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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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Do you know that if either of $\;\alpha\,,\,\beta\;$ exists then the sequence splits? In fact, this is an iff claim. – DonAntonio Oct 15 '13 at 4:45
up vote 3 down vote accepted

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 $1_B$.

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Very Big Hint: Proceed in two parts.

  1. Show that, if $y\in B$ satisfies the two equations $g(y) = 0$ and $\alpha(y)=0$, then $y=0$.
  2. Let $y=(f\alpha + \beta g)(x) - x$.

(I believe that Martin is right; this seems to require that $\alpha\beta=0$.)

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No need to use elements. The claim holds in arbitrary linear categories. – Martin Brandenburg Oct 15 '13 at 7:53

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