Questions about $eAe$-modules. Let $A$ be a $K$-algebra, $e$ be an idempotent, and $M$ be a right $A$-module. Let $f_M: \operatorname{Hom}_A(eA, M) \to Me$ be the map defined by $\varphi \mapsto \varphi(e)e$ for $\varphi \in \operatorname{Hom}_A(eA, M)$. We can show that $f_M$ is an isomorphism of right $eAe$-modules. By this result, take $M=eA$, we have $\operatorname{End}(eA)$ is isomorphic to $eAe$. It is said that the isomorphism $\operatorname{End}(eA) \to eAe$ induces an isomorphism of $K$-algebras. How to understand this? Is $\operatorname{End}(eA)$ isomorphic to $eAe$ as $K$-algebras? How to show that $f_M$ is functorial at $M$? Thank you very much.
 A: So there are two questions in there:

  
*
  
*Is $f_{eA}:\operatorname{End}(eA)\to eAe$ an isomorphism of $K$-algebras?
  
*Is $f_M$ functorial?
  

Let me answer them separately:


*

*So I think you believe that $f_{eA}$ is bijective (as it is an isomorphism of right $eAe$-modules). So we just have to prove that it is compatible with multiplication. The multiplication in $\operatorname{End}(eA)$ is composition, the multiplication in $eAe$ is just the restriction of the multiplication in $A$. So we have to prove that $f_{eA}(\varphi)\cdot f_{eA}(\psi)=f_{eA}(\varphi\circ \psi)$. By definition the left hand side is equal to $\varphi(e)e\cdot \psi(e)e$. Now the right hand side is $(\varphi\circ \psi)(e)\cdot e=\varphi(\psi(e))e$. Now since $\psi(e)\in eA$ we have $\psi(e)=e\psi(e)$, hence $$\varphi(\psi(e))e=\varphi(e\psi(e))e=\varphi(e)\psi(e)e,$$ where the last equality comes from the fact that $\psi(e)$ is an element of $A$ and $\varphi$ is a right $A$-module homomorphism. Now we can apply again $\psi(e)=e\psi(e)$ to get that the left hand side equals the right hand side.

*For the functoriality consider the following diagram for a right $A$-module homomorphism $r:M\to N$:
$$\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}
\begin{array}{c}
\operatorname{Hom}(eA,M) & \ra{f_M} & Me  \\
\da{r\circ -} & & \da{r|_{Me}} \\
\operatorname{Hom}(eA,N) & \ras{f_N} & Ne \\
\end{array}$$
So let $\varphi\in \operatorname{Hom}(eA,M)$. If you go first right then down you end up with $r(\varphi(e)e)$ and if you go first down then right you end up with $(r\circ \varphi)(e)e$. That both expressions are equal follows immediately from the fact that $r$ is a right $A$-module homomorphism.

