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?