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I am completely stuck on the following algebraic topology exercise:

Let $X$ and $Y$ be CW complexes and $\alpha \in H_p(X)$, $\beta \in H_q(Y)$, $p, q > 0$, homology classes such that the homology cross product $\alpha \times \beta \neq 0$ in $H_{p+q}(X \times Y).$ Show that $\alpha \times \beta$ is aspherical, i.e. there is no map $\phi \colon S^{p+q} \to X \times Y$ such that $\phi_*[S^{p+q}] = \alpha \times \beta$, where $[S^{p+q}]$ denotes the fundamental class of $S^{p+q}$.

I am quite sure that the Hurewicz theorem will play a prominent rule in the solution, but I just don't see how.

Maybe it could be useful to play around with the definition of the cellular homology of $X$ and $Y$, since it involves maps from disks and spheres to $X$ and $Y$, but again I don't see how this might actually lead to a solution.

(The exercise is taken from the book Stöcker/Zieschang - Algebraische Topologie and is not a homework problem.)

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I don't see how to use Hurewicz's theorem since we don't know anything about lower-dimensional classes. I would try to use the product structure in cohomology, e.g. if $a$ is a cohomology class dual to $\alpha$ and $b$ is a cohomology class dual to $\beta$ then $a\cup b$ will be a nonzero class. If it pairs nontrivially with a homology class represented by a sphere $S^{p+q}$ then you can pull back the classes $a$ and $b$ to the sphere and get nontrivial (co)homology classes there that are "forced" on you by the naturality of the cup product, yielding a contradiction. The technical details may not work out exactly as I stated but I am confident this direction has a better chance of succeeding than Hurewicz.

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  • $\begingroup$ Sorry, but I don't understand the details of your solution. I would be very glad, if you could try to give a more in-depth account of your argument. $\endgroup$ – AlgTop Aug 5 '13 at 6:19
  • $\begingroup$ The idea is to exhibit a $p+q$-dimensional cohomology class in $X\times Y$ which is a cup product of cohomology classes of dimensions $p$ and $q$. Since the product structure in cohomology is natural with respect to pullback of maps, the sphere $S^{p+q}$ would also inherit such a product decomposition for the pullback class, which is impossible. $\endgroup$ – Mikhail Katz Aug 5 '13 at 7:27

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