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I don't know why the following statement is true:

Theorem : In an abelian category, if the following diagram $\require{AMScd}$ \begin{CD} X' @>{f'}>> Y' \\ @V{g'}VV @VV{g}V\\ X @>>{f}> Y \end{CD} is a pullback and $f$ is an epimorphism, then the diagram is a pushout.

My attempt :

Assume that we are given the following commutative diagram: $\require{AMScd}$ \begin{CD} X' @>{f'}>> Y' \\ @V{g'}VV @VV{b}V\\ X @>>{a}> Z \end{CD} Since the following diagram $\require{AMScd}$ \begin{CD} \mathrm{ker}(f) @>{0}>> Y' \\ @V{i_f}VV @VV{g}V\\ X @>>{f}> Y \end{CD} is commutative, we have a morphism $k:\mathrm{ker}(f)\to X'$ such that $g'k=i_f$ and $f'k=0$.

Then, $ai_f=ag'k=bf'k=0b=0$. And since $f$ is epi, $(Y,f)$ is a cokernel of $i_f$. Then we can make $h:Y\to Z$ such that $hf=a$. But, how can i show that $hg=b$?

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A commutative diagram like yours induces a complex $$X'\stackrel{\iota}\to X\oplus Y'\stackrel{\pi}\to Y$$ where $\iota$ is $(g',f')$ and $\pi$ is $(-f,g)$. The commutative square is a pullback iff $\iota=\ker\pi$ and a pushout iff $\pi=\text{coker}\,\iota$.

As you have a pullback, $$0\to X'\stackrel{\iota}\to X\oplus Y'\stackrel{\pi}\to Y$$ is an exact sequence. If $f$ is epi, so is $\pi$ and so we have a short exact sequence $$0\to X'\stackrel{\iota}\to X\oplus Y'\stackrel{\pi}\to Y\to0.$$ Then $\pi=\text{coker}\,\iota$: the commutative square is a pushout.

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  • $\begingroup$ I checked all detail. I totally understood your answer. Thank you. $\endgroup$
    – PMJ
    Apr 29 '20 at 21:25

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