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

**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

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

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)

There are also: