# A certain post-composition is an isomorphism?

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 $$\operatorname{Ext}_{\mathsf{C}}^1(X, \operatorname{ker}(\pi)) = 0$$.

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?
• 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 $X \rightarrow \ker{\pi}$. Dec 25, 2022 at 21:02
• I don’t understand your comment: the exact sequence states that $\ker{\pi_{\ast}}$ is $\mathrm{Hom}(X,\ker{\pi})$? Dec 25, 2022 at 21:32
• 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? Dec 25, 2022 at 23:06
• 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 $\pi$. Dec 26, 2022 at 6:19