# A subobject allows $A\to B$ iff its cokernel kills $A\to B$

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 $S$?

• Try *allows* allows. :) Or $\textit{allows}$ $\textit{allows}$. Also, perhaps you can show something related to the kernel of the cokernel of $S\xrightarrow{s}B$. – awllower Oct 20 '16 at 15:00
• @awllower thanks!! I can find a map from $A$ to $ker(cok(s))$ but don't know what to do next.. – kousaka Oct 20 '16 at 15:14
• 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. – awllower Oct 20 '16 at 15:49
• @awllower thanks a lot!!! – kousaka Oct 20 '16 at 16:09
• It is my pleasure to help, and to find out such a good book. :D – 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 $\ker (\text{coker} (\ker f))=\ker f$.

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