We work in the setting of an abelian category.

We define a monomorphism $S\to B$ allows $A\to B$ if $A\to B$ factors through it.

The title is a lemma in Peter Freyd's abelian categories, P42. The only if part is easy to prove, for proving the if part, say the cokernel of $S\xrightarrow{s}B$ kills $A\xrightarrow{f}B$, then I can find a morphism from $cok(f)$ to $cok(s)$ by using the universal property of cokernel. How do I construct a map from $A$ to $S$?

  • $\begingroup$ Try *allows* allows. :) Or $\textit{allows}$ $\textit{allows}$. Also, perhaps you can show something related to the kernel of the cokernel of $S\xrightarrow{s}B$. $\endgroup$ – awllower Oct 20 '16 at 15:00
  • $\begingroup$ @awllower thanks!! I can find a map from $A$ to $ker(cok(s))$ but don't know what to do next.. $\endgroup$ – kousaka Oct 20 '16 at 15:14
  • $\begingroup$ Have a look at Theorem 2.11 in the referenced book. :D In fact $\operatorname{Ker}(\operatorname{Coker}(s))=S$, so basically you just did what you had to do. $\endgroup$ – awllower Oct 20 '16 at 15:49
  • $\begingroup$ @awllower thanks a lot!!! $\endgroup$ – kousaka Oct 20 '16 at 16:09
  • $\begingroup$ It is my pleasure to help, and to find out such a good book. :D $\endgroup$ – awllower Oct 20 '16 at 16:10

I would proceed in a different way: using the fact that $$\text{coker}(s) \circ f=0$$ we know that $f$ factors through $\ker (\text{coker} (s))$, but for any monomorphisms $m$, in an abelian category, $\ker(\text{coker}(m))=m$.

The fact on monomorphisms follows from the fact that in an abelian category every monomorphism $m=\ker h$ for some morphism $h$ and that in any additive category $\ker (\text{coker} (\ker f))=\ker f$.

  • $\begingroup$ I didn't add the other implication because how the OP said is trivial, and since I do not want to insult anyone........ :D $\endgroup$ – Giorgio Mossa Oct 20 '16 at 15:36
  • $\begingroup$ It turns out that Theorem 2.11 in the referenced book proves this fact on monos already. :D $\endgroup$ – awllower Oct 20 '16 at 15:41

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.