Let $A$ be a commutative ring, and $M$ be an $A$-module, and $M^\*$ be $\mathrm{Hom}\_A(M,A)$. Let $f$ be the map from $M \otimes\_A M^\*$ to $\mathrm{Hom}\_A(M,M)$, such that, for all $x=\sum\_i a\_i \otimes b\_i \in M \otimes\_A M^\*$, $f(x)$ is the homomorphism $y \in M \mapsto \sum\_ib\_i(y)a\_i \in M$.
Is it true that $f$ is always a monomorphism ?
If not, is there a necessary and sufficient condition on $M$ for $f$ to be a monomorphism ?



Tensor product and homomorphism
Here is how I would start.


Let $S$ be the class of all pairs of $A$-modules $(M,N)$ such that the canonical map
$$M \otimes N^\* \to \hom(N,M), ~ m \otimes \omega \mapsto (n \mapsto \omega(n) \cdot m)$$
is a monomorphism. Both sides are additive functors in both variables. It follows formally that $S$ is closed under finite direct sums (in both variables) as well as under direct summands (in both variables).


Clearly, $(A,N) \in S$ and $(M,A) \in S$. It follows that $(M,N) \in S$ whenever $M$ and $N$ are finitely generated and projective (in which case the map is actually an isomorphism).


We can also prove that $S$ is closed under infinite direct sums in the *first* variable: Let $I$ be any index set and $(M\_i,N) \in S$ for $i \in I$, then $(\bigoplus\_{i \in I} M\_i,N) \in S$: This is because the map
$$\textstyle (\bigoplus\_{i \in I} M\_i) \otimes N \to \hom(N,\bigoplus\_{i \in I} M\_i)$$ is mono iff the composition
$$\textstyle (\bigoplus\_{i \in I} M\_i) \otimes N \to \hom(N,\bigoplus\_{i \in I} M\_i) \hookrightarrow \hom(N,\prod\_{i \in I} M\_i)$$
is mono, which identifies with the mono
$$\textstyle \bigoplus\_i (M\_i \otimes N) \hookrightarrow \bigoplus\_{i \in I} \hom(N,M\_i) \hookrightarrow \prod\_{i \in I} \hom(N,M\_i).$$


In particular $(M,N) \in S$ whenever $M$ is a free module, and hence also when $M$ is a projective module. (This special case has been mentioned in the comments already, but I wanted to show a conceptual proof.)


Remark: $S$ is not closed under infinite direct sums in the second variable. In fact, it can happen that $M \otimes A^I \to M^I$ is not a mono, so that $(M,\bigoplus\_{i \in I} A) \notin S$.


The next case would be to look at two cyclic modules, say
$$M = A/I,~N = A/J$$
for two ideals $I,J$. Then
$$M \otimes N^\* \cong A/I \otimes \mathrm{Ann}(J) \cong \mathrm{Ann}(J)/(I \cdot \mathrm{Ann}(J))$$ and
$$\hom(N,M) \cong \{[a] \in A/I : aJ \subseteq I\} = (I:J)/I.$$
The canonical map then identifies with $[a] \mapsto [a]$. This is clearly not a monomorphism in general, even for $I=J$ in which case we just get the map $$\mathrm{Ann}(I) \to A/I,~a \mapsto [a]$$
with kernel $I \cap \mathrm{Ann}(I)$. This kernel can be non-trivial: take $A = k[X]/\langle X^2 \rangle$ and $I = \langle X \rangle$.


