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> [!definition] Definition. ([[category]])
> A **(locally small) category** $\mathsf{C}$ consists of
>- a class $\text{obj}(\mathsf{C})$ of **objects** of the category; and
>- or every two objects $A,B$ of $\mathsf{C}$, a set $\text{Hom}_{\mathsf{C}}(A,B)$ of **morphisms** or **arrows** satisfying:
>1. For every object $A$ of $\mathsf{C}$, there exists (at least) one morphism $1_{A} \in \text{Hom}_{\mathsf{C}}(A,A)$, the 'identity' on $A$;
>2. For every three objects $A,B,C$ there is defined a [[binary operation]] $\text{Hom}_{\mathsf{C}}(A,B) \times \text{Hom}_{\mathsf{C}}(B,C) \to \text{Hom}_{\mathsf{C}}(A,C)$ which determines from morphisms $f \in \text{Hom}_{\mathsf{C}}(A,B)$ and $g \in \text{Hom}_{\mathsf{C}}(B,C)$ a **composition** $gf \in \text{Hom}_{\text{C}}(A,C)$;
>3. This 'composition law' is [[associative]];
>4. The identity morphisms are identities with respect to composition: for all $f \in \text{Hom}_{\mathsf{C}}(A,B)$, we have $f1_{A}=f$ and $1_{B}f=f$;
>5. We always have $\text{Hom}_{\mathsf{C}}(A,B) \cap \text{Hom}_{\mathsf{C}}(C,D) = \emptyset$ unless $A=C$ and $B=D$.
>
When the category is understood, one may drop the index $\mathsf{C}$ — $f \in \text{Hom}(A,B)$ — or even use arrows as we do with set-functions: $f:A \to B$. $f \in \text{Hom}(A,A)=: \text{End}(A)$ is called an **endomorphism**.
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In general, $\text{Obj}(\mathsf{C})$ is 'too big' (in the sense of Russell's paradox) to be a set. When it *is* a set, we call $\mathsf{C}$ a **small category**.
>
Categories are objects of category $\mathsf{Cat}$, whose morphisms are [[covariant functor|functors]].
^definition
> [!intuition]
> As a prototype to keep it mind, it helps to think of the objects as 'sets' and of morphisms as 'functions'. This make properties 1-5 feel natural.
^intuition
> [!basicexample]
> Sets (as objects) together with set-functions (as morphisms) evidently form a category. It is denoted $\mathsf{Set}$.
^basic-example
> [!basicexample] Basic Example. (A pleasing small category)
> Suppose $S$ is a set and $\sim$ is a [[relation]] on $S$ satisfying reflexivity and transitivity. Then we may define a category thus:
>- Objects: the elements of $S$;
>- Morphisms: if $a,b$ are objects (that is, if $a,b \in S$), then let $\text{Hom}(a,b):= \ \begin{cases} \{(a,b) \in S \times S \} & \text{if } a \sim b \\ \emptyset & \text{else.} \end{cases}$.
>
Note that (unlike in $\mathsf{Set}$) the are very few morphisms: at most one for any pair of objects, and no morphisms at all between 'unrelated' objects.
>
We have to define 'composition of morphisms' and verify that the category definitions are satisfied. To begin, do we have 'identities'? Given an object $a \in S$, reflexivity implies $a \sim a$ and thus $\text{Hom}(a,a)=\{ (a,a) \}$. So we have no choice but to choose $1_{a}=\text{Hom}(a,a)$.
>
As for composition, let $a,b,c \in S$ be objects and $f \in \text{Hom}(a,b) , \ g \in \text{Hom}(b,c);$
we have to define a corresponding morphism $gf \in \text{Hom}(a,c)$. Of course $\text{Hom}(a,c)$ equals $(a,c)$ if $a \sim c$ and is empty otherwise. Since $a \sim b$, $b \sim c$ (their morphism sets are nonempty) it is indeed the case that $a \sim c$. And so we must define $gf=(a,c) \in \text{Hom}(a,c)$.
>
This operation is [[associative]]. Indeed, given additionally $h \in \text{Hom}(c,d)$, we have $hg=(b,d)$ and hence $h(gf)=(a,d)=(hg)f.$
It is immediate to show $1_{a}$ is an identity wrt composition.
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The most trivial case of this construction is when the [[equivalence relation]] is '$=