In the proof of split short exact sequence, when $\sigma \alpha = \text{id}_A$, then the sequence is split.

Question

I'm on the last part of the proof from Eisenbud's Commutative Algebra, page 16.

Given a short exact sequence $0 \to A \xrightarrow{\alpha} B \xrightarrow{\beta} C \to 0$, if there is a hom $\sigma: B \to A$ such that $\sigma \alpha = 1_A$, then again the sequence is split, ie. there is a map $\tau: C \to B$ such that $\beta \tau = 1_C$ or that is the route of proof I want to take.

So clearly, since $\beta$ is surjective, we do have a set map $\tau$ such that $\beta \tau = 1_C$. Now I need to prove that $\tau$ is indeed a hom. The book just says to follow a "dual path" to recover $\tau$. That is a little advanced, so I just want the tedious / direct way.

Let $x, y \in C$. $\tau(x+y) = w\in B \in $ domain of $\sigma$ given then $\sigma w \in A$, so define $\tau'= 1_B - \alpha\sigma$. Since the sequence is exact, $\beta \alpha = 0$ so $\beta \tau' = \beta$. Here I'm lost.

Answer 1

Hint: Consider $D\subset B$ generated $\{x-\alpha(\sigma(x)),x\in B\}$. Show that it is a submodule and the restriction of $\beta$ to $D$ is an isomorphism.