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

by Al Jebr   Last Updated August 11, 2019 16:20 PM

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

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