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?



Non-abelian Ext functor and non-abelian $H^2$
EDITED, taking into account the [comments](https://mathoverflow.net/questions/368093/non-abelian-ext-functor#comment928981_368204) of Donu Arapura.


As JLA wrote, a homomorphism $f\colon G\to N$ gives an extension
\begin{equation}\label{e:E}
1\to K\to E\to G\to 1.\tag{E}
\end{equation}
This extension defines a homomorphism
$$b\colon G\to \operatorname{Out} K$$
called the *band* (lien, kernel) of \eqref{e:E}.
By definition, $H^2(G,K,b)$ is the set of isomorphism classes of extensions \eqref{e:E} bound by $b$.


A cohomology class $\eta(E)\in H^2(G,K,b)$ is called *neutral* if the extension \eqref{e:E} splits, that is, there exists a homomorphism $G\to E$ such that the composite homomorphism $G\to E\to G$ is the identity automorphism of $G$. In this case we obtain an action $\varphi$ of $G$ on the normal subgroup $K$ of $E$, and we obtain an isomorphism $E\overset{\sim}{\to}K\rtimes\_\varphi G$ with the semidirect product.


There may be more that one neutral class in $H^2(G,K,b)$: they correspond to semidirect products with different actions $\varphi$ of $G$ on $K$. I have read that there may be no neutral elements, but I don't know examples. (In the Galois cohomology setting, for a connected reductive group, by Douai's theorem there always exists a neutral element in nonabelian $H^2$; see [2], Proposition 3.1).


If $K$ is abelian, then $\operatorname{Out} K = \operatorname{Aut} K$, so $b$ is just an action of $G$ on $K$, and $H^2(G,K,b)$ is the usual abelian group cohomology $H^2(G,K)$, where $G$ acts on $K$ via $b$.


The set $H^2(G,K,b)$ can be described in terms of cocycles. See Section 1.14 in Springer [1].


The band $b$ defines an action of $G$ on the center $Z=Z(K)$, and we may consider the usual (abelian) group cohomology $H^2(G,Z)$. From the cocyclic description of $H^2(G,K,b)$ it is clear that $H^2(G,Z)$ naturally acts on $H^2(G,K,b)$.


Moreover, if the set $H^2(G,K,b)$ is nonempty, then $H^2(G,Z)$ acts on it simply transitively; see Mac Lane, [Homology](https://link.springer.com/book/10.1007/978-3-642-62029-4), Theorem IV.8.8. The set $H^2(G,K,b)$ is nonempty if and only if a certain obstruction $\operatorname{Obs}(G,K,b)\in H^3(G,Z)$ vanishes; see Mac Lane, Theorem IV.8.7.


Note that we should not think that $H^2(G,K,b)$ "equals" $H^2(G,Z)$. First, $H^2(G,K,b)$ does not have a distinguished unit element. Secondly, $H^2(G,K,b)$ has a distinguished subset $N^2(G,K,b)$ of neutral elements. This is important because in many applications one uses nonabelian $H^2$ in order to determine whether a given extension \eqref{e:E} is split or not.


As far as I know, nonabelian $H^2$ is mostly used in the Galois cohomology setting.
Namely, if $k$ is an algebraic closure of a field $k\_0$ of characteristic 0, $G=\operatorname{Gal}(k/k\_0)$,
and $Y$ is a *quasi-projective* $k$-variety with additional structure (say, an algebraic group or a homogeneous space) such that for any $\sigma\in G=\operatorname{Gal}(k/k\_0)$ there exists an isomorphism $\alpha\colon\sigma Y\overset{\sim}{\to}Y$, then it defines an extension
$$1\to \operatorname{Aut} Y\to E\to G\to 1,$$
where $E$ is the set of such pairs $(\alpha,\sigma)$
with a suitably defined composition law.
We obtain the cohomology class $\eta(Y)\in H^2(k\_0,\operatorname{Aut} Y,b)$
of this extension for a suitable band $b$.
The variety $Y$ (with additional structure) admits a $k\_0$-model if and only if $\eta(Y)$ is neutral, that is, the extension splits;
see [this question](https://mathoverflow.net/questions/243312/nonabelian-h2-and-galois-descent).


For nonabelian $H^2$ in Galois cohomology see:


[1] T. A. Springer, Non-abelian $H^2$ in Galois cohomology, in: [Algebraic Groups and Discontinuous
Subgroups](http://www.ams.org/books/pspum/009/), Proc. Sympos. Pure Math. 9, Amer. Math. Soc., Providence, 1966, 164-182.


[2] M. Borovoi, Abelianization of the second nonabelian Galois cohomology. [Duke Math. J. 72 (1993), 217-239](https://doi.org/10.1215/S0012-7094-93-07209-2).


[3] Flicker, Scheiderer, Sujatha, Grothendieck's theorem on non-abelian $H^2$ and local–global principles.
[J. Amer. Math. Soc. 11 (1998), no. 3, 731–750](https://www.ams.org/journals/jams/1998-11-03/S0894-0347-98-00271-9/).


See also newer papers (they refer to these three), and [this preprint](https://arxiv.org/abs/1608.07366).


