Bifunctoriality stronger than functoriality in each variable?

Question

Is it correct to say that bifunctoriality is a stronger condition than "functoriality in each variable"?

More precisely, suppose that $\mathcal A, B, C$ are categories and I have $F \colon \mathcal A \times \mathcal B \to \mathcal C$. Bifunctoriality of $F$ means that $F$ is a functor from the product category $\mathcal A \times \mathcal B$ to $\mathcal C$. On the other hand, functoriality in each variable means that $F(-, B) \colon \mathcal A \to \mathcal C$ is a functor for each fixed $B \in \mathcal B$, and likewise $F(A, -) \colon \mathcal B \to \mathcal C$ is a functor for each fixed $A \in \mathcal A$. Given $A \to A'$ and $B \to B'$, bifunctoriality requires the following diagram to commute, but I don't see why functoriality in each variable should, too: $\require{AMScd}$ \begin{CD} F(A,B) @>>> F(A',B)\\ @VVV & @VVV\\ F(A,B') @>>> F(A',B') \end{CD} My question has the following motivation: to define adjoint functors $F$ and $G$, some authors (e.g. Vakil's algebraic geometry, Altman/Kleiman's commutative algebra) will state that there should be a "natural bijection" $$\operatorname{Hom}_{\mathcal B}(F(A),B) \to \operatorname{Hom}_{\mathcal A}(A, G(B))$$ for every $A \in \mathcal A$ and $B \in \mathcal B$. They then go on to explain that by "natural" they mean some commutative squares that essentially amount to naturality in each variable. I was wondering why they don't just come out and say that $\operatorname{Hom}_{\mathcal B}(F(-), -)$ and $\operatorname{Hom}_{\mathcal A}(-, G(-))$ should be naturally isomorphic bifunctors on $\mathcal A^{\text{op}} \times \mathcal B$ (which is the approach taken on the Wikipedia page for adjoints), and I thought perhaps this latter requirement is actually a stronger one that happens not to matter for $\operatorname{Hom}$.

Answer 1

Yes, bifunctoriality is a stronger condition than functoriality in each variable, and yes, the commutative diagram you write down need not commute in the second case. I don't actually know an example off the top of my head though.

Edit: We can find easy examples by thinking about one-object categories. If $G, H$ are two groups and $BG, BH$ the corresponding one-object categories, then a bifunctor $BG \times BH \to C$ is an action of $G \times H$ on an object of $C$, but a map which is a functor in each variable separately is an action of the free product $G \ast H$ on an object of $C$.

However, this is irrelevant to your motivation because $\text{Hom}(F(a), b)$ and $\text{Hom}(a, G(b))$ are already bifunctors. Since every morphism in the product category $A^{op} \times B$ is canonically a composite of a morphism in $A$ and a morphism in $B$, to check that a collection of maps $\eta_{a, b} : \text{Hom}(F(a), b) \to \text{Hom}(a, G(b))$ assemble into a natural isomorphism of bifunctors you only have to check it with respect to morphisms in $A$ and then morphisms in $B$.