# Does a cokernel in a exact sequence induce a monomorphism?

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.

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