I'm inexperienced with category theory, and studying adjoint functors using the hom-functor definition. This has led me to a very basic question about bifunctors. Given bifunctors $F,G: \mathcal{C×D} \to \mathcal{E}$, it is easy to show that $F \cong G \implies F(A,-)\cong G(A,-)$ and $F(-,B)\cong G(-,B)$ for all $A \in ob(\mathcal{C}$) and $B \in ob(\mathcal{D}$). My question is whether the converse holds, that is, whether this "argumentwise" isomorphism (for lack of knowledge of any better term) implies a genuine isomorphism of functors.
As far as I can see, the converse would follow providing each natural isomorphism $\eta:F(A,-) \to G(A,-)$ given agrees with each natural isomorphism $\mu:F(-,B) \to G(-,B)$ at $(A,B)$ (that is, $\eta_B = \mu_A$), and possibly only in such a case, and surely this is not always true. When I read about adjoint functors, though, the definition using the hom-bifunctor ($hom(F(-),-) \cong hom(-,G(-))$) is usually treated as equivalent to the definition using individual hom-functors ($hom(F(A),-) \cong hom(A,G(-))$ for all $A \in \mathcal{C}$ and $hom(F(-),B) \cong hom(-,G(B))$ for all $B \in \mathcal{D}$), which suggests that the former, an isomorphism of bifunctors, does indeed follow from the latter, an "argumentwise" isomorphism. If the result is not true in general, I would be curious if and why it is true in this case.
Apologies if this is overly obvious or easily researched; I have no idea where to search for an answer to this question. Thanks in advance :)
It indeed follows if $\eta_B=\mu_A$ i.e. there are isomorphisms $\gamma_{A,B}:F(A,B)\to G(A,B)$ which are natural in each variable separately. That's just because a family of maps are natural in each variable separately iff. they are 'jointly' natural i.e. natural between functors $C\times D\to E$, and a natural isomorphism is just a natural transformation consisting of isomorphisms.
If you cannot have $\eta_B=\mu_A$ then this is false, simply because if $\gamma:F\cong G$ then we may let $\eta_B=\gamma_{-,B}$ and $\mu_A=\gamma_{A,-}$. The only remaining case is if you know there exist isomorphisms $F(A,-)\cong G(A,-)$ and $F(-,B)\cong G(-,B)$ but you're unsure whether or not it is possible to make these isomorphisms agree. I imagine it's generally false in this case too.