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I know that there exist non-contractible spaces $X \not\simeq \ast$ with contractible suspension $\Sigma X \simeq \ast$. For instance the 2-skeleton of the Poincaré homology 3-sphere is such a space. But is there such a space which is itself a suspension ? In other words, is there a (non-connected) space $Y$ such that $\Sigma Y \not\simeq \ast$ but $\Sigma^2 Y \simeq \ast$ ?

toto
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1 Answers1

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Here I assume everything has a CW structure. If $X$ is path connected then $\sum X$ is simply connected. Now if we assume that $\sum^2 X$ is contractible then $\widetilde{H_n}(\sum ^2 X) = 0$ for all $n$. Again $\widetilde{H_{n+1}}(\sum^2 X)=\widetilde{H_n}(\sum X)$. This implies $\widetilde{H_n}(X)=0$ for all $n$. Thus Whitehead theorem implies that $\sum X$ is homotopically equivalent to a point, i.e contractible.

As @Eric Wofsey said (in the comment below) If $X$ is not path-connected, then $\widetilde{H_0}(X)\neq 0$ and hence no iterated suspension of $X$ can be contractible.

Anubhav Mukherjee
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  • can we say something in the non-connected case? i.e. without using Hurewicz on the suspension? – Riccardo Jun 20 '16 at 16:26
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    @Riccardo: If $X$ is not path-connected, then $\tilde{H}_0(X)$ is nontrivial and hence no iterated suspension of $X$ can be contractible. – Eric Wofsey Jun 20 '16 at 18:07
  • @EricWofsey Hope it's not trivial but why this shows that no suspension of $X$ can be contractible? at every step I'm increasing the connectivity right? – Riccardo Jun 20 '16 at 18:36
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    @Riccardo: For any $n$, $\tilde{H}_n(\Sigma^nX)\cong \tilde{H}_0(X)$ is nontrivial. – Eric Wofsey Jun 20 '16 at 18:37
  • oh my, I was so focused on $\tilde{H}_0(\Sigma^n X)$. My bad. Thanks for the clarification! – Riccardo Jun 20 '16 at 18:50