Let $R$ be a ring with identity and let $R^n=P⊕P'$ be a direct sum decomposition with right $R$-modules as its components. We take $e\in\operatorname{End}(R^n_R)$ as the projection of $R^n$ onto $P$, namely, $e_{|P'}=0$ and $e_{|P}=$ identity over $P$. Hereby, $e$ would be a so called "full idempotent" in $\operatorname{End}(R^n_R)=S $ in the sense that $e^2=e$ and $SeS=S$. My Question:
Why the endomorphisms in $eSe$ are "exactly" those sending $P'$ to $0$ and $P$ to $P$?
Thanks in advance!
Clearly everything in $eSe$ sends $P\to P$ and $P'\to 0$.
Conversely if $f$ is a homomorphism which sends $P\to P$ and has $P'$ in its kernel, then $ef=f$ (since the image is contained in $P$) and $fe=f$ (just compute $f(p+p')=f(p)+0 = fe(p)+fe(p')=fe(p+p')$.
So $f=ef=efe\in S$.