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Let $\mathsf{C}$ be a an abelian category. Suppose $\pi \colon T \rightarrow \nabla$ is an epimorphism in $\mathsf{C}$. Consider the short exact sequence $$0 \rightarrow \operatorname{ker}(\pi) \rightarrow T \xrightarrow{\pi} \nabla \rightarrow 0 \,.$$ Let $X$ be an object in $\mathsf C$. Assume that $\operatorname{Ext}_{\mathsf{C}}^1(X, \operatorname{ker}(\pi)) = 0$.

Since $\operatorname{Hom}_{\mathsf{C}}(X, -)$ is left exact, we obtain an exact sequence $$0\rightarrow \operatorname{Hom}_{\mathsf{C}}\big(X, \operatorname{ker}(\pi)\big) \rightarrow \operatorname{Hom}\big(X, T\big) \xrightarrow{\pi_*} \operatorname{Hom}\big(X, \nabla\big) \,.$$

I am trying to show that post-composition with $\pi$ (i.e., $\pi_*$) is an isomorphism. Why is this true?

Excuse me if this is a trivial question. I am all new to homological algebra. I know that I have to use that $\operatorname{Ext}_{\mathsf{C}}^1(X, \operatorname{ker}(\pi)) = 0$.

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If $0 \to A \to B \to C \to 0$ is a short exact sequence in an abelian category, there is an associated long exact sequence of abelian groups

$$0 \to \mathrm{Hom}(X,A) \to \mathrm{Hom}(X,B) \to \mathrm{Hom}(X,C)\\ \to \mathrm{Ext}^1(X,A) \to \mathrm{Ext}^1(X,B) \to \mathrm{Ext}^1(X,C)\\ \to \mathrm{Ext}^2(X,A) \to \cdots$$

This is explained in books about homological algebra. This immediately proves that $\pi_*$ is surjective. I do not think that you can prove that $\pi_*$ is injective. What is the context of your question?

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  • $\begingroup$ We can’t prove that $\pi_{\ast}$ is injective in this generality, but the exact sequence gives us a criterion: it is injective iff there is no nonzero map $X \rightarrow \ker{\pi}$. $\endgroup$
    – Aphelli
    Dec 25, 2022 at 21:02
  • $\begingroup$ I don’t understand your comment: the exact sequence states that $\ker{\pi_{\ast}}$ is $\mathrm{Hom}(X,\ker{\pi})$? $\endgroup$
    – Aphelli
    Dec 25, 2022 at 21:32
  • $\begingroup$ The kernel of $\pi_{\ast}$ is the image of $\mathrm{Hom}(X,\ker{pi})$ in $\mathrm{Hom}(X,T)$, no? And this last map is injective, isn’t it? $\endgroup$
    – Aphelli
    Dec 25, 2022 at 23:06
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    $\begingroup$ I deleted my stupid comments. Sorry! $\endgroup$
    – Martin Brandenburg
    Dec 25, 2022 at 23:25
  • $\begingroup$ Thank you! I've just realised that surjectivity is all I need. The context is: I wanted to argue that any morphism $f \in \operatorname{Hom}(X,\nabla)$ has a lifting $\tilde{f} \in \operatorname{Hom}(X,T)$ with respect to $\pi$. $\endgroup$
    – Margaret
    Dec 26, 2022 at 6:19

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