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Let $A$ and $B$ be rings, $M$ an $A$-module, $P$ an $B$-module, $N$ an $(A, B)$-bimodule. Then what is the meaning of $\operatorname{Hom}_A (M, \operatorname{Hom}_{B} (N, P ))$?

If $\operatorname{Hom}_A (M, \operatorname{Hom}_{B} (N, P ))$ is the $A$-linear maps from $M$ to $\operatorname{Hom}_{B} (N, P )$, doesn't $\operatorname{Hom}_{B} (N, P )$ have to be an $A$-module? Should it be made into one in some sense?

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It can indeed be made an $A$-module, in the following way: \begin{align} A\times \operatorname{Hom}_B(N,P)&\longrightarrow \operatorname{Hom}_B(N,P)\\ (a,f)&\longmapsto \bigl(af\colon n\mapsto f(an)\bigr) \end{align} You can check the axioms for the structure of $A$-module from the compatibility conditions for the structure of $(A,B)$-bimodule of $N$.

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