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Prove that $S^1\times \mathbb{R}^2$ is not homeomorphic to $S^2\times \mathbb{R}$.

So far our only approach to proving spaces are not homeomorphic is removing points and checking the number of path components.

Latest discussion was on pushouts and pullbacks, so another idea I though of was showing one is a pushout of some diagram that the other isn't, but while we've shown that two spaces satisfying the same pushout are homeomorphic, we have not shown two homeomorphic spaces satisfy the same pushout and I am not sure it is true.

Another approach was to try and prove that $S^1\times\mathbb{R}\not\cong\mathbb{R}^2$, but this has two problems - first is that I don't know how to do it either (without homotopies) and second is that I do not know if the canceling in the factorization is even legal. I know it is not legal in general (Taking any countable set $S$ with the discrete topology and $s\in S$ we have $S\times S \cong S\times\{s\}$), but am unable to prove or disprove the specific case of $\mathbb{R}$.

Are there any more elementary approaches then to prove this?

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    $\begingroup$ do you know about the fundamental group? $\endgroup$ – Pink Panther May 11 at 11:33
  • $\begingroup$ Take the loop $t\mapsto (e^{it},0)$. Its image in $S^2\times\mathbb{R}$ can be continuously contracted to a point. You can pullback that homotopy to $S^1\times\mathbb{R}^2$ and then project it down to to $S^1$. This would give you a homotopy from $t\mapsto e^{it}$ to a constant curve. $\endgroup$ – logarithm May 11 at 11:37
  • $\begingroup$ For $S^1\times \Bbb R\not\simeq\Bbb R^2$ you should take a look at this $\endgroup$ – Adam Chalumeau May 11 at 11:40
  • $\begingroup$ @PinkPanther Unfortunately not. We have only discussed $\pi_0$ (set of path components) so far. $\endgroup$ – Nescio May 11 at 11:59
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    $\begingroup$ (After you learn about the fundamental groups) $\pi_1(S_1\times \Bbb R^2)\cong \Bbb Z$ and $\pi_1(S_2\times \Bbb R)\cong \{e\}$, clearly they are non-isomorphic. $\endgroup$ – YuiTo Cheng May 11 at 12:01
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If you want to solve this with a topological invariant which isn't something like homotopy, you can use the theory of ends. I give the definition here:

Definition: Let $X$ be a topological space and let $\{K_n\}_{n\geq 1}$ be an increasing sequence of compact subsets such that $$X=\bigcup_{n\geq1}\stackrel{\circ}{K_n}.$$ An end of $X$ is a decreasing sequence $\{U_n\}_{n\geq 1}$ such that $U_n$ is a connected component of $X-K_n$.

Then you can prove that the number of ends of a topological space doesn't depend of the sequence $\{K_n\}_{n\geq 1}$ we have chosen (see here), and that is is a topological invariant (this is easy once you proved the first point).

Now using the sequences $K_n=S^1\times B(0,n)$ and ${K_n}^\prime=S^2\times [-n,n]$ for $S^1\times \Bbb R^2$ and $S^2\times \Bbb R$ respectively, you can see that $S^1\times \Bbb R^2$ has one end, whereas $S^2\times \Bbb R$ has two ends, which means that they are not homeomorphic.

Here the definition is very formal, you should try to do drawings in $\Bbb R^3$ to see what it represent, it's very intuitive!

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    $\begingroup$ To put this another way, in $X = S^2 \times \mathbb R$ there exists a compact subset $C$ (for example $C = S^2 \times \{0\}$) such that $\overline{X-C}$ has two noncompact components. However, in $X = S^1 \times \mathbb{R}^2$, for each compact subset $C$ there is only one noncompact component of $\overline{X-C}$. $\endgroup$ – Lee Mosher May 11 at 14:42
  • $\begingroup$ Yes, this argument has been given for the similar problem $\Bbb S^1\times \Bbb R\not\simeq \Bbb R^2$ in the link I gave in my comment. But when I first saw it it felt like it's a trick that you can't really think of when you want to solve such a problem. I wanted to give the general argument which can explain why we can think of your simpler argument. $\endgroup$ – Adam Chalumeau May 11 at 14:51
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    $\begingroup$ This is very interesting, and inspired me to do a lot more studying in the subject. Thank you very much $\endgroup$ – Nescio May 18 at 9:31

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