I am learning elementary category theory. My question was raised while I was reading a part about a characteristic of adjoint functors.
We adopt a policy for the notation $h_A(X) = \mathrm{Mor}(X, A)$ and $h^A(X) = \mathrm{Mor}(A, X)$. We use $C^\wedge$ to denote the category of presheaves on a category $C$. Thus, for $A$ being an object of a category $C$, the functor $h_A$, or $h^A$, is an object of $C^\wedge$, or $C^{\mathrm{op}\wedge}$, respectively.
Let $C, C'$ be categories. Let $F \colon C \to C'$ and $G \colon C' \to C$ be functors. We use the same notations for $F \colon C^{\mathrm{op}} \to C'^{\mathrm{op}}$ and $G \colon C'^{\mathrm{op}} \to C^{\mathrm{op}}$ We have a functor between presheaves $G^*: C^{\mathrm{op}\wedge} \to C'^{\mathrm{op}\wedge}$. Let $$ \psi \colon h_{C'^{\mathrm{op}}} \circ F \to G^* \circ h_{C^{\mathrm{op}}} $$ be a morphism between functors. Here, we used $h_C \colon C \to C^\wedge$ to denote a functor defined by $h_C(A) = h_A$ for an object $A$ and $h_C(f) \colon h_A \to h_B$ with $h_C(f)(X) = f_*$ for a morphism $f \colon A \to B$. We note that, actually, $h_{C^{\mathrm{op}}}(A) = h^A$ and $h_{C^{\mathrm{op}}}(f)(X) = f^*$.
For any $A$ being an object of $C$, for any $B$ being an object of $C'$, we have a function $$ \psi (A) (B) \colon \mathrm{Mor}_{C'} (F(A), B) \to \mathrm{Mor}_C (A, G(B)). $$
We want to define a morphism of functors $$ \varphi \colon \mathrm{Mor}_{C'} (F(\cdot), \cdot) \to \mathrm{Mor}_C (\cdot, G(\cdot)) $$ by $\varphi(A, B) = \psi(A)(B)$. My question is how to prove that $\varphi$ is really a morphism of functors.
Let me specify notations of proof. Let $f \colon A \to B$ be a morphism of $C$ and $f' \colon A' \to B'$ be a morphism of $C'$. We consider $f \times f' \colon A \times A' \to B \times B'$. Then, what we want to prove is that the following diagram is commutative. $$ \require{AMScd} \begin{CD} \mathrm{Mor}_{C'} (F(B), A') @> \mathrm{Mor}_{C'} (F(\cdot), \cdot)(f \times f') >> \mathrm{Mor}_{C'} (F(A), B') \\ @V \psi(B)(A') VV @V \psi(A)(B') VV\\ \mathrm{Mor}_{C} (B, G(A')) @> \mathrm{Mor}_C (\cdot, G(\cdot)) (f \times f') >> \mathrm{Mor}_{C} (A, G(B')). \end{CD} $$ Let $k \colon F(B) \to A'$ be a morphism of $C'$. This commutativity is equivalent to the following equation. $$ \psi(A)(B')(f' \circ k \circ F(f)) = G(f') \circ \psi(B)(A')(k) \circ f. $$ I can't prove this at the moment. Therefore I have decided to post this question. I would appreciate it if you give this a proof.