Does a cokernel in a exact sequence induce a monomorphism?

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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.

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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.