Is a map a homotopy equivalence if its suspension is so? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T17:47:59Z http://mathoverflow.net/feeds/question/46936 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/46936/is-a-map-a-homotopy-equivalence-if-its-suspension-is-so Is a map a homotopy equivalence if its suspension is so? emmet2 2010-11-22T12:12:49Z 2013-04-17T02:50:24Z <p>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?</p> http://mathoverflow.net/questions/46936/is-a-map-a-homotopy-equivalence-if-its-suspension-is-so/46937#46937 Answer by Andreas Thom for Is a map a homotopy equivalence if its suspension is so? Andreas Thom 2010-11-22T12:19:33Z 2013-02-19T18:37:54Z <p>Whitehead's Theorem (it is Corollary 4.33 in Allen Hatcher's <a href="http://www.math.cornell.edu/~hatcher/#ATI" rel="nofollow">book</a>) 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 $0$.</p> <p>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.</p> http://mathoverflow.net/questions/46936/is-a-map-a-homotopy-equivalence-if-its-suspension-is-so/46939#46939 Answer by rpotrie for Is a map a homotopy equivalence if its suspension is so? rpotrie 2010-11-22T12:25:40Z 2010-11-22T12:25:40Z <p>Look at the <a href="http://en.wikipedia.org/wiki/Double_suspension_theorem" rel="nofollow">double suspension theorem</a>, it asserts that the double suspension of the homology $3$-sphere of Poincare is homeomorphic to the 5-sphere. </p> <p>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.</p> http://mathoverflow.net/questions/46936/is-a-map-a-homotopy-equivalence-if-its-suspension-is-so/47007#47007 Answer by Jeffrey Giansiracusa for Is a map a homotopy equivalence if its suspension is so? Jeffrey Giansiracusa 2010-11-22T22:13:51Z 2013-02-19T10:49:26Z <p>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.</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/46936/is-a-map-a-homotopy-equivalence-if-its-suspension-is-so/122307#122307 Answer by Peter May for Is a map a homotopy equivalence if its suspension is so? Peter May 2013-02-19T14:10:19Z 2013-02-19T17:28:47Z <p>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.</p> http://mathoverflow.net/questions/46936/is-a-map-a-homotopy-equivalence-if-its-suspension-is-so/127782#127782 Answer by Fernando Lucatelli for Is a map a homotopy equivalence if its suspension is so? Fernando Lucatelli 2013-04-17T02:45:11Z 2013-04-17T02:50:24Z <p>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.</p> <p>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 $q&lt; 2n+1$. 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. </p> <p>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.</p>