# User:Michiexile/MATH198/Lecture 1

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## Revision as of 12:47, 3 September 2009

## Contents |

## 1 Welcome, administrativia

## 2 Introduction

Why this course? What will we cover? What do we require?

## 3 Category

A *graph* is a collection *G*_{0} of *vertices* and a collection *G*_{1} of *arrows*. The structure of the graph is captured in the existence of two functions, that we shall call *source* and *target*, both going from *G*_{1} to *G*_{1}. In other words, each arrow has a source and a target.

We denote by *[v,w]* the collection of arrows with source *v* and target *w*.

A *category* is a graph with some special structure:

- Each
*[v,w]*is a set and equipped with a composition operation . In other words, any two arrows, such that the target of one is the source of the other, can be composed to give a new arrow with target and source from the ones left out.

We write if .

=>

- The composition of arrows is associative.
- Each vertex
*v*has a dedicated arrow 1_{v}with source and target*v*, called the identity arrow. - Each identity arrow is a left- and right-identity for the composition operation.

The composition of with is denoted by . A mnemonic here is that you write things so associativity looks right. Hence, *(gf)(x) = g(f(x))*. This will make more sense once we get around to *generalized elements* later on.

### 3.1 Examples

- The empty category with no vertices and no arrows.
- The category
*1*with a single vertex and only its identity arrow. - The category
*2*with two objects, their identity arrows and the arrow . - For vertices take vector spaces. For arrows, take linear maps. This is a category, the identity arrow is just the identity map
*f*(*x*) =*x*and composition is just function composition. - For vertices take finite sets. For arrows, take functions.
- For vertices take logical propositions. For arrows take proofs in propositional logic. The identity arrow is the empty proof:
*P*proves*P*without an actual proof. And if you can prove*P*using*Q*and then*R*using*P*, then this composes to a proof of*R*using*Q*. - For vertices, take data types. For arrows take (computable) functions. This forms a category, in which we can discuss an abstraction that mirrors most of Haskell. There are issues making Haskell not quite a category on its own, but we get close enough to draw helpful conclusions and analogies.
- Suppose
*P*is a set equipped with a partial ordering relation*<*. Then we can form a category out of this set with elements for vertices and with a single element in*[v,w]*if and only if*v<w*. Then the transitivity and reflexivity of partial orderings show that this forms a category.

Some language we want settled:

A category is *concrete* if it is like the vector spaces and the sets among the examples - the collection of all sets-with-specific-additional-structure equipped with all functions-respecting-that-structure. We require already that *[v,w]* is always a set.

A category is *small* if the collection of all vertices, too, is a set.