Theorem: Let $R$ be (not necessarily commutative) ring with unity and $e$ a non-zero idempotent. Then $${\rm Hom}_R(eR,eR)\cong eRe \,\,\,\, (\mbox{isomorphic as abelian groups}).$$
I gave argument for the proof:
(1) Any $f\in {\rm Hom}_R(eR,eR)$ is determined by $f(e)$ [This is simply because the right $R$-module appearing in the domain of $f$ is generated by single element $e$.]
(2) Next, since $e$ is idempotent, it is easy to see that $f(e)$ lies in $eRe$. So to each $f$, we associate an element in $eRe$.
So from these remarks, it is easy to obtain desired map for isomorphism in the Theorem, then one can show that the map is actually an isomorphism.
Q. Beside this symbolic argument, I didn't understand the statement of the theorem and intuition behind it. Can one explain what the upshot of the theorem is?