# Let $0\to A\to B\to C\to 0$ be a short exact sequence of complexes in and Abelian category, and let $A$ and $B$ be exact. Then $C$ is exact.

Fix an arbitrary abelian category $\mathscr{A}$, and let $$0\to A\xrightarrow{f}B\xrightarrow{g}C\to 0$$ be a short exact sequence in the category of chains $\mathscr{A}_\bullet$, where $A$, $B$, and $C$ have chain maps $\varphi^A_n:A_n\to A_{n-1}$, $\varphi^B_n:B_n\to B_{n-1}$, $\varphi^C_n:C_n\to C_{n-1}$ respectively, and let $A$ and $B$ be exact. I claim that $C$ is exact.

My current approach is to try to show that $$\ker\varphi^C_{n-1} = \mathrm{coker\,}(\ker\varphi^C_n\hookrightarrow C_n) = \mathrm{coim\,}\varphi_n^C = \mathrm{im\,}\varphi_n^C.$$

So, for an arbitrary object $M\in\mathscr{A}$ and morphism $\psi:C_n\to M$ such that $$\left(\ker\varphi^C_n\hookrightarrow C_n\xrightarrow{\psi}M\right) = 0$$ I wish to show that there exists a unique $\ker\varphi^C_{n-1}\to M$ such that $$\left(C_n\twoheadrightarrow\mathrm{im\,}\varphi^C_n\hookrightarrow\ker\varphi^C_{n-1}\to M\right) = \left(C_n\xrightarrow{\psi}M\right).$$ However, despite playing around a lot with commutative diagrams, kernels, and cokernels, I haven't found a good way of doing this. What have I missed?

Then choose an element $c_n\in C_n$ which is in the kernel, and do some diagram chasing to show there's a $c_{n+1}$ such that $\varphi_n^C(c_{n+1})= c_n$.
For instance, to start off, $c_n$ lifts to some $b_n\in B_n$; then by commutativity, $g\varphi_n^B(b_n)=\varphi_n^C g(b_n)=0$, so by exactness we can write $\varphi_n^B(b_n)=f(a_{n-1})$ for some $a_{n-1}\in A_{n-1}$. Do something similar a few more times, using the exactness of $A$ and $B$, and you'll find your $c_{n+1}$.
• I figured the proof in elements would be doable, but in an Abelian category, it's not a given that we have elements (i.e. that there exists a forgetful functor that preserves isomorphisms). Consider the category of sheaves on a topological space $X$. – Monstrous Moonshine Jan 7 '17 at 17:07