On the Wikipedia page Hom functor it is pointed out that the two hom functors are `related in a natural manner'. That is, the square

$$ \require{AMScd} \begin{CD} Hom(A, B) @>{Hom(A, g)}>> Hom(A, B') \\ @V{Hom(f, B)}VV @VV{Hom(f, B')}V \\ Hom(A', B) @>>Hom(A', g)> Hom(A', B') \end{CD} $$

commutes for all morphisms $f : A' \rightarrow A$ and $g : B \rightarrow B'$.

Does this also hold for the internal hom of a closed symmetric monoidal category? I think it should, and I've been trying to show it using the adjunction $$ Hom(A \otimes B, C) \cong Hom(A, C^B) $$ that is natural in all three variables. Note that with $-^B$ and $C^-$, I denote the co- and contravariant internal hom functors, respectively.

So far my attempts have been unsuccesful. Can anyone point me in the right direction?


1 Answer 1


If I understand correctly, what you want to know is whether the square $$\begin{CD}B^{A}@>{g^{A}}>> B'^{A} \\@V{B^f}VV @VV{B'^f}V \\ B^{A'} @>>{g^{A'}}> B'^{A'} \end{CD}$$ commutes in $\mathcal{C}$ for all $f:A'\to A$ and $g:B\to B'$. To prove this, it is enough, by the Yoneda lemma, to prove that both sides induce the same natural transformation between the associated presheaves; in other words, it is enough to prove that the following commutative square of sets is commutative for all $C$ $$\begin{CD}Hom(C,B^{A})@>{Hom(C,g^{A})}>> Hom(C,B'^{A}) \\@V{Hom(C,B^f)}VV @VV{Hom(C,B'^f)}V \\ Hom(C,B^{A'}) @>>{Hom(C,g^{A'})}> Hom(C,B'^{A'}). \end{CD}$$ But since the isomorphism $Hom(C \otimes A, B) \cong Hom(C, B^A)$ is natural in $A$ and $B$, this square is isomorphic to the square $$\begin{CD}Hom(C\otimes A,B)@>{Hom(C\otimes A,g)}>> Hom(C\otimes A,B') \\@V{Hom(C\otimes f,B)}VV @VV{Hom(C\otimes f,B')}V \\ Hom(C\otimes A',B) @>>{Hom(C\otimes A',g)}> Hom(C\otimes A',B'), \end{CD}$$ and this one commutes since it is a particular case of your first one.

  • $\begingroup$ Just to be sure: you mean: 'But since the isomorphism $Hom(A \otimes B, C) \cong Hom(A, C^B)$ isomorphic in $B$ and $C$', right? $\endgroup$ Commented Apr 9, 2018 at 16:17
  • 1
    $\begingroup$ @ArnoudNijon Yes for the isomorphism and no for the diagram below... I initially wanted to write the tensor-exponential ajunction as you did, but it was a bit awkward to switch the variables, so now I've changed it so that at least my answer is coherent. Thanks for pointing out the issue, I hope I haven't complicated this too much! $\endgroup$
    – Arnaud D.
    Commented Apr 9, 2018 at 17:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .