1
$\begingroup$

In an abelian category we have the following exact sequence: $$0\rightarrow A^0 \xrightarrow{a^0} A^1\xrightarrow{a^1} A^2 \rightarrow \ldots$$ As part of a bigger proof I consider the cokernel of $a^0$ and its arrow to $A^2$: $$A^1\xrightarrow{c} \text{coker}(a^0)\xrightarrow{\alpha} A^2 $$ I want to prove that the morphism $\alpha:\text{coker}(a^0)\rightarrow A^2$ is a monomorphism.

I know that $\text{im}(a^0)=\ker(\text{coker}(a^0))$ by definition of image and $\text{im}(a^0)=\ker(a^1)$ by exactness, therefore I have $\ker(\text{coker}(a^0))=\ker(a^1)$. In most categories, that would mean that the cokernel of $a^0$ would be a subobject of $A^2$, but I don't know how to formalize that intuition.

$\endgroup$
1
$\begingroup$

Every epimorphism is the cokernel of its kernel in an abelian category, so $\operatorname{coker}(a^0)=\operatorname{coker(\ker(a^1))}$. Thus $\alpha$ is in fact the (co-)image of $a^1$, and thus a monomorphism.

$\endgroup$
1
  • $\begingroup$ I get it, thanks! I think I could have also deduced it from exactness but "dual": $\text{coim}(a^1)=\text{coker}(a^0)$. $\endgroup$ – Dani Dec 9 '20 at 20:34

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.