Covariant Hom Functor is Functor

Theorem

Let $\mathbf{Set}$ be the category of sets.

Let $C \in \mathbf C_0$ be an object of $\mathbf C$.

Let $\operatorname{Hom}_\mathbf C \paren {C, \cdot}: \mathbf C \to \mathbf{Set}$ be the covariant hom functor based at $C$.

Then $\operatorname{Hom}_\mathbf C \paren {C, \cdot}$ is a functor.

For brevity and readability, let us write $\operatorname{Hom}$ for $\operatorname{Hom}_\mathbf C$.

For any object $D$ of $\mathbf C$ and morphism $f: C \to D \in \operatorname{Hom} \paren {C, D}$, we have:

$\operatorname{Hom} \paren {C, \operatorname{id}_D} \paren f = \operatorname{id}_D \circ f = f$

and so:

$\operatorname{Hom} \paren {C, \operatorname{id}_D} = \operatorname{id}_{\operatorname{Hom} \paren {C, D} }$

Thus:

$\operatorname{Hom} \paren {C, \cdot}$ preserves identity morphisms

For composable morphisms $g$ and $f$ of $\mathbf C$ we furthermore have:

$\ds \operatorname{Hom} \paren {C, g \circ f} \paren x$

$=$

$\ds \paren {g \circ f} \circ x$

$\ds $

$=$

$\ds g \circ \paren {f \circ x}$

Composition of morphisms is associative

$\ds $

$=$

$\ds \operatorname{Hom} \paren {C, g} \circ \operatorname{Hom} \paren {C, f} \paren x$

for any $x \in \operatorname{Hom} \paren {C, \operatorname{dom} f}$ composable with $f$.

Hence the result, by Equality of Mappings.

$\blacksquare$