are loops each linking A once in the positive direction, then their sum B1 + B′ 1 is deformable to B2, linking A twice. Similarly, B1 + B−1 can be deformed to the loop B0, unlinked from A. More generally, we see that Bm + Bn can be deformed to Bm+n for arbitrary integers m and n. Note that in forming sums of loops we produce loops that pass through the base- point more than once. This is one reason why loops are deﬁned merely as continuous The Idea of the Fundamental Group 23 paths, which are allowed to pass through the same point many times. So if one is thinking of a loop as something made of stretchable string, one has to give the string the magical power of being able to pass through itself unharmed. However, we must be sure not to allow our loops to intersect the ﬁxed circle A at any time, otherwise we could always unlink them from A. Next we consider a slightly more complicated sort of linking, involving three cir- cles forming a conﬁguration known as the Borromean rings, shown at the left in the ﬁg- ure below. The interesting feature here is that if any one of the three circles is removed, the other two are not linked. In the same spirit as before, let us regard one of the cir- cles, say C, as a loop in the complement of the other
two, A and B, and we ask whether C can be continuously deformed to unlink it completely from A and B, always staying in the complement of A and B during the deformation. We can redraw the picture by pulling A and B apart, dragging C along, and then we see C winding back and forth between A and B as shown in the second ﬁgure above. In this new position, if we start at the point of C indicated by the dot and proceed in the direction given by the arrow, then we pass in sequence: (1) forward through A, (2) forward through B, (3) backward through A, and (4) backward through B. If we measure the linking of C with A and B by two integers, then the ‘forwards’ and ‘backwards’ cancel and both integers are zero. This reﬂects the fact that C is not linked with A or B individually. To get a more accurate measure of how C links with A and B together, we re- gard the four parts (1)–(4) of C as an ordered sequence. Taking into account the directions in which these segments of C pass through A and B, we may deform C to the sum a + b − a − b of four loops as in the ﬁgure. We write the third and fourth loops as the nega- tives of the ﬁrst two since they can be deformed to the ﬁrst two, but with the opposite orienta- tions, and as we saw in the preceding exam- ple, the sum of two oppositely oriented loops is deformable to a trivial loop, not linked with anything. We would like to view the expression a + b − a − b as lying in a nonabelian group, so that it is not automatically zero. Changing to the more usual multiplicative notation for nonabelian groups, it would be written aba−1b−1, the commutator of a and b. 24 Chapter 1 The Fundamental Group To shed further light on this example, suppose we modify it slightly so that the cir- cles A and B are now linked, as in the next ﬁgure. The circle C can then be deformed into the position shown at the right, where it again rep- resents the composite loop aba−1b−1, where a and b are loops linking A
and B. But from the picture on the left it is apparent that C can actually be unlinked completely from A and B. So in this case the product aba−1b−1 should be trivial. The fundamental group of a space X will be deﬁned so that its elements are loops in X starting and ending at a ﬁxed basepoint x0 ∈ X, but two such loops are regarded as determining the same element of the fundamental group if one loop can be continuously deformed to the other within the space X. (All loops that occur during deformations must also start and end at x0.) In the ﬁrst example above, X is the complement of the circle A, while in the other two examples X is the complement of the two circles A and B. In the second section in this chapter we will show: The fundamental group of the complement of the circle A in the ﬁrst example is inﬁnite cyclic with the loop B as a generator. This amounts to saying that every loop in the complement of A can be deformed to one of the loops Bn, and that Bn cannot be deformed to Bm if n ≠ m. The fundamental group of the complement of the two unlinked circles A and B in the second example is the nonabelian free group on two generators, represented by the loops a and b linking A and B. In particular, the commutator aba−1b−1 is a nontrivial element of this group. The fundamental group of the complement of the two linked circles A and B in the third example is the free abelian group on two generators, represented by the loops a and b linking A and B. As a result of these calculations, we have two ways to tell when a pair of circles A and B is linked. The direct approach is given by the ﬁrst example, where one circle is regarded as an element of the fundamental group of the complement of the other circle. An alternative and somewhat more subtle method is given by the second and third examples, where one distinguishes a pair of linked circles from a pair of unlinked circles by the fundamental group of their complement, which is abelian in one case and nonabelian in the other. This method is much more general: One can often show that two spaces are not homeomorphic by showing that their fundamental groups are not isomorphic, since it will be an easy
consequence of the deﬁnition of the fundamental group that homeomorphic spaces have isomorphic fundamental groups. Basic Constructions Section 1.1 25 This ﬁrst section begins with the basic deﬁnitions and constructions, and then proceeds quickly to an important calculation, the fundamental group of the circle, using notions developed more fully in §1.3. More systematic methods of calculation are given in §1.2. These are suﬃcient to show for example that every group is realized as the fundamental group of some space. This idea is exploited in the Additional Topics at the end of the chapter, which give some illustrations of how algebraic facts about groups can be derived topologically, such as the fact that every subgroup of a free group is free. Paths and Homotopy The fundamental group will be deﬁned in terms of loops and deformations of loops. Sometimes it will be useful to consider more generally paths and their defor- mations, so we begin with this slight extra generality. By a path in a space X we mean a continuous map f : I→X where I is the unit interval [0, 1]. The idea of continuously deforming a path, keeping its endpoints ﬁxed, is made precise by the following deﬁnition. A homotopy of paths in X is a family ft : I→X, 0 ≤ t ≤ 1, such that (1) The endpoints ft(0) = x0 and ft(1) = x1 are independent of t. (2) The associated map F : I × I→X deﬁned by F (s, t) = ft(s) is continuous. When two paths f0 and f1 are connected in this way by a homotopy ft, they are said to be homotopic. The notation for this is f0 ≃ f1. Example 1.1: Linear Homotopies. Any two paths f0 and f1 in Rn having the same endpoints x0 and x1 are homotopic via the homotopy ft(s) = (1 − t)f0(s) + tf1(s). During this homotopy each point f0(s) travels along the line segment to f1(s) at constant speed. This is because the line through f0(s) and f1(s) is
linearly parametrized as f0(s) + t[f1(s) − f0(s)] = (1 − t)f0(s) + tf1(s), with the segment from f0(s) to f1(s) covered by t values in the interval from 0 to 1. If f1(s) happens to equal f0(s) then this segment degenerates to a point and ft(s) = f0(s) for all t. This occurs in particular for s = 0 and s = 1, so each ft is a path from x0 to x1. Continuity of the homotopy ft as a map I × I→Rn follows from continuity of f0 and f1 since the algebraic operations of vector addition and scalar multiplication in the formula for ft are continuous. This construction shows more generally that for a convex subspace X ⊂ Rn, all paths in X with given endpoints x0 and x1 are homotopic, since if f0 and f1 lie in X then so does the homotopy ft. 26 Chapter 1 The Fundamental Group Before proceeding further we need to verify a technical property: Proposition 1.2. The relation of homotopy on paths with ﬁxed endpoints in any space is an equivalence relation. The equivalence class of a path f under the equivalence relation of homotopy will be denoted [f ] and called the homotopy class of f. Proof: Reﬂexivity is evident since f ≃ f by the constant homotopy ft = f. Symmetry is also easy since if f0 ≃ f1 via ft, then f1 ≃ f0 via the inverse homotopy f1−t. For transitivity, if f0 ≃ f1 via ft and if f1 = g0 with g0 ≃ g1 via gt, then f0 ≃ g1 via the homotopy ht that equals f2t for 0 ≤ t ≤ 1/2 and g2t−1 for 1/2 ≤ t ≤ 1. These two deﬁnitions agree for t = 1/2 since we assume f1 = g0. Continuity of the associated map H(s, t) = ht(s) comes from the elementary fact, which will be used frequently without explicit mention, that a function de
ﬁned on the union of two closed sets is continuous if it is continuous when restricted to each of the closed sets separately. In the case at hand we have H(s, t) = F (s, 2t) for 0 ≤ t ≤ 1/2 and H(s, t) = G(s, 2t − 1) for 1/2 ≤ t ≤ 1 where F and G are the maps I × I→X associated to the homotopies ft and gt. Since H is continuous on I × [0, 1/2] and on I × [1/2, 1], it is continuous on I × I. ⊔⊓ Given two paths f, g : I→X such that f (1) = g(0), there is a composition or product path f g that traverses ﬁrst f and then g, deﬁned by the formula f g(s) = f (2s), ( g(2s − 1), 0 ≤ s ≤ 1/2 1/2 ≤ s ≤ 1 Thus f and g are traversed twice as fast in order for f g to be traversed in unit time. This product operation respects homotopy classes since if f0 ≃ f1 and g0 ≃ g1 via homotopies ft and gt, and if f0(1) = g0(0) so that f0 g0 is deﬁned, then ft gt is deﬁned and provides a homotopy f0 g0 ≃ f1 g1. In particular, suppose we restrict attention to paths f : I→X with the same starting and ending point f (0) = f (1) = x0 ∈ X. Such paths are called loops, and the common starting and ending point x0 is referred to as the basepoint. The set of all homotopy classes [f ] of loops f : I→X at the basepoint x0 is denoted π1(X, x0). Proposition 1.3. π1(X, x0) is a group with respect to the product [f ][g] = [f g]. This group is called the fundamental group of X at the basepoint x0. We will see in Chapter 4 that π1(X, x0) is the ﬁrst
in a sequence of groups πn(X, x0), called homotopy groups, which are deﬁned in an entirely analogous fashion using the n dimensional cube In in place of I. Basic Constructions Section 1.1 27 Proof: By restricting attention to loops with a ﬁxed basepoint x0 ∈ X we guarantee that the product f g of any two such loops is deﬁned. We have already observed that the homotopy class of f g depends only on the homotopy classes of f and g, so the product [f ][g] = [f g] is well-deﬁned. It remains to verify the three axioms for a group. As a preliminary step, deﬁne a reparametrization of a path f to be a composition f ϕ where ϕ : I→I is any continuous map such that ϕ(0) = 0 and ϕ(1) = 1. Reparametrizing a path preserves its homotopy class since f ϕ ≃ f via the homotopy f ϕt where ϕt(s) = (1 − t)ϕ(s) + ts so that ϕ0 = ϕ and ϕ1(s) = s. Note that (1 − t)ϕ(s) + ts lies between ϕ(s) and s, hence is in I, so the composition f ϕt is deﬁned. If we are given paths f, g, h with f (1) = g(0) and g(1) = h(0), then both prod- ucts (f g) h and f (g h) are deﬁned, and f (g h) is a reparametrization of (f g) h by the piecewise linear function ϕ whose graph is shown in the ﬁgure at the right. Hence (f g) h ≃ f (g h). Restricting attention to loops at the basepoint x0, this says the product in π1(X, x0) is associative. Given a path f : I→X, let c be the constant path at f (1), deﬁned by c(s) = f (1) for all s ∈ I. Then f c is a
reparametrization of f via the function ϕ whose graph is shown in the ﬁrst ﬁgure at the right, so f c ≃ f. Similarly, c f ≃ f where c is now the constant path at f (0), using the reparametrization function in the second ﬁgure. Taking f to be a loop, we deduce that the homotopy class of the constant path at x0 is a two-sided identity in π1(X, x0). For a path f from x0 to x1, the inverse path f from x1 back to x0 is deﬁned by f (s) = f (1 − s). To see that f f is homotopic to a constant path we use the homotopy ht = ft gt where ft is the path that equals f on the interval [0, 1 − t] and that is stationary at f (1 − t) on the interval [1 − t, 1], and gt is the inverse path of ft. We could also describe ht in terms of the associated function H : I × I→X using the decomposition of I × I shown in the ﬁgure. On the bottom edge of the square H is given by f f and below the ‘V’ we let H(s, t) be independent of t, while above the ‘V’ we let H(s, t) be independent of s. Going back to the ﬁrst description of ht, we see that since f0 = f and f1 is the constant path c at x0, ht is a homotopy from f f to c c = c. Replacing f by f gives f f ≃ c for c the constant path at x1. Taking f to be a loop at the ⊔⊓ basepoint x0, we deduce that [ f ] is a two-sided inverse for [f ] in π1(X, x0). Example 1.4. For a convex set X in Rn with basepoint x0 ∈ X we have π1(X, x0) = 0, the trivial group, since any two loops f0 and f1 based at x0 are homotopic via the linear homotopy ft(s) = (
1 − t)f0(s) + tf1(s), as described in Example 1.1. 28 Chapter 1 The Fundamental Group It is not so easy to show that a space has a nontrivial fundamental group since one must somehow demonstrate the nonexistence of homotopies between certain loops. We will tackle the simplest example shortly, computing the fundamental group of the circle. It is natural to ask about the dependence of π1(X, x0) on the choice of the basepoint x0. Since π1(X, x0) involves only the path-component of X containing x0, it is clear that we can hope to ﬁnd a relation between π1(X, x0) and π1(X, x1) for two basepoints x0 and x1 only if x0 and x1 lie in the same path-component of X. So let h : I→X be a path from x0 to x1, with the inverse path h(s) = h(1 − s) from x1 back to x0. We can then associate to each loop f based at x1 the loop h f h based at x0. Strictly speaking, we should choose an order of forming the product h f h, either (h f ) h or h (f h), but the two choices are homotopic and we are only interested in homotopy classes here. Alternatively, to avoid any ambiguity we could deﬁne a general n fold product f1 ··· fn in which the path fi is traversed in the time interval. Either way, we deﬁne a change-of-basepoint map βh : π1(X, x1)→π1(X, x0) i−1 by βh[f ] = [h f h]. This is well-deﬁned since if ft is a homotopy of loops based at x1 then h ft h is a homotopy of loops based at x0. n, i n Proposition 1.5. The map βh : π1(X, x1)→π1(X, x0) is an isomorphism. Proof: We see ﬁrst that βh is a homomorphism since βh[f g] = [h f g h] = [h f h h
g h] = βh[f ]βh[g]. Further, βh is an isomorphism with inverse βh since βhβh[f ] = βh[h f h] = [h h f h h] = [f ], and similarly βhβh[f ] = [f ]. ⊔⊓ Thus if X is path-connected, the group π1(X, x0) is, up to isomorphism, independent of the choice of basepoint x0. In this case the notation π1(X, x0) is often abbreviated to π1(X), or one could go further and write just π1X. In general, a space is called simply-connected if it is path-connected and has trivial fundamental group. The following result explains the name. Proposition 1.6. A space X is simply-connected iﬀ there is a unique homotopy class of paths connecting any two points in X. Proof: Path-connectedness is the existence of paths connecting every pair of points, so we need be concerned only with the uniqueness of connecting paths. Suppose π1(X) = 0. If f and g are two paths from x0 to x1, then f ≃ f g g ≃ g since the loops g g and f g are each homotopic to constant loops, using the assumption π1(X, x0) = 0 in the latter case. Conversely, if there is only one homotopy class of paths connecting a basepoint x0 to itself, then all loops at x0 are homotopic to the ⊔⊓ constant loop and π1(X, x0) = 0. Basic Constructions Section 1.1 29 The Fundamental Group of the Circle Our ﬁrst real theorem will be the calculation π1(S 1) ≈ Z. Besides its intrinsic interest, this basic result will have several immediate applications of some substance, and it will be the starting point for many more calculations in the next section. It should be no surprise then that the proof will involve some genuine work. Theorem 1.7. π1(S 1) is an inﬁnite cyclic group generated by the homotopy class of the loop ω(s) = (cos 2π s, sin 2π s) based at (
1, 0). Note that [ω]n = [ωn] where ωn(s) = (cos 2π ns, sin 2π ns) for n ∈ Z. The theorem is therefore equivalent to the statement that every loop in S 1 based at (1, 0) is homotopic to ωn for a unique n ∈ Z. To prove this the idea will be to compare paths in S 1 with paths in R via the map p : R→S 1 given by p(s) = (cos 2π s, sin 2π s). This map can be visualized geometrically by embedding R in R3 as the helix parametrized by s ֏ (cos 2π s, sin 2π s, s), and then p is the restriction to the helix of the projection of R3 onto R2, (x, y, z) ֏ (x, y). Observe that ωn : I→R is the path the loop ωn is the composition p ωn(s) = ns, starting at 0 and ending at n, winding around the helix \|n\| times, upward if n > 0 and downward if n < 0. The relation f ωn = p ωn is expressed by saying that ωn is a lift of ωn. ωn where f f We will prove the theorem by studying how paths in S 1 lift to paths in R. Most of the arguments will apply in much greater generality, and it is both more eﬃcient f f and more enlightening to give them in the general context. The ﬁrst step will be to deﬁne this context. Given a space X, a covering space of X consists of a space X and a map p : satisfying the following condition: e X→X e (∗) For each point x ∈ X there is an open neighborhood U of x in X such that p−1(U) is a union of disjoint open sets each of which is mapped homeomorphically onto U by p. Such a U will be called evenly covered. For example, for the previously deﬁned map p : R→S 1 any open arc in S 1 is evenly covered. To prove the theorem we will need just the following two facts about covering X→X. e is a unique lift spaces
p : (a) For each path f : I→X starting at a point x0 ∈ X and each X starting at (b) For each homotopy ft : I→X of paths starting at x0 and each x0. e Before proving these facts, let us see how they imply the theorem. e e X of paths starting at is a unique lifted homotopy ft : I→ f : I→ x0. e e e e x0 ∈ p−1(x0) there e x0 ∈ p−1(x0) there 30 Chapter 1 The Fundamental Group Proof of Theorem 1.7: Let f : I→S 1 be a loop at the basepoint x0 = (1, 0), representing a given element of π1(S 1, x0). By (a) there is a lift f starting at 0. This path f (1) = f (1) = x0 and p−1(x0) = Z ⊂ R. Another f ends at some integer n since p e path in R from 0 to n is f ≃ ωn. ωn, and e e Composing this homotopy with p gives a homotopy f ≃ ωn so [f ] = [ωn]. e ωn via the linear homotopy (1 − t) f + t To show that n is uniquely determined by [f ], suppose that f ≃ ωn and f ≃ ωm, so ωm ≃ ωn. Let ft be a homotopy from ωm = f0 to ωn = f1. By (b) this ft of paths starting at 0. The uniqueness part of (a) homotopy lifts to a homotopy f0 = implies that ft is a homotopy of paths, the endpoint ft(1) is independent of t. For t = 0 this endpoint is m and for t = 1 it is n, so m = n. e f1 = ωn. Since e ωm and f f f f f e e e e It remains to prove (a) and (b). Both statements can be deduced from a more general assertion about covering spaces p : X→X : (c) Given a map F : Y × I→X and a map F : Y × {0}→ e X lifting
F \|Y × {0}, then there is a unique map F : Y × I→ X lifting F and restricting to the given F on Y × {0}. e e e Statement (a) is the special case that Y is a point, and (b) is obtained by applying (c) with Y = I in the following way. The homotopy ft in (b) gives a map F : I × I→X by setting F (s, t) = ft(s) as usual. A unique lift X is obtained by an F \|{0}× I application of (a). Then (c) gives a unique lift F : I × {0}→ F : I × I→ e e X. The restrictions e e and F \|{1}× I are paths lifting constant paths, hence they must also be constant by ft lifts ft e F (s, t) is a homotopy of paths, and the uniqueness part of (a). So ft(s) = e e e since → e e To prove (c) we will ﬁrst construct a lift e Nt × (at, bt) X for N some neighborhood in Y of a given point y0 ∈ Y. Since F is continuous, every point (y0, t) ∈ Y × I has a product neighborhood Nt × (at, bt) such that F is contained in an evenly covered neighborhood of F (y0, t). By compactness of {y0}× I, ﬁnitely many such products Nt × (at, bt) cover {y0}× I. This implies that we can choose a single neighborhood N of y0 and a partition 0 = t0 < t1 < ··· < tm = 1 of I so that for each i, F (N × [ti, ti+1]) is contained in an evenly covered neighborhood Ui. F has been constructed on N × [0, ti], starting with the given Assume inductively that F on N × {0}. We have F (N × [ti, ti+1]) ⊂ Ui, so since Ui is evenly covered there is X projecting homeomorphically onto Ui by p and containing the an open set e F (y0, ti). After replacing N by a smaller neighborhood of y0 we may assume point e Ui, namely, replace
N × {ti} by its intersection with F (N × {ti}) is contained in F on N × [ti, ti+1] to be the composition of F Ui. After a ﬁnite number of steps we eventually that e F \|\| N × {ti})−1( ( with the homeomorphism p−1 : Ui→ e get a lift Ui). Now we can deﬁne Ui ⊂ F : N × I→ X for some neighborhood N of y0. e Next we show the uniqueness part of (c) in the special case that Y is a point. In this are two lifts of F : I→X case we can omit Y from the notation. So suppose F and Basic Constructions Section 1.1 31 F is e ⊔⊓ e e e F F (0) = ′ (0). As before, choose a partition 0 = t0 < t1 < ··· < tm = 1 of such that I so that for each i, F ([ti, ti+1]) is contained in some evenly covered neighborhood on [0, ti]. Since [ti, ti+1] is connected, so is Ui. Assume inductively that F ([ti, ti+1]), which must therefore lie in a single one of the disjoint open sets Ui ′ projecting homeomorphically to Ui as in (∗). By the same token, ([ti, ti+1]) lies F e e ′ (ti) = F (ti). in a single F ([ti, ti+1]) since F e F = on [ti, ti+1], and Ui, in fact in the same one that contains Ui and p ′, it follows that ′ ′ Because p is injective on e e e e the induction step is ﬁnished. e The last step in the proof of (c) is to observe that since the F ’s constructed above on sets of the form N × I are unique when restricted to each segment {y}× I, they e e e e must agree whenever two such sets N × I overlap. So we obtain a well-deﬁned lift on all of Y × I. This F is continuous since it is continuous on each N × I. And F e unique since it is unique on each segment {y}×
I. e Now we turn to some applications of the calculation of π1(S 1), beginning with a proof of the Fundamental Theorem of Algebra. Theorem 1.8. Every nonconstant polynomial with coeﬃcients in C has a root in C. Proof: We may assume the polynomial is of the form p(z) = zn + a1zn−1 + ··· + an. If p(z) has no roots in C, then for each real number r ≥ 0 the formula fr (s) = p(r e2π is)/p(r ) \|p(r e2π is )/p(r )\| deﬁnes a loop in the unit circle S 1 ⊂ C based at 1. As r varies, fr is a homotopy of loops based at 1. Since f0 is the trivial loop, we deduce that the class [fr ] ∈ π1(S 1) is zero for all r. Now ﬁx a large value of r, bigger than \|a1\| + ··· + \|an\| and bigger than 1. Then for \|z\| = r we have \|zn\| > (\|a1\| + ··· + \|an\|)\|zn−1\| > \|a1zn−1\| + ··· + \|an\| ≥ \|a1zn−1 + ··· + an\| From the inequality \|zn\| > \|a1zn−1 + ··· + an\| it follows that the polynomial pt(z) = zn +t(a1zn−1 +··· +an) has no roots on the circle \|z\| = r when 0 ≤ t ≤ 1. Replacing p by pt in the formula for fr above and letting t go from 1 to 0, we obtain a homotopy from the loop fr to the loop ωn(s) = e2π ins. By Theorem 1.7, ωn represents n times a generator of the inﬁnite cyclic group π1(S 1). Since we have shown that [ωn] = [fr ] = 0, we conclude that n = 0. Thus the only polynomials without roots in C are constants. ⊔⊓ Our next application is the Brouwer ﬁxed point theorem in dimension 2. Theorem
1.9. Every continuous map h : D2→D2 has a ﬁxed point, that is, a point x ∈ D2 with h(x) = x. Here we are using the standard notation Dn for the closed unit disk in Rn, all vectors x of length \|x\| ≤ 1. Thus the boundary of Dn is the unit sphere S n−1. 32 Chapter 1 The Fundamental Group Proof: Suppose on the contrary that h(x) ≠ x for all x ∈ D2. Then we can deﬁne a map r : D2→S 1 by letting r (x) be the point of S 1 where the ray in R2 starting at h(x) and passing through x leaves D2. Continuity of r is clear since small perturbations of x produce small perturbations of h(x), hence also small perturbations of the ray through these two points. The crucial property of r, besides continuity, is that r (x) = x if x ∈ S 1. Thus r is a retraction of D2 onto S 1. We will show that no such retraction can exist. Let f0 be any loop in S 1. In D2 there is a homotopy of f0 to a constant loop, for example the linear homotopy ft(s) = (1 − t)f0(s) + tx0 where x0 is the basepoint of f0. Since the retraction r is the identity on S 1, the composition r ft is then a homotopy in S 1 from r f0 = f0 to the constant loop at x0. But this contradicts the fact that π1(S 1) is nonzero. ⊔⊓ This theorem was ﬁrst proved by Brouwer around 1910, quite early in the history of topology. Brouwer in fact proved the corresponding result for Dn, and we shall obtain this generalization in Corollary 2.15 using homology groups in place of π1. One could also use the higher homotopy group πn. Brouwer’s original proof used neither homology nor homotopy groups, which had not been invented at the time. Instead it used the notion of degree for maps S n→S n, which we shall deﬁne in §2.2 using homology but
which Brouwer deﬁned directly in more geometric terms. These proofs are all arguments by contradiction, and so they show just the exis- tence of ﬁxed points without giving any clue as to how to ﬁnd one in explicit cases. Our proof of the Fundamental Theorem of Algebra was similar in this regard. There exist other proofs of the Brouwer ﬁxed point theorem that are somewhat more con- structive, for example the elegant and quite elementary proof by Sperner in 1928, which is explained very nicely in [Aigner-Ziegler 1999]. The techniques used to calculate π1(S 1) can be applied to prove the Borsuk–Ulam theorem in dimension two: Theorem 1.10. For every continuous map f : S 2→R2 there exists a pair of antipodal points x and −x in S 2 with f (x) = f (−x). It may be that there is only one such pair of antipodal points x, −x, for example if f is simply orthogonal projection of the standard sphere S 2 ⊂ R3 onto a plane. The Borsuk–Ulam theorem holds more generally for maps S n→Rn, as we will show in Corollary 2B.7. The proof for n = 1 is easy since the diﬀerence f (x) − f (−x) changes sign as x goes halfway around the circle, hence this diﬀerence must be zero for some x. For n ≥ 2 the theorem is certainly less obvious. Is it apparent, for example, that at every instant there must be a pair of antipodal points on the surface of the earth having the same temperature and the same barometric pressure? Basic Constructions Section 1.1 33 The theorem says in particular that there is no one-to-one continuous map from S 2 to R2, so S 2 is not homeomorphic to a subspace of R2, an intuitively obvious fact that is not easy to prove directly. f (x) − f (−x) Proof: If the conclusion is false for f : S 2→R2, we can deﬁne a map g : S 2→S 1 by g(x) = /\|f (x) − f (−x)\|. Deﬁne a loop η circling the equator of S 2
⊂ R3 by η(s) = (cos 2π s, sin 2π s, 0), and let h : I→S 1 be the composed loop gη. Since g(−x) = −g(x), we have the relation h(s + 1/2) = −h(s) for all s in the interval [0, 1/2]. As we showed in the calculation of π1(S 1), the loop h can be lifted to a path h : I→R. The equation h(s + 1/2) = −h(s) implies that h(s) + q/2 for some odd integer q that might conceivably depend on s ∈ [0, 1/2]. But in fact q is e e e h(s)+ q/2 for q we see that independent of s since by solving the equation q depends continuously on s ∈ [0, 1/2], so q must be a constant since it is constrained h(0) + q. This means to integer values. In particular, we have that h represents q times a generator of π1(S 1). Since q is odd, we conclude that h e is not nullhomotopic. But h was the composition gη : I→S 2→S 1, and η is obviously nullhomotopic in S 2, so gη is nullhomotopic in S 1 by composing a nullhomotopy of ⊔⊓ η with g. Thus we have arrived at a contradiction. h(1/2) + q/2 = h(s + 1/2) = h(s + 1/2) = e h(1) = e e e Corollary 1.11. Whenever S 2 is expressed as the union of three closed sets A1, A2, and A3, then at least one of these sets must contain a pair of antipodal points {x, −x}. Proof: Let di : S 2→R measure distance to Ai, that is, di(x) = infy∈Ai is a continuous function, so we may apply the Borsuk–Ulam theorem to the map S 2→R2, x ֏, obtaining a pair of antipodal points x and −x with d1(x) = d1(−x) and d
2(x) = d2(−x). If either of these two distances is zero, then x and −x both lie in the same set A1 or A2 since these are closed sets. On the other hand, if the distances from x and −x to A1 and A2 are both strictly positive, then ⊔⊓ x and −x lie in neither A1 nor A2 so they must lie in A3. d1(x), d2(x) \|x − y\|. This To see that the number ‘three’ in this result is best possible, consider a sphere inscribed in a tetrahedron. Projecting the four faces of the tetrahedron radially onto the sphere, we obtain a cover of S 2 by four closed sets, none of which contains a pair of antipodal points. Assuming the higher-dimensional version of the Borsuk–Ulam theorem, the same arguments show that S n cannot be covered by n + 1 closed sets without antipodal pairs of points, though it can be covered by n+2 such sets, as the higher-dimensional analog of a tetrahedron shows. Even the case n = 1 is somewhat interesting: If the circle is covered by two closed sets, one of them must contain a pair of antipodal points. This is of course false for nonclosed sets since the circle is the union of two disjoint half-open semicircles. 34 Chapter 1 The Fundamental Group The relation between the fundamental group of a product space and the funda- mental groups of its factors is as simple as one could wish: Proposition 1.12. π1(X × Y ) is isomorphic to π1(X)× π1(Y ) if X and Y are pathconnected. Proof: A basic property of the product topology is that a map f : Z→X × Y is continuous iﬀ the maps g : Z→X and h : Z→Y deﬁned by f (z) = (g(z), h(z)) are both continuous. Hence a loop f in X × Y based at (x0, y0) is equivalent to a pair of loops g in X and h in Y based at x0 and y0 respectively. Similarly, a homotopy ft of a loop in X × Y is equivalent to a pair of homotopies gt and h
t of the corresponding loops ≈ π1(X, x0)× π1(Y, y0), in X and Y. Thus we obtain a bijection π1 [f ] ֏ ([g], [h]). This is obviously a group homomorphism, and hence an isomor⊔⊓ phism. X × Y, (x0, y0) Example 1.13: The Torus. By the proposition we have an isomorphism π1(S 1 × S 1) ≈ Z× Z. Under this isomorphism a pair (p, q) ∈ Z× Z corresponds to a loop that winds p times around one S 1 factor of the torus and q times around the other S 1 factor, for example the loop ωpq(s) = (ωp(s), ωq(s)). Interestingly, this loop can be knotted, as the ﬁgure shows for the case p = 3, q = 2. The knots that arise in this fashion, the so-called torus knots, are studied in Example 1.24. More generally, the n dimensional torus, which is the product of n circles, has fundamental group isomorphic to the product of n copies of Z. This follows by induction on n. Induced Homomorphisms Suppose ϕ : X→Y is a map taking the basepoint x0 ∈ X to the basepoint y0 ∈ Y. For brevity we write ϕ : (X, x0)→(Y, y0) in this situation. Then ϕ induces a homomorphism ϕ∗ : π1(X, x0)→π1(Y, y0), deﬁned by composing loops f : I→X based at x0 with ϕ, that is, ϕ∗[f ] = [ϕf ]. This induced map ϕ∗ is well-deﬁned since a homotopy ft of loops based at x0 yields a composed homotopy ϕft of loops based at y0, so ϕ∗[f0] = [ϕf0] = [ϕf1] = ϕ∗[f1]. Furthermore, ϕ∗ is a homomorphism since ϕ(f g) = (ϕ
f ) (ϕg), both functions having the value ϕf (2s) for 0 ≤ s ≤ 1/2 and the value ϕg(2s − 1) for 1/2 ≤ s ≤ 1. Two basic properties of induced homomorphisms are: ψ-----→ (Y, y0) (ϕψ)∗ = ϕ∗ψ∗ for a composition (X, x0) 11∗ = 11, which is a concise way of saying that the identity map 11 : X→X induces the identity map 11 : π1(X, x0)→π1(X, x0). ϕ-----→ (Z, z0). The ﬁrst of these follows from the fact that composition of maps is associative, so (ϕψ)f = ϕ(ψf ), and the second is obvious. These two properties of induced homo- morphisms are what makes the fundamental group a functor. The formal deﬁnition Basic Constructions Section 1.1 35 of a functor requires the introduction of certain other preliminary concepts, however, so we postpone this until it is needed in §2.3. As an application we can deduce easily that if ϕ is a homeomorphism with inverse ψ then ϕ∗ is an isomorphism with inverse ψ∗ since ϕ∗ψ∗ = (ϕψ)∗ = 11∗ = 11 and similarly ψ∗ϕ∗ = 11. We will use this fact in the following calculation of the fundamental groups of higher-dimensional spheres: Proposition 1.14. π1(S n) = 0 if n ≥ 2. The main step in the proof will be a general fact that will also play a key role in the next section: Lemma 1.15. If a space X is the union of a collection of path-connected open sets Aα each containing the basepoint x0 ∈ X and if each intersection Aα ∩ Aβ is pathconnected, then every loop in X at x0 is homotopic to a product of loops each of which is contained in a single Aα. Proof: Given a loop f : I→X at the basepoint x0, we claim there is a partition 0 = s0 < s1 < ··· < sm = 1
of I such that each subinterval [si−1, si] is mapped by f to a single Aα. Namely, since f is continuous, each s ∈ I has an open neighborhood Vs in I mapped by f to some Aα. We may in fact take Vs to be an interval whose closure is mapped to a single Aα. Compactness of I implies that a ﬁnite number of these intervals cover I. The endpoints of this ﬁnite set of intervals then deﬁne the desired partition of I. Denote the Aα containing f ([si−1, si]) by Ai, and let fi be the path obtained by restricting f to [si−1, si]. Then f is the composition f1 ··· fm with fi a path in Ai. Since we assume Ai ∩ Ai+1 is path-connected, we may choose a path gi in Ai ∩ Ai+1 from x0 to the point f (si) ∈ Ai ∩ Ai+1. Consider the loop (f1 g1) (g1 f2 g2) (g2 f3 g3) ··· (gm−1 fm) which is homotopic to f. This loop is a composition of loops each lying in a single Ai, the loops indicated by the parentheses. ⊔⊓ Proof of Proposition 1.14: We can express S n as the union of two open sets A1 and A2 each homeomorphic to Rn such that A1 ∩ A2 is homeomorphic to S n−1 × R, for example by taking A1 and A2 to be the complements of two antipodal points in S n. Choose a basepoint x0 in A1 ∩ A2. If n ≥ 2 then A1 ∩ A2 is path-connected. The lemma then applies to say that every loop in S n based at x0 is homotopic to a product of loops in A1 or A2. Both π1(A1) and π1(A2) are zero since A1 and A2 are homeomorphic to Rn. Hence every loop in S n is nullhomotopic. ⊔⊓ 36 Chapter 1 The Fundamental Group Corollary 1.16. R2 is not homeomorphic to Rn for n ≠ 2. Proof: Suppose f : R2→Rn
is a homeomorphism. The case n = 1 is easily disposed of since R2 − {0} is path-connected but the homeomorphic space Rn − {f (0)} is not path-connected when n = 1. When n > 2 we cannot distinguish R2 − {0} from Rn − {f (0)} by the number of path-components, but we can distinguish them by their fundamental groups. Namely, for a point x in Rn, the complement Rn − {x} is homeomorphic to S n−1 × R, so Proposition 1.12 implies that π1(Rn − {x}) is isomorphic to π1(S n−1)× π1(R) ≈ π1(S n−1). Hence π1(Rn − {x}) is Z for n = 2 and ⊔⊓ trivial for n > 2, using Proposition 1.14 in the latter case. The more general statement that Rm is not homeomorphic to Rn if m ≠ n can be proved in the same way using either the higher homotopy groups or homology In fact, nonempty open sets in Rm and Rn can be homeomorphic only if groups. m = n, as we will show in Theorem 2.26 using homology. Induced homomorphisms allow relations between spaces to be transformed into relations between their fundamental groups. Here is an illustration of this principle: Proposition 1.17. If a space X retracts onto a subspace A, then the homomorphism i∗ : π1(A, x0)→π1(X, x0) induced by the inclusion i : A ֓ X is injective. If A is a deformation retract of X, then i∗ is an isomorphism. Proof: If r : X→A is a retraction, then r i = 11, hence r∗i∗ = 11, which implies that i∗ is injective. If rt : X→X is a deformation retraction of X onto A, so r0 = 11, rt\|A = 11, and r1(X) ⊂ A, then for any loop f : I→X based at x0 ∈ A the composition rtf gives ⊔⊓ a homotopy of f to a loop in A,
so i∗ is also surjective. This gives another way of seeing that S 1 is not a retract of D2, a fact we showed earlier in the proof of the Brouwer ﬁxed point theorem, since the inclusion-induced map π1(S 1)→π1(D2) is a homomorphism Z→0 that cannot be injective. The exact group-theoretic analog of a retraction is a homomorphism ρ of a group G onto a subgroup H such that ρ restricts to the identity on H. In the notation above, if we identify π1(A) with its image under i∗, then r∗ is such a homomorphism from π1(X) onto the subgroup π1(A). The existence of a retracting homomorphism ρ : G→H is quite a strong condition on H. If H is a normal subgroup, it implies that G is the direct product of H and the kernel of ρ. If H is not normal, then G is what is called in group theory the semi-direct product of H and the kernel of ρ. t ∈ I, such that the associated map Recall from Chapter 0 the general deﬁnition of a homotopy as a family ϕt : X→Y, : X × I→Y, (x, t) = ϕt(x), is continuous. If ϕt takes a subspace A ⊂ X to a subspace B ⊂ Y for all t, then we speak of a homotopy of maps of pairs, ϕt : (X, A)→(Y, B). In particular, a basepoint-preserving homotopy Φ Φ Basic Constructions Section 1.1 37 ϕt : (X, x0)→(Y, y0) is the case that ϕt(x0) = y0 for all t. Another basic property of induced homomorphisms is their invariance under such homotopies: If ϕt : (X, x0)→(Y, y0) is a basepoint-preserving homotopy, then ϕ0∗ = ϕ1∗. This holds since ϕ0∗[f ] = [ϕ0f ] = [ϕ1f ]
= ϕ1∗[f ], the middle equality coming from the homotopy ϕtf. There is a notion of homotopy equivalence for spaces with basepoints. One says (X, x0) ≃ (Y, y0) if there are maps ϕ : (X, x0)→(Y, y0) and ψ : (Y, y0)→(X, x0) with homotopies ϕψ ≃ 11 and ψϕ ≃ 11 through maps ﬁxing the basepoints. In this case the induced maps on π1 satisfy ϕ∗ψ∗ = (ϕψ)∗ = 11∗ = 11 and likewise ψ∗ϕ∗ = 11, so ϕ∗ and ψ∗ are inverse isomorphisms π1(X, x0) ≈ π1(Y, y0). This somewhat formal argument gives another proof that a deformation retraction induces an isomorphism on fundamental groups, since if X deformation retracts onto A then (X, x0) ≃ (A, x0) for any choice of basepoint x0 ∈ A. Having to pay so much attention to basepoints when dealing with the fundamental group is something of a nuisance. For homotopy equivalences one does not have to be quite so careful, as the conditions on basepoints can actually be dropped: Proposition 1.18. If ϕ : X→Y is a homotopy equivalence, then the induced homomorphism ϕ∗ : π1(X, x0)→π1 is an isomorphism for all x0 ∈ X. The proof will use a simple fact about homotopies that do not ﬁx the basepoint: Y, ϕ(x0) Lemma 1.19. If ϕt : X→Y is a homotopy and h is the path ϕt(x0) formed by the images of a basepoint x0 ∈ X, then the three maps in the diagram at the right satisfy ϕ0∗ = βhϕ1∗. Proof: Let ht be the restriction of h to the interval [0, t], with a reparametrization so that the domain of
ht is still [0, 1]. Explicitly, we can take ht(s) = h(ts). Then if f is a loop in X at the basepoint x0, the product ht (ϕtf ) ht gives a homotopy of loops at ϕ0(x0). Restricting this homotopy to t = 0 and t = 1, we see that ϕ0∗([f ]) = ⊔⊓ βh. ϕ1∗([f ]) Proof of 1.18: Let ψ : Y →X be a homotopy-inverse for ϕ, so that ϕψ ≃ 11 and ψϕ ≃ 11. Consider the maps ϕ∗------------→ π1 ψ∗------------→ π1 π1(X, x0) X, ψϕ(x0) The composition of the ﬁrst two maps is an isomorphism since ψϕ ≃ 11 implies that ψ∗ϕ∗ = βh for some h, by the lemma. In particular, since ψ∗ϕ∗ is an isomorphism, Y, ϕψϕ(x0) Y, ϕ(x0) ϕ∗------------→ π1 38 Chapter 1 The Fundamental Group ϕ∗ is injective. The same reasoning with the second and third maps shows that ψ∗ is injective. Thus the ﬁrst two of the three maps are injections and their composition is an isomorphism, so the ﬁrst map ϕ∗ must be surjective as well as injective. ⊔⊓ Exercises 1. Show that composition of paths satisﬁes the following cancellation property: If f0 g0 ≃ f1 g1 and g0 ≃ g1 then f0 ≃ f1. 2. Show that the change-of-basepoint homomorphism βh depends only on the homotopy class of h. 3. For a path-connected space X, show that π1(X) is abelian iﬀ all basepoint-change homomorphisms βh depend only on the endpoints of the path h. 4. A subspace X ⊂ Rn
is said to be star-shaped if there is a point x0 ∈ X such that, for each x ∈ X, the line segment from x0 to x lies in X. Show that if a subspace X ⊂ Rn is locally star-shaped, in the sense that every point of X has a star-shaped neighborhood in X, then every path in X is homotopic in X to a piecewise linear path, that is, a path consisting of a ﬁnite number of straight line segments traversed at constant speed. Show this applies in particular when X is open or when X is a union of ﬁnitely many closed convex sets. 5. Show that for a space X, the following three conditions are equivalent: (a) Every map S 1→X is homotopic to a constant map, with image a point. (b) Every map S 1→X extends to a map D2→X. (c) π1(X, x0) = 0 for all x0 ∈ X. Deduce that a space X is simply-connected iﬀ all maps S 1→X are homotopic. this problem, ‘homotopic’ means ‘homotopic without regard to basepoints’.] [In 6. We can regard π1(X, x0) as the set of basepoint-preserving homotopy classes of maps (S 1, s0)→(X, x0). Let [S 1, X] be the set of homotopy classes of maps S 1→X, : π1(X, x0)→[S 1, X] with no conditions on basepoints. Thus there is a natural map is onto if X is path-connected, and that obtained by ignoring basepoints. Show that Φ Φ ([f ]) = ([g]) iﬀ [f ] and [g] are conjugate in π1(X, x0). Hence induces a oneto-one correspondence between [S 1, X] and the set of conjugacy classes in π1(X), Φ when X is path-connected. 7. Deﬁne f : S 1 × I→S 1 × I by f (θ, s) = (θ + 2π s, s), so
f restricts to the identity on the two boundary circles of S 1 × I. Show that f is homotopic to the identity by a homotopy ft that is stationary on one of the boundary circles, but not by any homotopy ft that is stationary on both boundary circles. [Consider what f does to the path s ֏ (θ0, s) for ﬁxed θ0 ∈ S 1.] 8. Does the Borsuk–Ulam theorem hold for the torus? In other words, for every map f : S 1 × S 1→R2 must there exist (x, y) ∈ S 1 × S 1 such that f (x, y) = f (−x, −y)? Φ Φ Basic Constructions Section 1.1 39 9. Let A1, A2, A3 be compact sets in R3. Use the Borsuk–Ulam theorem to show that there is one plane P ⊂ R3 that simultaneously divides each Ai into two pieces of equal measure. 10. From the isomorphism π1 loops in X × {y0} and {x0}× Y represent commuting elements of π1 Construct an explicit homotopy demonstrating this. X × Y, (x0, y0) ≈ π1(X, x0)× π1(Y, y0) it follows that. X × Y, (x0, y0) 11. If X0 is the path-component of a space X containing the basepoint x0, show that the inclusion X0 ֓ X induces an isomorphism π1(X0, x0)→π1(X, x0). 12. Show that every homomorphism π1(S 1)→π1(S 1) can be realized as the induced homomorphism ϕ∗ of a map ϕ : S 1→S 1. 13. Given a space X and a path-connected subspace A containing the basepoint x0, show that the map π1(A, x0)→π1(X, x0) induced by the inclusion A֓X is surjective iﬀ every path in X with endpoints in A is homotopic to a path in A. 14. Show that the isomorphism π1(X × Y ) ≈
π1(X)× π1(Y ) in Proposition 1.12 is given by [f ] ֏ (p1∗([f ]), p2∗([f ])) where p1 and p2 are the projections of X × Y onto its two factors. 15. Given a map f : X→Y and a path h : I→X from x0 to x1, show that f∗βh = βf hf∗ in the diagram at the right. 16. Show that there are no retractions r : X→A in the following cases: (a) X = R3 with A any subspace homeomorphic to S 1. (b) X = S 1 × D2 with A its boundary torus S 1 × S 1. (c) X = S 1 × D2 and A the circle shown in the ﬁgure. (d) X = D2 ∨ D2 with A its boundary S 1 ∨ S 1. (e) X a disk with two points on its boundary identiﬁed and A its boundary S 1 ∨ S 1. (f) X the M¨obius band and A its boundary circle. 17. Construct inﬁnitely many nonhomotopic retractions S 1 ∨ S 1→S 1. 18. Using Lemma 1.15, show that if a space X is obtained from a path-connected subspace A by attaching a cell en with n ≥ 2, then the inclusion A ֓ X induces a surjection on π1. Apply this to show: (a) The wedge sum S 1 ∨ S 2 has fundamental group Z. (b) For a path-connected CW complex X the inclusion map X 1 ֓ X of its 1 skeleton induces a surjection π1(X 1)→π1(X). [For the case that X has inﬁnitely many cells, see Proposition A.1 in the Appendix.] 19. Show that if X is a path-connected 1 dimensional CW complex with basepoint x0 a 0 cell, then every loop in X is homotopic to a loop consisting of a ﬁnite sequence of edges traversed monotonically. [See the proof of Lemma 1.15. This exercise gives an elementary proof that π1(S 1) is cyclic generated by the standard loop
winding once 40 Chapter 1 The Fundamental Group around the circle. The more diﬃcult part of the calculation of π1(S 1) is therefore the fact that no iterate of this loop is nullhomotopic.] 20. Suppose ft : X→X is a homotopy such that f0 and f1 are each the identity map. Use Lemma 1.19 to show that for any x0 ∈ X, the loop ft(x0) represents an element of the center of π1(X, x0). [One can interpret the result as saying that a loop represents an element of the center of π1(X) if it extends to a loop of maps X→X.] The van Kampen theorem gives a method for computing the fundamental groups of spaces that can be decomposed into simpler spaces whose fundamental groups are already known. By systematic use of this theorem one can compute the fundamental groups of a very large number of spaces. We shall see for example that for every group G there is a space XG whose fundamental group is isomorphic to G. To give some idea of how one might hope to compute fundamental groups by decomposing spaces into simpler pieces, let us look at an example. Consider the space X formed by two circles A and B intersecting in a single point, which we choose as the basepoint x0. By our preceding calculations we know that π1(A) is inﬁnite cyclic, generated by a loop a that goes once around A. Similarly, π1(B) is a copy of Z generated by a loop b going once around B. Each product of powers of a and b then gives an element of π1(X). For example, the product a5b2a−3ba2 is the loop that goes ﬁve times around A, then twice around B, then three times around A in the opposite direction, then once around B, then twice around A. The set of all words like this consisting of powers of a alternating with powers of b forms a group usually denoted Z ∗ Z. Multiplication in this group is deﬁned just as one would expect, for example (b4a5b2a−3)(a4b−1ab3) = b4a5b2ab−1ab3. The identity element is the empty word, and inverses are what they have to be
, for example (ab2a−3b−4)−1 = b4a3b−2a−1. It would be very nice if such words in a and b corresponded exactly to elements of π1(X), so that π1(X) was isomorphic to the group Z ∗ Z. The van Kampen theorem will imply that this is indeed the case. Similarly, if X is the union of three circles touching at a single point, the van Kampen theorem will imply that π1(X) is Z ∗ Z ∗ Z, the group consisting of words in powers of three letters a, b, c. The generalization to a union of any number of circles touching at one point will also follow. The group Z ∗ Z is an example of a general construction called the free product of groups. The statement of van Kampen’s theorem will be in terms of free products, so before stating the theorem we will make an algebraic digression to describe the construction of free products in some detail. Van Kampen’s Theorem Section 1.2 41 Free Products of Groups Q Suppose one is given a collection of groups Gα and one wishes to construct a single group containing all these groups as subgroups. One way to do this would be α Gα, whose elements can be regarded as the functions to take the product group α ֏ gα ∈ Gα. Or one could restrict to functions taking on nonidentity values at α Gα. Both these constructions produce most ﬁnitely often, forming the direct sum groups containing all the Gα ’s as subgroups, but with the property that elements of diﬀerent subgroups Gα commute with each other. In the realm of nonabelian groups this commutativity is unnatural, and so one would like a ‘nonabelian’ version of α Gα α Gα, it Q α Gα, and this is what the free α Gα is smaller and presumably simpler than α Gα. Since the sum should be easier to construct a nonabelian version of product ∗α Gα achieves. L L L L or Q and is not the identity element of Gαi Here is the precise deﬁnition. As a set, the free product ∗α Gα consists of all words g1g2 ··· gm of arbitrary ﬁnite length
m ≥ 0, where each letter gi belongs to, and adjacent letters gi and gi+1 a group Gαi belong to diﬀerent groups Gα, that is, αi ≠ αi+1. Words satisfying these conditions are called reduced, the idea being that unreduced words can always be simpliﬁed to reduced words by writing adjacent letters that lie in the same Gαi by canceling trivial letters. The empty word is allowed, and will be the identity element of ∗α Gα. The group operation in ∗α Gα is juxtaposition, (g1 ··· gm)(h1 ··· hn) = g1 ··· gmh1 ··· hn. This product may not be reduced, however: If gm and h1 belong to the same Gα, they should be combined into a single letter (gmh1) according to the multiplication in Gα, and if this new letter gmh1 happens to be the identity of Gα, it should be canceled from the product. This may allow gm−1 and h2 to be combined, and possibly canceled too. Repetition of this process eventually produces a reduced word. For example, in the product (g1 ··· gm)(g−1 1 ) everything cancels and we get the identity element of ∗α Gα, the empty word. as a single letter and m ··· g−1 Verifying directly that this multiplication is associative would be rather tedious, but there is an indirect approach that avoids most of the work. Let W be the set of reduced words g1 ··· gm as above, including the empty word. To each g ∈ Gα we associate the function Lg : W→W given by multiplication on the left, Lg(g1 ··· gm) = gg1 ··· gm where we combine g with g1 if g1 ∈ Gα to make gg1 ··· gm a reduced word. A key property of the association g ֏ Lg is the formula Lgg′ = LgLg′ for g, g′ ∈ Gα, that is, g(g′(g1 ··· gm)) = (gg′)(g1 ··· gm). This special case of associativity follows rather trivially from associativity in Gα. The formula Lgg′ = L
gLg′ implies that Lg is invertible with inverse Lg−1. Therefore the association g ֏ Lg deﬁnes a homomorphism from Gα to the group P (W ) of all permutations of W. More generally, we can deﬁne L : W→P (W ) by L(g1 ··· gm) = Lg1 for each reduced word g1 ··· gm. This function L is injective since the permutation L(g1 ··· gm) sends the empty word to g1 ··· gm. The product operation in W corresponds under L to ··· Lgm 42 Chapter 1 The Fundamental Group composition in P (W ), because of the relation Lgg′ = LgLg′. Since composition in P (W ) is associative, we conclude that the product in W is associative. In particular, we have the free product Z ∗ Z as described earlier. This is an example of a free group, the free product of any number of copies of Z, ﬁnite or inﬁnite. The elements of a free group are uniquely representable as reduced words in powers of generators for the various copies of Z, with one generator for each Z, just as in the case of Z ∗ Z. These generators are called a basis for the free group, and the number of basis elements is the rank of the free group. The abelianization of a free group is a free abelian group with basis the same set of generators, so since the rank of a free abelian group is well-deﬁned, independent of the choice of basis, the same is true for the rank of a free group. An interesting example of a free product that is not a free group is Z2 ∗ Z2. This is like Z ∗ Z but simpler since a2 = e = b2, so powers of a and b are not needed, and Z2 ∗ Z2 consists of just the alternating words in a and b : a, b, ab, ba, aba, bab, abab, baba, ababa, ···, together with the empty word. The structure of Z2 ∗ Z2 can be elucidated by looking at the homomorphism ϕ : Z2 ∗ Z2→Z2 associating to each word its length mod 2. Obviously �
� is surjective, and its kernel consists of the words of even length. These form an inﬁnite cyclic subgroup generated by ab since ba = (ab)−1 in Z2 ∗ Z2. In fact, Z2 ∗ Z2 is the semi-direct product of the subgroups Z and Z2 generated by ab and a, with the conjugation relation a(ab)a−1 = (ab)−1. This group is sometimes called the inﬁnite dihedral group. For a general free product ∗α Gα, each group Gα is naturally identiﬁed with a subgroup of ∗α Gα, the subgroup consisting of the empty word and the nonidentity one-letter words g ∈ Gα. From this viewpoint the empty word is the common identity element of all the subgroups Gα, which are otherwise disjoint. A consequence of associativity is that any product g1 ··· gm of elements gi in the groups Gα has a unique reduced form, the element of ∗α Gα obtained by performing the multiplications in any order. Any sequence of reduction operations on an unreduced product g1 ··· gm, combining adjacent letters gi and gi+1 that lie in the same Gα or canceling a gi that is the identity, can be viewed as a way of inserting parentheses into g1 ··· gm and performing the resulting sequence of multiplications. Thus associativity implies that any two sequences of reduction operations performed on the same unreduced word always yield the same reduced word. A basic property of the free product ∗α Gα is that any collection of homomorphisms ϕα : Gα→H extends uniquely to a homomorphism ϕ : ∗α Gα→H. Namely, the value of ϕ on a word g1 ··· gn with gi ∈ Gαi (gn), and using this formula to deﬁne ϕ gives a well-deﬁned homomorphism since the process of reducing an unreduced product in ∗α Gα does not aﬀect its image under ϕ. For example, for a free product G ∗ H the inclusions G ֓ G× H and H ֓ G× H induce a surjective homomorphism G ∗ H→
G× H. must be ϕα1 (g1) ··· ϕαn Van Kampen’s Theorem Section 1.2 43 The van Kampen Theorem Suppose a space X is decomposed as the union of a collection of path-connected open subsets Aα, each of which contains the basepoint x0 ∈ X. By the remarks in the preceding paragraph, the homomorphisms jα : π1(Aα)→π1(X) induced by the inclu: ∗α π1(Aα)→π1(X). The van Kampen sions Aα ֓ X extend to a homomorphism theorem will say that to have a nontrivial kernel in general. For if iαβ : π1(Aα ∩ Aβ)→π1(Aα) is the homomorphism induced by the inclusion Aα ∩ Aβ ֓ Aα then jαiαβ = jβiβα, both these compositions being induced by the inclusion Aα ∩ Aβ ֓ X, so the kernel of contains all the elements of the form iαβ(ω)iβα(ω)−1 for ω ∈ π1(Aα ∩ Aβ). Van Kampen’s theorem asserts that under fairly broad hypotheses this gives a full description of is very often surjective, but we can expect Φ Φ Φ Φ : Theorem 1.20. If X is the union of path-connected open sets Aα each containing the basepoint x0 ∈ X and if each intersection Aα ∩ Aβ is path-connected, then the : ∗α π1(Aα)→π1(X) is surjective. If in addition each intersection homomorphism Aα ∩ Aβ ∩ Aγ is path-connected, then the kernel of is the normal subgroup N generated by all elements of the form iαβ(ω)iβα(ω)−1 for ω ∈ π1(Aα ∩ Aβ), and hence induces an isomorphism π1(X) ≈ ∗α π1(Aα)/N. Φ Φ Φ Φ In Chapter 0 we deﬁned the wedge sum Example 1.21: Wedge Sum
s. α Xα of a collection of spaces Xα with basepoints xα ∈ Xα to be the quotient space of the α Xα in which all the basepoints xα are identiﬁed to a single point. disjoint union If each xα is a deformation retract of an open neighborhood Uα in Xα, then Xα is a deformation retract of its open neighborhood Aα = Xα β≠α Uβ. The intersection of two or more distinct Aα ’s is α Uα, which deformation retracts to a point. Van W Kampen’s theorem then implies that α Xα) is an isomorphism, assuming that each Xα is path-connected, hence also each Aα. : ∗α π1(Xα)→π1( ` W W W Thus for a wedge sum of copies of Z, one for each circle S 1 as in the example at the beginning of this section. α S 1 W Φ α of circles, π1( α) is a free group, the free product α. In particular, π1(S 1 ∨S 1) is the free group Z∗Z, α S 1 W It is true more generally that the fundamental group of any connected graph is free, as we show in §1.A. Here is an example illustrating the general technique. Example 1.22. Let X be the graph shown in the ﬁgure, consisting of the twelve edges of a cube. The seven heavily shaded edges form a maximal tree T ⊂ X, a contractible subgraph containing all the vertices of X. We claim that π1(X) is the free product of ﬁve copies of Z, one for each edge not in T. To deduce this from van Kampen’s theorem, choose for each edge eα of X − T an open neighborhood Aα of T ∪ eα in X that deformation retracts onto T ∪ eα. The intersection of two or more Aα ’s deformation retracts onto T, hence is contractible. The Aα ’s form a cover of 44 Chapter 1 The Fundamental Group X satisfying the hypotheses of van Kampen’s theorem, and since the intersection of any two of them is simply-connected we obtain an isomorphism π1(X)
≈ ∗α π1(Aα). Each Aα deformation retracts onto a circle, so π1(X) is free on ﬁve generators, as claimed. As explicit generators we can choose for each edge eα of X − T a loop fα that starts at a basepoint in T, travels in T to one end of eα, then across eα, then back to the basepoint along a path in T. Van Kampen’s theorem is often applied when there are just two sets Aα and Aβ in the cover of X, so the condition on triple intersections Aα∩Aβ∩Aγ is superﬂuous and one obtains an isomorphism π1(X) ≈ /N, under the assumption that Aα ∩ Aβ is path-connected. The proof in this special case is virtually identical with the proof in the general case, however. π1(Aα) ∗ π1(Aβ) One can see that the intersections Aα ∩ Aβ need to be path-connected by considering the example of S 1 decomposed as the union of two open arcs. In this case is not surjective. For an example showing that triple intersections Aα ∩ Aβ ∩ Aγ need to be path-connected, let X be the suspension of three points a, b, c, and let Φ Aα, Aβ, and Aγ be the complements of these three points. The theorem does apply to the covering {Aα, Aβ}, so there are isomorphisms π1(X) ≈ π1(Aα) ∗ π1(Aβ) ≈ Z ∗ Z since Aα ∩ Aβ is contractible. If we tried to use the covering {Aα, Aβ, Aγ}, which has each of the twofold intersections path-connected but not the triple intersection, then we would get π1(X) ≈ Z ∗ Z ∗ Z, but this is not isomorphic to Z ∗ Z since it has a diﬀerent abelianization. Proof of van Kampen’s theorem: We have already proved the ﬁrst part of the theorem in Lemma 1.15. To prove the harder part of the theorem, concerning surjectivity of that the
kernel of an element [f ] ∈ π1(X) we shall mean a formal product [f1] ··· [fk] where: is N, we ﬁrst introduce some terminology. By a factorization of Φ Φ Each fi is a loop in some Aα at the basepoint x0, and [fi] ∈ π1(Aα) is the homotopy class of fi. The loop f is homotopic to f1 ··· fk in X. A factorization of [f ] is thus a word in ∗α π1(Aα), possibly unreduced, that is is equivalent to saying that every [f ] ∈ π1(X) mapped to [f ] by. Surjectivity of has a factorization. We will be concerned with the uniqueness of factorizations. Call two factoriza- Φ Φ tions of [f ] equivalent if they are related by a sequence of the following two sorts of moves or their inverses: Combine adjacent terms [fi][fi+1] into a single term [fi fi+1] if [fi] and [fi+1] lie in the same group π1(Aα). Regard the term [fi] ∈ π1(Aα) as lying in the group π1(Aβ) rather than π1(Aα) if fi is a loop in Aα ∩ Aβ. Van Kampen’s Theorem Section 1.2 45 The ﬁrst move does not change the element of ∗α π1(Aα) deﬁned by the factorization. The second move does not change the image of this element in the quotient group Q = ∗α π1(Aα)/N, by the deﬁnition of N. So equivalent factorizations give the same element of Q. If we can show that any two factorizations of [f ] are equivalent, this will say that is injective, hence the kernel of is exactly N, and the map Q→π1(X) induced by the proof will be complete. Φ Φ 1] ··· [f ′ Let [f1] ··· [fk] and [f ′ 1 ··· f ′ 1 ··· f ′ ℓ] be two factorizations
of [f ]. The composed ℓ are then homotopic, so let F : I × I→X be a homopaths f1 ··· fk and f ′ topy from f1 ··· fk to f ′ ℓ. There exist partitions 0 = s0 < s1 < ··· < sm = 1 and 0 = t0 < t1 < ··· < tn = 1 such that each rectangle Rij = [si−1, si]× [tj−1, tj] is mapped by F into a single Aα, which we label Aij. These partitions may be obtained by covering I × I by ﬁnitely many rectangles [a, b]× [c, d] each mapping to a single Aα, using a compactness argument, then partitioning I × I by the union of all the horizontal and vertical lines containing edges of these rectangles. We may assume 1 ··· f ′ the s partition subdivides the partitions giving the products f1 ··· fk and f ′ ℓ. Since F maps a neighborhood of Rij to Aij, we may perturb the vertical sides of the rectangles Rij so that each point of I × I lies in at most three Rij ’s. We may assume there are at least three rows of rectangles, so we can do this perturbation just on the rectangles in the intermediate rows, leaving the top and bottom rows unchanged. Let us relabel the new rectangles R1, R2, ···, Rmn, ordering them as in the ﬁgure. If γ is a path in I × I from the left edge to the right edge, then the restriction F \|\| γ is a loop at the basepoint x0 since F maps both the left and right edges of I × I to x0. Let γr be the path separating the ﬁrst r rectangles R1, ···, Rr from the remaining rectangles. Thus γ0 is the bottom edge of I × I and γmn is the top edge. We pass from γr to γr +1 by pushing across the rectangle Rr +1. Let us call the corners of the Rr ’s vertices. For each vertex v with F (v) ≠ x0 we can choose a path gv from x0 to F
(v) that lies in the intersection of the two or three Aij ’s corresponding to the Rr ’s containing v, since we assume the intersection of any two or three Aij ’s is path-connected. Then we obtain a factorization of [F \|\| γr ] by inserting the appropriate paths gv gv into F \|\| γr at successive vertices, in Lemma 1.15. This factorization depends on as in the proof of surjectivity of Φ certain choices, since the loop corresponding to a segment between two successive vertices can lie in two diﬀerent Aij ’s when there are two diﬀerent rectangles Rij containing this edge. Diﬀerent choices of these Aij ’s change the factorization of [F \|\| γr ] to an equivalent factorization, however. Furthermore, the factorizations associated to successive paths γr and γr +1 are equivalent since pushing γr across Rr +1 to γr +1 changes F \|\| γr to F \|\| γr +1 by a homotopy within the Aij corresponding to Rr +1, and 46 Chapter 1 The Fundamental Group we can choose this Aij for all the segments of γr and γr +1 in Rr +1. We can arrange that the factorization associated to γ0 is equivalent to the factorization [f1] ··· [fk] by choosing the path gv for each vertex v along the lower edge of I × I to lie not just in the two Aij ’s corresponding to the Rs ’s containing v, but also to lie in the Aα for the fi containing v in its domain. In case v is the common endpoint of the domains of two consecutive fi ’s we have F (v) = x0, so there is no need to choose a gv for such v ’s. In similar fashion we may assume that the factorization associated to the ﬁnal γmn is equivalent to [f ′ ℓ]. Since the factorizations associated to all the γr ’s are equivalent, we conclude that the factorizations [f1] ··· [fk] and [f ′ ⊔⊓ 1] ··· [f ′ ℓ] are equivalent. 1] ··· [f
′ Example 1.23: Linking of Circles. We can apply van Kampen’s theorem to calculate the fundamental groups of three spaces discussed in the introduction to this chapter, the complements in R3 of a single circle, two unlinked circles, and two linked circles. The complement R3 −A of a single circle A deformation retracts onto a wedge sum S 1 ∨ S 2 embedded in R3 −A as shown in the ﬁrst of the two ﬁgures at the right. It may be easier to see that R3−A deformation retracts onto the union of S 2 with a diameter, as in the second ﬁgure, where points outside S 2 deformation retract onto S 2, and points inside S 2 and not in A can be pushed away from A toward S 2 or the diameter. Having this deformation retraction in mind, one can then see how it must be modiﬁed if the two endpoints of the diameter are gradually moved toward each other along the equator until they coincide, forming the S 1 summand of S 1 ∨S 2. Another way of seeing the deformation retraction of R3 − A onto S 1 ∨ S 2 is to note ﬁrst that an open ε neighborhood of S 1 ∨ S 2 obviously deformation retracts onto S 1 ∨ S 2 if ε is suﬃciently small. Then observe that this neighborhood is homeomorphic to R3 − A by a homeomorphism that is the identity on S 1 ∨ S 2. In fact, the neighborhood can be gradually enlarged by homeomorphisms until it becomes all of R3 − A. In any event, once we see that R3 − A deformation retracts to S 1 ∨ S 2, then we immediately obtain isomorphisms π1(R3 − A) ≈ π1(S 1 ∨ S 2) ≈ Z since π1(S 2) = 0. In similar fashion, the complement R3 − (A ∪ B) of two unlinked circles A and B deformation retracts onto S 1∨S 1∨S 2∨S 2, as in the ﬁgure to the right. From ≈ R3 − (A ∪ B) this we get π1 Z ∗ Z. On the other hand, if A
and B are linked, then R3 − (A ∪ B) deformation retracts onto the wedge sum of S 2 and a torus S 1 × S 1 separating A and B, ≈ as shown in the ﬁgure to the left, hence π1 π1(S 1 × S 1) ≈ Z× Z. R3 − (A ∪ B) Van Kampen’s Theorem Section 1.2 47 Example 1.24: Torus Knots. For relatively prime positive integers m and n, the torus knot K = Km,n ⊂ R3 is the image of the embedding f : S 1→S 1 × S 1 ⊂ R3, f (z) = (zm, zn), where the torus S 1 × S 1 is embedded in R3 in the standard way. The knot K winds around the torus a total of m times in the longitudinal direction and n times in the meridional direction, as shown in the ﬁgure for the cases (m, n) = (2, 3) and (3, 4). One needs to assume that m and n are relatively prime in order for the map f to be injective. Without this assumption f would be d –to–1 where d is the greatest common divisor of m and n, and the image of f would be the knot Km/d,n/d. One could also allow negative values for m or n, but this would only change K to a mirror-image knot. Let us compute π1(R3 − K). It is slightly easier to do the calculation with R3 replaced by its one-point compactiﬁcation S 3. An application of van Kampen’s theorem shows that this does not aﬀect π1. Namely, write S 3 − K as the union of R3 − K and an open ball B formed by the compactiﬁcation point together with the complement of a large closed ball in R3 containing K. Both B and B ∩ (R3 − K) are simply-connected, the latter space being homeomorphic to S 2 × R. Hence van Kampen’s theorem implies that the inclusion R3 − K ֓ S 3 − K induces an isomorphism on π1. We compute π1(S 3
− K) by showing that it deformation retracts onto a 2 dimensional complex X = Xm,n homeomorphic to the quotient space of a cylinder S 1 × I under the identiﬁcations (z, 0) ∼ (e2π i/mz, 0) and (z, 1) ∼ (e2π i/nz, 1). If we let Xm and Xn be the two halves of X formed by the quotients of S 1 × [0, 1/2] and S 1 × [1/2, 1], then Xm and Xn are the mapping cylinders of z ֏ zm and z ֏ zn. The intersection Xm ∩ Xn is the circle S 1 × {1/2}, the domain end of each mapping cylinder. To obtain an embedding of X in S 3 − K as a deformation retract we will use the standard decomposition of S 3 into two solid tori S 1 × D2 and D2 × S 1, the result of regarding S 3 as ∂D4 = ∂(D2 × D2) = ∂D2 × D2 ∪ D2 × ∂D2. Geometrically, the ﬁrst solid torus S 1 × D2 can be identiﬁed with the compact region in R3 bounded by the standard torus S 1 × S 1 containing K, and the second solid torus D2 × S 1 is then the closure of the complement of the ﬁrst solid torus, together with the compactiﬁcation point at inﬁnity. Notice that meridional circles in S 1 × S 1 bound disks in the ﬁrst solid torus, while it is longitudinal circles that bound disks in the second solid torus. In the ﬁrst solid torus, K intersects each of the meridian circles {x}× ∂D2 in m equally spaced points, as indicated in the ﬁgure at the right, which shows a meridian disk {x}× D2. These m points can be separated by a union of m radial line segments. Letting x vary, these radial segments then trace out a copy of the mapping cylinder Xm in the ﬁrst solid torus. Symmetrically, there is a copy of the other mapping cylinder X
n in the second solid torus. 48 Chapter 1 The Fundamental Group The complement of K in the ﬁrst solid torus deformation retracts onto Xm by ﬂowing within each meridian disk as shown. In similar fashion the complement of K in the second solid torus deformation retracts onto Xn. These two deformation retractions do not agree on their common domain of deﬁnition S 1 × S 1 − K, but this is easy to correct by distorting the ﬂows in the two solid tori so that in S 1 × S 1 − K both ﬂows are orthogonal to K. After this modiﬁcation we now have a well-deﬁned deformation retraction of S 3 − K onto X. Another way of describing the situation would be to say that for an open ε neighborhood N of K bounded by a torus T, the complement S 3 − N is the mapping cylinder of a map T→X. To compute π1(X) we apply van Kampen’s theorem to the decomposition of X as the union of Xm and Xn, or more properly, open neighborhoods of these two sets that deformation retract onto them. Both Xm and Xn are mapping cylinders that deformation retract onto circles, and Xm ∩ Xn is a circle, so all three of these spaces have fundamental group Z. A loop in Xm ∩ Xn representing a generator of π1(Xm ∩ Xn) is homotopic in Xm to a loop representing m times a generator, and in Xn to a loop representing n times a generator. Van Kampen’s theorem then says that π1(X) is the quotient of the free group on generators a and b obtained by factoring out the normal subgroup generated by the element amb−n. Let us denote by Gm,n this group π1(Xm,n) deﬁned by two generators a and b and one relation am = bn. If m or n is 1, then Gm,n is inﬁnite cyclic since in these cases the relation just expresses one generator as a power of the other. To describe the structure of Gm,n when m, n > 1 let us ﬁrst compute the center of Gm,n, the subgroup
consisting of elements that commute with all elements of Gm,n. The element am = bn commutes with a and b, so the cyclic subgroup C generated by this element lies in the center. In particular, C is a normal subgroup, so we can pass to the quotient group Gm,n/C, which is the free product Zm ∗ Zn. According to Exercise 1 at the end of this section, a free product of nontrivial groups has trivial center. From this it follows that C is exactly the center of Gm,n. As we will see in Example 1.44, the elements a and b have inﬁnite order in Gm,n, so C is inﬁnite cyclic, but we will not need this fact here. We will show now that the integers m and n are uniquely determined by the group Zm ∗ Zn, hence also by Gm,n. The abelianization of Zm ∗ Zn is Zm × Zn, of order mn, so the product mn is uniquely determined by Zm ∗ Zn. To determine m and n individually, we use another assertion from Exercise 1 at the end of the section, that all torsion elements of Zm ∗Zn are conjugate to elements of one of the subgroups Zm and Zn, hence have order dividing m or n. Thus the maximum order of torsion elements of Zm ∗ Zn is the larger of m and n. The larger of these two numbers is therefore uniquely determined by the group Zm ∗ Zn, hence also the smaller since the product is uniquely determined. The preceding analysis of π1(Xm,n) did not need the assumption that m and n Van Kampen’s Theorem Section 1.2 49 are relatively prime, which was used only to relate Xm,n to torus knots. An interesting fact is that Xm,n can be embedded in R3 only when m and n are relatively prime. This is shown in the remarks following Corollary 3.46. For example, X2,2 is the Klein bottle since it is the union of two copies of the M¨obius band X2 with their boundary circles identiﬁed, so this nonembeddability statement generalizes the fact that the Klein bottle cannot be embedded in R3. An algorithm
for computing a presentation for π1(R3−K) for an arbitrary smooth or piecewise linear knot K is described in the exercises, but the problem of determin- ing when two of these fundamental groups are isomorphic is generally much more diﬃcult than in the special case of torus knots. Example 1.25: The Shrinking Wedge of Circles. Consider the subspace X ⊂ R2 that is the union of the circles Cn of radius 1/n and center (1/n, 0) for n = 1, 2, ···. At ﬁrst glance one might confuse X with the wedge sum of an inﬁnite sequence of circles, but we will show that X has a much larger fundamental group than the wedge sum. Consider the retractions rn : X→Cn collapsing all Ci ’s except Cn to the origin. Each rn induces a surjection ρn : π1(X)→π1(Cn) ≈ Z, where we take the origin as the basepoint. The product of the ρn ’s is a homomorphism ρ : π1(X)→ ∞ Z to the direct product (not the direct sum) of inﬁnitely many copies of Z, and ρ is surjective since for every sequence of integers kn we can construct a loop f : I→X that wraps kn times around Cn in the time interval [1 − 1/n, 1 − 1/n+1]. This inﬁnite composition of loops is certainly continuous at each time less than 1, and it is continuous at time Q 1 since every neighborhood of the basepoint in X contains all but ﬁnitely many of the ∞ Z, it is uncountable. circles Cn. Since π1(X) maps onto the uncountable group On the other hand, the fundamental group of a wedge sum of countably many circles Q is countably generated, hence countable. The group π1(X) is actually far more complicated than ∞ Z. For one thing, it is nonabelian, since the retraction X→C1 ∪ ··· ∪ Cn that collapses all the circles smaller than Cn to the basepoint induces a surjection from π1(X) to a free group on n generators.
For a complete description of π1(X) see [Cannon & Conner 2000]. Q It is a theorem of [Shelah 1988] that for a path-connected, locally path-connected compact metric space X, π1(X) is either ﬁnitely generated or uncountable. Applications to Cell Complexes For the remainder of this section we shall be interested in cell complexes, and in particular in how the fundamental group is aﬀected by attaching 2 cells. Suppose we attach a collection of 2 cells e2 α to a path-connected space X via maps ϕα : S 1→X, producing a space Y. If s0 is a basepoint of S 1 then ϕα determines a loop at ϕα(s0) that we shall call ϕα, even though technically loops are maps I→X rather than S 1→X. For diﬀerent α ’s the basepoints ϕα(s0) of these loops ϕα may not all 50 Chapter 1 The Fundamental Group coincide. To remedy this, choose a basepoint x0 ∈ X and a path γα in X from x0 to ϕα(s0) for each α. Then γαϕαγα is a loop at x0. This loop may not be nullhomotopic in X, but it will certainly be nullhomotopic after the cell e2 α is attached. Thus the normal subgroup N ⊂ π1(X, x0) generated by all the loops γαϕαγα for varying α lies in the kernel of the map π1(X, x0)→π1(Y, x0) induced by the inclusion X ֓ Y. Proposition 1.26. (a) If Y is obtained from X by attaching 2 cells as described above, then the inclusion X ֓ Y induces a surjection π1(X, x0)→π1(Y, x0) whose kernel is N. Thus π1(Y ) ≈ π1(X)/N. (b) If Y is obtained from X by attaching n cells for a ﬁxed n > 2, then the inclusion X ֓ Y induces an isomorphism π1(X, x0) ≈ π1(Y, x0)
. (c) For a path-connected cell complex X the inclusion of the 2 skeleton X 2 ֓ X induces an isomorphism π1(X 2, x0) ≈ π1(X, x0). It follows from (a) that N is independent of the choice of the paths γα, but this can also be seen directly: If we replace γα by another path ηα having the same endpoints, then γαϕαγα changes to ηαϕαηα = (ηαγα)γαϕαγα(γαηα), so γαϕαγα and ηαϕαηα deﬁne conjugate elements of π1(X, x0). (a) Let us expand Y to a slightly larger space Z that deformation retracts Proof: onto Y and is more convenient for applying van Kampen’s theorem. The space Z is obtained from Y by attaching rectangular strips Sα = I × I, with the lower edge I × {0} attached along γα, the right edge {1}× I attached along an arc that starts at ϕα(s0) and goes radially into e2 α, and all the left edges {0}× I of the diﬀer- ent strips identiﬁed together. The top edges of the strips are not attached to anything, and this allows us to deformation retract Z onto Y. In each cell e2 S α choose a point yα not in the arc along which Sα is attached. Let α{yα} and let B = Z − X. Then A deformation retracts onto X, and B is A = Z − contractible. Since π1(B) = 0, van Kampen’s theorem applied to the cover {A, B} says that π1(Z) is isomorphic to the quotient of π1(A) by the normal subgroup generated by the image of the map π1(A ∩ B)→π1(A). More speciﬁcally, choose a basepoint z0 ∈ A ∩ B near x0 on the segment where all the strips Sα intersect, and choose loops δα in A ∩ B based
at z0 representing the elements of π1(A, z0) corresponding to [γαϕαγα] ∈ π1(A, x0) under the basepoint-change isomorphism βh for h the line segment connecting z0 to x0 in the intersection of the Sα ’s. To ﬁnish the proof of part (a) we just need to check that π1(A ∩ B, z0) is generated by the loops δα. This can be done by another application of van Kampen’s theorem, this time to the cover of A ∩ B by the open sets Aα = A ∩ B − β. Since Aα deformation retracts onto a circle in e2 α − {yα}, we have π1(Aα, z0) ≈ Z generated by δα. β≠α e2 S Van Kampen’s Theorem Section 1.2 51 The proof of (b) follows the same plan with cells en α. The only diﬀerence is that Aα deformation retracts onto a sphere S n−1 so π1(Aa) = 0 if n > 2 by Proposition 1.14. Hence π1(A ∩ B) = 0 and the result follows. α instead of e2 Part (c) follows from (b) by induction when X is ﬁnite-dimensional, so X = X n for some n. When X is not ﬁnite-dimensional we argue as follows. Let f : I→X be a loop at the basepoint x0 ∈ X 2. This has compact image, which must lie in X n for some n by Proposition A.1 in the Appendix. Part (b) then implies that f is homotopic to a loop in X 2. Thus π1(X 2, x0)→π1(X, x0) is surjective. To see that it is also injective, suppose that f is a loop in X 2 which is nullhomotopic in X via a homotopy F : I × I→X. This has compact image lying in some X n, and we can assume n > 2. Since π1(X 2, x0)→π1(X n, x0) is injective by (b
), we conclude that f is nullhomotopic in X 2. ⊔⊓ As a ﬁrst application we compute the fundamental group of the orientable surface Mg of genus g. This has a cell structure with one 0 cell, 2g 1 cells, and one 2 cell, as we saw in Chapter 0. The 1 skeleton is a wedge sum of 2g circles, with fundamental group free on 2g generators. The 2 cell is attached along the loop given by the product of the commutators of these generators, say [a1, b1] ··· [ag, bg]. Therefore π1(Mg) ≈ a1, b1, ···, ag, bg \|\|\|\| [a1, b1] ··· [ag, bg] \|\|\|\| rβ gα denotes the group with generators gα and relators rβ, in other where words, the free group on the generators gα modulo the normal subgroup generated by the words rβ in these generators. Corollary 1.27. The surface Mg is not homeomorphic, or even homotopy equivalent, to Mh if g ≠ h. Proof: The abelianization of π1(Mg) is the direct sum of 2g copies of Z. So if Mg ≃ Mh then π1(Mg) ≈ π1(Mh), hence the abelianizations of these groups are iso⊔⊓ morphic, which implies g = h. Nonorientable surfaces can be treated in the same way. If we attach a 2 cell to the wedge sum of g circles by the word a2 1 ··· a2 g we obtain a nonorientable surface Ng. For example, N1 is the projective plane RP2, the quotient of D2 with antipodal points of ∂D2 identiﬁed, and N2 is the Klein bottle, though the more usual representation of the Klein bottle is as a square with opposite sides identiﬁed via the word aba−1b. 52 Chapter 1 The Fundamental Group If one cuts the square along a diagonal and reassembles the resulting two triangles as shown in the ﬁgure, one obtains the other representation as a square with sides identiﬁed via the word a2c2. By the proposition, π
1(Ng) ≈. This abelianizes to the direct sum of Z2 with g − 1 copies of Z since in the abelian- ization we can rechoose the generators to be a1, ···, ag−1 and a1 + ··· + ag, with 2(a1 + ··· + ag) = 0. Hence Ng is not homotopy equivalent to Nh if g ≠ h, nor is Ng homotopy equivalent to any orientable surface Mh. a1, ···, ag 1 ··· a2 \|\|\|\| a2 g Here is another application of the preceding proposition: Corollary 1.28. For every group G there is a 2 dimensional cell complex XG with π1(XG) ≈ G. Proof: Choose a presentation G =. This exists since every group is a quotient of a free group, so the gα ’s can be taken to be the generators of this free group with the rβ ’s generators of the kernel of the map from the free group to G. α by attaching 2 cells e2 β by the loops speciﬁed by the Now construct XG from ⊔⊓ words rβ. α S 1 W \|\|\|\| rβ gα a \|\|\|\| an = Zn then XG is S 1 with a cell e2 attached by the map Example 1.29. If G = z ֏ zn, thinking of S 1 as the unit circle in C. When n = 2 we get XG = RP2, but for n > 2 the space XG is not a surface since there are n ‘sheets’ of e2 attached at each point of the circle S 1 ⊂ XG. For example, when n = 3 one can construct a neighborhood N of S 1 in XG by taking the product of the with the interval I, and then identifying graph the two ends of this product via a one-third twist as shown in the ﬁgure. The boundary of N consists of a single circle, formed by the three endpoints of cross section of N. To complete the construction of XG from N one attaches each a disk along the boundary circle of N. This cannot be done in R3, though it can in R4. For n = 4 one would use the graph, with a one-quarter twist instead of a one-third twist. For
larger n one would use an n pointed ‘asterisk’ and a 1/n twist. instead of Exercises 1. Show that the free product G ∗ H of nontrivial groups G and H has trivial center, and that the only elements of G ∗ H of ﬁnite order are the conjugates of ﬁnite-order elements of G and H. 2. Let X ⊂ Rm be the union of convex open sets X1, ···, Xn such that Xi ∩Xj ∩Xk ≠ ∅ for all i, j, k. Show that X is simply-connected. Van Kampen’s Theorem Section 1.2 53 3. Show that the complement of a ﬁnite set of points in Rn is simply-connected if n ≥ 3. 4. Let X ⊂ R3 be the union of n lines through the origin. Compute π1(R3 − X). 5. Let X ⊂ R2 be a connected graph that is the union of a ﬁnite number of straight line segments. Show that π1(X) is free with a basis consisting of loops formed by the boundaries of the bounded complementary regions of X, joined to a basepoint by suitably chosen paths in X. [Assume the Jordan curve theorem for polygonal simple closed curves, which is equivalent to the case that X is homeomorphic to S 1.] 6. Use Proposition 1.26 to show that the complement of a closed discrete subspace of Rn is simply-connected if n ≥ 3. 7. Let X be the quotient space of S 2 obtained by identifying the north and south poles to a single point. Put a cell complex structure on X and use this to compute π1(X). 8. Compute the fundamental group of the space obtained from two tori S 1 × S 1 by identifying a circle S 1 × {x0} in one torus with the corresponding circle S 1 × {x0} in the other torus. h does not retract onto its boundary circle C, and hence Mg does [Hint: abelianize π1.] But show that Mg does retract onto the 9. In the surface Mg of genus g, let C be a circle that separates Mg into two compact subsurfaces M ′ h and M
′ k obtained from the closed surfaces Mh and Mk by deleting an open disk from each. Show that M ′ not retract onto C. nonseparating circle C ′ in the ﬁgure. 10. Consider two arcs α and β embedded in D2 × I as shown in the ﬁgure. The loop γ is obviously nullhomotopic in D2 × I, but show that there is no nullhomotopy of γ in the complement of α ∪ β. 11. The mapping torus Tf of a map f : X→X is the quotient of X × I obtained In the case X = S 1 ∨ S 1 with f by identifying each point (x, 0) with (f (x), 1). basepoint-preserving, compute a presentation for π1(Tf ) in terms of the induced map f∗ : π1(X)→π1(X). Do the same when X = S 1 × S 1. [One way to do this is to regard Tf as built from X ∨ S 1 by attaching cells.] 12. The Klein bottle is usually pictured as a subspace of R3 like the subspace X ⊂ R3 shown in the ﬁrst ﬁgure at the right. If one wanted a model that could actually function as a bottle, one would delete the open disk bounded by the circle of selfintersection of X, producing a subspace Y ⊂ X. Show that π1(X) ≈ Z ∗ Z and that 54 Chapter 1 The Fundamental Group a, b, c \|\|\|\| aba−1b−1cbεc−1 for ε = ±1. (Changing the π1(Y ) has the presentation sign of ε gives an isomorphic group, as it happens.) Show also that π1(Y ) is isomorphic to π1(R3 −Z) for Z the graph shown in the ﬁgure. The groups π1(X) and π1(Y ) are not isomorphic, but this is not easy to prove; see the discussion in Example 1B.13. 13. The space Y in the preceding exercise can be obtained from a disk with two holes by identifying its three boundary circles. There are only two essentially diﬀerent ways of identifying the three boundary circles.
Show that the other way yields a space Z with π1(Z) not isomorphic to π1(Y ). [Abelianize the fundamental groups to show they are not isomorphic.] 14. Consider the quotient space of a cube I3 obtained by identifying each square face with the opposite square face via the right-handed screw motion consisting of a translation by one unit in the direction perpendicular to the face combined with a one-quarter twist of the face about its center point. Show this quotient space X is a cell complex with two 0 cells, four 1 cells, three 2 cells, and one 3 cell. Using this structure, show that π1(X) is the quaternion group {±1, ±i, ±j, ±k}, of order eight. 15. Given a space X with basepoint x0 ∈ X, we may construct a CW complex L(X) γ for each loop γ in X based at x0, and a 2 cell e2 having a single 0 cell, a 1 cell e1 τ for each map τ of a standard triangle P QR into X taking the three vertices P, Q, and R of the triangle to x0. The 2 cell e2 τ is attached to the three 1 cells that are the loops obtained by restricting τ to the three oriented edges P Q, P R, and QR. Show that the natural map L(X)→X induces an isomorphism π1 ≈ π1(X, x0). L(X) 16. Show that the fundamental group of the surface of inﬁnite genus shown below is free on an inﬁnite number of generators. 17. Show that π1(R2 − Q2) is uncountable. 18. In this problem we use the notions of suspension, reduced suspension, cone, and mapping cone deﬁned in Chapter 0. Let X be the subspace of R consisting of the sequence 1, 1/2, 1/3, 1/4, ··· together with its limit point 0. (a) For the suspension SX, show that π1(SX) is free on a countably inﬁnite set of generators, and deduce that π1(SX) is countable. In contrast to this, the reduced X, obtained from SX by collapsing the segment {0}× I to
a point, is suspension the shrinking wedge of circles in Example 1.25, with an uncountable fundamental group. Σ (b) Let C be the mapping cone of the quotient map SX→ countable by constructing a homomorphism from π1(C) onto X. Show that π1(C) is un∞ Z. Note ∞ Z/ Σ Q L Van Kampen’s Theorem Section 1.2 55 that C is the reduced suspension of the cone CX. Thus the reduced suspension of a contractible space need not be contractible, unlike the unreduced suspension. 19. Show that the subspace of R3 that is the union of the spheres of radius 1/n and center (1/n, 0, 0) for n = 1, 2, ··· is simply-connected. 20. Let X be the subspace of R2 that is the union of the circles Cn of radius n and center (n, 0) for n = 1, 2, ···. Show that π1(X) is the free group ∗n π1(Cn), the same ∞ S 1 are in fact homotopy as for the inﬁnite wedge sum W ∞ S 1. Show that X and W equivalent, but not homeomorphic. 21. Show that the join X ∗ Y of two nonempty spaces X and Y is simply-connected if X is path-connected. 22. In this exercise we describe an algorithm for computing a presentation of the fundamental group of the complement of a smooth or piecewise linear knot K in R3, called the Wirtinger presentation. To begin, we position the knot to lie almost ﬂat on a table, so that K consists of ﬁnitely many disjoint arcs αi where it intersects the table top together with ﬁnitely many disjoint arcs βℓ where K crosses over itself. The conﬁguration at such a crossing is shown in the ﬁrst ﬁgure below. We build a 2 dimensional complex X that is a deformation retract of R3 − K by the following three steps. First, start with the rectangle T formed by the table top. Next, just above each arc αi place a long, thin rectangular strip Ri, curved to run parallel to αi along the full length of αi and ar
ched so that the two long edges of Ri are identiﬁed with points of T, as in the second ﬁgure. Any arcs βℓ that cross over αi are positioned to lie in Ri. Finally, over each arc βℓ put a square Sℓ, bent downward along its four edges so that these edges are identiﬁed with points of three strips Ri, Rj, and Rk as in the third ﬁgure; namely, two opposite edges of Sℓ are identiﬁed with short edges of Rj and Rk and the other two opposite edges of Sℓ are identiﬁed with two arcs crossing the interior of Ri. The knot K is now a subspace of X, but after we lift K up slightly into the complement of X, it becomes evident that X is a deformation retract of R3 − K. (a) Assuming this bit of geometry, show that π1(R3 − K) has a presentation with one generator xi for each strip Ri and one relation of the form xixjx−1 i = xk for each square Sℓ, where the indices are as in the ﬁgures above. [To get the correct signs it is helpful to use an orientation of K.] (b) Use this presentation to show that the abelianization of π1(R3 − K) is Z. 56 Chapter 1 The Fundamental Group We come now to the second main topic of this chapter, covering spaces. We have already encountered these brieﬂy in our calculation of π1(S 1) which used the example of the projection R→S 1 of a helix onto a circle. As we will see, covering spaces can be used to calculate fundamental groups of other spaces as well. But the connection between the fundamental group and covering spaces runs much deeper than this, and in many ways they can be regarded as two viewpoints toward the same thing. Algebraic aspects of the fundamental group can often be translated into the geometric language of covering spaces. This is exempliﬁed in one of the main results in this section, an exact correspondence between connected covering spaces of a given space X and subgroups of π1(X). This is strikingly reminiscent of Galois theory, with its correspondence between ﬁeld extensions and subgroups of
the Galois group. Let us recall the deﬁnition. A covering space of a space X is a space X together X→X satisfying the following condition: Each point x ∈ X has an X, with a map p : open neighborhood U in X such that p−1(U) is a union of disjoint open sets in each of which is mapped homeomorphically onto U by p. Such a U is called evenly e X that project homeomorphically to U by p covered and the disjoint open sets in e e X over U. are called sheets of If U is connected these sheets are the connected components of p−1(U) so in this case they are uniquely determined by U, but when U is not connected the decomposition of p−1(U) into sheets may not be unique. We allow p−1(U) to be empty, the union of an empty collection of sheets over U, so p need not be surjective. The number of sheets over U is the cardinality of p−1(x) for x ∈ U. As x varies over X this number is locally constant, so it is constant if X is e e connected. An example related to the helix example is the helicoid surface S ⊂ R3 consisting of points of the form (s cos 2π t, s sin 2π t, t) for (s, t) ∈ (0, ∞)× R. This projects onto R2 − {0} via the map (x, y, z) ֏ (x, y), and this projection deﬁnes a covering space p : S→R2 − {0} since each point of R2 − {0} is contained in an open disk U in R2−{0} with p−1(U) consisting of countably many disjoint open disks in S projecting homeomorphically onto U. Another example is the map p : S 1→S 1, p(z) = zn where we view z as a complex number with \|z\| = 1 and n is any positive integer. The closest one can come to realizing this covering space as a linear projection in 3 space analogous to the projection of the helix is to draw a circle wrapping around a cylinder n times and intersecting itself in n − 1 points that one has to imagine are not really intersections. For an alternative picture without this defect,
embed S 1 in the boundary torus of a solid torus S 1 × D2 so that it winds n times Covering Spaces Section 1.3 57 monotonically around the S 1 factor without self-intersections, then restrict the projection S 1 × D2→S 1 × {0} to this embedded circle. The ﬁgure for Example 1.29 in the preceding section illustrates the case n = 3. These n sheeted covering spaces S 1→S 1 for n ≥ 1 together with the inﬁnitesheeted helix example exhaust all the connected coverings spaces of S 1, as our general theory will show. There are many other disconnected covering spaces of S 1, such as n disjoint circles each mapped homeomorphically onto S 1, but these disconnected covering spaces are just disjoint unions of connected ones. We will usually restrict our attention to connected covering spaces as these contain most of the interesting features of covering spaces. The covering spaces of S 1 ∨ S 1 form a remarkably rich family illustrating most of the general theory very concretely, so let us look at a few of these covering spaces to get an idea of what is going on. To abbreviate notation, set X = S 1 ∨ S 1. We view this as a graph with one vertex and two edges. We label the edges a and b and we choose orientations for a and b. Now let X be any other graph with four ends of edges at each vertex, as e in X, and suppose each edge of X has been assigned a label a or b and an orientation in such a way that the local picture near each vertex is the same as in X, so there is an a edge end oriented toward the vertex, an a edge end oriented away from the vertex, a b edge end oriented toward the vertex, and a b edge end oriented away from the vertex. To give a name to this structure, let us call X a 2 oriented graph. The table on the next page shows just a small sample of the inﬁnite variety of possible examples. Given a 2 oriented graph of X to the vertex of X and sending each edge of X we can construct a map p : X→X sending all vertices X to the edge of X with the same e e label by a map that is a homeomorphism on the interior of the edge and preserves orientation. It is clear that the covering space condition is satisﬁed for p.
Conversely, every covering space of X is a graph that inherits a 2 orientation from X. As the reader will discover by experimentation, it seems that every graph having four edge ends at each vertex can be 2 oriented. This can be proved for ﬁnite graphs as follows. A very classical and easily shown fact is that every ﬁnite connected graph with an even number of edge ends at each vertex has an Eulerian circuit, a loop traversing each edge exactly once. If there are four edge ends at each vertex, then labeling the edges of an Eulerian circuit alternately a and b produces a labeling with two a edge ends and two b edge ends at each vertex. The union of the a edges is then a collection of disjoint circles, as is the union of the b edges. Choosing orientations for all these circles gives a 2 orientation. It is a theorem in graph theory that inﬁnite graphs with four edge ends at each vertex can also be 2 oriented; see Chapter 13 of [K¨onig 1990] for a proof. There is also a generalization to n oriented graphs, which are covering spaces of the wedge sum of n circles. e e e e 58 Chapter 1 The Fundamental Group Covering Spaces Section 1.3 59 A simply-connected covering space of X = S 1 ∨ S 1 can be constructed in the following way. Start with the open intervals (−1, 1) in the coordinate axes of R2. Next, for a ﬁxed number λ, 0 < λ < 1/2, for example λ = 1/3, adjoin four open segments of length 2λ, at distance λ from the ends of the previous segments and perpendicular to them, the new shorter segments being bisected by the older ones. For the third stage, add perpendicular open segments of length 2λ2 at distance λ2 from the endpoints of all the previous segments and bisected by them. The process is now repeated indeﬁnitely, at the nth stage adding open segments of length 2λn−1 at distance λn−1 from all the previous endpoints. The union of all these open segments is a graph, with vertices the intersection points of horizontal and vertical segments, and edges the subsegments between adjacent vertices. We label all the horizontal edges a, oriented to the right, and all the vertical edges b, oriented upward. This covering space
is called the universal cover of X because, as our general theory will show, it is a covering space of every other connected covering space of X. The covering spaces (1)–(14) in the table are all nonsimply-connected. Their fun- e e words in a and b, starting at the basepoint damental groups are free with bases represented by the loops speciﬁed by the listed x0 indicated by the heavily shaded vertex. This can be proved in each case by applying van Kampen’s theorem. One can also interpret the list of words as generators of the image subgroup p∗ x0) in π1(X, x0) =. A general fact we shall prove about covering spaces is that x0)→π1(X, x0) is always injective. Thus we have the atthe induced map p∗ : π1( ﬁrst-glance paradoxical fact that the free group on two generators can contain as a π1( a, b X, X, e e e subgroup a free group on any ﬁnite number of generators, or even on a countably inﬁnite set of generators as in examples (10) and (11). Changing the basepoint vertex changes the subgroup p∗ to a conjugate subgroup in π1(X, x0). The conjugating element of π1(X, x0) is represented by X joining one basepoint to the other. For any loop that is the projection of a path in x0) π1( X, e e example, the covering spaces (3) and (4) diﬀer only in the choice of basepoints, and the corresponding subgroups of π1(X, x0) diﬀer by conjugation by b. e X, x0) π1( to the covering space p : The main classiﬁcation theorem for covering spaces says that by associating the X→X, we obtain a one-to-one subgroup p∗ correspondence between all the diﬀerent connected covering spaces of X and the conjugacy classes of subgroups of π1(X, x0). vertex x0 ∈ X, then this is a one-to-one correspondence between covering spaces x0
)→(X, x0) and actual subgroups of π1(X, x0), not just conjugacy classes. e Of course, for these statements to make sense one has to have a precise notion of If one keeps track of the basepoint p : ( X, e e e e when two covering spaces are the same, or ‘isomorphic’. In the case at hand, an iso- e e 60 Chapter 1 The Fundamental Group morphism between covering spaces of X is just a graph isomorphism that preserves the labeling and orientations of edges. Thus the covering spaces in (3) and (4) are isomorphic, but not by an isomorphism preserving basepoints, so the two subgroups of π1(X, x0) corresponding to these covering spaces are distinct but conjugate. On the other hand, the two covering spaces in (5) and (6) are not isomorphic, though the graphs are homeomorphic, so the corresponding subgroups of π1(X, x0) are isomorphic but not conjugate. Some of the covering spaces (1)–(14) are more symmetric than others, where by a ‘symmetry’ we mean an automorphism of the graph preserving the labeling and orientations. The most symmetric covering spaces are those having symmetries taking any one vertex onto any other. The examples (1), (2), (5)–(8), and (11) are the ones with this property. We shall see that a covering space of X has maximal symmetry exactly when the corresponding subgroup of π1(X, x0) is a normal subgroup, and in this case the symmetries form a group isomorphic to the quotient group of π1(X, x0) by the normal subgroup. Since every group generated by two elements is a quotient group of Z ∗ Z, this implies that every two-generator group is the symmetry group of some covering space of X. Lifting Properties e Covering spaces are deﬁned in fairly geometric terms, as maps p : X→X that are local homeomorphisms in a rather strong sense. But from the viewpoint of algebraic topology, the distinctive feature of covering spaces is their behavior with respect to lifting of maps. Recall the terminology from the proof of Theorem 1.7: A lift of a map
f : Y →X is a map properties of covering spaces and derive a few applications of these. f = f. We will describe three special lifting X such that p f : Y → e e e First we have the homotopy lifting property, also known as the covering homo- topy property: Proposition 1.30. Given a covering space p : map lifts ft. e f0 : Y → e e X lifting f0, then there exists a unique homotopy X→X, a homotopy ft : Y →X, and a f0 that ft : Y → X of e e e Proof: This was proved as property (c) in the proof of Theorem 1.7. ⊔⊓ X→X, which says that for each path f : I→X and each lift Taking Y to be a point gives the path lifting property for a covering space x0 of the starting p : f : I→ point f (0) = x0 there is a unique path x0. In particular, the uniqueness of lifts implies that every lift of a constant path is constant, but this could be deduced more simply from the fact that p−1(x0) has the discrete topology, by the deﬁnition of a covering space. X lifting f starting at e e e e e Covering Spaces Section 1.3 61 homotopy Taking Y to be I, we see that every homotopy ft of a path f0 in X lifts to a ft is a homotopy of paths, ft traces out a path lifting a f0 of f0. The lifted homotopy ﬁxing the endpoints, since as t varies each endpoint of e ft of each lift e e constant path, which must therefore be constant. e Here is a simple application: X, x0)→(X, x0) is injective. The image subgroup p∗ Proposition 1.31. The map p∗ : π1( p : ( consists of the homotopy classes of loops in X based at x0 whose lifts to at x0)→π1(X, x0) induced by a covering space in π1(X, x0) π1( X starting x0) X, X, e e e e e e x0 are loops. e Proof: An element of the kernel of p∗ is represented by a loop homotopy
ft : I→X of f0 = p proposition, there is a lifted homotopy of loops X with a f0 to the trivial loop f1. By the remarks preceding the ft starting with f0 and ending with x0) and p∗ is injective. e f0] = 0 in π1( a constant loop. Hence [ X, e e e e For the second statement of the proposition, loops at x0 lifting to loops at e x0 e x0)→π1(X, x0). Conversely, certainly represent elements of the image of p∗ : π1( e a loop representing an element of the image of p∗ is homotopic to a loop having such ⊔⊓ a lift, so by homotopy lifting, the loop itself must have such a lift. X, e e e e f0 : I→ Proposition 1.32. The number of sheets of a covering space p : ( X path-connected equals the index of p∗ with X and X, X, x0)→(X, x0) π1( g be its lift to x0) in π1(X, x0). e e e g ending at the same point as X starting at e x0. A product g from cosets H[g] to p−1(x0) e is surjective X implies that e e e Proof: For a loop g in X based at x0, let h g with [h] ∈ H = p∗ since π1( h is a loop. Thus we may deﬁne a function e has the lift e x0) X, h e by sending H[g] to e g(1). The path-connectedness of Φ e x0 can be joined to any point in p−1(x0) by a path e is injective, observe that (H[g1]) = e e since x0. To see that e lifts to a loop in x0, so [g1][g2]−1 ∈ H and hence H[g1] = H[g2]. Φ Φ g projecting to a loop g at Φ (H[g2]) implies that g1 g2 e ⊔⊓ X based at Φ e e It is important also to know about the existence and uniqueness of
lifts of general maps, not just lifts of homotopies. For the existence question an answer is provided by the following lifting criterion: Proposition 1.33. Suppose given a covering space p : ( x0)→(X, x0) and a map f : (Y, y0)→(X, x0) with Y path-connected and locally path-connected. Then a lift f : (Y, y0)→( π1(Y, y0) x0) of f exists iﬀ f∗ When we say a space has a certain property locally, such as being locally path- e e X, π1( x0) ⊂ p∗ X, X, e e e. e e connected, we usually mean that each point has arbitrarily small open neighborhoods with this property. Thus for Y to be locally path-connected means that for each point 62 Chapter 1 The Fundamental Group y ∈ Y and each neighborhood U of y there is an open neighborhood V ⊂ U of y that is path-connected. e g f (y) = x0. Deﬁne f γ starting at Proof: The ‘only if’ statement is obvious since f∗ = p∗ f∗. For the converse, let y ∈ Y and let γ be a path in Y from y0 to y. The path f γ in X starting at x0 f γ(1). To show this is wellhas a unique lift deﬁned, independent of the choice of γ, let γ′ be another path from y0 to y. Then e (f γ′) (f γ) is a loop h0 at x0 with [h0] ∈ f∗. This means there is a homotopy ht of h0 to a loop h1 that lifts to a loop property to ht to get a lifting x0, so is the ﬁrst half of e half is h0. By the uniqueness of lifted paths, x0. Apply the covering homotopy π1(Y, y0) f γ traversed backwards, with f γ′ and the second h1 is a loop at π1( x0) X based at ht. Since h1 in ⊂ p∗ h0 is g X the common midpoint f γ′
(1). This shows that g well-deﬁned. g g f γ(1) = f is g e e e U ⊂ To see that f (y) such that p : X containing e f is continuous, let U ⊂ X be an open neighborhood of f (y) having U→U is a homeomorphism. Choose a a lift path-connected open neighborhood V of y with f (V ) ⊂ U. For paths from y0 to points y ′ ∈ V we can take a ﬁxed path γ from y0 to y followed by paths η in V from y to the points y ′. Then the paths (f γ) (f η) in X have lifts ( f η) where f \|V = p−1f, hence g f η = p−1f η and p−1 : U→ U is the inverse of p : f is continuous at y. U→U. Thus U and g ⊔⊓ f (V ) ⊂ g f γ) ( e e e e e e e Without the local path-connectedness assumption on Y the lifting criterion in the e preceding proposition can fail, as shown by an example in Exercise 7 at the end of this section. Next we have the unique lifting property: Proposition 1.34. Given a covering space p : f2 : Y → f1, on all of Y. e e e e X of f agree at one point of Y and Y is connected, then X→X and a map f : Y →X, if two lifts f2 agree f1 and e e By continuity of U1 and f1 and Proof: For a point y ∈ Y, let U be an evenly covered open neighborhood of f (y) in X, so p−1(U) is decomposed into disjoint sheets each mapped homeomorphically f2(y), respectively. onto U by p. Let f1 e U2 are disjoint and e f1(y) = f2(y) then e U1 = U2. Thus the e ⊔⊓ f2 agree is both open and closed in Y. U1 and and into f1 ≠ f2 throughout the neighborhood N. On the other hand, if e U1 = U2 so e e set of points where e e U
2 be the sheets containing f2 there is a neighborhood N of y mapped into f2 and p is injective on e f1(y) ≠ e f2 on N since p f2(y) then e f2. If e f1(y) and U2, hence e f1 = f1 = p f1 and U1 by U2 by U1 ≠ Covering Spaces Section 1.3 63 The Classiﬁcation of Covering Spaces We consider next the problem of classifying all the diﬀerent covering spaces of a ﬁxed space X. Since the whole chapter is about paths, it should not be surprising that we will restrict attention to spaces X that are at least locally path-connected. Path-components of X are then the same as components, and for the purpose of clas- sifying the covering spaces of X there is no loss in assuming that X is connected, or equivalently, path-connected. Local path-connectedness is inherited by covering spaces, so connected covering spaces of X are the same as path-connected covering spaces. The main thrust of the classiﬁcation will be a correspondence between connected covering spaces of X and subgroups of π1(X). This is often called the Galois correspondence because of its surprising similarity to another basic correspondence in the purely algebraic subject of Galois theory. We will also describe a diﬀerent method of classiﬁcation that includes disconnected covering spaces as well. X, space p : ( x0)→(X, x0) the subgroup p∗ The Galois correspondence arises from the function that assigns to each covering of π1(X, x0). First we consider whether this function is surjective. That is, we ask whether every subgroup of e x0)→(X, x0). π1(X, x0) is realized as p∗ In particular we can ask whether the trivial subgroup is realized. Since p∗ is always injective, this amounts to asking whether X has a simply-connected covering space. for some covering space p : ( π1( x0) x0) π1( X, X, X, e e e e e e e Answering this will take some work. A necessary condition for X to have a simply-connected covering space is the following: Each point x
∈ X has a neighborhood U such that the inclusion-induced map π1(U, x)→π1(X, x) is trivial; one says X is semilocally simply-connected if X→X is a covering this holds. To see the necessity of this condition, suppose p : X simply-connected. Every point x ∈ X has a neighborhood U having a space with U ⊂ lift e e e e X projecting homeomorphically to U by p. Each loop in U lifts to a loop in U, and the lifted loop is nullhomotopic in X since π1( X) = 0. So, composing this nullhomotopy with p, the original loop in U is nullhomotopic in X. e e A locally simply-connected space is certainly semilocally simply-connected. For example, CW complexes have the much stronger property of being locally contractible, as we show in the Appendix. An example of a space that is not semilocally simplyconnected is the shrinking wedge of circles, the subspace X ⊂ R2 consisting of the circles of radius 1/n centered at the point (1/n, 0) for n = 1, 2, ···, introduced in Example 1.25. On the other hand, if we take the cone CX = (X × I)/(X × {0}) on the shrink- ing wedge of circles, this is semilocally simply-connected since it is contractible, but it is not locally simply-connected. We shall now show how to construct a simply-connected covering space of X if motivate the construction, suppose p : ( X is path-connected, locally path-connected, and semilocally simply-connected. To x0)→(X, x0) is a simply-connected coverx0 by a unique homotopy class of X can then be joined to ing space. Each point x ∈ X, e e e e e e 64 Chapter 1 The Fundamental Group starting at paths, by Proposition 1.6, so we can view points of X as homotopy classes of paths x0. The advantage of this is that, by the homotopy lifting property, homox0 are the same as homotopy classes of paths X starting at e topy classes of paths in e in X starting at x0. This gives a way of describing e e Given a path-connected, locally path-connected,
semilocally simply-connected X purely in terms of X. space X with a basepoint x0 ∈ X, we are therefore led to deﬁne X = [γ] \|\|\|\| γ is a path in X starting at x0 that ﬁx the endpoints γ(0) and γ(1). The function p : where, as usual, [γ] denotes the homotopy class of γ with respect to homotopies X→X sending [γ] to γ(1) is then well-deﬁned. Since X is path-connected, the endpoint γ(1) can be any point of e e e X, so p is surjective. Before we deﬁne a topology on X we make a few preliminary observations. Let U be the collection of path-connected open sets U ⊂ X such that π1(U)→π1(X) is trivial. Note that if the map π1(U)→π1(X) is trivial for one choice of basepoint in U, it is trivial for all choices of basepoint since U is path-connected. A path-connected open subset V ⊂ U ∈ U is also in U since the composition π1(V )→π1(U)→π1(X) will also be trivial. It follows that U is a basis for the topology on X if X is locally path-connected and semilocally simply-connected. e Given a set U ∈ U and a path γ in X from x0 to a point in U, let U[γ] = [γ η] \|\|\|\| η is a path in U with η(0) = γ(1) As the notation indicates, U[γ] depends only on the homotopy class [γ]. Observe that p : U[γ]→U is surjective since U is path-connected and injective since diﬀerent choices of η joining γ(1) to a ﬁxed x ∈ U are all homotopic in X, the map π1(U)→π1(X) being trivial. Another property is (∗) U[γ] = U[γ′] if [γ′] ∈ U[γ]. For if �
�′ = γ η then elements of U[γ′] have the form [γ η µ] and hence lie in U[γ], while elements of U[γ] have the form [γ µ] = [γ η η µ] = [γ′ η µ] and hence lie in U[γ′]. X. For if This can be used to show that the sets U[γ] form a basis for a topology on we are given two such sets U[γ], V[γ′] and an element [γ′′] ∈ U[γ] ∩ V[γ′], we have U[γ] = U[γ′′] and V[γ′] = V[γ′′] by (∗). So if W ∈ U is contained in U ∩ V and contains γ′′(1) then W[γ′′] ⊂ U[γ′′] ∩ V[γ′′] and [γ′′] ∈ W[γ′′]. e The bijection p : U[γ]→U is a homeomorphism since it gives a bijection between the subsets V[γ′] ⊂ U[γ] and the sets V ∈ U contained in U. Namely, in one direction we have p(V[γ′]) = V and in the other direction we have p−1(V ) ∩ U[γ] = V[γ′] for any [γ′] ∈ U[γ] with endpoint in V, since V[γ′] ⊂ U[γ′] = U[γ] and V[γ′] maps onto V by the bijection p. The preceding paragraph implies that p : X→X is continuous. We can also deduce that this is a covering space since for ﬁxed U ∈ U, the sets U[γ] for varying [γ] partition p−1(U) because if [γ′′] ∈ U[γ] ∩ U[γ′] then U[γ] = U[γ′′] = U[γ′] by (∗). e Covering Spaces Section 1.3 65 e It remains only to show that X is simply-connected. For a point [γ] ∈ e X is path-connected. To show that
π1( X let γt be the path in X that equals γ on [0, t] and is stationary at γ(t) on [t, 1]. Then the function t ֏ [γt] is a path in X lifting γ that starts at [x0], the homotopy class of X, this the constant path at x0, and ends at [γ]. Since [γ] was an arbitrary point in X, [x0]) = 0 it suﬃces to show shows that that the image of this group under p∗ is trivial since p∗ is injective. Elements in the X at [x0]. We have image of p∗ are represented by loops γ at x0 that lift to loops in observed that the path t ֏ [γt] lifts γ starting at [x0], and for this lifted path to be a loop means that [γ1] = [x0]. Since γ1 = γ, this says that [γ] = [x0], so γ is nullhomotopic and the image of p∗ is trivial. e e e e e This completes the construction of a simply-connected covering space X→X. In concrete cases one usually constructs a simply-connected covering space by e which simply-connected covering spaces one can attempt to build a simply-connected covering space more direct methods. For example, suppose X is the union of subspaces A and B for B→B are already known. Then X→X by assembling B. For example, for X = S 1 ∨ S 1, if we take A and B to be the two B are each R, and we can build the simply-connected cover X A→A and copies of A and A and e e e circles, then e e described earlier in this section by glueing together inﬁnitely many copies of A and e X. Here is another illustration of this method: B, the horizontal and vertical lines in e e e e e Example 1.35. For integers m, n ≥ 2, let Xm,n be the quotient space of a cylinder S 1 × I under the identiﬁcations (z, 0) ∼ (e2π i/mz, 0) and (z, 1) ∼ (e2π i/nz, 1). Let A
⊂ X and B ⊂ X be the quotients of S 1 × [0, 1/2] and S 1 × [1/2, 1], so A and B are the mapping cylinders of z ֏ zm and z ֏ zn, with A ∩ B = S 1. The simplest case is m = n = 2, when A and B are M¨obius bands and X2,2 is the Klein bottle. We encountered the complexes Xm,n previously in analyzing torus knot complements in Example 1.24. The ﬁgure for Example 1.29 at the end of the preceding section shows what A looks like in the typical case m = 3. We have π1(A) ≈ Z, A is homeomorphic to a product Cm × R where and the universal cover Cm is the graph that is a cone on m points, as shown in the ﬁgure to B is homeomorphic to the right. The situation for B is similar, and Cn × R. Now we attempt to build the universal cover Xm,n from copies of A. Its boundary, the outer edges of its ﬁns, consists of m copies of R. Along each of these m boundary e B. Start with a copy of A and e e e e e lines we attach a copy of B. Each of these copies of B has one of its boundary lines attached to the initial copy of A, leaving n − 1 boundary lines free, and we attach a A to each of these free boundary lines. Thus we now have m(n − 1) + 1 e e new copy of copies of A. Each of the newly attached copies of and to each of these lines we attach a new copy of e e e A has m − 1 free boundary lines, B. The process is now repeated ad e e 66 Chapter 1 The Fundamental Group inﬁnitum in the evident way. Let Xm,n be the resulting space. e e give A = Cm × R and B = Cn × R The product structures Xm,n the structure of a product Tm,n × R where Tm,n e is an inﬁnite graph constructed by an inductive scheme Xm,n. Thus Tm,n is the union of a sequence of ﬁnite subgraphs, each obtained from the e preceding by attaching new copies of
Cm or Cn. Each of these ﬁnite subgraphs deformation retracts onto the just like the construction of e preceding one. The inﬁnite concatenation of these deformation retractions, with the k th graph deformation retracting to the previous one during the time interval [1/2k, 1/2k−1], gives a deformation retraction of Tm,n onto the initial stage Cm. Since Cm is contractible, this means Tm,n is contractible, hence also Xm,n is simply-connected. Xm,n, which is the product Tm,n × R. In particular, The map that projects each copy of Xm,n to A and e each copy of B to B is a covering space. To deﬁne this map precisely, choose a point x0 ∈ S 1, and then the image of the line segment {x0}× I in Xm,n meets A in a line segment whose A consists of an inﬁnite number of line segments, preimage in A in e e e e appearing in the earlier ﬁgure as the horizontal segments spi- raling around the central vertical axis. The picture in B is e similar, and when we glue together all the copies of A and e B to form exactly. This decomposes Xm,n, we do so in such a way that these horizontal segments always line up Xm,n into inﬁnitely many rectangles, each formed from a Xm,n→Xm,n is the B. The covering projection A and a rectangle in a e e e rectangle in an e quotient map that identiﬁes all these rectangles. e e e Now we return to the general theory. The hypotheses for constructing a simply- connected covering space of X in fact suﬃce for constructing covering spaces realizing arbitrary subgroups of π1(X) : Proposition 1.36. Suppose X is path-connected, locally path-connected, and semilocally simply-connected. Then for every subgroup H ⊂ π1(X, x0) there is a covering space p : XH→X such that p∗ = H for a suitably chosen basepoint x0) x0 ∈ XH. π1(XH, e e Proof:
For points [γ], [γ′] in the simply-connected covering space X constructed above, deﬁne [γ] ∼ [γ′] to mean γ(1) = γ′(1) and [γ γ′] ∈ H. It is easy to see that this is an equivalence relation since H is a subgroup: it is reﬂexive since H contains e the identity element, symmetric since H is closed under inverses, and transitive since X obtained by H is closed under multiplication. Let XH be the quotient space of identifying [γ] with [γ′] if [γ] ∼ [γ′]. Note that if γ(1) = γ′(1), then [γ] ∼ [γ′] iﬀ [γ η] ∼ [γ′ η]. This means that if any two points in basic neighborhoods U[γ] e Covering Spaces Section 1.3 67 and U[γ′] are identiﬁed in XH then the whole neighborhoods are identiﬁed. Hence the natural projection XH→X induced by [γ] ֏ γ(1) is a covering space. If we choose for the basepoint c at x0, then the image of p∗ : π1(XH, for a loop γ in X based at x0, its lift to e of this lifted path in XH is a loop iﬀ [γ] ∼ [c], or equivalently, [γ] ∈ H. x0 ∈ XH the equivalence class of the constant path x0)→π1(X, x0) is exactly H. This is because X starting at [c] ends at [γ], so the image ⊔⊓ e e Having taken care of the existence of covering spaces of X corresponding to all subgroups of π1(X), we turn now to the question of uniqueness. More speciﬁcally, we are interested in uniqueness up to isomorphism, where an isomorphism between covering spaces p1 : X2 such that p1 = p2f. This condition means exactly that f preserves the covering space e 2 (x) for each x ∈ X. The inverse f −1 is then also an 1 (x) to p
−1 structures, taking p−1 isomorphism, and the composition of two isomorphisms is an isomorphism, so we X2→X is a homeomorphism f : X1→X and p2 : X1→ e e e have an equivalence relation. e e e X2, = p2∗ e x2) 1 (x0) to a basepoint X1→ x1) X1, e e X1→X and p2 : x1 ∈ p−1 X2 taking a basepoint. π1( Proposition 1.37. If X is path-connected and locally path-connected, then two pathX2→X are isomorphic via an isomorconnected covering spaces p1 : phism f : 2 (x0) iﬀ π1( p1∗ Proof: If there is an isomorphism f : ( p1 = p2f and p2 = p1f −1 it follows that p1∗ e = p2∗ versely, suppose that p1∗ we may lift p1 to a map X2, x2)→( obtain X1, X2, p2 = 11 and p1 p2 e e e p2 are inverse isomorphisms. e e = p2∗ X2,. By the lifting criterion, x1) X1, π1( e x1)→( p1 = p1. Symmetrically, we p1 : ( X1, e e p2 = p2. Then by the unique lifting property, x1) with p1 e e p1 and ⊔⊓ e x2), then from the two relations. Conπ1( p1 = 11 since these composed lifts ﬁx the basepoints. Thus x1) x2) e x2) with p2 e X2, π1( e π1( x2 ∈ p−1 x2) x1)→( X1, e X2, e e X1, p2 : ( We have proved the ﬁrst half of the following classiﬁcation theorem: Theorem 1.38. Let X be path-connected, locally path-connected, and semilocally simply-connected. Then there is a bijection between the set of basepoint-preserving x
0)→(X, x0) and the x0) x0). If basepoints are ignored, this correspondence gives a e e X→X isomorphism classes of path-connected covering spaces p : ( set of subgroups of π1(X, x0), obtained by associating the subgroup p∗ to the covering space ( bijection between isomorphism classes of path-connected covering spaces p : e and conjugacy classes of subgroups of π1(X, x0). π1( X, X, X, e e e e X, Proof: It remains only to prove the last statement. We show that for a covering space x0 within p−1(x0) corresponds exactly p : ( to changing p∗ x1 X, e is another basepoint in p−1(x0), and let γ projects e x0)→(X, x0), changing the basepoint x0) to a conjugate subgroup of π1(X, x0). Suppose that γ be a path from x1. Then x0 to π1( e e e e e e e e 68 Chapter 1 The Fundamental Group to a loop γ in X representing some element g ∈ π1(X, x0). Set Hi = p∗ for i = 0, 1. We have an inclusion g−1H0g ⊂ H1 since for x0, a loop at gives H1 ⊂ g−1H0g, so g−1H0g = H1. Thus, changing the basepoint from changes H0 to the conjugate subgroup H1 = g−1H0g. xi) γ is e x1. Similarly we have gH1g−1 ⊂ H0. Conjugating the latter relation by g−1 x1 π1( γ f e e e e x0 to e Conversely, to change H0 to a conjugate subgroup H1 = g−1H0g, choose a loop γ(1). The preceding γ representing g, lift this to a path argument then shows that we have the desired relation H1 = g−1H0g. e x0, and let γ starting at f a loop at x1 = X, ⊔�
�� e e e e e e e A consequence of the lifting criterion is that a simply-connected covering space of a path-connected, locally path-connected space X is a covering space of every other path-connected covering space of X. A simply-connected covering space of X is therefore called a universal cover. It is unique up to isomorphism, so one is justiﬁed in calling it the universal cover. More generally, there is a partial ordering on the various path-connected covering spaces of X, according to which ones cover which others. This corresponds to the partial ordering by inclusion of the corresponding subgroups of π1(X), or conjugacy classes of subgroups if basepoints are ignored. Representing Covering Spaces by Permutations We wish to describe now another way of classifying the diﬀerent covering spaces of a connected, locally path-connected, semilocally simply-connected space X, with- e X2, and out restricting just to connected covering spaces. To give the idea, consider the 3 sheeted covering spaces of S 1. There are three of these, X3, with the subscript indicating the number of compoX1, Xi→S 1 the three diﬀerent nents. For each of these covering spaces p : e lifts of a loop in S 1 generating π1(S 1, x0) determine a permutation of p−1(x0) sending the starting point of the lift to the ending point of the X2 it is a transposition of lift. For X3 it is the identity permutation. These permutations obviously determine the covering spaces X1 this is a cyclic permutation, for two points ﬁxing the third point, and for e e e e uniquely, up to isomorphism. The same would be true for n sheeted covering spaces of S 1 for arbitrary n, even for n inﬁnite. e The covering spaces of S 1 ∨ S 1 can be encoded using the same idea. Referring back to the large table of examples near the beginning of this section, we see in the covering space (1) that the loop a lifts to the identity permutation of the two vertices and b lifts to the permutation that transposes the two vertices. In (2), both a and b lift to transpositions of the two vertices. In (3) and (4), a
and b lift to transpositions of diﬀerent pairs of the three vertices, while in (5) and (6) they lift to cyclic permutations of the vertices. In (11) the vertices can be labeled by Z, with a lifting to the identity permutation and b lifting to the shift n ֏ n + 1. Indeed, one can see from these Covering Spaces Section 1.3 69 examples that a covering space of S 1 ∨ S 1 is nothing more than an eﬃcient graphical representation of a pair of permutations of a given set. e γ(0) of each lift e γ to its ending point This idea of lifting loops to permutations generalizes to arbitrary covering spaces. X→X, a path γ in X has a unique lift For a covering space p : γ starting at a given point of p−1(γ(0)), so we obtain a well-deﬁned map Lγ : p−1(γ(0))→p−1(γ(1)) by e sending the starting point γ(1). It is evident that Lγ is a bijection since Lγ is its inverse. For a composition of paths γ η we have Lγ·η = LηLγ, rather than LγLη, since composition of paths is written from left to right while composition of functions is written from right to left. To compensate for this, let us modify the deﬁnition by replacing Lγ by its inverse. Thus the new Lγ is a bijection p−1(γ(1))→p−1(γ(0)), and Lγ·η = LγLη. Since Lγ depends only on the homotopy class of γ, this means that if we restrict attention to loops at a basepoint x0 ∈ X, then the association γ ֏ Lγ gives a homomorphism from π1(X, x0) to the group of permutations of p−1(x0). This is called the action of π1(X, x0) on the ﬁber p−1(x0). e e Let us see how the covering space p : X→X can be reconstructed from the associated action of π1(X, x0) on the ﬁber
F = p−1(x0), assuming that X is path-connected, e locally path-connected, and semilocally simply-connected, so it has a universal cover X0→X. We can take the points of X0 to be homotopy classes of paths in X starting X0× F→ at x0, as in the general construction of a universal cover. Deﬁne a map h : X e e x0. Then h sending a pair ([γ], e is continuous, and in fact a local homeomorphism, since a neighborhood of ([γ], x0) e e x0) with η a path in a suitable neighborhood of γ(1). It is obvious that h is surjective since X is path-connected. If h were injec- e X0 × F consists of the pairs ([γ η], γ is the lift of γ to X starting at γ(1) where x0) to in e e e e e e tive as well, it would be a homeomorphism, which is unlikely since X is probably not X0 × F. Even if h is not injective, it will induce a homeomorphism X. To see what this quotient space is, e homeomorphic to e e x′ from some quotient space of x0) = h([γ′], X0 × F onto 0). Then γ and γ′ are both x′ suppose h([γ], e e paths from x0 to the same endpoint, and from the ﬁgure e x0). Letting λ be the loop γ′ γ, this we see that x0)). Conversely, for x0)). Thus h X from the quotient space of x0) with ([λ γ], Lλ( x0)) e any loop λ we have h([γ], x0) = h([λ γ], Lλ( x0) = h([λ γ], Lλ( means that h([γ], 0 = Lγ′·γ( e e e e induces a well-deﬁned map to X0 × F obtained by identifying ([γ], e for each [λ] ∈ π1(X, x0). Let this quotient space be denoted Xρ where ρ is the hoe momorphism from π1(X, x0
) to the permutation group of F speciﬁed by the action. e e e e Notice that the deﬁnition of Xρ makes sense whenever we are given an action ρ of π1(X, x0) on a set F regarded as a space with the discrete topology. There is a x0) to γ(1), and this is a covering space since natural projection X0 is a product U × π1(X, x0), if U ⊂ X is an open set over which the universal cover Xρ→X sending ([γ], e e e e 70 Chapter 1 The Fundamental Group then the identiﬁcations deﬁning Xρ simply collapse U × π1(X, x0)× F to U × F. Returning to our given covering space X→X with associated action ρ, the map X induced by h is a bijection and therefore a homeomorphism since h was a e Xρ to Xρ→ local homeomorphism. Since this homeomorphism e the corresponding ﬁber of X, it is an isomorphism of covering spaces. X takes each ﬁber of Xρ→ e e e e e e X1→X and p2 : If two covering spaces p1 : X2→X are isomorphic, one may ask how the corresponding actions of π1(X, x0) on the ﬁbers F1 and F2 over x0 are X2 restricts to a bijection F1→F2, and evidently related. An isomorphism h : x0), this x0 for Lγ( Lγ(h( x0). A bijection F1→F2 with relation can be written more concisely as γh( this property is what one would naturally call an isomorphism of sets with π1(X, x0) action. Thus isomorphic covering spaces have isomorphic actions on ﬁbers. The x0)). Using the less cumbersome notation γ x0)) = h(Lγ( x0) = h(γ e X1 converse is also true, and easy to prove. One just observes that for isomorphic actions ρ1 and ρ2, an isomorphism h : F1→F2 induces a map and h−1 induces a similar map in the opposite direction,
such that the compositions of these two maps, Xρ1→ Xρ2 e e in either order, are the identity. This shows that n sheeted covering spaces of X are classiﬁed by equivalence classes of homomorphisms π1(X, x0)→ n is the symmetric group on n symbols and the equivalence relation identiﬁes a homomorphism ρ with each of its Σ conjugates h−1ρh by elements h ∈ n. The study of the various homomorphisms n is a very classical topic in group theory, so we see that this from a given group to n, where Σ algebraic question has a nice geometric interpretation. Σ Σ Deck Transformations and Group Actions For a covering space p : X→X the isomorphisms X→ X are called deck transfor- e X) under composition. mations or covering transformations. These form a group G( For example, for the covering space p : R→S 1 projecting a vertical helix onto a circle, the deck transformations are the vertical translations taking the helix onto itself, so X) ≈ Z in this case. For the n sheeted covering space S 1→S 1, z ֏ zn, the deck G( transformations are the rotations of S 1 through angles that are multiples of 2π /n, e so G( e e e X) = Zn. By the unique lifting property, a deck transformation is completely determined e by where it sends a single point, assuming X is path-connected. In particular, only the identity deck transformation can ﬁx a point of X. e A covering space p : x′ of x there is a deck transformation taking e X→X is called normal if for each x ∈ X and each pair of lifts x′. For example, the covering x, space R→S 1 and the n sheeted covering spaces S 1→S 1 are normal. Intuitively, a e normal covering space is one with maximal symmetry. This can be seen in the covering spaces of S 1 ∨ S 1 shown in the table earlier in this section, where the normal covering e x to e e e Covering Spaces Section 1.3 71 spaces are (1), (2), (5)–(8), and (11). Note that in (7) the group of deck transformations is Z4 while in (8) it is
Z2 × Z2. Sometimes normal covering spaces are called regular covering spaces. The term ‘normal’ is motivated by the following result. X, Proposition 1.39. Let p : ( x0)→(X, x0) be a path-connected covering space of the path-connected, locally path-connected space X, and let H be the subgroup p∗ π1( (a) This covering space is normal iﬀ H is a normal subgroup of π1(X, x0). (b) G( e X) is isomorphic to the quotient N(H)/H where N(H) is the normalizer of ⊂ π1(X, x0). Then : x0) X, e e e H in π1(X, x0). e In particular, G( for the universal cover X) is isomorphic to π1(X, x0)/H if X→X we have G( X) ≈ π1(X). e e X is a normal covering. Hence e e x0 ∈ p−1(x0) to Proof: We observed earlier in the proof of the classiﬁcation theorem that changing x1 ∈ p−1(x0) corresponds precisely to conjugating the basepoint H by an element [γ] ∈ π1(X, x0) where γ lifts to a path x1. Thus [γ] e, which by the lifting is in the normalizer N(H) iﬀ p∗ π1( π1( e criterion is equivalent to the existence of a deck transformation taking x1. e Hence the covering space is normal iﬀ N(H) = π1(X, x0), that is, iﬀ H is a normal subgroup of π1(X, x0). γ from x1) x0) X, e e x0 to x0 to = p∗ X, e e e e e e Deﬁne ϕ : N(H)→G( X) sending [γ] to the deck transformation τ taking x0 to x1, in the notation above. Then ϕ is a homomorphism, for if γ′ is another loop corree γ′)), sponding to the
deck transformation τ ′ taking γ (τ( e x0), so ττ ′ is the deck transformation corresponding a path from to [γ][γ′]. By the preceding paragraph ϕ is surjective. Its kernel consists of classes e = H. ⊔⊓ X. These are exactly the elements of p∗ [γ] lifting to loops in 1 then γ γ′ lifts to x′ 1) = ττ ′( x′ x0 to τ( x0 to X, e e e e e e e The group of deck transformations is a special case of the general notion of e ‘groups acting on spaces’. Given a group G and a space Y, then an action of G on Y is a homomorphism ρ from G to the group Homeo(Y ) of all homeomorphisms from Y to itself. Thus to each g ∈ G is associated a homeomorphism ρ(g) : Y →Y, which for notational simplicity we write simply as g : Y →Y. For ρ to be a homomorphism amounts to requiring that g1(g2(y)) = (g1g2)(y) for all g1, g2 ∈ G and y ∈ Y. If ρ is injective then it identiﬁes G with a subgroup of Homeo(Y ), and in practice not much is lost in assuming ρ is an inclusion G ֓ Homeo(Y ) since in any case the subgroup ρ(G) ⊂ Homeo(Y ) contains all the topological information about the action. π1( x0) e e 72 Chapter 1 The Fundamental Group We shall be interested in actions satisfying the following condition: (∗) Each y ∈ Y has a neighborhood U such that all the images g(U) for varying g ∈ G are disjoint. In other words, g1(U) ∩ g2(U) ≠ ∅ implies g1 = g2. X), then g1( The action of the deck transformation group G( X) on X project homeomorphically to U ⊂ X. e x1) = g2( x1, U ⊂ let g1, g2 ∈ G( e e in the same set p−1(x), which intersects e 1 g2
ﬁxes this point, so g−1 Then g−1 1 g2 = 11 and g1 = g2. Note that in (∗) it suﬃces to take g1 to be the identity since g1(U) ∩ g2(U) ≠ ∅ 1 g2(U) ≠ ∅. Thus we have the equivalent condition that e U in only one point, we must have e U ) ∩ g2( If g1( e x2 ∈ U. Since X satisﬁes (∗). To see this, U ) ≠ ∅ for some x2 must lie x1 and x1 = x2. x2) for some is equivalent to U ∩ g−1 U ∩ g(U) ≠ ∅ only when g is the identity Given an action of a group G on a space Y, we can form a space Y /G, the quotient space of Y in which each point y is identiﬁed with all its images g(y) as g ranges over G. The points of Y /G are thus the orbits Gy = { g(y) \| g ∈ G } in Y, and Y /G is called the orbit space of the action. For example, for a normal covering space X→X, the orbit space X) is just X. X/G( e e Proposition 1.40. If an action of a group G on a space Y satisﬁes (∗), then : (a) The quotient map p : Y →Y /G, p(y) = Gy, is a normal covering space. (b) G is the group of deck transformations of this covering space Y →Y /G if Y is e path-connected. (c) G is isomorphic to π1(Y /G)/p∗ connected. π1(Y ) if Y is path-connected and locally path- Proof: Given an open set U ⊂ Y as in condition (∗), the quotient map p simply identiﬁes all the disjoint homeomorphic sets { g(U) \| g ∈ G } to a single open set p(U) in Y /G. By the deﬁnition of the quotient topology on Y /G, p restricts to a homeomorphism
from g(U) onto p(U) for each g ∈ G so we have a covering space. Each element of G acts as a deck transformation, and the covering space is normal since g2g−1 takes g1(U) to g2(U). The deck transformation group contains G as a subgroup, and equals this subgroup if Y is path-connected, since if f is any deck transformation, then for an arbitrarily chosen point y ∈ Y, y and f (y) are 1 in the same orbit and there is a g ∈ G with g(y) = f (y), hence f = g since deck transformations of a path-connected covering space are uniquely determined by where they send a point. The ﬁnal statement of the proposition is immediate from part (b) of Proposition 1.39. ⊔⊓ In view of the preceding proposition, we shall call an action satisfying (∗) a covering space action. This is not standard terminology, but there does not seem to be a universally accepted name for actions satisfying (∗). Sometimes these are called ‘properly discontinuous’ actions, but more often this rather unattractive term means Covering Spaces Section 1.3 73 something weaker: Every point x ∈ X has a neighborhood U such that U ∩ g(U) is nonempty for only ﬁnitely many g ∈ G. Many symmetry groups have this proper discontinuity property without satisfying (∗), for example the group of symmetries of the familiar tiling of R2 by regular hexagons. The reason why the action of this group on R2 fails to satisfy (∗) is that there are ﬁxed points: points y for which there is a nontrivial element g ∈ G with g(y) = y. For example, the vertices of the hexagons are ﬁxed by the 120 degree rotations about these points, and the midpoints of edges are ﬁxed by 180 degree rotations. An action without ﬁxed points is called a free action. Thus for a free action of G on Y, only the identity element of G ﬁxes any point of Y. This is equivalent to requiring that all the images g(y) of each y ∈ Y are distinct, or in other words g1(y) = g2(y)
only when g1 = g2, since g1(y) = g2(y) is equivalent to g−1 1 g2(y) = y. Though condition (∗) implies freeness, the converse is not always true. An example is the action of Z on S 1 in which a generator of Z acts by rotation through an angle α that is an irrational multiple of 2π. In this case each orbit Zy is dense in S 1, so condition (∗) cannot hold since it implies that orbits are discrete subspaces. An exercise at the end of the section is to show that for actions on Hausdorﬀ spaces, freeness plus proper discontinuity implies condition (∗). Note that proper discontinuity is automatic for actions by a ﬁnite group. Example 1.41. Let Y be the closed orientable surface of genus 11, an ‘11 hole torus’ as shown in the ﬁgure. This has a 5 fold rotational symme- try, generated by a rotation of angle 2π /5. Thus we have the cyclic group Z5 acting on Y, and the condition (∗) is obviously satisﬁed. The quotient space Y /Z5 is a surface of genus 3, obtained from one of the ﬁve subsurfaces of Y cut oﬀ by the circles C1, ···, C5 by identifying its two boundary circles Ci and Ci+1 to form the circle C as shown. Thus we have a covering space M11→M3 where Mg denotes the closed orientable surface of genus g. In particular, we see that π1(M3) contains the ‘larger’ group π1(M11) as a normal subgroup of index 5, with quotient Z5. This example obviously generalizes by replacing the two holes in each ‘arm’ of M11 by m holes and the 5 fold symmetry by n fold symmetry. This gives a covering space Mmn+1→Mm+1. An exercise in §2.2 is to show by an Euler characteristic argument that if there is a covering space Mg→Mh then g = mn + 1 and h = m + 1 for some m and n. As a special case of the ﬁnal statement of the preceding proposition we see that for a covering space action
of a group G on a simply-connected locally path-connected space Y, the orbit space Y /G has fundamental group isomorphic to G. Under this isomorphism an element g ∈ G corresponds to a loop in Y /G that is the projection of 74 Chapter 1 The Fundamental Group a path in Y from a chosen basepoint y0 to g(y0). Any two such paths are homotopic since Y is simply-connected, so we get a well-deﬁned element of π1(Y /G) associated to g. This method for computing fundamental groups via group actions on simplyconnected spaces is essentially how we computed π1(S 1) in §1.1, via the covering space R→S 1 arising from the action of Z on R by translations. This is a useful general technique for computing fundamental groups, in fact. Here are some examples illustrating this idea. Example 1.42. Consider the grid in R2 formed by the horizontal and vertical lines through points in Z2. Let us decorate this grid with arrows in either of the two ways shown in the ﬁgure, the diﬀerence between the two cases being that in the second case the horizontal arrows in adjacent lines point in opposite direc- tions. The group G consisting of all symmetries of the ﬁrst decorated grid is isomorphic to Z× Z since it consists of all translations (x, y) ֏ (x + m, y + n) for m, n ∈ Z. For the second grid the symmetry group G contains a subgroup of translations of the form (x, y) ֏ (x + m, y + 2n) for m, n ∈ Z, but there are also glide-reﬂection symmetries consisting of vertical translation by an odd integer distance followed by reﬂection across a vertical line, either a vertical line of the grid or a vertical line halfway between two adjacent grid lines. For both decorated grids there are elements of G taking any square to any other, but only the identity element of G takes a square to itself. The minimum distance any point is moved by a nontrivial element of G is 1, which easily implies the covering space condition (∗). The orbit space R2/G is the quotient space of a square in the grid with opposite edges identiﬁed according to the arrows
. Thus we see that the fundamental groups of the torus and the Klein bottle are the symme- try groups G in the two cases. In the second case the subgroup of G formed by the translations has index two, and the orbit space for this subgroup is a torus forming a two-sheeted covering space of the Klein bottle. Example 1.43: RPn. The antipodal map of S n, x ֏ −x, generates an action of Z2 on S n with orbit space RPn, real projective n space, as deﬁned in Example 0.4. The action is a covering space action since each open hemisphere in S n is disjoint from its antipodal image. As we saw in Proposition 1.14, S n is simply-connected if n ≥ 2, so from the covering space S n→RPn we deduce that π1(RPn) ≈ Z2 for n ≥ 2. A generator for π1(RPn) is any loop obtained by projecting a path in S n connecting two antipodal points. One can see explicitly that such a loop γ has order two in π1(RPn) if n ≥ 2 since the composition γ γ lifts to a loop in S n, and this can be homotoped to the trivial loop since π1(S n) = 0, so the projection of this homotopy into RPn gives a nullhomotopy of γ γ. Covering Spaces Section 1.3 75 One may ask whether there are other ﬁnite groups that act freely on S n, deﬁning covering spaces S n→S n/G. We will show in Proposition 2.29 that Z2 is the only possibility when n is even, but for odd n the question is much more diﬃcult. It is easy to construct a free action of any cyclic group Zm on S 2k−1, the action generated by the rotation v ֏ e2π i/mv of the unit sphere S 2k−1 in Ck = R2k. This action is free since an equation v = e2π iℓ/mv with 0 < ℓ < m implies v = 0, but 0 is not a point of S 2k−1. The orbit space S 2k−1/Zm is one of a family
of spaces called lens spaces deﬁned in Example 2.43. There are also noncyclic ﬁnite groups that act freely as rotations of S n for odd n > 1. These actions are classiﬁed quite explicitly in [Wolf 1984]. Examples in the simplest case n = 3 can be produced as follows. View R4 as the quaternion algebra H. Multiplication of quaternions satisﬁes \|ab\| = \|a\|\|b\| where \|a\| denotes the usual Euclidean length of a vector a ∈ R4. Thus if a and b are unit vectors, so is ab, and hence quaternion multiplication deﬁnes a map S 3 × S 3→S 3. This in fact makes S 3 into a group, though associativity is all we need now since associativity implies that any subgroup G of S 3 acts on S 3 by left-multiplication, g(x) = gx. This action is free since an equation x = gx in the division algebra H implies g = 1 or x = 0. As a concrete example, G could be the familiar quaternion group Q8 = {±1, ±i, ±j, ±k} from group theory. More generally, for a positive integer m, let Q4m be the subgroup of S 3 generated by the two quaternions a = eπ i/m and b = j. Thus a has order 2m and b has order 4. The easily veriﬁed relations am = b2 = −1 and bab−1 = a−1 imply that the subgroup Z2m generated by a is normal and of index 2 in Q4m. Hence Q4m is a group of order 4m, called the generalized quaternion group. Another common name for this group is the binary dihedral group D∗ 4m since its quotient by the subgroup {±1} is the ordinary dihedral group D2m of order 2m. Besides the groups Q4m = D∗ 4m there are just three other noncyclic ﬁnite sub24, O∗ groups of S 3 : the binary tetrahedral, octahedral, and icosahedral groups T ∗ 48, and I∗ 120, of orders indicated by the subscripts. These project two-to-one onto the groups of rotational
symmetries of a regular tetrahedron, octahedron (or cube), and icosahedron (or dodecahedron). In fact, it is not hard to see that the homomorphism S 3→SO(3) sending u ∈ S 3 ⊂ H to the isometry v→u−1vu of R3, viewing R3 as the ‘pure imaginary’ quaternions v = ai + bj + ck, is surjective with kernel {±1}. Then 48, I∗ the groups D∗ 120 are the preimages in S 3 of the groups of rotational symmetries of a regular polygon or polyhedron in R3. 4m, T ∗ 24, O∗ There are two conditions that a ﬁnite group G acting freely on S n must satisfy: (a) Every abelian subgroup of G is cyclic. This is equivalent to saying that G contains no subgroup Zp × Zp with p prime. (b) G contains at most one element of order 2. A proof of (a) is sketched in an exercise for §4.2. For a proof of (b) the original source [Milnor 1957] is recommended reading. The groups satisfying (a) have been 76 Chapter 1 The Fundamental Group completely classiﬁed; see [Brown 1982], section VI.9, for details. An example of a group satisfying (a) but not (b) is the dihedral group D2m for odd m > 1. There is also a much more diﬃcult converse: A ﬁnite group satisfying (a) and (b) acts freely on S n for some n. References for this are [Madsen, Thomas, & Wall 1976] and [Davis & Milgram 1985]. There is also almost complete information about which n ’s are possible for a given group. e In Example 1.35 we constructed a contractible 2 complex Example 1.44. Xm,n = Tm,n × R as the universal cover of a ﬁnite 2 complex Xm,n that was the union of the mapping cylinders of the two maps S 1→S 1, z ֏ zm and z ֏ zn. The group of deck transformations of this covering space is therefore the fundamental group π1(Xm
,n). From van Kampen’s theorem applied to the decomposition of Xm,n into the two mapping cylinders we have the presentation for this group Gm,n = π1(Xm,n). It is interesting to look at the action of Gm,n on Xm,n more closely. Xm,n into rectangles, with Xm,n the quotient of We described a decomposition of e Xm,n lifting a cell one rectangle. These rectangles in fact deﬁne a cell structure on structure on Xm,n with two vertices, three edges, and one 2 cell. The group Gm,n is e thus a group of symmetries of this cell structure on Xm,n. If we orient the three edges Xm,n, then Gm,n is the group of all of Xm,n and lift these orientations to the edges of symmetries of Xm,n preserving the orientations of edges. For example, the element a acts as a ‘screw motion’ about an axis that is a vertical line {va}× R with va a vertex of Tm,n, and b acts similarly for a vertex vb. a, b \|\|\|\| amb−n e e e e Since the action of Gm,n on Xm,n preserves the cell structure, it also preserves the product structure Tm,n × R. This means that there are actions of Gm,n on Tm,n and R such that the action on the product Xm,n = Tm,n × R is the diagonal action g(x, y) = for g ∈ Gm,n. If we make the rectangles of unit height in the R coordinate, then the element am = bn acts on R as unit translation, while a acts by 1/m translation and b by 1/n translation. The translation actions of a and b on R generate a group of translations of R that is inﬁnite cyclic, generated by translation g(x), g(y) e by the reciprocal of the least common multiple of m and n. The action of Gm,n on Tm,n has kernel consisting of the powers of the element am = bn. This inﬁnite cyclic subgroup is precisely the center of Gm,n, as we saw in Example 1.24
. There is an induced action of the quotient group Zm ∗ Zn on Tm,n, but this is not a free action since the elements a and b and all their conjugates ﬁx vertices of Tm,n. On the other hand, if we restrict the action of Gm,n on Tm,n to the kernel K of the map Gm,n→Z given by the action of Gm,n on the R factor of Xm,n, then we do obtain a free action of K on Tm,n. Since this action takes vertices to vertices and edges to edges, it is a covering space action, so K is a free group, the fundamental group of the graph Tm,n/K. An exercise at the end of the section is to determine Tm,n/K explicitly and compute the number of generators of K. Covering Spaces Section 1.3 77 Cayley Complexes Covering spaces can be used to describe a very classical method for viewing e e e gα \|\|\|\| rβ XG/G = XG in the following way. Let the vertices of groups geometrically as graphs. Recall from Corollary 1.28 how we associated to each group presentation G = a 2 dimensional cell complex XG with π1(XG) ≈ G by taking a wedge-sum of circles, one for each generator gα, and then attaching a XG with a covering space 2 cell for each relator rβ. We can construct a cell complex XG be action of G such that the elements of G themselves. Then, at each vertex g ∈ G, insert an edge joining g to ggα for each of the chosen generators gα. The resulting graph is known as the Cayley graph of G with respect to the generators gα. This graph is connected since every element of G is a product of gα ’s, so there is a path in the graph joining each vertex to the identity vertex e. Each relation rβ determines a loop in the graph, starting at any vertex g, and we attach a 2 cell for each such loop. The resulting cell complex XG by multiplication on the left. Thus, an element g ∈ G sends a vertex g′ ∈ G to the vertex gg′, and the edge from g′ to g′gα is sent to the edge from gg′ to gg′
gα. The action extends to 2 cells in the obvious way. This is clearly a covering space action, and the orbit space is just XG. In fact XG is the universal cover of XG since it is simply-connected. This can be seen by considering the homomorphism ϕ : π1(XG)→G deﬁned in the proof of Proposition 1.39. For an edge eα in XG corresponding to a generator gα of G, it is clear from the deﬁnition of ϕ that ϕ([eα]) = gα, so ϕ is an isomorphism. In particular, is zero, hence also π1( the kernel of ϕ, p∗ XG is the Cayley complex of G. The group G acts on XG) since p∗ is injective. e e e Let us look at some examples of Cayley complexes. e e π1( XG) Example 1.45. When G is the free group on two generators a and b, XG is S 1 ∨ S 1 and XG is the Cayley graph of Z ∗ Z pictured at the right. The action of a on this graph is a e rightward shift along the central horizontal axis, while b acts by an upward shift along the central vertical axis. The composition ab of these two shifts then takes the vertex e to the vertex ab. Similarly, the action of any w ∈ Z ∗ Z takes e to the vertex w. Example 1.46. The group G = Z× Z with presentation has XG XG is R2 with vertices the integer lattice Z2 ⊂ R2 and edges the torus S 1 × S 1, and the horizontal and vertical segments between these lattice points. The action of G is by translations (x, y) ֏ (x + m, y + n). x, y \|\|\|\| xyx−1y −1 e 78 Chapter 1 The Fundamental Group x \|\|\|\| x2 x \|\|\|\| xn XG = S 2. More generally, for Example 1.47. For G = Z2 =, XG is S 1 with a disk attached by the map z ֏ zn and Zn = XG consists of n disks D1, ···, Dn with their boundary circles identiﬁed. A generator of Zn acts on this union
of disks by sending Di to Di+1 via a 2π /n rotation, the subscript i being taken mod n. The common boundary circle of the disks is rotated by 2π /n., XG is RP2 and e e Example 1.48. If G = Z2 ∗ Z2 = an inﬁnite sequence of circles each tangent to its two neighbors. a, b \|\|\|\| a2, b2 then the Cayley graph is a union of e We obtain XG from this graph by making each circle the equator of a 2 sphere, yielding an inﬁnite sequence of tangent 2 spheres. Elements of the index-two normal subgroup Z ⊂ Z2 ∗ Z2 generated by ab act on XG as translations by an even number of units, while each of the remaining elements of Z2 ∗ Z2 acts as the antipodal map on one of the spheres and ﬂips the whole chain of spheres end-for-end about this sphere. The orbit space XG is RP2 ∨ RP2. e It is not hard to see the generalization of this example to Zm ∗ Zn with the presentation XG consists of an inﬁnite union of copies of the Cayley complexes for Zm and Zn constructed in Example 1.47, arranged in a tree-like pattern. The case of Z2 ∗ Z3 is pictured below. a, b \|\|\|\| am, bn, so that e Covering Spaces Section 1.3 79 Exercises 1. For a covering space p : X→X and a subspace A ⊂ X, let A = p−1(A). Show that the restriction p : A→A is a covering space. e X1→X1 and p2 : 2. Show that if p1 : e X2→X1 × X2. X1 × p1 × p2 : e X→X be a covering space with p−1(x) ﬁnite and nonempty for all x ∈ X. 3. Let p : e X is compact Hausdorﬀ iﬀ X is compact Hausdorﬀ. Show that e X2→X2 are covering spaces, so is their product 4. Construct a simply-connected covering space of the space X ⊂ R3 that is the union of a sphere and
a diameter. Do the same when X is the union of a sphere and a circle e e e e intersecting it in two points. 5. Let X be the subspace of R2 consisting of the four sides of the square [0, 1]× [0, 1] together with the segments of the vertical lines x = 1/2, 1/3, 1/4, ··· inside the square. X→X there is some neighborhood of the left Show that for every covering space X. Deduce that X has no simply-connected edge of X that lifts homeomorphically to covering space. e e 6. Let X be the shrinking wedge of circles in Example 1.25, and let X be its covering space shown in the ﬁgure below. e X→X Construct a two-sheeted covering space Y → of the two covering spaces is not a covering space. Note that a composition of two X such that the composition Y → e e covering spaces does have the unique path lifting property, however. 7. Let Y be the quasi-circle shown in the ﬁgure, a closed subspace of R2 consisting of a portion of the graph of y = sin(1/x), the segment [−1, 1] in the y axis, and an arc connecting these two pieces. Collapsing the segment of Y in the y axis to a point gives a quotient map f : Y →S 1. Show that f does not lift to the covering space R→S 1, even though π1(Y ) = 0. Thus local path-connectedness of Y is a necessary hypothesis in the lifting criterion. X and 8. Let path-connected spaces X and Y. Show that if X ≃ Y then Y be simply-connected covering spaces of the path-connected, locally e e Chapter 0 may be helpful.] X ≃ Y. [Exercise 11 in e e 9. Show that if a path-connected, locally path-connected space X has π1(X) ﬁnite, then every map X→S 1 is nullhomotopic. [Use the covering space R→S 1.] 10. Find all the connected 2 sheeted and 3 sheeted covering spaces of S 1 ∨ S 1, up to isomorphism of covering spaces without basepoints. 80 Chapter 1 The Fundamental Group 11. Construct ﬁnite graphs X1 and X2 having a
common ﬁnite-sheeted covering space X1 = X2, but such that there is no space having both X1 and X2 as covering spaces. 12. Let a and b be the generators of π1(S 1 ∨ S 1) corresponding to the two S 1 e summands. Draw a picture of the covering space of S 1 ∨ S 1 corresponding to the normal subgroup generated by a2, b2, and (ab)4, and prove that this covering space is indeed the correct one. e 13. Determine the covering space of S 1 ∨ S 1 corresponding to the subgroup of π1(S 1 ∨ S 1) generated by the cubes of all elements. The covering space is 27 sheeted and can be drawn on a torus so that the complementary regions are nine triangles with edges labeled aaa, nine triangles with edges labeled bbb, and nine hexagons with edges labeled ababab. [For the analogous problem with sixth powers instead of cubes, the resulting covering space would have 228325 sheets! And for k th powers with k suﬃciently large, the covering space would have inﬁnitely many sheets. The underlying group theory question here, whether the quotient of Z ∗ Z obtained by factoring out all k th powers is ﬁnite, is known as Burnside’s problem. It can also be asked for a free group on n generators.] 14. Find all the connected covering spaces of RP2 ∨ RP2. e 15. Let p : X→X be a simply-connected covering space of X and let A ⊂ X be a X a path-component of A→A is the covering space corresponding to the kernel of the path-connected, locally path-connected subspace, with p−1(A). Show that p : map π1(A)→π1(X). 16. Given maps X→Y →Z such that both Y →Z and the composition X→Z are covering spaces, show that X→Y is a covering space if Z is locally path-connected, and show that this covering space is normal if X→Z is a normal covering space. A ⊂ e e e 17. Given a group G and a normal subgroup N, show that there exists a normal X) ≈ N, and deck transformation group X→X with π1(X)
≈ G, π1( covering space X) ≈ G/N. G( e e e e 18. For a path-connected, locally path-connected, and semilocally simply-connected X→X abelian if it is normal and has abelian deck transformation group. Show that X has an abelian covering space that is space X, call a path-connected covering space a covering space of every other abelian covering space of X, and that such a ‘universal’ abelian covering space is unique up to isomorphism. Describe this covering space explicitly for X = S 1 ∨ S 1 and. 19. Use the preceding problem to show that a closed orientable surface Mg of genus g has a connected normal covering space with deck transformation group isomorphic to Zn (the product of n copies of Z ) iﬀ n ≤ 2g. For n = 3 and g ≥ 3, describe such a covering space explicitly as a subspace of R3 with translations of R3 as deck transformations. Show that such a covering space in R3 exists iﬀ there is an embedding Covering Spaces Section 1.3 81 of Mg in the 3 torus such that the induced map π1(Mg)→π1(T 3) is surjective. 20. Construct nonnormal covering spaces of the Klein bottle by a Klein bottle and by a torus. 21. Let X be the space obtained from a torus S 1 × S 1 by attaching a M¨obius band via a homeomorphism from the boundary circle of the M¨obius band to the circle S 1 × {x0} in the torus. Compute π1(X), describe the universal cover of X, and describe the action of π1(X) on the universal cover. Do the same for the space Y obtained by attaching a M¨obius band to RP2 via a homeomorphism from its boundary circle to the circle in RP2 formed by the 1 skeleton of the usual CW structure on RP2. 22. Given covering space actions of groups G1 on X1 and G2 on X2, show that the action of G1 × G2 on X1 × X2 deﬁned by (g1, g2)(x1, x2) = (g1(x1), g2(x2)) is a covering space action
, and that (X1 × X2)/(G1 × G2) is homeomorphic to X1/G1 × X2/G2. 23. Show that if a group G acts freely and properly discontinuously on a Hausdorﬀ space X, then the action is a covering space action. (Here ‘properly discontinuously’ means that each x ∈ X has a neighborhood U such that { g ∈ G \| U ∩ g(U) ≠ ∅ } is ﬁnite.) In particular, a free action of a ﬁnite group on a Hausdorﬀ space is a covering space action. 24. Given a covering space action of a group G on a path-connected, locally path- connected space X, then each subgroup H ⊂ G determines a composition of covering spaces X→X/H→X/G. Show: (a) Every path-connected covering space between X and X/G is isomorphic to X/H for some subgroup H ⊂ G. (b) Two such covering spaces X/H1 and X/H2 of X/G are isomorphic iﬀ H1 and H2 are conjugate subgroups of G. (c) The covering space X/H→X/G is normal iﬀ H is a normal subgroup of G, in which case the group of deck transformations of this cover is G/H. 25. Let ϕ : R2→R2 be the linear transformation ϕ(x, y) = (2x, y/2). This generates an action of Z on X = R2 − {0}. Show this action is a covering space action and compute π1(X/Z). Show the orbit space X/Z is non-Hausdorﬀ, and describe how it is a union of four subspaces homeomorphic to S 1 × R, coming from the complementary components of the x axis and the y axis. 26. For a covering space p : X→X with X connected, locally path-connected, and semilocally simply-connected, show: e (a) The components of X are in one-to-one correspondence with the orbits of the action of π1(X, x0) on the ﬁber p−1(x0). e (b)
Under the Galois correspondence between connected covering spaces of X and subgroups of π1(X, x0), the subgroup corresponding to the component of X e 82 Chapter 1 The Fundamental Group containing a given lift x0 of x0 is the stabilizer of x0, the subgroup consisting of elements whose action on the ﬁber leaves x0 ﬁxed. e e X→X there are two actions of π1(X, x0) on the ﬁber 27. For a universal cover p : p−1(x0). The ﬁrst is the action deﬁned on page 69 in which the element of π1(X, x0) X, and the second is determined by a loop γ sends γ(0) for each lift e γ(1) to γ of γ to e the action given by restricting deck transformations to the ﬁber (see Proposition 1.39). Show that these two actions are diﬀerent when X = S 1 ∨ S 1 and when X = S 1 × S 1 and determine when the two actions are the same. [This is a revised version of the e e e e original form of this exercise.] 28. Show that for a covering space action of a group G on a simply-connected space Y, π1(Y /G) is isomorphic to G. [If Y is locally path-connected, this is a special case of part (c) of Proposition 1.40.] 29. Let Y be path-connected, locally path-connected, and simply-connected, and let G1 and G2 be subgroups of Homeo(Y ) deﬁning covering space actions on Y. Show that the orbit spaces Y /G1 and Y /G2 are homeomorphic iﬀ G1 and G2 are conjugate subgroups of Homeo(Y ). 30. Draw the Cayley graph of the group Z ∗ Z2 = 31. Show that the normal covering spaces of S 1 ∨ S 1 are precisely the graphs that are Cayley graphs of groups with two generators. More generally, the normal cov-. a, b \|\|\|\| b2 ering spaces of the wedge sum of n circles are the Cayley graphs of groups with n generators. 32. Consider covering spaces p : X→X with X and X connected CW complexes, the

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