What exactly is an identity morphism?

Question

I come from a Physics/CS background and I just picked up a book on Category Theory for programmers. The first chapters talks about considering the $\leq$ as a morphism and seeing if we have a category.

I get the first requirement of associativity, but I'm having trouble with the identity morphism requirement. As I understand it, the identity morphism must satisfy $f\circ id_{A} = f$ and $id_{B}\circ f = f$ for some $f::A\rightarrow B$. I'm confused here because it seems that $id$ simply returns its input so wouldn't $id$ always exist?

My textbook says that because "every object is less than or equal to itself" then we have an identity morphism. I don't follow the reasoning here; why does the fact that $a\leq a$ being always true imply an identity morphism?

Answer 1

The morphisms of a general category need not be functions in any sense.
We can regard a category as a (possibly huge) directed graph, equipped with a(n associative) composition operation on the edges.

If we have a partial order relation $R$ on a set $X$, then taking elements of $X$ for objects and taking exactly one arrow (morphism) $x\to y$ if $x\,R\,y$ defines a category, in which the single arrow $x\to x$ is the identity morphism of object $x$.

Note also that ${\rm id}_x$ needs to satisfy ${\rm id}_x\circ f=f$ and $g\circ {\rm id}_x=g$ for all morphisms $f$ ending at $x$ and for all $g$ starting at $x$.

Answer 2

If you consider the category induced by the $\leqslant$ relation, you have a unique morphism from $A$ to $B$ if and only if $A\leqslant B$. That means that there is a unique morphism from $A$ to $A$, since $A\leqslant A$. Then you should also check that it is the identity morphism, which follows from the trivial computation that $A\leqslant A\leqslant B$ implies $A\leqslant B$ and $C\leqslant A\leqslant A$ implies $C\leqslant A$.

If you consider the category induced by the $<$ relation, in which you have a unique morphism from $A$ to $B$ if and only if $A < B$, you will find that there are no morphisms from $A$ to $A$, since $A\nless A$. You can introduce a new identity morphism for each object, and this will make this "oriented graph" into a category, but this will be the category we studied earlier, the one induced by the $\leqslant$ relation, not by $<$.