Problem: Let $(1)\quad 0\to \mathfrak a\to R\to R/\mathfrak a\to 0$ be an exact sequence. Show that if there exists $e\in\mathfrak a$ such that $e^2=e$ and $\mathfrak a=Re$, then $(1)$ splits.
Question: The proof is straightforward: let $k:R\to R/\mathfrak a$ be the canonical isomorphism. Consider $\sigma:R/\mathfrak a\to R$; $r+\mathfrak a\mapsto (1-r)e$. Then $\sigma$ is a section of $k$. Thus, $(1)$ splits.
But aren't we supposed to have $k\sigma=1_{R/\mathfrak a}$? I don't see why this proof works. Any insight? Thanks in advance.
Let $\varphi : R \to \mathfrak a$ such that $\varphi(r) = re$. This homomorphism is well defined since $\mathfrak a = Re$. Let $\iota : \mathfrak a \to R$ be the inclusion homomorphism.
Consider $a \in \mathfrak a$. Since $\mathfrak a = Re$, we can find $r \in R$ such that $a = re$. We have: $$ (\varphi \circ \iota)(a) = \varphi(re) = re^2 = re = a. $$
Thus, $\varphi \circ \iota = 1_\mathfrak a$ as desired.