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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?

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Madara Uchiha
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1 Answers1

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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.

Giorgio Mossa
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    I had quite a hard time seeing that such $\operatorname{Ker}u\to\operatorname{Im}f$ exists. I am writing this comment for future people like me. To see why it exists, note that $\operatorname{Im}f = \operatorname{Ker}\operatorname{Cok}f$. So $\operatorname{Ker}(X\to\operatorname{Coker}f)$ in the answer is just $\operatorname{Im}f\to X$. Since $\operatorname{Ker}u\to\operatorname{Ker}g\to X\to\operatorname{Cok}f$ is zero, we get a desired morphism. I don't know if this was what the answerer meant, but it works anyway. – zxcv Aug 22 '21 at 10:04
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    I think you mean to say the morphism $\textrm{im} f\rightarrow \ker u$ is a right inverse to $\ker u \rightarrow \textrm{im} f$. – subrosar Jul 20 '22 at 21:47
  • They probably wrote left inverse instead of right inverse because if you write your function compositions as arrows, then the right inverse is on the left. – Sha Vuklia Jun 25 '23 at 16:49