Let $G$ be a group and $$0\rightarrow K\rightarrow M\rightarrow N\rightarrow 0$$ a short exact sequence of groups. Now these are abelian groups, if I want to show that $\text{Hom}(G,M)\rightarrow \text{Hom}(G,N)$ is surjective, I would show that $\text{Ext}^1(G,K)=0$. However, if I'm studying the same question for non-abelian groups, then I do not have the tool of derived categories at my disposal. Can this be overcome with (non-abelian) cohomology?