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For a cylinder as the way $S^1 \times [0,1]$, both surfaces are orientable and with Euler characteristic 0. So they are homeomorphic, yeah?

But they have different fundamental group. So they are not homeomorphic??

Please, help me.

Thanks!

LH8
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  • I'm not an expert in algebraic topology so I don't know why your first statement is wrong (perhaps homeomorphic spaces have the same Euler characteristic but not vice versa?). But definitely those two spaces are not homeomorphic (yes they have different fundamental group, in fact $S^1\times [0,1]$ is homotopic to $S^1$). – Yanko Mar 01 '19 at 12:22
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    A topological invariant is a property of topological spaces that is invariant under homeomorphisms. So homeomorphic $\Rightarrow$ same Eurler characteristic. However the opposite is not true. – FWE Mar 01 '19 at 12:28
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    An important question to ask yourself is: do they look homeomorphic? – Santana Afton Mar 01 '19 at 12:53
  • A remember a result: if two surfaces are orientable (or not orientable) and have same Euler characterstic, they are homemorphic. This is a counter-example?? And the converse also its true (of course). – LH8 Mar 01 '19 at 13:09
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    The classification theorem you're thinking of is for compact, connected surfaces without boundary. It doesn't apply to a cylinder. (The closed cylinder has a boundary, and the open cylinder is not compact.) – William Mar 01 '19 at 13:28

4 Answers4

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We can handle this at a more basic level than fundamental groups, orientation, and Euler characteristic. The torus is a 2-manifold (without boundary). The closed cylinder $S^1\times [0,1]$ isn't; neighborhoods of points with second coordinate $0$ or $1$ aren't homeomorphic to the plane. As such, they're not homeomorphic.

And if we cut out those points and looked at $S^1\times (0,1)$ instead? That's a 2-manifold, but it's not compact. It can't be homeomorphic to the torus (which is compact) either.

jmerry
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The cyclinder is not a CLOSED surface.

https://en.wikipedia.org/wiki/Surface_(topology)#Classification_of_closed_surfaces

Tsemo Aristide
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Well $\pi_1(S^1 \times [0, 1]) \cong \pi_1(S^1) \times \pi_1([0, 1]) = \mathbb{Z} \times \{1\} = \mathbb{Z}$. But $\pi_1(\mathbb{T}^2) = \mathbb{Z} \times \mathbb{Z}$, and since homeomorphic spaces have isomorphic fundamental groups the spaces are not homeomorphic.

Perturbative
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  • The conclusion you made at last is not correct. You have said homeomorphic spaces have isomorphic fundamental groups. In this case the fundamental groups of these two soaces are not isomorphic and hence the spaces should not be homeomorphic as well. – little o Mar 01 '19 at 13:25
  • @Dbchatto67 Whoops forgot to add in the "not" there. Thanks for letting me know – Perturbative Mar 01 '19 at 13:41
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The following would be a heuristic way to see that the two are not homeomorphic.

The cylinder has boundary, i.e. near the edges it will locally look like a closed half-disc, there is no way to continually deform the cylinder so to remove these, whereas torus is locally Euclidean (i.e. every point has a neighbourhood where it locally looks like the open 2-disc).

Eugaurie
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