I am following a course on category theory, using the notes Algebra and Topology of Shapira, and I am working on the following exercise. Let $A$ be a commutative ring and consider $\mathcal{C} = Mod(A)$. Prove that $Ext^1_A(N,M) = 0$ if and only if any short exact sequence $0 \to M \to E \to N \to 0$ splits, where $Ext^j_\mathcal{C}(X,Y) = R^n Hom_{\mathcal{C}}(-,Y)(X) = R^n Hom_{\mathcal{C}}(X,-)(Y).$
The direction "$\implies$" works out:
If $0 \to M \to G \to N \to 0$ is an arbitrary short exact sequence, then apply $Hom_A(N,-)$. This gives a short exact sequence $0 \to Hom_A(N,M) \to Hom_A(N, E) \to Hom_A(N,N) \to Ext_A^1(N,M)$. The last term being zero, this means that the map from $Hom_A(N, E) \to Hom_A(N,N)$ is an epimorphism, so there is a map $s: N \to E$, such that $g \circ s = id_N$, which is equivalent to saying that the original sequence splits (for example proposition 4.1.7 of the notes of Shapira).
I don't know how to prove the other direction.
Firstly, find a partial projective resolution $$\require{AMScd} \begin{CD} P_2 @>{d_2}>>P_1 @>{d_1}>> P_0 @>{\varepsilon}>> N. \end{CD}$$
The module $\text{Ext}^1_A(N,M)$ is calculated via applying $\text{Hom}_A(-,M)$ to this complex and taking the cohomology group at $\text{Hom}_A(P_1,M)$. Pick a class in this module represented by some $f\colon P_1\to M$ with $(d_2)^*f = f\circ d_2 = 0$. Forming the pushout $X$ of the two maps $d_1$ and $f$, we obtain a map of complexes. $$\require{AMScd} \begin{CD} P_2 @>{d_2}>> P_1 @>{d_1}>> P_0 @>{\varepsilon}>> N @>{}>> 0;\\ & @V{f}VV @V{g}VV @V{\text{id}}VV\\ 0 @>>> M @>{i}>> X @>{}>> N @>{}>> 0; \end{CD}$$ Even better, one can verify that the bottom complex is a short exact sequence, due to the fact that $f\circ d_2= 0$.
We now use the assumption, to obtain a splitting of this sequence, a map $r\colon X \to M$ such that $r\circ i = \text{id}_M$, a so called "retraction" of $i$. Since $f = r\circ g \circ d_1$, the element $r\circ g$ witnesses that $(d_1)^* [r\circ g] = [f]$ is a boundary in the cohomology group $\text{Ext}^1_A(N,M)$, and hence equal to 0.