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Given a short exact sequence for some finite groups $A,B,C$, $$1\to A \to B \to C \to 1,$$

how could we construct an exact sequence of their cohomology group out of it?

One version of the story I found is here,

$$1\to H^0(G,A) \to H^0(G,B) \to H^0(G,C)\to H^1(G,A) \to H^1(G,B) \to H^1(G,C)$$

but I am looking for a relation of an exact sequence relating the following cohomology groups by group homomorphism: $H^d(A,\mathbb{R}/\mathbb{Z})$, $H^d(B,\mathbb{R}/\mathbb{Z})$, $H^d(C,\mathbb{R}/\mathbb{Z})$, $H^{d-1}(A,\mathbb{R}/\mathbb{Z})$, $H^{d-1}(B,\mathbb{R}/\mathbb{Z})$, $H^{d-1}(C,\mathbb{R}/\mathbb{Z})$.

Is there a relation like

$$\to H^d(A,\mathbb{R}/\mathbb{Z}) \to H^d(B,\mathbb{R}/\mathbb{Z}) \to H^d(C,\mathbb{R}/\mathbb{Z})\to ?$$

or the other way around

$$\to H^d(C,\mathbb{R}/\mathbb{Z}) \to H^d(B,\mathbb{R}/\mathbb{Z}) \to H^d(A,\mathbb{R}/\mathbb{Z})\to ?$$

and how do we proceed the sequence?

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    $\begingroup$ Perhaps the Lyndon-Hochschild-Serre spectral sequence is what you're looking for? In low degrees you get a short-ish exact sequence, which is usually called the inflation-restriction sequence. $\endgroup$ – JHF May 1 '17 at 16:48
  • $\begingroup$ Thanks JHF, +1, It looks to me that in LHS $1\to H^1(C,\mathbb{R}/\mathbb{Z}) \to H^1(B,\mathbb{R}/\mathbb{Z}) \to H^1(A,\mathbb{R}/\mathbb{Z})\to H^2(C,\mathbb{R}/\mathbb{Z}) \to H^2(B,\mathbb{R}/\mathbb{Z})$, but does this continue go on for $H^d$? $\endgroup$ – wonderich May 1 '17 at 17:34
  • $\begingroup$ In general, no. You only get an exact sequence at the beginning because they aren't that many terms in the associated graded of the $E_\infty$-page of the spectral sequence. However, if you know that there is a vanishing region in the spectral spectral for low degrees, you may get a corresponding exact sequence for the first nonvanishing degrees. Second, I want to point out that in the inflation-restriction sequence, the coefficient modules may change depending on the group action, so you should be careful with the sequence you've written . $\endgroup$ – JHF May 1 '17 at 18:04
  • $\begingroup$ +1, JHF, Thanks, can you write a much complete answer? For example, when do I have $H^d(C,\mathbb{R}/\mathbb{Z}) \to H^d(B,\mathbb{R}/\mathbb{Z})$ and $H^{d-1}(C,\mathbb{R}/\mathbb{Z}) \to H^{d-1}(B,\mathbb{R}/\mathbb{Z})$? $\endgroup$ – wonderich May 1 '17 at 18:09
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As requested, I am turning my comments into an answer.

First, given a short exact sequence $$1 \to A \to B \to C \to 1$$ of finite discrete groups, one can think of them as a sequence of topological spaces $$K(A,1) \to K(B,1) \to K(C,1).$$ If this were a cofiber sequence, then we would get a long exact sequence in cohomology as usual, but unfortunately this is not the case. Instead, this is a fibration, so we have the (Lyndon-Hochschild-)Serre spectral sequence. More precisely, suppose we are given compatible systems of local coefficients, which in this case amounts to a $B$-module $M$. Then we have a spectral sequence $$E_2^{p,q} = H^p(C, H^q(A, M)) \Rightarrow H^{p+q}(B,M).$$

This is a first-quadrant spectral sequence, and some of the terms at the lower left corner of he $E_2$ page are unaffected by diffentials and hence survives intact to the $E_\infty$ page. Then, when reconstructing $H^*(B, M)$ from the associated graded on the $E_\infty$ page, we find that we get a short "long" exact sequence $$0 \to H^1(C, M^A) \to H^1(B, M) \to H^1(A, M)^C \xrightarrow{\delta} H^2(C, M^A) \to H^2(B, M).$$ The map $\delta$ is given by the $d_2$ differential $E_2^{0,1} \to E_2^{2,0}$. In group cohomology, this sequence is called the inflation-restriction sequence. However, this sequence can be extended only for a couple more terms, but even then one of the terms would be less explicit. For example, the next term in the sequence would be $H^1(C, H^1(A,M))$.

On the other hand, if you can show that $E_2^{p,q} = 0$ for $0 < p \leq d$ and $0 < q < d$, then you would have a five-term exact sequence starting in a higher degree $$0 \to H^d(C, M^A) \to H^d(B, M) \to H^d(A, M)^C \to H^{d+1}(C, M^A) \to H^{d+1}(B,M).$$

Finally, you asked when you have maps $H^d(C, M) \to H^d(B, M)$. The answer is always, and this map is called the restriction. (There is also sometimes a wrong-way map $H^d(B,M) \to H^d(C,M)$ called the transfer.) The problem is that they don't knit together to form an exact sequence, and there certainly isn't going to be a "connecting homomorphism" in general, as we see from the spectral sequence.

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  • $\begingroup$ thanks, JHF, +1, looks a very serious one. $\endgroup$ – wonderich May 1 '17 at 19:16
  • $\begingroup$ I think there is always a group homomorphism map $H^d(A,M) \to H^d(B,M)$ is that correct? $\endgroup$ – annie heart May 1 '17 at 19:22
  • $\begingroup$ @annieheart In this case, yes, since $A$ and $B$ are finite. $\endgroup$ – JHF May 1 '17 at 19:24
  • $\begingroup$ How do you call the name of the space $K(G,n)$ and how is it defined? $\endgroup$ – annie heart May 14 '17 at 1:04
  • $\begingroup$ $K(G,n)$ are called Eilenberg-Mac Lane spaces. Their homotopy type is defined by the property that they only have one nontrivial homotopy group $G$ in degree $n$. There are various constructions for point-set models for them, for example via the bar construction. Of course, any two such CW models of them will be homotopy equivalent. $\endgroup$ – JHF May 14 '17 at 14:38

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