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User:Michiexile/MATH198/Lecture 4

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1 Product

Recall the construction of a cartesian product: for sets S,T, the set S\times T=\{(s,t) : s\in S, t\in T\}.

The cartesian product is one of the canonical ways to combine sets with each other. This is how we build binary operations, and higher ones - as well as how we formally define functions, partial functions and relations in the first place.

This, too, is how we construct vector spaces: recall that \mathbb R^n is built out of tuples of elements from \mathbb R, with pointwise operations. This constructions reoccurs all over the place - sets with structure almost always have the structure carry over to products by pointwise operations.

Given the cartesian product in sets, the important thing about the product is that we can extract both parts, and doing so preserves any structure present, since the structure is defined pointwise.

This is what we use to define what we want to mean by products in a categorical setting.

Definition Let C be a category. The product of two objects A,B is an object A\times B equipped with maps p_1: A\times B\to A and p_2: A\times B\to B such that any other object V with maps A\leftarrow^{q_1} V\rightarrow^{q_2} B has a unique map V\to A\times B such that both maps from V factor through the p1,p2.

In the category of Set, the unique map from V to A\times B would be given by q(v) = (q1(v),q2(v)).

The uniqueness requirement is what, in the theoretical setting, forces the product to be what we expect it to be - pairing of elements with no additional changes, preserving as much of the structure as we possibly can make it preserve.

In the Haskell category, the product is simply the Pair type:

Product a b = (a,b)
and the projection maps p1,p2 are just
fst, snd

Recall from the first lecture, the product construction on categories: objects are pairs of objects, morphisms are pairs of morphisms, identity morphisms are pairs of identity morphisms, and composition is componentwise.

This is, in fact, the product construction applied to Cat - or even to CAT: we get functors P1,P2 picking out the first and second components, and everything works out exactly as in the cases above.

2 Coproduct

The other thing you can do in a Haskell data type declaration looks like this:

Coproduct a b = A a | B b
and the corresponding library type is
Either a b = Left a | Right b

This type provides us with functions

A :: a -> Coproduct a b
B :: b -> Coproduct a b

and hence looks quite like a dual to the product construction, in that the guaranteed functions the type brings are in the reverse directions from the arrows that the product projection arrows.

So, maybe what we want to do is to simply dualize the entire definition?

Definition Let C be a category. The coproduct of two objects A,B is an object A + B equipped with maps i_1:A\to A+B and i_2:B\to A+B such that any other object V with maps A\rightarrow_{v_1} V \leftarrow_{v_2} B has a unique map v:A+B\to V such that v1 = vi1 and v2 = vi2.

In the Haskell case, the maps i1,i2 are the type constructors A,B. And indeed, this Coproduct, the union type construction, is the type which guarantees inclusion of source types, but with minimal additional assumptions on the type.

In the category of sets, the coproduct construction is one where we can embed both sets into the coproduct, faithfully, and the result has no additional structure beyond that. Thus, the coproduct in set, is the disjoint union of the included sets: both sets are included without identifications made, and no extra elements are introduced.

  • Diagram definition
  • Disjoint union in Set
  • Coproduct of categories construction
  • Union types

3 Algebra of datatypes

Recall from last week that we can consider endofunctors as container datatypes. A few nice endofunctors we may have around include (with some abuse of notation):

data 0 a = Boring
data 1 a = Singleton a

with the obvious Functor implementations. From these, we can start building new container types, such as:

Bool = 0 + 0
Maybe = 1 + 0
Pair = 1 * 1

and we can use this to produce recursive definitions

List = 0 + 1*List
NonEmptyList = 1 + NonEmptyList

and to argue in a highly algebraic manner about data type definitions

List = 0 + 1*List
     = 0 + 1*(0 + 1*List)
     = 0 + 1*0 + 1*1*List
     = 0 + 1 + 1*1*List