$\DeclareMathOperator\Ext{Ext}\DeclareMathOperator\Hom{Hom}$I'm self-studying homological algebra and got stuck in my attempt to prove the following:
Let $A$ be a fixed abelian group. Then A is a free abelian group if $\Ext^{1}_{\mathbf Z}(A, F) \cong 0$ for every free abelian group F.
I've attempted to solve the exercise by making long exact sequences from suitable short exact ones (for example we know that there is a short exact sequence $ 0 \to F_1 \to F_0 \to A \to 0$ with $F_1$ and $F_0$ free), and by using that for every abelian group $B$ we know that $\Ext^{2}_{\mathbf Z}(A,B) \cong 0 $ and $\Ext^{0}_{\mathbf Z}(A,B ) \cong \Hom_{\mathbf Z}(A,B)$... but without luck so far.
$\DeclareMathOperator\Ker{Ker}$Let $F=\Bbb Z^{\oplus A}$ and $\pi:F\to A$ be the canonical homomorphism. Then we have an exact sequence $$\{0\}\to\Ker\pi\to F\to A\to\{0\}$$ By assumption this splits, hence $A$ is a projective (because it's a direct summand of the free abelian group $F$), hence free, abelian group.