About equivalent definitions of cohomology in abelian categories

Question

I was trying to understand why certain definitions of cohomology in abelian categories are equivalent (isomorphic) and I came to the proof below, but I don't see how he concludes that $\mathrm{Ker}\ u\simeq \mathrm{Im} f$. Can you please help?

Answer 1

Probably this is just one of many possible solutions.

You have that $\text{Ker} u$ is the pullback of $\ker (X \to \text{Coker f})$ along the map $\text{Ker} g \to X$. So we have the following pullback diagram $$ \require{AMScd} \begin{CD} \text{Ker} u @>>> \text{Ker} g \\ @VVV @VVV \\ \text{Im} f @>>> X \end{CD} $$

From the pullback property we get a morphism $\text{Ker} u \to \text{Im}f$ which is a monomorphism.

From the commutative diagram $$ \begin{CD} \text{Im} f @>>> \text{Ker} g \\ @V{\text{id}}VV @VVV \\ \text{Im}f @>>> X \end{CD} $$ and the pullback property of $\text{Ker} u$ we get a morphism $\text{Im}f \to \text{Ker} u$. This morphism is a left inverse to the map $\text{Ker} u \to \text{Im}f$.

From this it follows that the morphism $\text{Ker} u \to \text{Im}f$ is an isomorphism being a monomorphism with left inverse.

I hope this helps.