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Exist simply connected CW complexes $X$, $Y$ and a mapping $f:X\to Y$ with the property that the reduced suspension $\Sigma f:\Sigma X\to\Sigma Y$ is a homotopy equivalence but $f$ is not?

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As said below, the answer is yes. It is interesting to note that this fails if you drop the simply connectedness assumption on $X$ or $Y$. Let's do $X$ not simply connected. Let $X$ be any connected acyclic CW-complexes with non-trivial fundamental group (these exist, see e.g. Hatcher). Then $X \to pt$ is a homology equivalence, but not a homotopy equivalence. Since suspension raises the connectedness, the induced map $\Sigma X \to \Sigma pt$ is a homology equivalence of simply connected CW-complexes and using Whitehead's theorem from Andreas' answer, this is a homotopy equivalence. – skupers Nov 22 '10 at 23:19
up vote 17 down vote accepted

Whitehead's Theorem (it is Corollary 4.33 in Allen Hatcher's book) says that a map between simply connected CW-complexes is a homotopy equivalence if and only if the induced map on homology (with $\mathbb Z$-coefficients) is an isomorphism. If $\Sigma f : \Sigma X \to \Sigma Y$ is a homotopy equivalence, then this is clearly the case, since suspension just shifts dimensions and the spaces are connected, so that there is no problem in dimension $0$.

It would be interesting to have an argument which does not use all the machinery that goes into Whitehead's Theorem, since your assumption is rather strong.

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Andreas, which Whitehead's theorem is that? The only one I can find is the one where "a weak equivalence between CW complexes (in more generality, fibrant-cofibrant objects) admits a homotopy inverse." – Harry Gindi Nov 22 '10 at 12:26
Alan Hatcher calls it a version of Whitehead's Theorem. It follows from a combination of the relative version of Whitehead's Theorem and a relative version of the Hurewicz Theorem about Isomorphism of homology groups and homotopy groups in the lowest non-vanishing degree. – Andreas Thom Nov 22 '10 at 12:33
Ultimately, what is needed is a way to compare the connectivities of the fiber and cofiber of a map. It can be shown without homology that for simply-connected spaces (you can do slightly better by imposing conditions on $\pi_1$), the connectivity of the fiber is precisely one more than the connectivity of the cofiber. You can measure the connectivity of a map using cofibers rather than fibers in this case. – Jeff Strom Nov 22 '10 at 12:52

Since adequate answers to the question asked have been given, I will address the relevant underlying theorems. There are two relevant theorems, both called ``Whitehead's theorem''. One says that a weak homotopy equivalence between CW complexes is a homotopy equivalence. The other says that an integral homology isomorphism between suitable spaces, say nilpotent but it actually holds a little more generally, is a weak homotopy equivalence; I say weak because that is what comes naturally out of the proof. I observed in "The dual Whitehead theorems'', #47 on my web page, that these two theorems are word for word Eckmann-Hilton dual to each other when thought of in the right homotopical way (this was presented at a birthday conference for Hilton). The idea is that you study $[X,Y]$ by decomposing $X$ using cells to get the first theorem and decompose $Y$ using "cocells'', via (generalized) Postnikov towers, to get the second. This point of view is explained more leisurely in "More concise algebraic topology", by Kate Ponto and myself: it dominates our treatment of localizations and completions of spaces. It is an especially precise application of the intuition of model category theory, but it is best understood when worked out directly, without invoking that language.

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If the spaces are simply connected, or somewhat more generally, if they are simple (meaning that $\pi_1$ is abelian and acts trivially on the higher homotopy groups) then as Andreas points out, there is Whitehead's theorem that a homology isomorphism between simple spaces is a weak equivalence, and also Whitehead's other theorem that a weak equivalence between CW complexes is a homotopy equivalence.

However, as rpotrie's example indicates, with non-simple spaces the question is more interesting, and the answer is that there are certainly examples where the suspension is a homotopy equivalence but the map itself is not.

Here's a way to construct such a map. Let $G$ be group containing a nontrivial perfect normal subgroup $H$ (i.e., $H= [H,H]$) let $X=BG$, and let $Y=BG^+$ - Quillen's plus construction. The plus construction attaches 2 cells and 3 cells to a space to produce a new space with the same homology but with fundamental group now the quotient of the original fundamental group by $H$. The inclusion $BG \subset BG^+$ is a homology isomorphism, but the spaces have different fundamental groups so the inclusion is not a homotopy equivalence. However, if $H$ happens to be the whole commutator subgroup and you suspend the map once then $\Sigma BG$ and $\Sigma (BG^+)$ are both simply connected, and so Whitehead's theorems tell you that the map is a homotopy equivalence.

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Actually, I think it's enough that the space be nilpotent. If I remember well, a theorem of Drozd generalizing Whitehead's says that if $f\colon X\rightarrow Y$ is a map of connected nilpotent CW-complexes inducing isomorphism on $\pi_1$ and $H_n$, $n\geq 2$, then it is a homotopy equivalence. – Fernando Muro Feb 19 '13 at 13:04
Sorry, I meant Dror, Emmanuel Dror Farjoun. – Fernando Muro Feb 19 '13 at 17:50

Look at the double suspension theorem, it asserts that the double suspension of the homology $3$-sphere of Poincare is homeomorphic to the 5-sphere.

I believe that there is a map from $S^3$ to this homology sphere, I guess that the double suspension of this map may be a homotopy equivalence, I am not sure about this, but this can help.

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Your homology sphere will not be simply connected. – Andreas Thom Nov 22 '10 at 12:31
There is a map from $S^3$ to the Poincare homology sphere $P$ ($S^3$ is its universal covering) and there is a map $P\rightarrow S^3$ ($P$ can be constructed usin a dodecahedron identifying some faces. Collapsing the whole boundary gives $S^3$ and the map). I guess it is better to consider the latter map (but it is just a feeling). – HenrikRüping Nov 22 '10 at 12:33
Alternatively, you can delete a point the homology sphere to get a space with trivial homology and nontrivial fundamental group. – Ben Wieland Nov 22 '10 at 12:38
@Andreas: You are right, I've missed that hypothesis. I hadn't seen your answer when I wrote mine, it is clear from your answer the need for simple conectedeness. @Henrik: I think you are right, with the first map, the induced map of the one I say will have degree greater than one, yours seems to have degree one (after taking double suspension) which is enough to be homotopic to the identity by Hopf theorem. – rpotrie Nov 22 '10 at 12:47

I liked very much the comment/answer of Jeff Strom. And this result seems to be, indeed, essentialy, a consequence of what he mentioned. At first glance, the result seems to be a consequence of the relative Hurewicz isomorphism (and, obviously, Whitehead theorem). But we can prove the result using homotopy excision without passing to homology.

Assuming $\Sigma f $ is a weak equivalence between simply connected spaces, we get, by the homotopy excision (pag 81, May's Concise Course), that $\Sigma C_f\equiv C_{\Sigma f} $ is weakly equivalent to a point. Now, we complete by induction. We already know that $f$ is a $1$-equivalence. By induction, we assume that $ f $ is a $n$-equivalence. And, again, by homotopy excision, we know that this hypothesis implies that $(M_f,X)\to C_f $ is $(n+2)$-equivalence. So $C_f$ is $n$ connected. And, then, by Freudenthal theorem, $\Sigma :\pi_q( C_f)\to \pi _{q+1}(\Sigma C_f) $ is an isomorphism for $q< 2n+1 $. In particular, $C_f$ is $2n$-connected. Therefore, by the $(n+2)$-equivalence, we conclude that $f$ is a $(n+2)$-connected space. And this concludes our induction.

Concisely, this proof is about two lemmas: one is that commented by Jeff Strom (which can be proved using excision). The other lemma is a consequence of Freudenthal/excision: If $\sum X $ is $n$-connected and $X$ is simply connected, then $X$ is $n-1$ connected.

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