Lemma 10.12.8. For any three $R$-modules $M, N, P$,

$\mathop{\mathrm{Hom}}\nolimits _ R(M \otimes _ R N, P) \cong \mathop{\mathrm{Hom}}\nolimits _ R(M, \mathop{\mathrm{Hom}}\nolimits _ R(N, P))$

Proof. An $R$-linear map $\hat{f}\in \mathop{\mathrm{Hom}}\nolimits _ R(M \otimes _ R N, P)$ corresponds to an $R$-bilinear map $f : M \times N \to P$. For each $x\in M$ the mapping $y\mapsto f(x, y)$ is $R$-linear by the universal property. Thus $f$ corresponds to a map $\phi _ f : M \to \mathop{\mathrm{Hom}}\nolimits _ R(N, P)$. This map is $R$-linear since

$\phi _ f(ax + y)(z) = f(ax + y, z) = af(x, z)+f(y, z) = (a\phi _ f(x)+\phi _ f(y))(z),$

for all $a \in R$, $x \in M$, $y \in M$ and $z \in N$. Conversely, any $f \in \mathop{\mathrm{Hom}}\nolimits _ R(M, \mathop{\mathrm{Hom}}\nolimits _ R(N, P))$ defines an $R$-bilinear map $M \times N \to P$, namely $(x, y)\mapsto f(x)(y)$. So this is a natural one-to-one correspondence between the two modules $\mathop{\mathrm{Hom}}\nolimits _ R(M \otimes _ R N, P)$ and $\mathop{\mathrm{Hom}}\nolimits _ R(M, \mathop{\mathrm{Hom}}\nolimits _ R(N, P))$. $\square$

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