Short exact sequence is split iff contractible Let $0\rightarrow A\overset{f}{\rightarrow} B \overset{g}{\rightarrow} C\rightarrow 0$ be a short exact sequence in an abelian category. I am trying to prove this SES is contractible iff it is split.
I managed the $\implies$ direction but I am having trouble with the converse.
First, the diagram:

The chain homotopy relations in the left and right diamonds are respectively $s_n\circ f+0\circ s_{n+1}=1_A$ and $s_{n+2}\circ 0+g\circ s_{n-1}=1_C$. By assumption, the sequence is split, so we can (and must) take $s_n$ to be the left inverse of $f$ and $s_{n-1}$ to be the right inverse of $g$. But now, in the middle diamond we are left with the relation $s_{n-1}g+fs_n=1_{A\oplus C}$. I can't seem to show why this arrow must be the identity.
 A: $\DeclareMathOperator{\id}{id}$I think you mixed up the indices a bit. What you need to prove is that $f \circ s_n + s_{n-1} \circ g = \id_{A \oplus C}$. For simplicity I will go back to $B = A \oplus C$ for the middle term, and I will denote $r = s_n$, $s = s_{n-1}$. We have $r \circ f = \id_A$ and $g \circ s = \id_C$.
So we want to be that $\id_B = fr + sg$. Consider $\varphi = \id_B - sg$; then $g\varphi = g - gsg = g - g = 0_B$. Thus by exactness $\varphi$ factorizes through $A$ (by definition of the kernel), say $\varphi = fh$ for some $h : B \to A$. Then $fr\varphi = frfh = fh = \varphi$. Given the expression of $\varphi$, this translates to:
$$\begin{align}
& fr - frsg = \id_B - sg \\
\iff & \id_B = fr + sg - frsg.
\end{align}$$
So we're left with proving that $frsg = 0 \iff rsg = 0$ (because $f$ is monic). We thus have to prove that $rs = 0$ (since $g$ is epic).
But now this is because of how you constructed the (left/right) inverses of $f$ and $g$. Typically you start with either $r$ or $s$, and you construct the other out of it (by following the procedure described there for example); and if you follow this procedure, you'll see that it will satisfy $r \circ s = 0$. Another way to see this is that, to write $B = A \oplus C$, you implicitly assume that the left inverse of $f$ is the projection $A \oplus C \to A$ and the right inverse of $g$ is the injective $C \to A \oplus C$; it's clear that the composition of these two maps is zero.
