Understanding how group cohomology classifies extensions using the derived functor point of view

Question

I am rereading some material about group extensions, in particular because I needed to recall the formula $$H^2(G;A)\cong \mathcal{E}(G;A).$$ We have that $G$ is some group acting on an abelian group $A$, the group on the left is the second cohomology group of $G$ with coefficients in $A$ and the group on the right is the group of equivalence classes of extensions $0\to A\to E\to G\to 0$ under Baer sum.

This isomorphism is well known, and the proof of it which I have seen is a bit hands on but far from impossible using factor sets and 2-cocycles. However, understanding group cohomology this way is one of several ways to do this. In particular, one could try to understand the above isomorphism pretending to only know about the derived functor definition of group cohomology. My question is does anybody have a proof with this perspective?

Thank you in advance to anyone who has an answer to my querry.

Answer 1

There is actually not so direct a connection with the derived functor point of view, at least as far as I know, and for a pretty simple reason: the derived functor POV concerns "abelian" stuff, abelian groups or chain complexes being acted on by $G$. But extensions are nonabelian groups! So how are they supposed to appear?

Actually the connection is with the homotopy-theoretic point of view on group cohomology (if you like this has to do with "nonabelian derived functors"). This concerns the classifying space $BG$, also known as the Eilenberg-MacLane space $K(G, 1)$. This is a space with $\pi_1(BG) \cong G$ and all higher homotopy groups trivial, and group cohomology

$$H^n(G, A) \cong H^n(BG, A) \cong [BG, B^n A]$$

is the cohomology of this space, which can in turn be identified with homotopy classes of maps from $BG$ to spaces of the form $B^n A$, which can either be thought of as iterated classifying spaces or as Eilenberg-MacLane spaces $K(A, n)$, with $\pi_n(B^n A) \cong A$ and all other homotopy groups trivial.

So what does this have to do with group extensions? Given a homotopy class of maps $f : BG \to B^n A$ representing a class in group cohomology we can take its homotopy fiber, which is a "nonabelian derived kernel," if you like. This produces a space $X$ fitting into a fiber sequence (a homotopical version of a long exact sequence)

$$\dots \to B^{n-1} A \to X \to BG \xrightarrow{f} B^n A$$

and this space $X$ has the following properties, which can be shown using the long exact sequence in homotopy (which is the result of applying $\pi_0$ to the fiber sequence):

• If $n = 2$, then $\pi_1(X)$ is a group fitting into a short exact sequence $1 \to A \to \pi_1(X) \to G \to 1$ and all higher homotopy is trivial. This turns out to be the central extension classified by $f$.
• If $n \ge 3$, then $\pi_1(X) \cong G$ and $\pi_{n-1}(X) \cong A$, and all other homotopy groups are trivial. Furthermore every space with only these two nontrivial homotopy groups and trivial $\pi_1$ action on $\pi_{n-1}$ arises in this way. $f$ here is one of the simplest examples of a $k$-invariant and this becomes one of the simplest examples of a Postnikov tower.

For more on this you can check out, for example, John Baez's Lectures on $n$-categories and cohomology. This version of the story only covers central extensions and trivial $\pi_1$ action on $\pi_{n-1}$; non-central extensions and arbitrary action of $\pi_1$ on $\pi_{n-1}$ can also be classified this way but one needs to work with cohomology with local / twisted coefficients. Instead of working with homotopy types you can tell this story in terms of, equivalently, either homotopy types with a $G$-action or homotopy types equipped with a map to $BG$, which is like a "nonabelian derived category of $G$-modules."