# Tag Info

14

Here's a counterexample. Set $X'=S^1\vee S^2$. Consider the following map $F':S^2\vee S^2\vee S^2\rightarrow X'$: It maps the first $S^2$ summand to the $S^2$ summand of $X'$ via a map that represents $2\in\pi_{2}S^2$; it maps the second summand once around the $S^1$ factor of $X'$, and maps the third $S^2$ summand to the $S^2$ summand in $X'$ by a map ...

5

Let $\tilde{X}_k$ be the normalization of the closed $k$-codimension stratum, so $\tilde{X}_0$ is the normalization of $X$. Then there is a diagram of pullback functors between the categories $\text{Perf}(\tilde{X}_k),$ such that $\text{Perf}(X)$ is the pullback of this diagram in the $\infty$-category of derived categories (for example perfect derived ...

5

In your situation, the Serre spectral sequence looks as follows $H_p(C,H_q(A,K))\Rightarrow H_{p+q}(B,K).$ The left hand side is the $E^2$ term. There, $H_q(A,K)$ carries an action of $\pi_1(C)$ induced by the fibration, and $H_p(C,H_q(A,K))$ is the homology with local coefficients. Now, if $A$ has the homology of a point then then $H_q(A,K)=0$ for $q>... 5 First, let me fix generators for$H^*(SU(n))$and$H^*(SU(n)/\mathbb T)$: For the first, consider the vector bundle on$\Sigma SU(n)$with clutching map$\operatorname{id}_{SU(n)}$, i.e. with classifying map$f_n\colon\Sigma SU(n)\simeq \Sigma\Omega BSU(n)\to BSU(n)$, and let$\Sigma x_{2i-1} = f^*c_i$. For the second, let$\pi_k\colon \mathbb T\subset U(1)^...

4

Here is a compromise between Fernando's answer and S. Carmeli's comment. Also, we may as well use $E_*$ any homology theory and there's no need to assume $C$ is nilpotent, just that every fiber is $E_*$-acyclic (so, e.g., if $C$ is connected and you have your assumption then we're okay.) The statement is okay with filtered colimits and equivalences, so we ...

1

Bertram already mentioned this in the comments but I thought I'd write an answer for completness's sake. The Leray-Hirsch theorem says that $H^{*}P(E)\cong H^{*}M\otimes H^{*}(Fiber)\ \$ $\textit{as$H^{*}M$modules}$. So if $x$ is the first chern class of the tautological line bundle over $P(E)$, there's no reason to expect $x^{n}=0$, even if its ...

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