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Is that true that in an abelian category $\mathcal{C}$ if I have the pullback diagram:

$$ \require{AMScd} \begin{CD} P @>{p_1}>> C\\ @V{p_2}VV @V{g}VV \\ A @>{f}>> B \end{CD} $$

with $f$ and $p_1$ monomorphisms, then $Coker(f) = Coker(p_1)$?

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There's no reason for that to be true. Take $C = 0$, for example. –  Zhen Lin Dec 4 '13 at 9:34

1 Answer 1

Nope! Let $C = 0$, for example.

What is true is that $\text{Coker}(p_1) \to \text{Coker}(f)$ is a monomorphism.

(it doesn't matter whether or not $f$ and $p_1$ are monomorphisms)

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Yeah sure, it was just an adjoint that I wrote, 'cause is in the case that I'm studyng, is not so important. Thank you so much! –  freedfromthereal Dec 4 '13 at 9:43
    
What if I add the hypothesis that $g$ and $p_2$ are epimorphisms? –  freedfromthereal Dec 4 '13 at 10:06
    
If $g$ is an epimorphism, then the map of cokernels is also an epimorphism. –  Hurkyl Dec 4 '13 at 19:46

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