Let $X$ be a CW complex such that for all extraordinary homology theories, if you plug $X$ into them you get the same value as plugging in a point. Must $X$ be contractible?
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2 Answers
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First of all I claim that asking that $X$ is acyclic for ordinary homology with integral coefficients is the same as asking that it is acyclic with respect to every homology theory. This is because of the Atiyah-Hirzebruch spectral sequence
$H_p(X;E_q) \Rightarrow E_{p+q}X$
So if $X$ is acylic with respect to ordinary homology we have that the $E_2$ page of the spectral sequence is zero outside of the 0th column and so it collapses. Hence we have that the map $E_*X\to E_*$ is an isomorphism. However it is easy to find examples of acyclic but noncontractible spaces.
As a small light of hope let me mention that an acyclic space, while not necessarily contractible, it is not far from being contractible. In fact it is stably contractible.
Since $H_*X$ is trivial, $X$ is connected and in particular nonempty so we can choose a basepoint. I claim that $\tilde H_*X=0$ if and only if $\Sigma X$ is contractible (where by $\Sigma X$ I mean the reduced suspension $X\wedge S^1$). In fact $\tilde H_*(\Sigma X)=\tilde H_{*-1}(X)$, so if $\Sigma X$ is contractible then $X$ is clearly acyclic. On the other hand if $X$ is acyclic it is connected, so $\Sigma X$ is simply connected. Moreover the map $\Sigma X\to *$ is an isomorphism in homology between simply connected spaces and so it is an homotopy equivalence by the Whitehead theorem.
answered May 28, 2015 at 2:42
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$\begingroup$ Maybe this is easy to fix, but it seems like the AHSS argument ignores the difference between reduced and unreduced homology. $\endgroup$ Commented May 28, 2015 at 3:27
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$\begingroup$ It's not quite correct to say the $E_2$ page is zero: you have the coefficients of your homology theory (i.e., the value of $E_\ast$ on a point) in the $p=0$ column. Hence the spectral sequence collapses and $X$ has the $E$-homology of a point. So I think a simple change of wording will make Qiaochu happier. $\endgroup$ Commented May 28, 2015 at 7:18
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$\begingroup$ Ok, sorry. Now it should be ok (in any case the second paragraph could be seen as another proof of the fact). $\endgroup$ Commented May 28, 2015 at 12:19
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$\begingroup$ @DenisNardin: do you really need the AHSS here? That's too much of a gun IMO. Isn't it just the relative Hurewicz theorem? The point is that if if $X$ is acyclic with respect to singular theory then $\Sigma X$ is contratible. But this shows that any exotic homology theory $E$ applied to $\Sigma X$ is trivial. Finally, the $E$-homology of $\Sigma X$ is always the shifted $E$-homology of $X$. $\endgroup$ Commented May 28, 2015 at 21:56
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$\begingroup$ Well, we may disagree about whether the AHSS is more or less advanced than the relative Hurewicz theorem (it's just cellular homology!). Also the AHSS proves the slightly stronger statement that an homology equivalence between bounded below spectra is an equivalence.. $\endgroup$ Commented May 28, 2015 at 22:06
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If you go to exotic cohomology with twisted coefficients, then the answer is yes.
Alternatively, one can state the condition on the level of a universal cover and then the answer is yes.
It is certainly not true without passage to a covering: for example, a non-trivial knot complement has the cohomology of a circle with respect to all exotic cohomology theories.
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2$\begingroup$ If you introduce twisted coefficients you do not even need exotic homology. In fact it is enough to consider ordinary homology with coefficients in $\mathbb{Z}[\pi_1X]$. I think that the role played by extraordinary homology theories is a red herring here. $\endgroup$ Commented May 28, 2015 at 12:39
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1$\begingroup$ @Denis Nardin: Absolutely: it's just a version of the relative Hurewicz theorem. $\endgroup$ Commented May 28, 2015 at 20:21
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$\begingroup$ I think the use of the AHSS also demonstrates the extent to which using extraordinary homology theories are a red herring. Of course, you are very familiar with my personal biases John, :). $\endgroup$ Commented May 30, 2015 at 21:03
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$\begingroup$ @SeanTilson: Come on Sean: The relative Hurewicz Theorem produces the same demonstration. And the relative Hurewicz Theorem dates back to the 1950s. I don't why people want to throw spectral sequences at every problem that arises. It's like using an AK47 to to attack a problem that only requires a pea-shooter. $\endgroup$ Commented May 31, 2015 at 22:15
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$\begingroup$ @JohnKlein: you really like that analogy. I understand that some things predate others historically. As ... everyone seems to agree the argument is essentially the relative Hurewicz theorem that gives you the appropriate identification of the $E_2$-page of the AHSS. I think certain arguments, just as different language frequently does, can highlight certain aspects to some people. For me, I find certain spectral sequence proofs very illuminating. I know this sounds shocking for you, but for me SSs are a tool for understanding filtered objects. (I hope you aren't offended by my comment.) $\endgroup$ Commented May 31, 2015 at 22:42