Is a torus homeomorphic to a cylinder? 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!
 A: 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.
A: The cyclinder is not a CLOSED surface.
https://en.wikipedia.org/wiki/Surface_(topology)#Classification_of_closed_surfaces
A: 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.
A: 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).
