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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 …
Oscar Randal-Williams's user avatar
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 …
Oscar Randal-Williams's user avatar
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 …
Oscar Randal-Williams's user avatar
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 …
Oscar Randal-Williams's user avatar
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 …
Oscar Randal-Williams's user avatar