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12 votes
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Are long exact sequences in homology a special case of spectral sequences?

To answer the question in the title : yes, long exact sequences can be derived from the degeneracy of some spectral sequences. In fact, there are several spectral sequences leading to the long exact ...
Roland's user avatar
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12 votes
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Wang Sequence for the circle $S^1$

Here is the Mayer-Vietoris argument : Let $N,W,S,E$ be the north, west, south and east pole on the circle. Let $U=S^1\setminus\{E\}$ and $V=S^1\setminus \{W\}$. This is an open covering of $S^1$ such ...
Roland's user avatar
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8 votes
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Understanding Adams spectral sequence and Pontryagin-Thom isomorphism intuitively

Your questions seem very broad, and would probably be better served by a textbook reference. I'll try to give some perspectives that might help. Given two spectra $X$ and $Y$, we would compute the ...
JHF's user avatar
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8 votes
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Question about isomorphism given by the Serre Spectral sequence

In the Serre spectral sequence the edge homomorphism $$H_p(X;M)\to E^\infty_{p,0}\subset E^2_{p,0}=H_p(Y;M)$$ is the map induced by $X\to Y$. If $F$ is acyclic then the Serre spectral sequence ...
Vincent Boelens's user avatar
7 votes
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Grading of Cech-de Rham cohomology

You are quite right that $H_D^n$ is single-graded. However, there is a filtration$$H_D^n = F_0 \supset F_1 \supset F_2 \supset \ldots \supset F_n \supset 0$$on $H_D^n$, and $H_D^{p, n - p}$ is defined ...
Brian Ng's user avatar
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7 votes

Calculating the extraordinary cohomology of $\mathbb{C}P^n$

Your condition should be that the orientation is an element of the reduced $E$ cohomology of $CP^\infty$, restricting to a generator of $\pi_0(E)$. Look at the reduced AHSS. The elements with ...
Michael Andrews's user avatar
7 votes
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How can I determine the Steenrod Square $Sq^2$ for complex projective space?

The cohomology ring $H^*(\mathbb{C}P^n;R)\cong R[x]/(x^{n+1})$, $|x|=2$, can be calculated for any coeffient ring $R$ using the Gysin sequence of the fibration $S^1\hookrightarrow S^{2n+1}\rightarrow \...
Tyrone's user avatar
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7 votes

Leray spectral sequence

Do you mean the five term exact sequence? That would be the following. $$0 \to H^1(Y,f_*A) \to H^1(X,A) \to H^0(Y,R^1f_*A) \to H^2(Y,f_*A) \to H^2(X,A).$$
RCT's user avatar
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7 votes
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Computing $\pi_4(SU(3))$ using Serre spectral sequence

The problem is your conclusion that $H^5X$ is isomorphic to $\mathbb{Z}\oplus\mathbb{Z}_2$. The spectral sequence converges to the the associated graded module. To extract $H^5X$ from the resulting ...
Tyrone's user avatar
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5 votes
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Vanishing terms in Leray spectral sequence

Here is a heavy machinery proof for the vanishing $R^if_*V$. It uses the notion of coherent $\mathcal{O}_X$-modules and complex spaces, since we have to work on the fibres which may not be manifolds. ...
Ben's user avatar
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5 votes

Bockstein homomorphism and Steenrod square

This is proven in Hatcher's book as Theorem 4L.12. It would be way too long to give a full proof here, because I would basically need to redefine the Steenrod squares. But the basic idea is as follows....
Najib Idrissi's user avatar
5 votes
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Hochschild-Serre spectral sequence

When you have a convergent spectral sequence $$ E_2^{pq}\Rightarrow H^{p+q} $$ you indeed have a filtration on $H^{pq}$, but its graded part are $E_{\infty}^{pq}$ which are subquotients of $E_2^{pq}$. ...
Roland's user avatar
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5 votes
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Künneth formula and Leray spectral sequence

The projection $pr_1:X\times Y\rightarrow X$ induces a map of fibrations from $X\rightarrow X\times Y\xrightarrow{pr_2} Y$ to $X\xrightarrow{=}X\rightarrow\ast$, which in particular is an isomorphism ...
Tyrone's user avatar
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5 votes
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Homology groups of unit tangent bundle

Here's a proof that $d = \pm \chi(M)$. Consider the universal orientable $n$-plane bundle $ESO(n)\rightarrow BSO(n)$. It turns out that the unit sphere bundle in $ESO(n)$ is homotopy equivalent to $...
Jason DeVito - on hiatus's user avatar
5 votes
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Computing induced map on homology

No, the subgroup $E^\infty_{2n, 0}$ is $2^n \mathbb{Z} \subseteq \mathbb{Z} = E^2_{2n, 0}$. The idea is that none of the $\mathbb{Z}/2$'s on the $E^2$ page can survive, since the homology of $K(\...
JHF's user avatar
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5 votes

Question about isomorphism given by the Serre Spectral sequence

$\require{AMScd}$ I think you can use the fact that the Serre spectral sequence converges naturally; here is a rough argument: Consider the following commutative diagram of fibrations \begin{CD} F @&...
jasnee's user avatar
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4 votes
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Computing the "limit" of a SSeq with $E_{\infty}^{*,*}$ a free graded $\Gamma$-module.

I think I was able to solve my doubt. I wrote the answer here, but I'll type it down here: So first of all the difficult point is to show that the AHSS collapses at the second page. After that it is ...
Riccardo's user avatar
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4 votes
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An Exercise in Allen Hatcher's book on Spectral Sequences

If $A$, $B$, $C$ are finite groups that fit into an exact sequence $0 \to A \hookrightarrow B \xrightarrow{\phi} C \to 0$ with $A$ abelian, we will show that there is a well defined action of $\pi_1 K(...
Hari Rau-Murthy's user avatar
4 votes

Examples of group extension $G/N=Q$ with continuous $G$ and $Q$, but finite $N$

In general, there is a way to classify extensions of group $1 \to N \to G \to Q \to 1$ for given $Q$ and $N$. You need a group morphism $\phi$ from $Q$ to $Out(N)$ (the outer automorphism group of ...
Clément Guérin's user avatar
4 votes

Examples of group extension $G/N=Q$ with continuous $G$ and $N$, but finite $Q$

Just choose an open, normal subgroup of a compact group, eg. $$1\to p\Bbb Z_p\to \Bbb Z_p\to \Bbb Z/p\to 1$$ So long as the upstairs group is Hausdorff, the open subgroups are exactly the closed ...
Adam Hughes's user avatar
4 votes
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fibrations and map of spectral sequences.

You do get an induced map between the $E_r$ pages for $2 \leq r \leq \infty$, but a priori the map on the $E_\infty$ page coincides with $f^*$ only up to filtration. Here's a (somewhat silly) ...
JHF's user avatar
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4 votes
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Serre classes and the Serre spectral sequence

What about the bundle $\mathbb{Z}/2 \to S^n \to \mathbb{R}P^n$ where $n>0$ is even? Then for all $k>0$ the groups $H_k(\mathbb{Z}/2)$ and $H_k(\mathbb{R}P^n)$ are all in the Serre class of ...
William's user avatar
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4 votes

Derived proofs of elementary homological algebra theorems?

I have not read it too much, but in the beggining (1.3) of this the author relates derived stuff, K√ľnneth and spectral sequences. Thus it seems that at least this short paragraph might be something ...
Con's user avatar
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4 votes
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Why is this map from Leray spectral sequence zero?

I believe Tag 0BUT implies something stronger, namely that $R^{1}f_{\ast}\mathcal{O}_{X}^{\ast} = 0$. One can see this as follows: Let $\mathcal{F}$ be the presheaf on $Y$ given by $V \mapsto H^{1}(f^{...
Minseon Shin's user avatar
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4 votes
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Motivation for construction of Spectral Sequence from a Biregularly Filtrated Cochain Complex

My comment was perhaps a bit too vague/incomplete. I think there are two steps to trying to motivate/gain intuition for the spectral sequence of a filtered cochain complex. The first is to understand ...
Thorgott's user avatar
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4 votes
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Question about spectral sequences

Here is a simple example demonstrating that degeneration of the classical mod $p$ Adams spectral sequence does not imply degeneration of the corresponding Atiyah-Hirzebruch spectral sequence. Let's ...
JHF's user avatar
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4 votes
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cohomology of free loop space on $S^3$

The fiber sequence $\Omega S^3\to LS^3\to S^3$ leads to the long exact sequence of homotopy groups. Since any map $S^1\to S^3$ is nullhomotopic, $LS^3$ is path connected and the long exact sequence ...
Vincent Boelens's user avatar

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