# Is $\mathscr F(\mathrm{Hom}_{R\text{-mod}}(M,N))=\mathrm{Hom}_{S\text{-mod}}(\mathscr F(M),\mathscr F(N))$?

Let $$R$$ and $$S$$ be commutative rings.

Let $$\mathscr F$$ is a functor from $$R$$-mod to $$S$$-mod.

Let $$M,N$$ be two $$R$$-modules. Then $$\mathrm{Hom}_{R\text{-mod}}(M,N)$$ is an $$R$$-module and $$\mathrm{Hom}_{S\text{-mod}}(\mathscr F(M), \mathscr F (N))$$ is an $$S$$-module.

Since $$\mathscr F: R\text{-mod} \to S\text{-mod}$$ is a functor, then this defines a function $$\mathrm{Hom}_{R\text{-mod}}(M,N) \to \mathrm{Hom}_{S\text{-mod}}(\mathscr F(M),\mathscr F(N)).$$

However, since $$\mathrm{Hom}_{R\text{-mod}}(M,N)$$ is itself an $$R$$-module, do we know what the object $$\mathscr F(\mathrm{Hom}_{R\text{-mod}}(M,N))$$ must be, based on $$\mathscr F$$?

Must it be the $$S$$-module $$\mathrm{Hom}_{S\text{-mod}}(\mathscr F(M),\mathscr F(N))$$?

• How about the functor that sends all $R$-modules to the $0$ $S$-module? – RghtHndSd Aug 11 at 16:17
• @RghtHndSd That would make both objects be zero. – k.stm Aug 11 at 16:19

First of all, they are not the same in general.

Let $$S = ℤ$$ and $$\mathscr F$$ be the forgetful functor $$R\mathrm{-Mod} → \mathrm{Ab}$$. Then $$\operatorname{Hom}_S (\mathscr F M, \mathscr F N)$$ is the set of abelian group homomorphisms $$M → N$$ and $$\mathscr F \operatorname{Hom}_R (M, N)$$ is the abelian group of $$R$$-module homomorphisms $$M → N$$. Clearly, they are not the same in general, as not every abelian group homomorphism $$M → N$$ has to respect the $$R$$-structure of the modules. Take $$M = N = ℂ$$ as $$ℂ$$-modules, for example.

The counterexample of course generalises to other scalar restrictions.

Then, in general, the concept you are looking for are functors of closed categories. These are functors of closed categories that respect the internal hom-functor. The category of $$R$$- or $$S$$-modules is not only a closed category, but a cartesian closed category.

So we have seen that not every functor $$R\mathrm{-Mod} → S\mathrm{-Mod}$$ respects the internal hom-functor, but there are functors that do. Boring examples include the null functor and the identity for $$R = S$$. If I find a nonboring example, I’ll let you know.

In general, there may be some connection. Taking again the scalar restriction example from above, we have for instance at least a natural inclusion map $$\mathscr F \operatorname{Hom}_R (M, N) → \operatorname{Hom}_S (\mathscr F M, \mathscr F N),$$ which just happens to almost never be an isomorphism. But at least it’s a natural inclusion.

A similar phenomenon occurs for $$R = S$$ when $$R$$ is an integral domain and $$\mathscr F = T$$ is taking torsion modules. For $$R$$-modules $$M$$, $$N$$, we have a map $$T \operatorname{Hom}_R (M,N) → \operatorname{Hom}_S (TM, TN),~f ↦ f|_{TM → TN}$$ but it doesn’t even have to be injective now, for example if $$M$$ happens to be nontrivial torsion-free and $$N$$ happens to be a nontrivial torsion module, with $$M$$ mapping nontrivially into $$N$$ somehow. So, $$M = ℤ$$ and $$N = ℤ/nℤ$$ over $$R = ℤ$$ for example.

If I can think of a nonboring example with a natural map going in the other direction, I’ll let you know.

Here is another kind of example: take $$R = S$$ and let $$\mathscr{F}$$ be the functor sending $$M \mapsto M^{\oplus 2}$$. Then $$\mathscr F(\mathrm{Hom}_{R\text{-mod}}(M,N)) \simeq (\mathrm{Hom}_{R\text{-mod}}(M,N))^{\oplus 2}$$ while $$\mathrm{Hom}_{S\text{-mod}}(\mathscr F(M),\mathscr F(N)) \simeq (\mathrm{Hom}_{R\text{-mod}}(M,N))^{\oplus 4}$$ so they would not be the same in general.