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Problem

(Exercise 15.12 in Bott & Tu's Differential Forms in Algebraic Topology) If $X$ is a space having a good cover, e.g., a triangularizable space, and $Y$ is any topological space, prove using the spectral sequence of the fiber bundle $\pi\colon X\times Y\to X$ that $H^n(X\times Y)=\bigoplus_{p+q=n}H^p(X,H^q(Y))$.

Remark

Cochain complexes and cohomology groups are with integer coefficients.

Thoughts

Suppose $\mathfrak U$ is a good cover of $X$, and $\pi^{-1}\mathfrak U$ is a cover of $X\times Y$. Consider the double complex $K^{\bullet,\bullet}=C^\bullet(\pi^{-1}\mathfrak U,S^\bullet(-\times Y))$, where $S^\bullet$ is the singular cochain complex. On one hand, consider the spectral sequence $E_r'$ associated to filtration by rows of $K^{\bullet,\bullet}$ which is $E_2'$ degenerate, we have $$G'H_D^n(K)=\bigoplus_{p+q=n}E_2^{\prime p,q}=H_S^n(X\times Y)$$ to be the singular cohomology group.

Note that the filtration of $H_D^n$ should be $\underbrace{H_D^n\supseteq0}_{E_2^{0,n}}$, therefore $H_D^n(K)=H_S^n(X\times Y)$.

On the other hand, consider the spectral sequence $E_r$ associated to filtration by columns of $K^{\bullet,\bullet}$, we have $E_2^{p,q}=H^p(\pi^{-1}\mathfrak U,\mathscr H^q)=H^p(X,H^q(Y))$ where $\mathscr H^q$ is the presheaf $U\mapsto H^q(\pi^{-1}U)$.

However, I cannot see any reason either that $E_r$ is $E_2$ degenerate or that $GH_D^n(K)=H_D^n(K)$ for filtration of columns. I need to know what to do next.

Any idea? Thanks!

Update

It seems that $E_2$-degeneracy is easier, although for the moment I don't have a rigorous argument for this.

Roughly speaking, an element $e^{pq}\in E_2^{pq}\cong H^p(\mathfrak U,H^q(Y))$, represented by some $c^{pq}\in C^p(\mathfrak U,S^q(-\times Y))$. We note that for every "good" $U$, the morphism $\phi_U\colon S^\bullet(Y)\to S^\bullet(U\times Y)$ induced by the projection $U\times Y\to Y$ is a quasi-isomorphism. On the other hand, the morphisms $\phi_U$ is natural in $U$, i.e. they constitute a natural transformation between the constant functor $U\mapsto S^q(Y)$ and the contravariant functor $U\mapsto S^\bullet(U\times Y)$. We can identify $c^{pq}$ with an element in $C^p(\mathfrak U,S^q(Y))$ with some effort. We can then extend $c^{pq}$ throgh the zig-zag path indefinitely. For example, the image of $\delta c^{pq}\in C^{p+1}(\mathfrak U,S^q(Y))$ in $C^{p+1}(\mathfrak U,H^q(Y))$ is zero, hence there exists a $c_2^{p+1,q-1}\in C^{p+1}(\mathfrak U,C^{q-1}(Y))$ such that $dc_2^{p+1,q-1}=c^{p+1,q}$, and we can further extend $c_2^{p+1,q-1}\in C^{p+1}(\mathfrak U,C^{q-1}(Y))$ to some $c_3^{p+2,q-2}\in C^{p+2}(\mathfrak U,C^{q-2}(Y))$, etc.

I still have no idea about the extension problem.

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  • $\begingroup$ Integral or real coefficients? $\endgroup$ – Qiaochu Yuan Feb 4 '15 at 5:55
  • $\begingroup$ @QiaochuYuan Integral. $\endgroup$ – Yai0Phah Feb 4 '15 at 14:48
  • $\begingroup$ Then I don't think I believe this. Even if the $E_2$ page of the spectral sequence degenerates, at best you get that the LHS and the RHS can be identified up to extension problems. Probably you can check that this doesn't hold using an example where the usual Kunneth theorem has some nontrivial stuff going on in it. $\endgroup$ – Qiaochu Yuan Feb 4 '15 at 19:26
  • $\begingroup$ @QiaochuYuan The usual Künneth formula for PIDs seems to coincide with the formula given above, if we rewrite the term $H^p(X,H^q(Y))$ by the universal coefficient theorem. The usual Künneth formula could be proved by a spectral sequence argument, but that is using the Cartan-Eilenberg resolution of a chain complex, which does not seem to correspond to the Postnikov tower. $\endgroup$ – Yai0Phah Nov 28 '18 at 13:40

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