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 $Coker(f) = Coker(p_1)$?

  • 2
    $\begingroup$ There's no reason for that to be true. Take $C = 0$, for example. $\endgroup$ – Zhen Lin Dec 4 '13 at 9:34

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)

  • $\begingroup$ 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! $\endgroup$ – freedfromthereal Dec 4 '13 at 9:43
  • $\begingroup$ What if I add the hypothesis that $g$ and $p_2$ are epimorphisms? $\endgroup$ – freedfromthereal Dec 4 '13 at 10:06
  • $\begingroup$ If $g$ is an epimorphism, then the map of cokernels is also an epimorphism. $\endgroup$ – Hurkyl Dec 4 '13 at 19:46
  • $\begingroup$ I was looking again at this question and I can't get, at the moment, why the map between the two cokernels is a monomorphism. Can you explain? $\endgroup$ – freedfromthereal Apr 12 '17 at 10:17
  • $\begingroup$ @freedfromthereal: $\operatorname{coker}(p_1)$ is isomorphic to the image of the composite $C \xrightarrow{g} B \to \operatorname{coker}(f)$. I don't recall an easy way to see this; I only know an argument with elements (construct the inverse map), or maybe spectral sequence argument. If it helps, note that $\ker(p_1) \cong \ker(f)$ as well. $\endgroup$ – Hurkyl Apr 12 '17 at 13:59

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