Let $0 \to A \stackrel f\to B \stackrel g\to C \to 0$ be short exact sequence of modules ($f:A \to B$, $g:B \to C$). Suppose that there exists $\alpha: B \to A$ and $\beta: C \to B$ such that $g \beta = id_C$, $\alpha f = id_A$. How can I prove that $f \alpha + \beta g = id_B$?
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The claim is not correct (see below). However, the following is true: Assume that we have the short exact sequence, as well as some $\beta : C \to B$ with $g \beta = 1_C$. Then there is a unique $\alpha : B \to A$ with the desired properties:
The morphism $\beta g - 1_B$ has the property that $g(\beta g - 1_B) = g - g = 0$. Hence, by exactness, there is a unique $\alpha : B \to A$ such that $\beta g - 1_B = f \alpha$, i.e. $f \alpha + \beta g = 1_B$. It also follows that $f \alpha f = f$, hence $\alpha f = 1_A$ since $f$ is monic.
If $\alpha' f = 1_A$, this means that $\alpha' = \alpha + hg$ for some $h : C \to A$, and then $f \alpha' + \beta g = 1_B + f h g$ has no reason to coincide with $1_B$.
Very Big Hint: Proceed in two parts.
(I believe that Martin is right; this seems to require that $\alpha\beta=0$.)