Tensor product and idempotents

Question

Given an associative unital ring $R$ and an idempotent $e^2=e\in R$ one knows that $\operatorname{End}_R(Re)\simeq eRe \simeq \operatorname{End}_R(eR)$ as rings. Also we have a surjective $R-R$-bimodule map $Re \otimes_{eRe} eR \to ReR$ sending $ae \otimes eb$ to $aeb$. My question is whether this map is always an isomorphism. (In the case I am interested, $eRe$ is a commutative domain and $Re$ (resp. $eR$) is a free right (resp. left) $eRe$-modules of rank, say $n$.)

Answer 1

Let $k$ be a field, let $R$ be the path algebra over $k$ of the quiver $$1\xrightarrow{\gamma}2\xrightarrow{\delta}3$$ modulo the relation $\gamma\delta=0$, and let $e=e_2$ be the idempotent associated to the middle vertex.

Then $eRe\cong k$, $\dim_k(ReR)=3$, $\dim_k(eR)=2=\dim_k(Re)$, so the map $Re \otimes_{eRe} eR \to ReR$ is not an isomorphism. Specifically, $\gamma\otimes\delta$ is in the kernel.