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(This question is originally from Math.SE, where it didn't receive any answers.) $\DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\im}{im} \DeclareMathOperator{\id}{id} \DeclareMathOperator{\ext}{Ext} \newcommand{\Z}{\mathbb{Z}}$

Let $G$ be a group, let $A$ be a $G$-module, and let $P_3\to P_2\to P_1\to P_0\to\Z\to0$ be the start of a projective resolution of the $G$-module $\mathbb{Z}$. Consider the cohomology group

$$H^2(G,A)=\frac{\ker(\Hom_{\Z G}(P_2,A)\to\Hom_{\Z G}(P_3,A))}{\im(\Hom_{\Z G}(P_1,A)\to\Hom_{\Z G}(P_2,A))}.$$

It can be shown that $\lvert H^2(G,A)\rvert$ counts the number of equivalence classes of group extensions $0\to A\to E\to G\to0$. The only proof that I know of this result involves choosing a specific projective resolution (namely, the bar resolution).

Is there a proof of this result that does not require choosing a specific projective resolution?

For context, $\lvert\ext_R^n(M,N)\rvert$ counts the number of equivalence classes of extensions $0\to N\to X_n\to\ldots\to X_1\to M\to0$. The proof of this result is fairly abstract and does not require picking a specific projective resolution of $M$ or a specific injective resolution of $N$.

Also, I am aware that we actually have isomorphisms in both of these results but I am more interested in the existence of an explicit bijection.

Here is one approach for constructing an element of $H^2(G,A)$ from an extension $0\to A\to E\to G\to0$: Treat $A$ as an $E$-module and consider the transgression map $H^1(A,A)^{E/A}\to H^2(E/A,A^A)$. Rewriting this gives a homomorphism $\Hom(A,A)^G\to H^2(G,A)$. The image of $\id_A$ under this map will be an element of $H^2(G,A)$.

To make this work, this map would need to be a bijection from equivalence classes of group extensions and elements of $H^2(G,A)$.

Another approach that I considered was to work directly with the arbitrary projective resolution (similar to the proof of the Yoneda Ext result). Suppose that we are given a group extension $0\to A\to E\to G\to0$. We want to construct an element of $\ker(\Hom_{\Z G}(P_2,A)\to\Hom_{\Z G}(P_3,A))$. Equivalently, we want to construct a $\Z G$-module homomorphism $P_2/\im(P_3\to P_2)\to A$. However, $\im(P_3\to P_2)=\ker(P_2\to P_1)$ and $P_2/\ker(P_2\to P_1)\cong\im(P_2\to P_1)=\ker(P_1\to P_0)$. Thus, we want to construct a $\Z G$-module homomorphism $f\colon\ker(P_1\to P_0)\to A$. Furthermore, if we unwind some more definitions, we see that we only need to construct $f$ up to the restriction of a $\Z G$-module homomorphism $P_1\to A$.

Unfortunately, the only information we have about $A$ is the short exact sequence $0\to A\to E\to G\to0$ which makes it hard to define a $\Z G$-module homomorphism to $A$.

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  • $\begingroup$ I like this question! I wonder why you specify just thinking of $\lvert\operatorname H^2(G, A)\rvert$ as counting extensions, rather than of $\operatorname H^2(G, A)$ as parameterising extensions. $\endgroup$
    – LSpice
    Commented May 10, 2019 at 18:00
  • $\begingroup$ See also math.stanford.edu/~conrad/210BPage/handouts/gpext.pdf $\endgroup$
    – user2831784
    Commented May 10, 2019 at 20:02
  • $\begingroup$ Actually your last method works. Since $P_1$ is projective, we get a commutative square $P_1 \to G$ with maps via $E$ and $P_0$. This gives a map $Ker (P_1\to P_0) \to A \cong Ker (E\to G)$ and you are done. $\endgroup$
    – user43326
    Commented May 11, 2019 at 14:28
  • $\begingroup$ @user43326, could you elaborate a bit more? Are you claiming that we have maps $P_1\to E$ and $P_0\to G$? $\endgroup$
    – Thomas Browning
    Commented May 11, 2019 at 18:55
  • $\begingroup$ Sorry, that comment was a crap. $\endgroup$
    – user43326
    Commented May 12, 2019 at 6:03

1 Answer 1

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Here is a simple way. The extension $A \to E \to G$ induces a map of classifying spaces $BA\to BE \to BG$, which is a principal fibration, so classified by (homotopy class of) a map $BG \to BBA=K(A,2)$, i.e., an element of $H^2(BG,A)$.

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    $\begingroup$ This is interesting, but is there a way to convert between $H^2(BG,M)$ and $H^2(G,M)$ without the choice of a resolution? $\endgroup$
    – Thomas Browning
    Commented May 10, 2019 at 20:21
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    $\begingroup$ @ThomasBrowning Personally, my definition of group cohomology is just cohomology of $BG$... I think it's going to be hard to get off the ground if you don't at least give yourself that. $\endgroup$
    – Kevin Casto
    Commented May 10, 2019 at 23:22
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    $\begingroup$ @ThomasBrowning Any model for $EG$ gives such a projective resolution of $\mathbb{Z}$. $\endgroup$
    – J Cameron
    Commented May 15, 2019 at 16:21
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    $\begingroup$ @ThomasBrowning No you can't get all resolutions that way, for example you'll always get a free resolution from $EG$. Are you trying to avoid using the fact that cohomology doesn't depend on which resolution you use? $\endgroup$
    – J Cameron
    Commented May 15, 2019 at 16:40
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    $\begingroup$ @ThomasBrowning the algebraic topological correlate of an algebraic resolution is a CW approximation to the space $BG$. More precisely, a pair $(Y,h)$ consisting of a CW complex and a homotopy equivalence $Y \to BG$. To each such pair, there is an associated $\Bbb Z[G]$ resolution of $\Bbb Z$ (namely, the cellular chain complex of the universal cover $\tilde Y$). These approximations form a category in an evident way. No, not every algebraic resolution will arise in this fashion. But what you have is a functor (CW approximations of BG) $\mapsto$ ($\Bbb Z[G]$ resolutions of $\Bbb Z$). $\endgroup$
    – John Klein
    Commented May 17, 2019 at 14:43

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