# $F$ free $\implies$ $\mathrm{Hom}_R(M,R)\otimes F\to\mathrm{Hom}_R(M,F)$ is a monomorphism?

Suppose $M$ and $F$ are modules over a commutative ring $R$. In general, it is not true that the natural (functorial) homomorphism $\mathrm{Hom}_R(M,R)\otimes F\to\mathrm{Hom}_R(M,F)$ is a monomorphism.

However, if $F$ is actually a free module, then we actually do have a monomorphism on our hands. Is there a clear explanation of why is this so? Thank you.

-
Think about what happens if $F=\bigoplus F_i$, the $F_i$ being arbitrary. – Pierre-Yves Gaillard Mar 12 '12 at 5:54

This fact can be proved by simply appling a little of abstract non-sense. First of all you have that you module $F$ can be written as a direct sum (i.e. coproduct in the category of $R$-modules) $$F= \bigoplus_{i \in I} R$$ for some set $I$. So we have the following isomorphisms $$\hom(M,R) \otimes F \cong \hom(M,R) \otimes \bigoplus_{i \in I} R$$ because tensor product commute with direct sum you have $$\hom(M,R) \otimes F \cong \bigoplus_{i \in I} \hom(M,R) \otimes R$$ and then $$\hom(M,R) \otimes F \cong \bigoplus_{i \in I} \hom(M,R)\ \text{.}$$
Now there's the natural embedding $$\bigoplus_{i \in I} \hom(M,R) \hookrightarrow \prod_{i \in I} \hom(M,R) \cong \hom \left(M,R^I\right)$$ gives us the embedding $$f \colon \hom(M,R) \otimes F \hookrightarrow \hom\left(M,R^I\right)$$ for each $\varphi \in \hom(M,R)$ and $\sum_{i \in I} a_i e_i \in F$ (where $\{e_i | i \in I\}$ is a basis for $F$ and all but a finite number of $a_i$ is $0$) we have that $f(\varphi \otimes (\sum_{i \in I}a_i e_i))$ is the element of $\hom(M,R^I)$ such that for each $m \in M$ $$f\left(\varphi \otimes \left(\sum_{i \in I}a_i e_i\right)\right)(m)= (a_i \varphi(m))_{i \in I} \in R^I$$ Because just a finite number of $a_i$ is not null we have the $(a_i \varphi(m))_{i \in I} \in \bigoplus_{i \in I}R=F \subseteq R^I$, this said to us that $f(\varphi \otimes (\sum_{i \in I}a_i e_i)) \in \hom(M,F)$ for every $(\varphi \otimes (\sum_{i \in I}a_i e_i)) \in \hom(M,R) \otimes F$ and because this is a set of generators for $\hom(M,R) \otimes F$ this also says to us that $f(v) \in \hom(M,F)$ for every $v \in \hom(M,R) \otimes F$. So the $f$ give rise to a morphism (by restriction of the target of the morphism $f$) $$\tilde f \colon \hom(M,R) \otimes F \hookrightarrow \hom(M,F)$$ that I think it's your functorial homomorphism, which is a monomorphism.
Infinite direct sums do not commute with hom: in general, $\textrm{Hom}(\bigoplus M, N)$ gets turned into $\prod \textrm{Hom}(M, N)$, which gets turned into $\textrm{Hom}(M, \prod N)$. – Zhen Lin Mar 12 '12 at 8:57