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Mar
26
comment Classification of $O(2)$-bundles in terms of characteristic classes.
Aha! Thank you @OscarRandal-Williams!
Mar
25
comment Classification of $O(2)$-bundles in terms of characteristic classes.
(I guess one possibility is that this correspondence is surjective but not bijective.)
Mar
25
comment Classification of $O(2)$-bundles in terms of characteristic classes.
I'm confused... it seems like an unoriented bundle $E$ is the same as the knowledge of (1) $\text{det}(E)$ and (2) $E \otimes \text{det}(E)$ (or maybe duals thereof, etc.) In other words, an oriented rank 2 bundle is the same as line bundle and an oriented rank 2 bundle with the correspondence given by $(V, L) \mapsto V \otimes L^{-1}$. I think that's what Chanler's answer is saying, but it feels like there's a contradiction with what Mike is saying... what's going on?
Mar
23
comment Surjective homomorphism of groups.
What have you tried?
Mar
23
comment Smash and join products of spheres
As for joins one definition of these is as the suspension of the smash product... so there's that... Another way to think of them is as coming from a natural operation on simplices. If you have a $p$ simplex (on vertices $0, ..., p$) and a $q$-simplex (on vertices $0, ..., q$) it seems resonable to make a new guy by concatenating their ordered vertices to get a $p+q+1$-simplex $0, ..., p, 0,..., q$. Extending this operation of concatenation formally from simplices to simplicial complexes gives the join.
Mar
23
comment Smash and join products of spheres
What have you tried? As for intuition: spheres are one-point compactifications of vector spaces, i.e. $S^n$ is the one-point compactification of $\mathbb{R}^n$. A natural way to add vector spaces is to take their direct sum. Show that the one-point compactification of $\mathbb{R}^n \oplus \mathbb{R}^m$ is the smash product of one point compactifications.
Mar
17
awarded  Good Answer
Mar
9
comment A CW complex for $\mathbb{CP}^n$ so that $\mathbb{RP}^n$ is a subcomplex
This shouldn't be so bad: you just need a C_2-CW decomposition where C_2 acts on complex projective space by conjugation. This exists, the question is can you find a small/pretty one and I think the answer is yes...
Feb
26
comment Spectrum of spectrum in the stable homotopy category
The answer is yes to both, if you make sense of things correctly. There are a couple of ways to do that: (1) you could interpret that list of spaces as presenting a spectrum as a homotopy colimit of suspension spectra, and if you compute the corresponding homotopy colimit of spectra you get with your E-tilde construction you get an equivalent thing, (2) you could take the spanier whitehead category, and the spanier whitehead category of a stable category doesn't spits out what you started with.
Feb
25
comment Clarification about Borel $G$-homology theory
It's in any standard algebraic topology book. Here's the proof: fibrations give long exact sequences in homotopy groups, and maps of fibrations induce maps of long exact sequences. Now use the five lemma. (With a bit of fuss near pi_1 and pi_0, I guess)
Feb
25
comment Clarification about Borel $G$-homology theory
Hmm, perhaps I spoke too soon- I still think it's homotopy fiber is G but maybe that's not so elementary, so a different tactic: There is a fiber bundle $EG\times X \rightarrow EG\times_GX$ and this maps to the quotient map of X by G. The fiber of both maps over a point is just G and a map of fibrations which is an equivalence on fiber and total space also induces an equivalence on bases.
Feb
25
comment Clarification about Borel $G$-homology theory
If G does not act freely that map is not an equivalence (modding out by G doesn't preserve equivalences, which is basically the reason for doing the borel construction anyway). Instead, note that when G acts freely the map from X to the orbits is a fiber bundle with fiber G. It follows that your map is a fibration with fiber $EG\times_G G \cong EG$ which is contractible, so the map is an equivalence in this case.
Feb
4
comment Cohomology of geometric realization of a simplicial topological space
I believe the answer is "yes" (I'm more confident for the corresponding statement about homology, but I think this is ok too). Non-elementary proof: C_* commutes with homotopy colimits. Elementary proof: maybe use that there is a map between both objects and a spectral sequence computing their homology that has the same E_2 term?
Feb
1
awarded  Nice Answer
Jan
24
comment Why is symmetric group action needed for symmetric spectra?
We know that connective spectra are spaces with an action of the E_infty operad, and you can explicitly write down deloopings. I wonder if you can reinvent this formula by writing down the formula for the deloopings of a smash product of E_infty spaces? The symmetric group appears because it acts on the spaces of the operad
Jan
24
comment Fibration and induced mapping
And the definition of Serre fibration via the homotopy lifting property
Jan
24
comment Fibration and induced mapping
You can do it without that theorem, straight from the definition of fibration... Use the adjunction between maps and products/smash products
Jan
18
comment $d$ operator for Mayer Vietoris sequence in De Rahm cohomology
Use exactness- in the case in question the term after the target of delta is zero.
Jan
14
comment Surjective Homomorphisms of Isomorphic Abelian Groups
Hint: use the classification of finitely generated abelian groups. Try taking a quotient and seeing whether it's possible to end up with what you started with...
Jan
6
comment If s is a (global) section of a quasi coherent sheaf F on a scheme X, is there a reasonable scheme structure to gives its vanishing locus?
If you had a global section of the dual of a quasicoherent sheaf then that would give a morphism from X to $\mathbf{V}(\mathcal{E})$ (relative spec of sym on your bundle). You also have the zero section, always, and then you can form the fiber product of schemes to put a scheme structure on the zero locus.