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Suppose that a path-connected CW complex $X$ is homotopy equivalent to its own suspension $SX$. Is it true that $X$ must be contractible?

I have tried to show that since $X \times I \simeq SX = X \times I /\! \sim$, the projection $X \times I \to X \times \lbrace 0 \rbrace$ induces a projection $X \times I /\! \sim\; \to \lbrace x_0 \rbrace \times \lbrace 0 \rbrace$ such that $X \times \lbrace 0 \rbrace \simeq \lbrace x_0 \rbrace \times \lbrace 0 \rbrace$ as desired, but this does not seem to be successful or true in general.

Edit: Following freakish's comment, by the Freudenthal suspension theorem and Whitehead's theorem, this is true. Might a more elementary proof be possible? (I don't think so, but I'd be pleasantly surprised if there was one.)

Jake Lai
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    It is true for CW complex by combination of Freudenthal (all homotopy groups have to be trivial) and Whitehead (so it is contractible). But outside of CW I wouldn't expect much. – freakish Mar 22 '24 at 17:08
  • @freakish Just to be clear: Suppose $X$ is $n$-connected. By Freudenthal, $\pi_{n+1}(X) \cong \pi_{n+1}(SX) \cong \pi_n(X) = 0$. We can then argue inductively, assuming $X$ is $0$-connected in the first place, correct? – Jake Lai Mar 22 '24 at 17:23
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    Yes. And it is $0$-connected, because every suspension is. – freakish Mar 22 '24 at 17:25
  • My original motivation for this question was an exercise in Hatcher: Show that $S^\infty$, with its usual CW structure, is contractible. It seems that this was not the intended method of proof! – Jake Lai Mar 22 '24 at 17:37
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    Since $X$ is naturally a subspace of $SX$, a less general but perhaps more answerable question could be: "Is $X$ contractible if $X\hookrightarrow SX$ is a homotopy equivalence?" As for the Hatcher exercise... (several of the "proofs" there are wrong, see the end for correct stuff with precise hypotheses) – FShrike Mar 22 '24 at 17:37
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    The intended solution by Hatcher is likely the one I refer to somewhere in my link, where we inductively define a deformation. A lot of the fluff in the link is about trying to generalise it, but in the specific case we're good – FShrike Mar 22 '24 at 17:43
  • $X\simeq SX\simeq S^2X\simeq ...S^\infty X=\star$ – Bob Dobbs Mar 22 '24 at 18:02
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    @BobDobbs do you have a reference for the last step? Also I would question jf there are any hypotheses needed to promote a homotopy equivalence to a one on the colimit – FShrike Mar 22 '24 at 19:04
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    @FShrike (I work in the reduced setting.) It is enough to suppose that ${\ast}\rightarrow X$ is a closed cofibration and $X\subseteq SX$ a (pointed) h-equivalence. You obtain a "ladder" with top row $SX\subseteq S^2X\subseteq\dotsc$ and bottom row $X=X=\dotsc$ and vertical maps all (pointed) h-equivalences and horizontal maps all cofibrations, which implies the colimit $X\subseteq S^{\infty}X$ is a (pointed) h-equivalence. On the other hand, it is not too hard to show that a colimit of pointed nullhomotopic cofibrations such as $S^{\infty}X$ deformation retracts to its basepoint. – Thorgott Mar 22 '24 at 21:11
  • If, however, we merely assume an abstract h-equivalence $X\simeq SX$, I'm not sure if that allows to deduce much. If $X$ is well-pointed as before, appealing to Freudenthal still yields that it is weakly contractible (and for that it would suffice to assume that $X$ and $SX$ are weakly equivalent, even), but I don't see contractibility lurking in sight. – Thorgott Mar 22 '24 at 21:13
  • @FShrike No reference. Oops It is not equal.... I dont know colinit. – Bob Dobbs Mar 22 '24 at 21:34

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