Suppose $e$ is an idempotent element of a unital ring $R$. I'm trying to show $Re$ is a projective left $R$-module. I've seen a few answers out there about showing $Re \oplus R(1-e) = R$ to finish the proof, but I'd like to show it directly using the definition I know about lifting. Let me lay out what I have so far.
Let $M$ and $N$ be $R$-modules and $h: Re \rightarrow N$ be a $R$-module homomorphism, and $g: M \rightarrow N$ be a surjective $R$-module homomorphism. We want to define a map $f$ so that $h = g \circ f$.
Since $Re = \{ re \mid r \in R\}$ is principally generated and $h$ is a $R$-module homomorphism, we have that $h$ is uniquely defined by $h(e)$, i.e. $h(re) = rh(e)$ for any $r \in R$. Let $h(e) = a \in N$. Then since $g$ is onto, we have a $b \in M$ with $g(b) = a$. Define $f: Re \rightarrow M$ by $f(re) = rb$. This is a $R$-module homomorphism.
Since we made a choice of $b \in M$, we should show $f$ is well-defined. But I'm not sure exactly how to do it. If $re=se$, then the fact that $f(re)=f(se)$ is immediate, so I feel like I should be showing something about $b$. That is, if $c \in M$ with $g(c) = a$ too, then the there is some $r \in R$ such that $c = rb$ so that the definition is consistent.
Up to this point, I don't think I've used idempotence at all, which of course should be necessary. Any advice on how to proceed would be appreciated.