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.
Best Answer
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.