Artificial intelligent assistant

For a ring with idempotent $e$, is $r \mapsto ere$ a homomorphism? Let $R$ be a ring with idempotent $e$, then $eRe$ (the corner ring) is also a ring. Then is the map $\varphi(r) := ere$ a homomorphism? Surely if $e$ is central, then we have $e(rs)e = e^2(rs)e = erese = (ere)(ese)$, but what if $e$ is not neccessarily central?

For a concrete counterexample, consider $R=M_2(K)$ and the idempotent $e=\begin{pmatrix} 1 & 0 \cr 0 & 0 \end{pmatrix}$. Let $r=\begin{pmatrix} a & b \cr c & d \end{pmatrix}$ and $s=\begin{pmatrix} a' & b' \cr c' & d' \end{pmatrix}$, then we have $$ \phi(r)\phi(s)=\begin{pmatrix} aa' & 0 \cr 0 & 0 \end{pmatrix}\
eq \begin{pmatrix} aa'+bc' & 0 \cr 0 & 0 \end{pmatrix}=\phi(rs). $$

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