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Let $\mathscr{A}$ be an Abelian category, and let $$A\xrightarrow{f} B\xrightarrow{g} C\to 0$$ be an exact sequence in $\mathscr{A}$, and let $h:X\to B$ be any morphism, which gives us a sequence $$A\xrightarrow{f}B\twoheadrightarrow\mathrm{coker\,}h\to\mathrm{coker\,}gh\to 0$$ where the map $\mathrm{coker\,}h\to\mathrm{coker\,}gh$ is given by universal properties. If we collapse the first two maps, I claim that the resulting sequence $$A\to\mathrm{coker\,}h\to\mathrm{coker\,}gh\to 0$$ is exact.

Using relatively elementary methods, I've managed to prove that this is a complex, and that the map $\mathrm{coker\,}h\to\mathrm{coker\,}gh$ is surjective. But when I try to prove that $$\mathrm{im}\left(A\xrightarrow{f}B\twoheadrightarrow\mathrm{coker\,}h\right) = \ker\left(\mathrm{coker\,}\to\mathrm{coker\,}gh\right)$$ I can't seem to line up any of the arrows to use the universal properties of the kernel and cokernel.

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This isn't too hard to prove with spectral sequences. Even if you are unfamiliar with them, maybe they'll still be inspirational.


Consider the bicomplex

$$ \begin{matrix} 0 &\to& X &\to& X \\ \downarrow & & \downarrow & & \downarrow & & \\ A &\to& B &\to& C \\ \downarrow & & \downarrow & & \downarrow & & \\ A &\to& \operatorname{coker}{h} &\to& \operatorname{coker}(gh) \end{matrix} $$

You can compute the homology of its total complex in two different ways. If we start by taking the horizontal homology, the next page of the spectral sequence is

$$ \begin{matrix} 0 & 0 & 0 \\ \bullet & 0 & 0 \\ \bullet & U & V \end{matrix} $$ where $\bullet$ indicates groups we don't care about. This implies the bottom two homology groups are $U$ and $V$.

If we compute by taking vertical homology first, the next page of the spectral sequence is $$ \begin{matrix} 0 & \bullet & \bullet \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} $$ from which we can determine the bottom two homology groups are both zero.

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