I was wondering whether the following conjecture is true and, if so, how one would proof this. All spaces are assumed to be pointed spaces but we drop the base point from notation.

Conjecture: Let $f: X \rightarrow Y$ be a Hurewicz-fibration between associative $H$-spaces and let $F$ denote the fiber. If $f$ induces an isomorphism $f_*: H_k(X;\mathbb{Z}) \rightarrow H_k(Y;\mathbb{Z})$ for all $k$ then $H_k(F;\mathbb{Z}) = 0$ for $k>0$ and $H_0(F;\mathbb{Z}) = \mathbb{Z}$.

My idea for proving this is as follows. The case $k=0$ is straightforward. Since $f$ induces an isomorphism on $H_0$ it induces a bijection of pointed sets on path components. It then follows from the long exact sequence of a fibration in homotopy that $F$ is pathconnected hence that $H_0(F) = \mathbb{Z}$. I run into problems however with the case $k>0$. The obvious thing to do would be to consider the Serre spectral sequence arising from $f$ but I am not sure how to proceed with this.

To give some context to my question: my main goal is to prove the following

Theorem: Let $f: X \rightarrow Y$ be a Hurewicz-fibration between associative $H$-spaces and let $F$ denote the fiber. If $f$ induces an isomorphism $f_*: H_k(X;\mathbb{Z}) \rightarrow H_k(Y;\mathbb{Z})$ for all $k$ then $f$ is a weak homotopy equivalence.

Edit: I forgot to mention that $f$ is assumed to preserve to $H$-structure.

My idea for the proof consists of three steps. (1) Prove the above conjecture. (2) Apply Hurewicz Theorem to conclude that $\pi_k(F)$ is trivial for all $k$. (3) Apply the long exact sequence of homotopy groups of a fibration.

So, my question is two-fold. (I) Is the above conjecture true and, if so, how should I proceed with my proof. (II) If the conjecture is false then how would one go about proving the theorem?

Any help would be greatly appreciated.

  • $\begingroup$ In the theorem, does $f$ preserve the associative H-space structure? $\endgroup$ – Baby Dragon Feb 13 '13 at 16:44
  • $\begingroup$ Yes, $f$ is assumed to be an $H$-map but I forgot to mention that. $\endgroup$ – user62163 Feb 13 '13 at 17:49
  • $\begingroup$ Nice question. I think your theorem has a good chance of being true. In Hatcher it is an exercise to show that $f:X \to Y$ is a w.e. if it is an isomorphism on $\pi_1$ and if the induced map on universal covers induces an isomoprhism on homology. You have the $\pi_1$ part, since the fundamental group of a $H$-space is abelian. $\endgroup$ – Juan S Feb 14 '13 at 0:11

Let me show how to do something related.

If the map $f$ gives an isomorphism in homology, using the naturality of the universal coefficient theorem for cohomology and the $5$-lemma, we see that $f$ induces an isomorphism in cohomology too. This means that $H^*(Y)\cong H^*(X)$, so that $H^*(X)$ is a free $H^*(Y)$-module. Now there Eilenberg-Moore spectral sequence is a spectral sequence going from $Tor_{H^*(Y)}(\mathbb Z,H^*(X))$ and going to $H^*(F)$, with $F$ the fiber. The freeness of $H^*(X)$ implies that this $Tor$ is trivial in positive degrees, and we conclude at once that $H^p(F)=0$ for all $p>0$.

From this it follows (again using universal coefficients for cohomology) that for all $p>0$ we have $Ext^1_{\mathbb Z}(H_p(F),\mathbb Z)=0$ and $\hom_{\mathbb Z}(H_p(F),\mathbb Z)=0$. These imply that $H_p(F)=0$ if $H_p(F)$ is a countable group, by the positive solution to Whitehead's problem for countable groups (notice that this applies if $F$ has the homotopy type of a countable CW-complex, for example) but the general problem is complicated by the famous Shelah's independence result for the Whitehead problem.

As I noted in a comment below, there is an Eilenberg-Moore spectral sequence, looking like $E^2=Cotor^{H_*(Y)}(\mathbb Z. H_*(X))$ converging to $H_*(F)$ —here we see $H_*(Y)$ as a coalgebra and $H_*(X))$ as an $H_*(Y)$-comodule. Since $H_*(X)\cong H_*(Y)$, $H_*(X)$ is a coflat $H_*(Y)$-comodule, and exactly the same reasoning applies directy, sidestepping the silly detour through the Whitehead problem.

  • $\begingroup$ (Maybe one needs the fibration to give a simple local system on $Y$ to get the E-M spectral sequence, I don't remember; if $Y$ is simply connected this is no limitation...) (Maybe the fact that $f$ is a map of $H$-spaces implies tat the local system given by $f$ on $Y$ is simple automatically.) $\endgroup$ – Mariano Suárez-Álvarez Feb 14 '13 at 1:21
  • $\begingroup$ There is a E-M spectral sequence for homology, which involves a $Cotor$ functor, which should eliminate the silly need to solve Whitehead's problem! I guess this is discussed in McCleary's book on spectral sequences or in the original paper by E-M. $\endgroup$ – Mariano Suárez-Álvarez Feb 14 '13 at 1:23

I believe your conjecture is true. The following has been known as dual Whitehead theorem:

Theorem. A homology isomorphism between simple spaces is a weak homotopy equivalence.

A space is simple when it is path connected, its fundamental group is abelian and it acts trivially on higher homotopy groups. You may restrict your map to the path-connected components of the identity on both sides, this does not change the fiber nor the condition about isomorphism on homology. This way you reduce to the case of path-connected H-spaces and these are known to be simple. Applying dual Whitehead theorem you see that this is an equivalence, so the homotopy fiber - in this case just the fiber - must be contractible.

The simplicity of H-spaces is proved, for example, as Theorem 4.18 in Chapter III of Whitehead's "Elements of Homotopy theory". As for the dual Whitehead theorem, I was not able to find any reference, though this is discussed in the following Mathoverflow question. Peter May's article mentioned can be also freely found on the internet and contains a proof, although it is probably not the standard proof.

Note that this may very well be possible that the statement you're after admits a much easier proof, because you also assume much more (not only simplicity, but that your spaces are actual H-spaces and the map between them respects that structure).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.