Prove that a "representable functor" is indeed a functor

Question

I guess I'm missing some obvious step here, but my goal is to prove that the definition of a representable functor indeed provides a functor. Now, a representable functor is $H^A \equiv Hom(A, -)$.

First, to prove that it takes identities to identities, I have that for $g:C \to B$, then $H^A(id_B) (g) = id_B \circ g = g$. All good. But this only shows that $H^A(id_B)$ is a left identity. I'd still have to prove that it is a right identity... Right? Or is the proof done and I'm missing something?

Answer 1

Let C and D be categories. A functor $F$ from $C$ to $D$ is a mapping that

1. Associates to each object $X$ in $C$ an object $F(X)$ in $D$,
2. Associates to each morphism $f: X → Y$ in $C$ a morphism $F(f): F ( X ) → F ( Y )$ in $D$ such that the following two conditions hold:

\begin{align} (a)& \quad \quad \quad \quad \quad \quad F(\mathrm {id} _{X})=\mathrm {id} _{F(X)}\,\! \text{ for every object } X \in C \\ (b)& \quad \quad \quad \quad \quad \quad F(g\circ f)=F(g)\circ F(f) \text{ for all morphisms } f : X → Y \text{ and } g : Y → Z \text{ in } C \end{align}

The term representable functor is usually used as an adjective, to describe any functor naturally isomorphic to $\mathrm{Hom}(A, -)$ functor for some fixed $A$ (or the natural contravariant version). The $\mathrm{Hom}(A, -)$ functor is a functor $F$ from some (typically locally small) category $C$ to the category Set, which associates to each object $X$ in $C$ the object $F(X) = \mathrm{Hom}(A, X)$ for some fixed $A$, and to each morphism $f:X → Y$ in the category $C$ the morphism $(h \mapsto f\circ h): \mathrm{Hom}(A, X) \to \mathrm{Hom}(A, Y)$. Here $\mathrm{Hom}(A, X)$ denotes the set of morphisms from $X$ to $A$. Saying that the category $C$ is locally small, is simply saying that the set $\mathrm{Hom}(Y, X)$ is a set for all $Y$ and $X$.

Now to prove that this is indeed a functor means to verify that condition (a) and (b) holds;

For property (a) $F(\mathrm{id}_{X})(g)=(g \circ \mathrm{id} _{X}) = g$ for all $g$. Therefore $F(\mathrm{id}_{X}) = \mathrm{id}_{Hom(A, X)}$. Notice that \mathrm{id}_{X} is a morphism in the category $C$, while $\mathrm{id}_{Hom(A, X)}$ is a morphism in the category of sets. That the identity element is a two-sided identity is something you have to check, when verifying that something is a category. When verifying that something is a functor you only have to give the assignments (1) and (2) and check conditions (a) and (b).

For property (b) we have that $$F(g\circ f)=(g\circ f) \circ h = g \circ (f \circ h) = F(g)\circ F(f)$$ Here the second equality follows from the associativity axiom, which holds by the assumption that $C$ is a category.