Hom functor

Remarks

Associativity of composition gives $\Hom(A, f' \circ f) = \Hom(A, f') \circ \Hom(A, f)$, and $\Hom(A, \id_X)$ is the identity, so clause 1 defines a functor. Clause 2 defines a contravariant functor, that is, a functor on the opposite category $\mathcal{C}^{\mathrm{op}}$. The two together form the bifunctor $\Hom(-, -) \colon \mathcal{C}^{\mathrm{op}} \times \mathcal{C} \to \mathbf{Set}$ on the product category. Other notations are $\mathcal{C}(A, X)$, $h^A = \Hom(A, -)$ and $h_A = \Hom(-, A)$.

Example: in the category of real vector spaces, take $A = \R$. A linear map $\R \to V$ is determined by the image of $1$, so $\Hom(\R, V)$ is in bijection with $V$, and $\Hom(\R, f)$ corresponds to $f$ itself.

Non-example: the assignment $X \mapsto \Hom(A, X)$ with $f \mapsto (g \mapsto g \circ f)$ is not a covariant hom functor. For $f \colon X \to Y$ and $g \colon A \to X$, the composite $g \circ f$ is in general not defined. So it does not give a function $\Hom(A, X) \to \Hom(A, Y)$.