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I have a probably basic question concerning Hilbert cube manifolds. There are already many questions asking very similar things in this forum, but I could not find the particular answer I am looking for.

Let $Q=[0,1]^\omega$ be the Hilbert cube and let $X=Q\setminus\{0\}$ the locally compact space obtained by deleting one point of $Q$. It is clear that $X$ is a contractible non-compact $Q$-manifold, as is $X\times X$.

My question is: Is $X$ homeomorphic to $X\times X$ ?

I know that compact contractible $Q$-manifolds are all homeomorphic, and that the situation is more complicated in the non-compact case. However, since the space $X$ seems like a very "easy" non-compact Hilbert cube manifold, I would expect the answer to be yes, I just could not find the correct reference.

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  • $\begingroup$ By $\{0\}$ do you mean the point all of whose coordinates are 0? Because it's not trvially obvious that the structure of $X$ shouldn't depend on how many of the coordinates of the deleted point are endpoints of $[0,1]$. $\endgroup$ Commented Sep 21 at 13:56
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    $\begingroup$ Yes, I meant the point with all coordinates equal to zero. However, it does not matter which point you cut out as the Hilbert cube is homogeneous (which is far from obvious) $\endgroup$ Commented Sep 21 at 14:48

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There is an easy answer, based on a non-trivial theorem from Infinite-dimensional Topology. The Hilbert cube $Q$ is homeomorphic to its own cone $C(Q)$, which is $Q\times[0,1]$ with $Q\times\{1\}$ shrunk to a point, $p$ say. By homogeneity your $X$ is the same as $C(Q)\setminus\{p\}$, which is homeomorphic to $Q\times[0,1)$. Now $[0,1)\times[0,1)$ is homeomorphic to $[0,1]\times[0,1)$ (both are a disc with a closed seqment taken out of the boundary circle; draw a picture). It follows that $\bigl(Q\times[0,1)\bigr)^2$ is homeomorphic to $Q^2\times[0,1]\times[0,1)$, which is (homeomorphic to) $Q\times[0,1)$.

For a proof that $C(Q)$ is homeomorphic to $Q$ see section 6.1 of Jan van Mill's book Infinite-dimensional topology. Prerequisites and introduction (pages 254-258).

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  • $\begingroup$ Perfect! Thank you very much for this wonderful answer! $\endgroup$ Commented Sep 25 at 18:55
  • $\begingroup$ I just realized that $[0,1)\times Q$ being homeomorphic to $[0,1)^2\times Q$ shows that $[0,1)$ is very different from $\mathbb{R}$ as $\mathbb{R}\times Q$ is not homeomorphic to $\mathbb{R}^2\times Q$ ... (see mathoverflow.net/questions/482483/…) ... So, the argument for $\mathbb{R}$ cannot carry over to $[0,1)$ ... Fascinating... $\endgroup$ Commented Sep 27 at 9:37

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