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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