2
$\begingroup$

Here is the question I encountered:

$$\require{AMScd} \begin{CD} s @>{f^\prime}>> a @>{\varphi^\prime}>> \bar a\\ @V{g^\prime}VV @V{g}VV @V{\bar g}VV\\ b @>{f}>> c @>{\varphi}>> \bar c \end{CD}$$

Given a commutative diagram as above in an abelian category such that $(f^\prime,g^\prime)$ is a pullback of $(f,g)$ and that $\varphi$, $\varphi^\prime$ are cokernel of $f$, $f^\prime$ respectively. Then $\bar g$ is monic.

I proved it using diagram chasing but I wonder if there is any simpler and more straightforward proof. I view this as a small generalization of the fact that a pullback of an epimorphism is epic in an abelian category, I think it's highly possible to find one, but I haven't succeeded yet. Does anyone have any idea?

$\endgroup$
  • $\begingroup$ Hi, can you post your diagram chasing? Because I can't get the proof. Thank you! $\endgroup$ – freedfromthereal Apr 12 '17 at 9:24
  • $\begingroup$ @freedfromthereal Never mind, the diagram chasing method is not so illustrative. My newest thought is that I can prove the case when $f$ is epic or monic at first, and then combine these two special cases to yield the general one. $\endgroup$ – Censi LI Apr 12 '17 at 9:38
  • $\begingroup$ And what if $f$ is monic? This can be interesting for me. $\endgroup$ – freedfromthereal Apr 12 '17 at 9:57
  • $\begingroup$ @freedfromthereal When $f$ is monic, it is the kernel of $\varphi$, and in this situation one can prove that $f'$ is the kernel of of $\varphi g$. In fact the monic case holds more generally than epic case, for in any category, the pullback of equalizer is again an equalizer. $\endgroup$ – Censi LI Apr 12 '17 at 10:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.