# If a topological space has vanishing $n$th homology for every possible homology theory, does it have vanishing $n$th homotopy?

I don't have any strong preference as to whether or not the homology theories are required to be ordinary.

Also, if this does not hold in general, does it hold for some nice category of spaces, like CW-complexes?

Finally, in a more general context, does anyone have a reference for where to find information about things one can deduce from properties common to all homology theories evaluated on a given space?

• This is very unlikely: a punctured homology $3$-sphere would have all homology group (including extraordinary, I think) isomorphic to those of a point, but still a nontrivial $\pi_1$. And this is a finite polyhedron. Jun 19, 2015 at 20:23
The first answer to your question is no. There are many acyclic spaces with nonvanishing $\pi_1$. Since generalized homology is a stable invariant, and the suspension of an acyclic space is contractible (exercise), this also means that any generalized cohomology theory vanishes on such a space. For an explicit example, take the classifying space of a perfect group.
But now you'll say it's a $\pi_1$ issue, and that's sort of true. If $E_n(X) =0$ for every generalized homology theory, then that means that $E_*(X) = 0$ for any generalized homology theory, including ordinary homology. Indeed: if $E_*(-)$ is a homology theory, so is $E_{*+n}$. But if you're willing to stick to, say, connective cohomology theories (i.e. $E_k(pt) = 0$ for $k<0$), then there are simply connected counterexamples. Indeed: $\pi_3S^2 = \mathbb{Z}$ but $H_3(S^2) = 0$. So let $X = S^2_{\mathbb{Q}}$ denote the rationalization of the 2-sphere, then for any homology theory $E$ with $E_{-1} \otimes \mathbb{Q} = 0$ we get $E_3(X) = 0$ but $\pi_3X = \mathbb{Q}$. Indeed, $\Sigma^{\infty}_+X$ is just $\Sigma^2H\mathbb{Q}$ so $E_3(X) = \pi_{-1}(H\mathbb{Q} \wedge E) = 0$.