Hot answers tagged cohomology
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There are many different things going on here. One is that to invert the suspension functor is to stabilize. The Freudenthal suspension theorem tells you that the system of suspension maps $[X,Y] \to [SX,SY] \to [S^2X,S^2Y] \to \cdots$ eventually stabilizes (at least for finite CW-complexes), and so you can view this as a simplification of the usual ...
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Is there any finite simply-connected CW complex on which $\mathbb{Z}/2$ acts freely? No, since then the classifying space $B(\mathbb{Z}/2)$ would have finite homological dimension, which would imply that the group $\mathbb{Z}/2$ has finite homological dimension. But one can compute $H_p(\mathbb{Z}/2,\mathbb{Z})=\mathbb{Z}/2$ for odd $p$.
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A more or less tautological example that shows that group cohomology is "needed" is the fact that taking invariants under finite groups in short exact sequences does not preserve exactness.
It is easy to find examples of surjections $f:M\to N$ of $G$-modules such that the induced map $M^G\to N^G$ on the invariant subspaces is not surjective.
For example, ...
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Another application is called "Galois descent". Roughly, for a Galois extension $K/k$, if two structures are isomorphic over $K$, we may ask whether they are isomorphic over $k$. Galois descent provides an answer in terms of Galois cohomology.
As an example, let $k$ be a field and let $M$ be a matrix with entries in $k$ and let $K/k$ be a Galois extension. ...
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Well, if the group homomorphisms $\gamma_n^*$ are one-to-one, so you're essentially forming a directed union, then this will be infinitely generated, because any finitely many generators would be in one of the groups of your direct system and therefore can't generate the additional elements in the next group of the system.
Presumably, you want sufficient ...
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I am also learning all this, and I had the same crisis of asking what the hell are spectra for, so maybe I can help you with the idea that I have of spectra now. As Aaron said, there are many different things going on here, and his two reasons are already enough to motivate such a construction. Here are a few more (1reason and 2 good consequences).
I will ...
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Which class are you taking? This material is not easy. You should ask your professor to ask for a proof or some hints.
There is a "simple" proof not using spectral sequences at here and is quite readable. Notice there is an obvious mistake in the proof.
I hope David Speyer or someone else can give an answer on the spectral sequence part(which I do not ...
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Of course!
For homology: We have that $X^k \subset X^{k+1} \subset X$, so by functoriality we get $H_n(X^k)\rightarrow H_n(X^{k+1}) \rightarrow H_n(X)$. So if I'm in the image of the composite, certainly I'm in the image of the second map. Indeed, denote the first map by $i$, then we have $i_{k} = i_{k+1} \circ i$, so if $i_k(x) = y$ then $i_{k+1}(i(x)) = ...
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If a group $G$ acts on an abelian group $N$, we can form the semidirect product $N\rtimes G$ and there is a canonical surjection $p:N\rtimes G\to G$.
This surjection is split, and in fact split surjections are of this form with abelian kernel.
Now, a split surjection admits many different splittings. As soon as you try to classify them, you end up with ...
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The construction of cross product algebras is a very natural problem.
It is very easy to arrive at the $2$-cocycle condition for assocativity and to the condition that such cocycles be cohomologous for the algebras to be isomoorphic.
With sufficient hand waving, this example can be concluded by mentioning the Brauer group of fields and the amazing fact ...
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