Acyclic complexes for extraordinary cohomology theories 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?
 A: 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.
A: 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.
