Conditions for injectivity of the multiplication map $Re\otimes_{eRe} eM\to M$ Let $R$ be a noncommutative ring with $1$. Consider a left $R$-module $M$ and an idempotent element $e\in R$. Then $eRe$ is a ring with identity $e$, and there is an obvious multiplication map $\varphi:Re\otimes_{eRe} eM\to M$, which sends $re\otimes_{eRe} em$ to $rem$.
I would like to know when $\varphi$ is injective, but embarrasingly enough, I haven't even been able to cook up even one example where this fails.
This is what I have so far: the kernel of $\varphi$ is killed by everything in $ReR$, so in particular $\varphi$ is injective if $ReR=R$ (in fact, requiring that $ReR$ admit a multiplicative identity is enough, because such an element would necessarily fix all of $Re$ and therefore $Re\otimes_{eRe} eM$ as well). This type of argument comes up often in discussions about Morita equivalence and the like.
I guess my question is twofold:


*

*Can $\varphi$ fail to be injective?

*If so, under which other conditions (preferably on $R$) is it injective?


I would also be happy with an answer in the special case $R=M$.
A somewhat more general follow-up: if $N$ is a right $R$-module, can we say anything meaningful about the injectivity of the "inclusion" map $Ne\otimes_{eRe} eM\to N\otimes_R M$?
 A: If you are familiar with representations of quivers, here is an example where $\varphi$ is not injective.  Let $K$ be a field, and take $R$ to be the path $K$-algebra of the quiver
$$
1 \stackrel{a}{\to} 2
$$
I compose arrows from right to left.
Consider the semisimple representation given by
$$
M = K \stackrel{0}{\to} K.
$$
As a vector space, $M$ is isomorphic to $K^2$, so I'll write elements of $M$ as elements of $K^2$.
Let $e = e_1$ be the idempotent at vertex $1$.  Then $e_1 R e_1$ is isomorphic to the field $K$. Now, consider the element $a\otimes (1,0)$ of the tensor product $R e_1 \otimes_{e_1 R e_1} e_1 M$.  Then $\varphi(a\otimes (1,0)) = 0$.
However, the element $a\otimes (1,0)$ is non-zero.  This is because the tensor product is taken over $e_1 R e_1$, so the element $a$ cannot be passed to the right in $a\otimes (1,0)$.  Therefore, $\varphi$ is not injective.
In view of this relatively simple example, it seems that asking for $\varphi$ to be injective for any choice of idempotent $e$ is a pretty strong condition.  For instance, in the case where $R$ is, as above, a finite-dimensional path $K$-algebra of a quiver modulo some relations, this would only happen when there are no arrows between any pair of distinct vertices.
Added
Even in the case where $R=M$, $\varphi$ might not be injective.  Take $R$ to be the path algebra of
$$
1 \stackrel{a}{\to} 2 \stackrel{b}{\to} 3,
$$
modulo the relation $ba=0$.  Take $e=e_2$.  Then in $R e_2 \otimes_{e_2 R e_2} e_2 R$, the element $b \otimes a$ is non-zero, even though its image by $\varphi$ is zero.
