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14
votes
To compare the total, base and fiber spaces of two fiber bundles
No. Consider the map from the fibre bundle
$$B\mathbb{Z} \to BD_\infty \to B\mathbb{Z}/2$$
to $* \to * \to *$. Here $D_\infty = \mathbb{Z} \rtimes \mathbb{Z}/2$ is the infinite dihedral group.
You ca …
2
votes
fibrations of classifying spaces - Leray Hirsch Theorem converse
If you work with coefficients in a field $\mathbb{F}$, assume that $H^*(BH;\mathbb{F})$ is a free $H^*(BG;\mathbb{F})$-module, and add the assumption that the Serre spectral sequence has a product str …
15
votes
Accepted
Computation of stable homotopy groups of $RP^2$
The version of the AHSS you wrote down converges to $\pi_*^s(\mathbb{RP}^2_+)$, i.e. with an extra basepoint. This splits canonically as $\pi_*^s(\mathbb{RP}^2) \oplus \pi_*^s(S^0)$, and the left hand …
1
vote
Accepted
Transgression in terms of k-invariant for chain complexes
I think the difficulty is that you are assuming that $X$ only has homology in two degrees, but are then looking at the cohomology spectral sequence. (To get sensible answers I seem to have to take coh …
9
votes
Accepted
Calculate the group cohomology classes $H^d[U(1)\rtimes Z_2, Z]$ and $H^d[U(1)\rtimes Z_2...
The group $U(1) \rtimes \mathbb{Z}/2$ you describe is nothing but the group $O(2)$ (as $U(1) = SO(2)$).
As such I think one can see the spectral sequence for the extension does collapse, and one obta …