Duality in the homology of 5-manifolds

W.H. Mannan

###### Abstract

We show that the homological properties of a 5-manifold $M$ with fundamental
group $G$ are encapsulated in a $G$–invariant stable form on the dual of the
third syzygy of $\mathbb{Z}$. In this notation one may express an even
stronger version of Poincaré duality for $M$. However we find an obstruction
to this duality.

Department of Mathematics, City University London, Northampton Square, London,
EC1V 0HB Tel: 07951825081 email<EMAIL_ADDRESS>

Keywords: Manifold, Poincaré duality, Chain complex

Subject Classification: 57P10, 57M60, 55U15, 55N91, 55N45

##  1 Introduction

The purpose of this article is to consider a 5-manifold $M$ with finite
fundamental group $G$, and examine the interaction between the module theory
over $\mathbb{Z}[G]$ with the homology theory of the simply connected
5-manifold $\tilde{M}$. Much of what we do holds more generally for
$2n+1$-manifolds with $n-1$-connected universal cover. However we focus on
5-manifolds for clarity and in order to avoid such additional assumptions, as
well as to highlight that low-dimensional phenomena to not impinge on our
results.

We associate to $M$ an algebraic complex $C(M)$; the cellular chain complex
for $\tilde{M}$ which one may regard as an algebraic complex over
$\mathbb{Z}[G]$. This algebraic object (defined up to chain homotopy
equivalence) is universal in the sense that all twisted homology/cohomology of
$M$ may be extracted from it (by taking coefficients in the appropriate module
over $\mathbb{Z}[G]$.

Using purely algebraic methods from our earlier paper [5], in §2 we show that
we may take $C(M)$ to satisfy a certain duality, so that its modules and those
of its dual $C(M)^{*}$ may be identified with four of the five differentials
the same in both cases. We call this dual form. Note this does not imply
$C(M)\cong C(M)^{*}$.

A self-dual (up to signs on boundary maps) polyhedral cell description for $M$
itself could be obtained from Lefschetz’ classical proof of Poincaré Duality
cf. [6, Theorem 2.1] or via geometric arguments cf [4, Theorem 6.5]. Our
result differs in that we are working with the universal cover of $M$ and
modules over $\mathbb{Z}[G]$. As mentioned earlier this object more completely
represents the (co)homology of $M$.

In §3 we observe that when $C(M)$ is in dual form, the equivalence induced by
Poincaré duality may be taken to be plus/minus the identity on four of the six
modules. However some redundancy in the proof of this, together with evidence
on the level of the derived category of $\mathbb{Z}[G]$ on the one hand, and
the homology of $\tilde{M}$ on the other, suggest that we may extend this to
all six modules.

In this case $C(M)$ would satisfy a stronger duality and we say $M$ satisfies
anti-self-duality. Here the modules in $C(M)$ and $C(M)^{*}$ are again
identified, with four of the five differentials the same, but now the
remaining differential differs by a sign in the two cases. Finally we find an
obstruction to anti-self-duality and show that not all 5-manifolds satisfy it,
whilst providing a class of examples which do.

Notation: Fix a finite group $G$. All modules are right modules unless
otherwise stated and assumed to be over $\mathbb{Z}[G]$, as are linear maps
between modules. Bilinear forms on modules are assumed to be $G$ invariant. We
abbreviate finitely generated to fg.

By algebraic complex we mean a sequence of modules and maps (referred to as
differentials) where the composition of successive terms is 0. The maps will
be represented by solid arrows. Dashed arrows will denote identifications of
the kernel/cokernel of the maps at either end with a module. Dashed arrows
should not be regarded as part of the complex (for homotopies between chain
maps etc.).

By a _simple homotopy equivalence_ we mean interchanging an algebraic complex
with one which differs from it in only two consecutive modules, where
isomorphic free modules have been added to both modules, and the differential
between them has been extended by the isomorphism. The preceding/following
differentials are composed with the natural inclusion/projection respectively.
If two algebraic complexes are related in this way, then we have a projection
chain map and an inclusion chain map between them, which form a pair of
mutually inverse chain homotopy equivalences.

######  Definition 1.1.

For a finite algebraic complex $\mathcal{A}$, the $n^{\rm th}$ _stabilization_
of $\mathcal{A}$, denoted $\mathcal{A}^{n}$, is the algebraic complex obtained
by adding a free rank $n$ module $F$ to the leftmost module in $\mathcal{A}$
(with maps written going from left to right). The differential from this
module is extended to $0$ on $F$.

We define the dual of a right $\mathbb{Z}[G]$ module $X$ to be the right
$\mathbb{Z}[G]$ module $X^{*}$ obtained by taking the left $\mathbb{Z}[G]$
module ${\rm Hom}_{\mathbb{Z}}(X,\mathbb{Z})$ and defining the right action of
$g\in G$ to be left multiplication by $g^{-1}$.

We extend this definition in the natural way to linear maps and algebraic
complexes. Note that dualization preserves exactness at a module in an
algebraic complex of torsion free (over $\mathbb{Z}$) fg $\mathbb{Z}[G]$
modules. We do not introduce signs when we dualize differentials. The price
for this is a sign factor in the statement of Poincaré duality (see (11)).
Define $H^{i}(\mathcal{A})=H_{i}(\mathcal{A}^{*})$ for an algebraic complex
$\mathcal{A}$.

The integers $\mathbb{Z}$ are a module over $\mathbb{Z}[G]$ with trivial $G$
action. A bilinear form $X\times X\to\mathbb{Z}$ will be referred to
interchangeably as a map $X\to X^{*}$.

Let ${\rm TOP}^{5}$ denote the category of closed connected oriented five
dimensional topological manifolds with base point, equipped with an
identification of their fundamental group with $G$. Let the morphisms in ${\rm
TOP}^{5}$ be the continuous maps which preserve the base point and induce the
identity on $G$.

Given an object $M\in{\rm TOP}^{5}$ we may find a finite CW complex
$M^{\prime}$ which is homotopy equivalent to it [3, Ann. 2]. Let
$C_{*}(\tilde{M^{\prime}})$ be the (cellular) chain complex of
$\tilde{M^{\prime}}$:

$\displaystyle
C_{*}(\tilde{M^{\prime}})\\!:=C_{5}\stackrel{{\scriptstyle\partial_{5}}}{{\longrightarrow}}C_{4}\stackrel{{\scriptstyle\partial_{4}}}{{\longrightarrow}}C_{3}\stackrel{{\scriptstyle\partial_{3}}}{{\longrightarrow}}C_{2}\stackrel{{\scriptstyle\partial_{2}}}{{\longrightarrow}}C_{1}\stackrel{{\scriptstyle\partial_{1}}}{{\longrightarrow}}C_{0}.$

This is an algebraic complex of fg free $\mathbb{Z}[G]$ modules and
$\mathbb{Z}[G]$–linear maps. It is exact at $C_{1}$ and
coker$(\partial_{1})\cong\mathbb{Z}$ (as $M$ connected). We fix this
isomorphism by fixing the homology class represented by a point to correspond
to $1\in\mathbb{Z}$.

Poincaré duality states that $C_{*}(\tilde{M^{\prime}})$ is chain homotopy
equivalent to its dual $C_{*}(\tilde{M^{\prime}})^{*}$. Therefore it is exact
at $C_{4}$ and ker$(\partial_{5})\cong\mathbb{Z}$. We fix this isomorphism by
demanding the generator of ker$(\partial_{5})$ preferred by the orientation,
maps to $1\in\mathbb{Z}$.

The Euler characteristic (alternating sum of torsion free ranks over
$\mathbb{Z}$) of $C_{*}(\tilde{M^{\prime}})$ is minus that of
$C_{*}(\tilde{M^{\prime}})^{*}$ (as dualization preserves the torsion free
$\mathbb{Z}$ rank of an fg $\mathbb{Z}[G]$ module). However if they are chain
homotopy equivalent they must also have equal Euler characteristic, which must
then be $0$.

######  Definition 1.2.

Let ${\rm ALG}^{5}$ denote the category of algebraic 5–complexes of fg free
$\mathbb{Z}[G]$ modules:

$\mathbb{Z}\dashrightarrow
F_{5}\stackrel{{\scriptstyle\partial_{5}}}{{\longrightarrow}}F_{4}\stackrel{{\scriptstyle\partial_{4}}}{{\longrightarrow}}F_{3}\stackrel{{\scriptstyle\partial_{3}}}{{\longrightarrow}}F_{2}\stackrel{{\scriptstyle\partial_{2}}}{{\longrightarrow}}F_{1}\stackrel{{\scriptstyle\partial_{1}}}{{\longrightarrow}}F_{0}\dashrightarrow\mathbb{Z},$

satisfying exactness at $F_{4}$ and $F_{1}$, and with Euler characteristic
$0$. Define the morphisms in this category to be homotopy classes of chain
maps.

We may define a functor $C\colon{\rm TOP}^{5}\to{\rm ALG}^{5}$ by choosing a
homotopy equivalence, $h_{M}\colon M\to M^{\prime}$ with $M^{\prime}$ a finite
CW complex, for each $M\in{\rm TOP}^{5}$. In each case fix also
$h_{M}^{\prime}$ a homotopy inverse of $h_{M}$. Then define $C(M)$ to be
$C_{*}(\tilde{M^{\prime}})$.

Given a morphism in ${\rm TOP}^{5}$, $f\colon M_{1}\to M_{2}$, we may select a
cellular map $f^{\prime}\colon M_{1}^{\prime}\to M_{2}^{\prime}$, which is
homotopic to $(h_{M_{2}}\circ f\circ h_{M_{1}}^{\prime})$. Define $C(f)$ to be
the equivalence class of the chain map $f^{\prime}_{*}\colon
C_{*}(\tilde{M_{1}^{\prime})}\to C_{*}(\tilde{M_{2}^{\prime})}$.

The isomorphism class of $C(M)$ in ${\rm ALG}^{5}$ is an invariant of $M$ as
by construction different choices of $M^{\prime}$ must be homotopy equivalent
to each other.

##  2 Dual form

Consider an algebraic 2–complex $\mathcal{B}$ of free fg $\mathbb{Z}[G]$
modules, where the cokernel of the last map is identified with $\mathbb{Z}$,
we have exactness at the middle term and the dual of the kernel of the first
map is denoted $J$:

$\mathcal{B}:=\qquad
J^{*}\stackrel{{\scriptstyle\iota}}{{\dashrightarrow}}F_{2}\stackrel{{\scriptstyle
d_{2}}}{{\to}}F_{1}\stackrel{{\scriptstyle
d_{1}}}{{\to}}F_{0}\dashrightarrow\mathbb{Z}.$

Thus $J^{*}\in\Omega_{3}(\mathbb{Z})$ the third syzygy of $\mathbb{Z}$ [2,
p6]. Equip $\mathcal{B}$ with a $G$ invariant bilinear form $\beta$ on $J$. We
can associate to the pair $(\mathcal{B},\beta)$ an algebraic 5–complex:

$\displaystyle i(\mathcal{B},\beta):=\qquad\mathbb{Z}\dashrightarrow
F_{0}^{*}\stackrel{{\scriptstyle
d_{1}^{*}}}{{\to}}F_{1}^{*}\stackrel{{\scriptstyle
d_{2}^{*}}}{{\to}}F_{2}^{*}\,\,\,\stackrel{{\scriptstyle
d_{3}}}{{\longrightarrow}}\,\,\,\,F_{2}\stackrel{{\scriptstyle
d_{2}}}{{\to}}F_{1}\stackrel{{\scriptstyle
d_{1}}}{{\to}}F_{0}\dashrightarrow\mathbb{Z}.\quad\,$
$\displaystyle\iota^{*}\searrow\quad\,\,\,\,\quad\nearrow\iota\qquad\qquad\qquad\,\,\quad\quad\,\,\,$
$\displaystyle
J\stackrel{{\scriptstyle\beta}}{{\to}}J^{*}\qquad\qquad\qquad\,\,\,\,\,\quad\quad\,\,\quad$

Let ${\rm DUAL}^{2}$ denote the category whose objects are pairs
$(\mathcal{B},\beta)$ as above.

We define the morphisms of ${\rm DUAL}^{2}$ to be homotopy equivalence classes
of chain maps between the associated algebraic 5–complexes. If an object of
${\rm ALG}^{5}$ is of the form $i(\mathcal{B},\beta)$, for some
$(\mathcal{B},\beta)\in{\rm DUAL}^{2}$ then we say it is in _dual form_.

######  Theorem 2.1.

The functor $i$ is a natural equivalence of categories

By construction $i\colon{\rm DUAL}^{2}\to{\rm ALG}^{5}$ is a full faithful
functor. It remains to show that every object in ${\rm ALG}^{5}$ is isomorphic
to an object in the image of $i$. Our proof will follow stages analogous to
the proof of [5, Theorem 1.1].

Consider an arbitrary element of ${\rm ALG}^{5}$:

$\displaystyle\mathbb{Z}\dashrightarrow
C_{5}\stackrel{{\scriptstyle\partial_{5}}}{{\longrightarrow}}C_{4}\stackrel{{\scriptstyle\partial_{4}}}{{\longrightarrow}}C_{3}\stackrel{{\scriptstyle\partial_{3}}}{{\longrightarrow}}C_{2}\stackrel{{\scriptstyle\partial_{2}}}{{\longrightarrow}}C_{1}\stackrel{{\scriptstyle\partial_{1}}}{{\longrightarrow}}C_{0}\dashrightarrow\mathbb{Z}.$
(1)

We perform three pairs of simple homotopy equivalences. Firstly, the complex
(1) is chain homotopy equivalent to:

$\displaystyle C_{5}\oplus
C_{0}^{*}\stackrel{{\scriptstyle\delta_{5}}}{{\longrightarrow}}C_{4}\oplus
C_{0}^{*}\stackrel{{\scriptstyle\partial_{4}\oplus
0}}{{\longrightarrow}}C_{3}\stackrel{{\scriptstyle\partial_{3}}}{{\longrightarrow}}C_{2}\stackrel{{\scriptstyle\partial_{2}}}{{\longrightarrow}}C_{1}\oplus
C_{5}^{*}\stackrel{{\scriptstyle\delta_{1}}}{{\longrightarrow}}C_{0}\oplus
C_{5}^{*},$ (2)

where

$\delta_{1}=\left(\begin{array}[]{cc}\partial_{1}&0\\\
0&1\end{array}\right),\qquad\qquad\delta_{5}=\left(\begin{array}[]{cc}\partial_{5}&0\\\
0&1\end{array}\right).$

Let $R_{0}=C_{0}$, $R_{5}=C_{5}$, and $R_{1}=C_{1}\oplus C_{5}^{*}$,
$R_{4}=C_{4}\oplus C_{0}^{*}$. Rewrite (2):

$\displaystyle R_{5}\oplus
R_{0}^{*}\stackrel{{\scriptstyle\delta_{5}}}{{\longrightarrow}}R_{4}\stackrel{{\scriptstyle\partial_{4}\oplus
0}}{{\longrightarrow}}C_{3}\stackrel{{\scriptstyle\partial_{3}}}{{\longrightarrow}}C_{2}\stackrel{{\scriptstyle\partial_{2}}}{{\longrightarrow}}R_{1}\stackrel{{\scriptstyle\delta_{1}}}{{\longrightarrow}}R_{0}\oplus
R_{5}^{*}.$ (3)

Again we perform a pair of simple homotopy equivalences. The complex (3) is
chain homotopy equivalent to:

$\displaystyle R_{5}\oplus
R_{0}^{*}\stackrel{{\scriptstyle\delta_{5}}}{{\longrightarrow}}R_{4}\oplus
R_{1}^{*}\stackrel{{\scriptstyle\delta_{4}}}{{\longrightarrow}}C_{3}\oplus
R_{1}^{*}\stackrel{{\scriptstyle\partial_{3}\oplus
0}}{{\longrightarrow}}C_{2}\oplus
R_{4}^{*}\stackrel{{\scriptstyle\delta_{2}}}{{\longrightarrow}}R_{1}\oplus
R_{4}^{*}\stackrel{{\scriptstyle\delta_{1}\oplus
0}}{{\longrightarrow}}R_{0}\oplus R_{5}^{*},$ (4)

where

$\delta_{2}=\left(\begin{array}[]{cc}\partial_{2}&0\\\
0&1\end{array}\right),\qquad\qquad\delta_{4}=\left(\begin{array}[]{cc}\partial_{4}\oplus
0&0\\\ 0&1\end{array}\right).$

Let $R_{2}=C_{2}\oplus R_{4}^{*}$, $R_{3}=C_{3}\oplus R_{1}^{*}$. Then (4) can
be written:

$\displaystyle R_{5}\oplus
R_{0}^{*}\stackrel{{\scriptstyle\delta_{5}}}{{\longrightarrow}}R_{4}\oplus
R_{1}^{*}\stackrel{{\scriptstyle\delta_{4}}}{{\longrightarrow}}R_{3}\stackrel{{\scriptstyle\partial_{3}\oplus
0}}{{\longrightarrow}}R_{2}\stackrel{{\scriptstyle\delta_{2}}}{{\longrightarrow}}R_{1}\oplus
R_{4}^{*}\stackrel{{\scriptstyle\delta_{1}\oplus
0}}{{\longrightarrow}}R_{0}\oplus R_{5}^{*}.$ (5)

As all the modules in this complex are free and the Euler characteristic is 0,
we can assume the existence of some isomorphism $\theta\colon R_{2}^{*}\to
R_{3}^{*}$.

We perform a final simple homotopy equivalence to get:

$\displaystyle R_{5}\oplus
R_{0}^{*}\stackrel{{\scriptstyle\delta_{5}}}{{\longrightarrow}}R_{4}\oplus
R_{1}^{*}\stackrel{{\scriptstyle\delta_{4}}}{{\longrightarrow}}R_{3}\oplus
R_{2}^{*}\stackrel{{\scriptstyle\delta_{3}}}{{\longrightarrow}}R_{2}\oplus
R_{3}^{*}\stackrel{{\scriptstyle\delta_{2}\oplus
0}}{{\longrightarrow}}R_{1}\oplus
R_{4}^{*}\stackrel{{\scriptstyle\delta_{1}\oplus
0}}{{\longrightarrow}}R_{0}\oplus R_{5}^{*},$ (6)

where:

$\delta_{3}=\left(\begin{array}[]{cc}\partial_{3}\oplus 0&0\\\
0&\theta\end{array}\right).\qquad\qquad$

Our initial algebraic complex (1) is therefore chain homotopy equivalent to
(6). We will show that (6) is chain isomorphic to an algebraic 5–complex in
dual form.

######  Lemma 2.2.

There exist a pair of inverse chain isomorphisms $h,k$:

$\displaystyle R_{2}\oplus R_{3}^{*}\stackrel{{\scriptstyle\delta_{2}\oplus
0}}{{\longrightarrow}}R_{1}\oplus
R_{4}^{*}\stackrel{{\scriptstyle\delta_{1}\oplus
0}}{{\longrightarrow}}R_{0}\oplus
R_{5}^{*}\stackrel{{\scriptstyle\epsilon\oplus
0}}{{\dashrightarrow}}\mathbb{Z}\to 0$ (7) $\displaystyle
h_{2}\downarrow\uparrow{k_{2}}\qquad{h_{1}}\downarrow\uparrow{k_{1}}\qquad\quad
h_{0}\downarrow\uparrow{k_{0}}\qquad||1\quad\,\,\,$ $\displaystyle
R_{3}^{*}\oplus R_{2}\stackrel{{\scriptstyle\delta_{4}^{*}\oplus
0}}{{\longrightarrow}}R_{4}^{*}\oplus
R_{1}\stackrel{{\scriptstyle\delta_{5}^{*}\oplus
0}}{{\longrightarrow}}R_{5}^{*}\oplus
R_{0}\stackrel{{\scriptstyle\epsilon^{\prime}\oplus
0}}{{\dashrightarrow}}\mathbb{Z}\to 0$ (8)

###### Proof.

This is just a case of [5, Lemma 2.1]. ∎

Let $d_{3}=\delta_{3}k_{2}^{*}$ and write (7) as:

$S_{2}\stackrel{{\scriptstyle d_{2}}}{{\to}}S_{1}\stackrel{{\scriptstyle
d_{1}}}{{\to}}S_{0}.$

######  Lemma 2.3.

The complex (6) is chain isomorphic to:

$S_{0}^{*}\stackrel{{\scriptstyle
d_{1}^{*}}}{{\longrightarrow}}S_{1}^{*}\stackrel{{\scriptstyle
d_{2}^{*}}}{{\longrightarrow}}S_{2}^{*}\stackrel{{\scriptstyle
d_{3}}}{{\longrightarrow}}S_{2}\stackrel{{\scriptstyle
d_{2}}}{{\longrightarrow}}S_{1}\stackrel{{\scriptstyle
d_{1}}}{{\longrightarrow}}S_{0}.$

###### Proof.

We have the following chain isomorphism:

$\displaystyle R_{5}\oplus
R_{0}^{*}\stackrel{{\scriptstyle\delta_{5}}}{{\longrightarrow}}R_{4}\oplus
R_{1}^{*}\stackrel{{\scriptstyle\delta_{4}}}{{\longrightarrow}}R_{3}\oplus
R_{2}^{*}\stackrel{{\scriptstyle\delta_{3}}}{{\longrightarrow}}R_{2}\oplus
R_{3}^{*}\stackrel{{\scriptstyle\delta_{2}\oplus
0}}{{\longrightarrow}}R_{1}\oplus
R_{4}^{*}\stackrel{{\scriptstyle\delta_{1}\oplus
0}}{{\longrightarrow}}R_{0}\oplus R_{5}^{*}\quad$ $\displaystyle\downarrow
h_{0}^{*}\qquad\qquad\downarrow h_{1}^{*}\,\,\,\quad\qquad\downarrow
h_{2}^{*}\qquad\qquad\downarrow 1\qquad\qquad\downarrow
1\qquad\qquad\downarrow 1\qquad$ $\displaystyle\quad
S_{0}^{*}\quad\stackrel{{\scriptstyle d_{1}^{*}}}{{\longrightarrow}}\quad\quad
S_{1}^{*}\quad\stackrel{{\scriptstyle d_{2}^{*}}}{{\longrightarrow}}\quad
S_{2}^{*}\quad\stackrel{{\scriptstyle d_{3}}}{{\longrightarrow}}\quad\quad
S_{2}\quad\stackrel{{\scriptstyle d_{2}}}{{\longrightarrow}}\quad
S_{1}\quad\stackrel{{\scriptstyle d_{1}}}{{\longrightarrow}}\quad\quad
S_{0}\quad\quad$

We may verify that the central square commutes:
$d_{3}h_{2}^{*}=\delta_{3}k_{2}^{*}h_{2}^{*}=\delta_{3}$. ∎

Clearly $d_{3}$ is induced by a $G$ invariant bilinear form on the dual of
ker$(d_{2})$. This then completes the proof of Theorem 2.1.

Thus when using the functor $C$ to provide an invariant (up to isomorphism) of
an element of ${\rm TOP}^{5}$, we may work in the category ${\rm DUAL}^{2}$.
From now on we will suppress the functor $i$ and regard ${\rm DUAL}^{2}$ as a
full subcategory of ${\rm ALG}^{5}$.

Note that if $(\mathcal{B},\beta)\in{\rm DUAL}^{2}$, and
$\mathcal{B^{\prime}}$ is related to $\mathcal{B}$ via a simple homotopy
equivalence (recall definition from §1) then $(\mathcal{B},\beta)$ is related
to $(\mathcal{B}^{\prime},\beta)$ by a pair of simple homotopy equivalences.

Recall Definition 1.1. Then $(\mathcal{B},\beta)$ and
$(\mathcal{B}^{n},\beta)$ are related by a simple homotopy equivalence, (where
in the latter $\beta$ is understood to extend to the standard inner product on
the free module which we regard as a direct sum of copies of $\mathbb{Z}[G]$).

In some sense $\mathcal{B}$ is not important. That is we may fix an algebraic
2–complex $\mathcal{B}$, so that up to isomorphism any element of ${\rm
DUAL}^{2}$ has the form $(\mathcal{B}^{n},\beta)$ for some integer $n$ and
bilinear form $\beta$. To that end fix an algebraic 2–complex:

$\mathcal{B}:=\quad
J^{*}\stackrel{{\scriptstyle}}{{\dashrightarrow}}F_{2}\stackrel{{\scriptstyle
d_{2}}}{{\to}}F_{1}\stackrel{{\scriptstyle
d_{1}}}{{\to}}F_{0}\dashrightarrow\mathbb{Z},$

with $F_{0},F_{1},F_{2}$ fg free and exactness at $F_{1}$. Let $J_{n}$ denote
$J\oplus\mathbb{Z}[G]^{n}$.

######  Theorem 2.4.

Given $(\mathcal{C},\gamma)\in{\rm DUAL}^{2}$ there exists an integer $n$ and
bilinear form $\beta$ on $J_{n}$ such that $(\mathcal{C},\gamma)$ is chain
homotopy equivalent to $(\mathcal{B}^{n},\beta)$.

###### Proof.

By [5, Lemma 2.1] there exists an integer $n$ such that we may apply a pair of
simple homotopy equivalences to $\mathcal{B}^{n}$ to get a complex:

$L_{2}\stackrel{{\scriptstyle
D_{2}}}{{\longrightarrow}}L_{1}\stackrel{{\scriptstyle
D_{1}}}{{\longrightarrow}}L_{0},$

and a pair of simple homotopy equivalences to some stabilization of
$\mathcal{C}$ to get:

$S_{2}\stackrel{{\scriptstyle\partial_{2}}}{{\longrightarrow}}S_{1}\stackrel{{\scriptstyle\partial_{1}}}{{\longrightarrow}}S_{0},$

such that there is a chain isomorphism:

$\displaystyle
S_{2}\stackrel{{\scriptstyle\partial_{2}}}{{\longrightarrow}}S_{1}\stackrel{{\scriptstyle\partial_{1}}}{{\longrightarrow}}S_{0}\,\,$
$\displaystyle\downarrow\theta_{2}\quad\downarrow\theta_{1}\quad\downarrow\theta_{0}$
$\displaystyle L_{2}\stackrel{{\scriptstyle
D_{2}}}{{\longrightarrow}}L_{1}\stackrel{{\scriptstyle
D_{1}}}{{\longrightarrow}}L_{0}\,\,$

As noted above, $(\mathcal{C},\gamma)$ is chain homotopy equivalent to:

$S_{0}^{*}\stackrel{{\scriptstyle\partial_{1}^{*}}}{{\longrightarrow}}S_{1}^{*}\stackrel{{\scriptstyle\partial_{2}^{*}}}{{\longrightarrow}}S_{2}^{*}\stackrel{{\scriptstyle\partial_{3}}}{{\longrightarrow}}S_{2}\stackrel{{\scriptstyle\partial_{2}}}{{\longrightarrow}}S_{1}\stackrel{{\scriptstyle\partial_{1}}}{{\longrightarrow}}S_{0},$

for some map $\partial_{3}$. Let
$D_{3}=\theta_{2}\circ\partial_{3}\circ\theta_{2}^{*}$. We have a chain
isomorphism:

$\displaystyle
S_{0}^{*}\stackrel{{\scriptstyle\partial_{1}^{*}}}{{\longrightarrow}}S_{1}^{*}\stackrel{{\scriptstyle\partial_{2}^{*}}}{{\longrightarrow}}\,\,\,S_{2}^{*}\,\,\,\stackrel{{\scriptstyle\partial_{3}}}{{\longrightarrow}}\,\,S_{2}\,\,\,\stackrel{{\scriptstyle\partial_{2}}}{{\longrightarrow}}\,\,\,S_{1}\stackrel{{\scriptstyle\partial_{1}}}{{\longrightarrow}}S_{0}\,\,\,\,$
(9)
$\displaystyle\quad\downarrow\theta_{0}^{*-1}\,\,\,\downarrow\theta_{1}^{*-1}\,\,\,\downarrow\theta_{2}^{*-1}\qquad\downarrow\theta_{2}\quad\quad\downarrow\theta_{1}\quad\downarrow\theta_{0}$
$\displaystyle L_{0}^{*}\stackrel{{\scriptstyle
D_{1}^{*}}}{{\longrightarrow}}\,\,L_{1}^{*}\stackrel{{\scriptstyle
D_{2}^{*}}}{{\longrightarrow}}\,\,L_{2}^{*}\,\,\,\stackrel{{\scriptstyle
D_{3}}}{{\longrightarrow}}\,\,L_{2}\,\,\,\stackrel{{\scriptstyle
D_{2}}}{{\longrightarrow}}\,\,\,L_{1}\stackrel{{\scriptstyle
D_{1}}}{{\longrightarrow}}L_{0}\,\,\,\,$ (10)

Finally note that (10) is obtained from $(\mathcal{B}^{n},\,\beta)$ for some
$\beta$, by two pairs of simple homotopy equivalences. Hence
$(\mathcal{B}^{n},\,\beta)$ is chain homotopy equivalent to
$(\mathcal{C},\gamma)$. ∎

So with $\mathcal{B}$ fixed, for any $M\in{\rm TOP}^{5}$ we have an integer
$n$ and bilinear form $\beta$ on $J_{n}$, such that $C(M)$ is isomorphic to
$(\mathcal{B}^{n},\beta)$ in ${\rm ALG}^{5}$. In particular the form $\beta$
encapsulates all the twisted (co)homology of $M$. A benefit of this is that
one may attempt to classify objects in ${\rm ALG}^{5}$ and ${\rm TOP}^{5}$
without having to classify algebraic 2–complexes over $\mathbb{Z}[G]$ (an
abstruse problem in its own right [2]).

##  3 Anti-self-duality

Pick $\eta\in H_{5}(C(M))$ a generator. Poincaré Duality may be stated: There
is a chain homotopy equivalence $\phi\colon C(M)^{*}\to C(M)$, such that for
$\alpha\in H^{p}(C(M))$ we have:

$\displaystyle\phi_{*}(\alpha)=-1^{p(p+1)/2}\eta\,\,\,\widehat{}\,\,\,\alpha.$
(11)

Given $\alpha\in H^{p}(C(M))$, $\gamma\in H^{5-p}(C(M))$ the graded symmetry
of the cup product implies that
$\gamma(\eta\,\,\,\widehat{}\,\,\,\alpha)=\alpha(\eta\,\,\,\widehat{}\,\,\,\gamma)$.
Substituting (11) into this we get:

$\displaystyle\gamma(\phi_{*}(\alpha))=-\alpha(\phi_{*}(\gamma)).$ (12)

Suppose, that we have a chain homotopy equivalence $f\colon
C(M)\to\mathcal{A}$, for some $\mathcal{A}\in{\rm ALG}^{5}$. We have a chain
homotopy equivalence $\phi^{\prime}=f\circ\phi\circ
f^{*}\colon\mathcal{A}^{*}\to\mathcal{A}$.

######  Definition 3.1.

If a homotopy equivalence $\mathcal{A}^{*}\to\mathcal{A}$ is chain homotopy
equivalent to one constructed in this way for some $f$, we say it is a
_duality equivalence_.

Given $\alpha\in H^{p}(\mathcal{A})$, $\gamma\in H^{5-p}(\mathcal{A})$ we have
$\gamma(\phi^{\prime}_{*}(\alpha))=f^{*}(\gamma)(\phi_{*}f^{*}(\alpha))$ which
is antisymmetric in $\alpha,\gamma$ (by (12)), so we may conclude
$\gamma(\phi^{\prime}_{*}(\alpha))=-\alpha(\phi^{\prime}_{*}(\gamma))$.

By Theorem 2.1 we may choose $\mathcal{A}$ to be of the form
$(\mathcal{B},\beta)$ for some:

$\mathcal{B}=J^{*}\stackrel{{\scriptstyle\iota}}{{\dashrightarrow}}F_{2}\stackrel{{\scriptstyle
d_{2}}}{{\to}}F_{1}\stackrel{{\scriptstyle d_{1}}}{{\to}}F_{0}\dashrightarrow
Z,$

and bilinear form $\beta\colon J\times J\to\mathbb{Z}$. Let
$d_{3}=\iota\beta\iota^{*}$.

The duality equivalence
$\phi^{\prime}\colon(\mathcal{B},\beta)^{*}\to(\mathcal{B},\beta)$ may be
written:

$\displaystyle\mathbb{Z}\dashrightarrow F_{0}^{*}\stackrel{{\scriptstyle
d_{1}^{*}}}{{\to}}F_{1}^{*}\stackrel{{\scriptstyle
d_{2}^{*}}}{{\to}}F_{2}^{*}\stackrel{{\scriptstyle
d_{3}^{*}}}{{\to}}F_{2}\stackrel{{\scriptstyle
d_{2}}}{{\to}}F_{1}\stackrel{{\scriptstyle
d_{1}}}{{\to}}F_{0}\dashrightarrow\mathbb{Z}\,\,\,$ $\displaystyle
y\downarrow\,\,\,\,\,\,\,\phi^{\prime}_{5}\downarrow\,\,\,\,\phi^{\prime}_{4}\downarrow\,\,\,\,\phi^{\prime}_{3}\downarrow\,\,\,\phi^{\prime}_{2}\downarrow\,\,\,\phi^{\prime}_{1}\downarrow\,\,\,\phi^{\prime}_{0}\downarrow\quad\,x\downarrow\,\,\,$
(13) $\displaystyle\mathbb{Z}\dashrightarrow F_{0}^{*}\stackrel{{\scriptstyle
d_{1}^{*}}}{{\to}}F_{1}^{*}\stackrel{{\scriptstyle
d_{2}^{*}}}{{\to}}F_{2}^{*}\stackrel{{\scriptstyle
d_{3}}}{{\to}}F_{2}\stackrel{{\scriptstyle
d_{2}}}{{\to}}F_{1}\stackrel{{\scriptstyle
d_{1}}}{{\to}}F_{0}\dashrightarrow\mathbb{Z}\,\,\,$

The natural pairing $F_{0}^{*}\times F_{0}\to\mathbb{Z}$ induces a pairing on
the copies of $\mathbb{Z}$ on the left of this diagram, with those on the
right. We set the generators $1\in\mathbb{Z}$ on the right and left so that
this pairing is given by multiplication. Let $\alpha\in
H^{5}((\mathcal{B},\beta)),\,\gamma\in H^{0}((\mathcal{B},\beta))$ correspond
to these generators.

We have
$x=\gamma(\phi^{\prime}_{*}(\alpha))=-\alpha(\phi^{\prime}_{*}(\gamma))=-y$.
Without loss of generality we may assume (by for example replacing $d_{3}$
with $-d_{3}$), that $x=1,y=-1$.

######  Theorem 3.2.

For some maps $\theta_{1}$, $\theta_{2}$ we have a duality equivalence:

$\displaystyle F_{0}^{*}\stackrel{{\scriptstyle
d_{1}^{*}}}{{\longrightarrow}}F_{1}^{*}\stackrel{{\scriptstyle
d_{2}^{*}}}{{\longrightarrow}}F_{2}^{*}\stackrel{{\scriptstyle
d_{3}^{*}}}{{\longrightarrow}}F_{2}\stackrel{{\scriptstyle
d_{2}}}{{\longrightarrow}}F_{1}\stackrel{{\scriptstyle
d_{1}}}{{\longrightarrow}}F_{0}\,\,\,\,\,$
$\displaystyle\downarrow-1\quad\downarrow-1\,\,\,\,\,\,\downarrow\theta_{2}\,\,\,\quad\downarrow\theta_{1}\,\,\quad\downarrow
1\qquad\downarrow 1$ (14) $\displaystyle F_{0}^{*}\stackrel{{\scriptstyle
d_{1}^{*}}}{{\longrightarrow}}F_{1}^{*}\stackrel{{\scriptstyle
d_{2}^{*}}}{{\longrightarrow}}F_{2}^{*}\stackrel{{\scriptstyle
d_{3}}}{{\longrightarrow}}F_{2}\stackrel{{\scriptstyle
d_{2}}}{{\longrightarrow}}F_{1}\stackrel{{\scriptstyle
d_{1}}}{{\longrightarrow}}F_{0}\,\,\,\,\,$

###### Proof.

By the projectivity of the $F_{i}$ we may define maps
$I_{0},I_{1},I_{3},I_{4}$:

$\textstyle{J^{*}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{F_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{1-\phi_{2}^{\prime}}$$\scriptstyle{d_{2}}$$\textstyle{F_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{I_{1}}$$\scriptstyle{1}$$\scriptstyle{-\phi_{1}^{\prime}}$$\scriptstyle{d_{1}}$$\textstyle{F_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{I_{0}}$$\scriptstyle{1-\phi_{0}^{\prime}}$$\textstyle{\mathbb{Z}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{0}$$\textstyle{J^{*}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{F_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{-1-\phi_{3}^{\prime*}}$$\scriptstyle{d_{2}}$$\textstyle{F_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{I_{3}^{*}}$$\scriptstyle{-1}$$\scriptstyle{-\phi_{4}^{\prime*}}$$\scriptstyle{d_{1}}$$\textstyle{F_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{I_{4}^{*}}$$\scriptstyle{-1-\phi_{5}^{\prime*}}$$\textstyle{\mathbb{Z}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{0}$$\textstyle{J^{*}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{F_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{d_{2}}$$\textstyle{F_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{d_{1}}$$\textstyle{F_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathbb{Z}}$$\textstyle{J^{*}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{F_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{d_{2}}$$\textstyle{F_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{d_{1}}$$\textstyle{F_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathbb{Z}}$

such that:

$\displaystyle d_{1}I_{0}=1-\phi_{0}^{\prime}$ , $\displaystyle\qquad
I_{0}d_{1}+d_{2}I_{1}=1-\phi_{1}^{\prime},$ $\displaystyle
d_{1}I_{4}^{*}=-1-\phi_{5}^{\prime*}$ , $\displaystyle\qquad
I_{4}^{*}d_{1}+d_{2}I_{3}^{*}=-1-\phi_{4}^{\prime*}.$

Set

$\displaystyle\theta_{1}=\phi_{2}^{\prime}+I_{1}d_{2},\qquad\theta_{2}=\phi_{3}^{\prime}+d_{2}^{*}I_{3},\qquad
I_{2}=0.$ (15)

Then the $I_{i}$, $i=0,1,2,3,4$ form the required chain homotopy (13) to (14).
∎

Theorem 2.1 tells us that we may always pick a representative
$\mathcal{A}=(\mathcal{B},\beta)$ of the homotopy type of $C(M)$ such that the
modules of $\mathcal{A}$ and $\mathcal{A}^{*}$ may be identified, with four of
the five differentials identical. Theorem 3.2 tells us that Poincaré duality
induces a homotopy equivalence $\mathcal{A}^{*}\to\mathcal{A}$ which is
plus/minus the identity on four of the six maps. The natural completion of
these results would be that we could also take $\theta_{1}=1,\,\theta_{2}=-1$
(which would imply $d_{3}^{*}=-d_{3}$).

######  Definition 3.3.

Say that $M\in{\rm TOP}^{5}$ is _anti-self-dual_ if we can pick $\mathcal{A}$
as above with $\theta_{1}=1,\,\theta_{2}=-1$.

Indeed in (15) we arbitrarily set $I_{2}=0$, so it may seem that a more
careful choice of $I_{2}$ could lead to $\theta_{1}=1,\,\theta_{2}=-1$. There
are two further reasons to conjecture that all $M\in{\rm TOP}^{5}$ are anti-
self dual. These arise in the two themes which we seek to marry in this paper:
the derived category of the fundamental group of $M$ and the homology of the
simply connected manifold $\tilde{M}$.

Let $\lambda_{1}\colon J\to J$ be the dual of the map induced by $\theta_{1}$
and let $\lambda_{2}\colon J\to J$ be the map induced by $\theta_{2}$.
Commutativity in the central square of (14) implies that for all $u,v\in J$:
$\beta(\lambda_{1}u,v)=\beta(\lambda_{2}v,u).$

Thus if we had $\lambda_{1}=1_{J},\,\lambda_{2}=-1_{J}$ then $\beta$ would be
antisymmetric. What we do know from Theorem 3.2 is that this is true at level
of the derived category of $G$ (in the sense of [2, Chapter 4]);
$\lambda_{1},\lambda_{2}$ augment to $1,-1\in\mathbb{Z}/|G|$ respectively.

On the other hand Tor$(H^{3}(\tilde{M}))$ may be naturally identified with a
subquotient of $J\otimes\mathbb{Q}$. With respect to this identification, the
linking form on Tor$(H^{3}(\tilde{M}))$ (taking values in
$\mathbb{Q}/\mathbb{Z}$) is induced by $\beta(\lambda_{1}\\_,\,\\_)$ [1, 7].
From [1, Lemma D(ii)] we know that this is antisymmetric.

Despite all this, it transpires that not all manifolds are anti-self-dual. An
obstruction arises from a combination of a condition on $G$ and a condition on
$H_{*}(\tilde{M})$.

######  Theorem 3.4.

Let $M\in{\rm TOP}^{5}$. If $G$ has even order and $H_{3}(\tilde{M})$ has even
rank over $\mathbb{Z}$, then $M$ is not anti-self-dual.

###### Proof.

Suppose $M$ is anti-self-dual. Then we have that $C(M)$ is chain homotopy
equivalent to $(\mathcal{B},\beta)$, with $\beta$ an antisymmetric form on a
module $J$ with $J^{*}\in\Omega_{3}(\mathbb{Z})$. Thus ${\rm
rk}_{\mathbb{Z}}(J)\equiv-1\mod|G|$, and in particular is odd. Now
$H_{3}(\tilde{M})$ may be identified naturally with the kernel of $\beta$.

Thus we may split (over $\mathbb{Z}$) $J=H_{3}(\tilde{M})\oplus J^{\prime}$
and extend $\beta$ to the tensor product $V=J^{\prime}\otimes\mathbb{Q}$. Then
$V$ is an odd dimensional vector space over $\mathbb{Q}$, equipped with a non-
degenerate anti-symmetric bilinear form. This is impossible. ∎

As an example let $M$ be the Lens space $L(n;1,1)$. This has fundamental group
$C_{n}=\langle t|\,\,t^{n}=1\rangle$. Let $\Sigma$ denote the sum of the
$t^{i}$. We have $\mathcal{A}$ in dual form, with a chain homotopy equivalence
$C(M)\to\mathcal{A}$ yielding the duality morphism $\phi$:

$\textstyle{\mathcal{A^{*}}\\!:=\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi}$$\textstyle{\mathbb{Z}[t]\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{-1}$$\scriptstyle{\times(1-t^{-1})}$$\textstyle{\mathbb{Z}[t]\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{-1}$$\scriptstyle{\times\Sigma}$$\textstyle{\mathbb{Z}[t]\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{-t}$$\scriptstyle{\times(1-t)}$$\textstyle{\mathbb{Z}[t]\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{1}$$\scriptstyle{\times\Sigma}$$\textstyle{\mathbb{Z}[t]\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{1}$$\scriptstyle{\times(1-t)}$$\textstyle{\mathbb{Z}[t]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{1}$$\textstyle{\mathcal{A}\,:=}$$\textstyle{\mathbb{Z}[t]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\times(1-t^{-1})}$$\textstyle{\mathbb{Z}[t]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\times\Sigma}$$\textstyle{\mathbb{Z}[t]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\times(1-t^{-1})}$$\textstyle{\mathbb{Z}[t]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\times\Sigma}$$\textstyle{\mathbb{Z}[t]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\times(1-t)}$$\textstyle{\mathbb{Z}[t]}$

Note that dualizing multiplication by an element of $\mathbb{Z}[t]$
corresponds to involuting the element (e.g. substituting $t^{-1}$ for $t$).
This results in the change in sign between the components of $\phi$ on the
left and right sides, implied by Theorem 3.2.

Note that the fourth component (from the right) of $\phi$ is $-t$, not $-1$.
If $n$ is even then this is unavoidable; $\tilde{M}=S^{5}$ so we know
$H_{3}(\tilde{M})=0$ and Theorem 3.4 implies that $M$ is not anti-self-dual.
Conversely:

######  Proposition 3.5.

If $n=4k+1$ then $M$ is anti-self-dual.

###### Proof.

$\,\,\,\,{\rm
Let}\qquad\alpha=t^{k+1}+t^{k}-t^{-k}-t^{-(k+1)},\qquad\beta=\sum_{r=-k+1}^{k}t^{r}-\sum_{r=k+2}^{3k}t^{r}.$

Then $\alpha$ generates the same ideal as $t-1$, as
$\alpha(t^{1+k}+t^{1-k})=t^{2}-1$, which when multiplied by
$1+t^{2}+t^{4}+\cdots+t^{4k}$ returns $t-1$.

Now $\beta(1-t^{-1})=\alpha$ and $\Sigma\beta=\Sigma$. Thus
$\gamma\beta(1-t^{-1})=(1-t^{-1})$ for some $\gamma$, and
$\gamma\beta=1+s\Sigma$ for some integer $s$. Then $(\gamma-s\Sigma)\beta=1$
and $\beta$ is a unit. Thus we have a chain homotopy equivalence
$f\colon\mathcal{A}\to\mathcal{A^{\prime}}$:

$\textstyle{\mathcal{A}\,:=\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\textstyle{\mathbb{Z}[t]\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{1}$$\scriptstyle{\times(1-t^{-1})}$$\textstyle{\mathbb{Z}[t]\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{1}$$\scriptstyle{\times\Sigma}$$\textstyle{\mathbb{Z}[t]\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{1}$$\scriptstyle{\times(1-t^{-1})}$$\textstyle{\mathbb{Z}[t]\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\beta}$$\scriptstyle{\times\Sigma}$$\textstyle{\mathbb{Z}[t]\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{1}$$\scriptstyle{\times(1-t)}$$\textstyle{\mathbb{Z}[t]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{1}$$\textstyle{\mathcal{A^{\prime}}\,:=}$$\textstyle{\mathbb{Z}[t]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\times(1-t^{-1})}$$\textstyle{\mathbb{Z}[t]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\times\Sigma}$$\textstyle{\mathbb{Z}[t]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\times\alpha}$$\textstyle{\mathbb{Z}[t]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\times\Sigma}$$\textstyle{\mathbb{Z}[t]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\times(1-t)}$$\textstyle{\mathbb{Z}[t]}$

The chain homotopy equivalence $f\phi f^{*}$ is chain homotopic to the chain
map whose first three components are 1 and whose last three components are -1.
The required homotopy has only one non-zero component; $I_{2}=x$, where
$\alpha x=\beta-1$. ∎

## References

  * [1] D. Barden, Simply connected five-manifolds, _The Annals of Mathematics_ 2nd Series Vol. 82, No. 3 (Nov., 1965), 365–385.
  * [2] F.E.A.Johnson, _Stable modules and the D(2) problem, LMS 301_ , Cambridge University Press, Cambridge, 2003.
  * [3] R Kirby and L Siebenmann, _Foundational essays on topological manifolds, smoothings and triangulations, Annals of Mathematics Studies 88_ , Princeton University Press, Princeton, NJ, 1977.
  * [4] J. Klein, Poincaré duality spaces, _Surveys on surgery theory_ Vol 1, Annals of Mathematics Studies 145 (2000), 135–165.
  * [5] W.H. Mannan, Homotopy types of truncated projective resolutions, _Homology, Homotopy and Applications_ Vol. 9, No. 2 (2007), 445–449.
  * [6] C.T.C. Wall, _Surgery on compact manifolds, AMS Mathematical Surveys and Monographs 69_ , $2^{\rm nd}$ edition, Providence, RI, 1999.
  * [7] C.T.C. Wall, Quadratic forms on finite groups and related topics, _Topology_ 2 (1963), 281–298.