are two the following functors isomorphic

Question

Consider the following two functors that go $C^{op}\times C\to Set$, where $X$ is an object in $C$

1. $(A,B)\mapsto Set(C(X,A)\times A,B)$

2. $(A,B)\mapsto Set(C(X,A),Set(A,B))$

Are these two functors isomorphic? (i.e. is there an natural transformation between them that is an isomorphism?) I want to think that they are due to $Set(A\times B,C)\cong Set(A,Set(B,C))$, but I haven't grasped all the details.

Answer 1

Yes, in the end the two functors are isomorphic, but you need some care in specifying in what sense $Set(X,Y)\times Z$ is an object of $\cal C$ for every $Z$ (instead of $Set(C(X,A)\times A,B)$, you want to write ${\cal C(\cal C}(X,A)\times A,B)$, or the only way to make $Set(C(X,A)\times A,B)$ type-check is to impose ${\cal C} = Set$).

The correct statement involves tensors over the category of sets.

Now, the adjunction given by the fact that $\cal C$ is tensored $$ {\cal C}({\cal C}(X,A)\times A,B)\cong Set({\cal C}(X,A), {\cal C}(A,B)) $$ is a natural isomorphism in all arguments, and the claim follows.

Answer 2

If we have a fixed functor $U:C\to Set$ (typically the 'underlying set' functor) and a fixed object $X\in C$, then the cited natural isomorphism only implies $$Set(C(X,A)\times UA,\, UB)\ \cong\ Set(C(X,A),\, Set(UA,UB))\,.$$ In case $U$ admits a left adjoint $F$, we get that $Set(UA,UB)$ is further naturally isomorphic to $C(FUA,\, B)$, and by the way, if the unit $1_C\to FU$ of this adjunction is invertible (which is rarely the case, at least in categories of algebraic structures), then it will also be naturally isomorphic to $C(A,B)$.