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Let $A$ be a ring (which might or might not be commutative), and let $M,N$ and $K$ be three bi-modules over $A$.

There are two Tensor-Hom adjunctions. One says that

$\operatorname{Hom}_A(M\otimes_A N, K) \cong \operatorname{Hom}_A(M,Hom_A(N,K))$.

The other says that

$\operatorname{Hom}_A(M\otimes_A N, K) \cong \operatorname{Hom}_A(N,\operatorname{Hom}_A(M,K))$.

Are these isomorphisms of bimodules?

If so, does this mean that the two bimodules $\operatorname{Hom}_A(N,\operatorname{Hom}_A(M,K))$ and $\operatorname{Hom}_A(M,\operatorname{Hom}_A(N,K))$ are isomorphic?

Hank Scorpio
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the L
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    Have you even tried simply writing down the isomorphism and checking whether it preserves the module structure? Category theory is not magic... – Najib Idrissi Nov 21 '15 at 07:41

2 Answers2

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Be careful. It's cleanest to describe the tensor-hom adjunction with three different rings instead of one, to make it as hard as possible to accidentally write down the wrong thing, so let $A, B, C$ be three different rings, let $_A M_B$ be an $(A, B)$-bimodule, let $_B N_C$ be a $(B, C)$-bimodule, and let $_A K_C$ be an $(A, C)$-bimodule. Then

$$\text{Hom}_C(M \otimes_B N, K) \cong \text{Hom}_B(M, \text{Hom}_C(N, K))$$

as $(A, A)$-bimodules, and

$$\text{Hom}_A(M \otimes_B N, K) \cong \text{Hom}_B(N, \text{Hom}_A(M, K))$$

as $(C, C)$-bimodules.

Specializing to the case that $A = B = C$ shows that your notation is sloppy (to be fair, so is mine): when you write $\text{Hom}_A$ you haven't been careful about whether this means left $A$-module or right $A$-module homomorphisms, and it has different meanings in the different parts of your adjunctions unless $A$ is commutative and $M, N, K$ are plain $A$-modules, in which case there's no need to make left/right distinctions.

(Specifically, $\text{Hom}_A$ means left the second, fifth, and sixth times you used it, but right the first, third, and fourth times.)

Qiaochu Yuan
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  • Do you mean the first two hom modules are isomorphic as left $A$-modules? – Jaehyeon Seo Jun 13 '21 at 20:24
  • @The Great Seo: hmm, okay, I thought about it for a bit and actually I mean it's an isomorphism of $(A, A)$-bimodules. There are two $A$-module structures available, one coming from acting on $M$ (contravariantly because it's in the first slot of the hom, so it switches to a right action) and one coming from acting on $K$ (this stays a left action), and the tensor-hom adjunction respects both. – Qiaochu Yuan Jun 25 '21 at 01:42
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It turns out the two bimodules you mention are isomorphic. Adjunction in general gives you the bijection you described. However, in the proof of the Hom/Tensor adjunction, the map that you define for the bijection can be seen to also be a homomorphism. Really you have to write out the proof in detail, and observe that you are dealing with homomorphisms. More information can be found here:

Adjointness of Hom and Tensor

In fact the exact statement you are asking about is mentioned here:

http://en.wikipedia.org/wiki/Tensor-hom_adjunction

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Sergio Da Silva
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