# Linear map between $M^*\otimes_RN\rightarrow \text{Hom}_R(M,N)$

For $R$-module $M$ and $N$, there is a linear map $M^*\otimes_RN\rightarrow \text{Hom}_R(M,N)$ sending each elements tensor $\varphi\otimes n$ in $M^*\otimes_RN$ to the linear map $M\rightarrow N$ defined by $m\mapsto \varphi(m)n$.

I think it is obvious and argue like this:

We have a bilinear map from $M^*\times N$ to $\text{Hom}_R(M,N)$ defined by $(\varphi,n)\mapsto \varphi(\cdot)n$. So by universal property of tensors, we get a linear map from $M^*\otimes_RN$ to $\text{Hom}_R(M,N)$.

However, the author of the note does not prove like this and argues seems more complicated:

The function $M^*\times M\times N\rightarrow N$ given by $(\varphi,m,n)\mapsto \varphi(m)n$ is trilinear, so (by a mentioned theorem) this trilinear map induces a bilinear map $B:(M^*\otimes_RN)\times M\rightarrow N$ where $B(\varphi\otimes m,n)=\varphi(m)n$. For fixed $t\in M^*\otimes_RN$, $B(t,\cdot)$ is in $\text{Hom}_R(M,N)$, so we have a linear map $f:M^*\times N\rightarrow\text{Hom}_R(M,N)$ defined by $t\mapsto B(t,\cdot)$.

I wonder why the author argued like this? Is there any problem in my simple proof?

-
 you're thing sounds legit to me – uncookedfalcon Nov 1 '12 at 4:33

He's essentially doing the same thing in a more complicated way for no apparent reason. Your proof is fine. The details of his "why" probably lie in the structure of the document itself. Maybe it's just a bad proof, or maybe he wasn't okay writing $\varphi( \cdot )n$ or something, but without the book I can't tell.