I have seen several proofs for $e$ idempotent $\Rightarrow$ $eA$ projective where $A$ is an algebra. I tried something different and produced a proof without using the fact that $e$ is idempotent (so it's obviously wrong). Can someone point out where my mistake is? Here is my (fake) proof:
Let $\psi:M \longrightarrow N$ be an epimorphism, where $M,N$ are $A$-modules, and let $\varphi: eA\longrightarrow N$ be a morphism of $A$-modules. I want to find and $A$-module morphism $\kappa: eA\longrightarrow M$ such that $\phi\circ \kappa = \varphi$. Since $eA$ is generated by $\{e\}$ as an $A$-module, I only have to determine how $\kappa$ acts on $e$. Let $u\in \psi^{-1}(\varphi(e))$. Then by defining $\kappa(e)=u$, I have my desired morphism.
Edit: When I talk about modules, I mean right modules.
Given a $A$-module $B$ that is generated by a single element $b\in B$, and given another $A$-module $C$ with a given element $c\in C$, you cannot be sure that the assignment $b\mapsto c$ extends to an $A$-linear map $f\colon B\to C$. In full generality, an $A$-module $B$ for which this would hold is a free $A$-module of rank 1. For instance, taking $A=\mathbb{Z}$ and $B=\mathbb{Z}/2$ with generator $1$, a general map of abelian groups $\mathbb{Z}/2\to C$ is specified by an assignment $1\mapsto c\in C$ for which $2c=0$, i.e. you would need to make sure that $c$ satisfies extra relations. If we take $A=k[x]/(x^2)$ and $e=x\in A$, then a map $xA\to C$ (with $C$ an $A$-module) is specified by choosing an element $c\in C$ for which $xc=0$. All in all, in your case you don't know that your choice $u\in\psi^{-1}(\varphi(e))$ satisfies the necessary relations for it to define a map $eA\to M$.