Wiki states that:
The pair of functors $ \DeclareMathOperator{\Hom}{Hom}\Hom(A, –)$ and $ \Hom(–, B)$ are related in a natural manner.
Than there is a commuting diagram, which I can't understand. Let's take one of its paths: $ \Hom(A, B) \xrightarrow{\Hom(A, f: B \rightarrow B')} \Hom(A, B') \xrightarrow{\Hom(h: A \rightarrow A', B')} \Hom(A', B')$
So, $\Hom(A, B) \xrightarrow{\Hom(A, f: B \rightarrow B')} \Hom(A, B')$ maps $\{ A \rightarrow B \}$ arrows to the $\{ A \rightarrow B' \}$ arrows by a known way. Here things getting unclear to me: according to the path above, next morphism is $\Hom(h: A \rightarrow A', B')$. The problem is that it does not work with arrows of type $\{ A \rightarrow B' \}$, it instead expects arrows of type $\{ A \leftarrow B' \}$.
Hence, I got stuck.
I assume that this is some kind of common "abuse of notation".
I would like to know how to read abovementioned diagram and define proper natural transformation (or natural isomorphism?) between the pair of functors $\Hom(A, –)$ and $\Hom(–, B)$?
In short: the commutativity of the diagram tell us two things at the same time and both things involve natural transformations:
If $\tau_B:=\mathsf{Hom}(h,B):\mathsf{Hom}(A,B)\to\mathsf{Hom}(A',B)$ for every object $B$ then $\tau$ is a natural transformation $\mathsf{Hom}(A,-)\to\mathsf{Hom}(A',-)$.
If $\rho_A:=\mathsf{Hom}(A,f):\mathsf{Hom}(A,B)\to\mathsf{Hom}(A,B')$ for every object $A$ then $\rho$ is a natural transformation $\mathsf{Hom}(-,B)\to\mathsf{Hom}(-,B')$.
Let $\mathcal C$ denotes the category that is involved.
Then $h$ denotes an arrow in homset $\mathcal C^{\text{op}}(A,A')=\mathcal C(A',A)$ and $\tau_B=\mathsf{Hom}(h,B)$ is prescribed by: $$u\mapsto u\circ h$$
Further $f$ denotes an arrow in homset $\mathcal C(B,B')$ and $\rho_B=\mathsf{Hom}(A,f)$ is prescribed by: $$u\mapsto f\circ u$$