1

I want to prove that $[0,1] \times X \cong [0,1] \times Y$ where $[0,1] \subset \mathbb R$ has the usual Euclidean topology, $X$ is a Möbius strip and $Y$ is the curved surface of a cylinder. Here, $\cong$ denotes that there exists a homeomorphism between the two spaces.

I know how to express $X$ and $Y$ as quotient spaces of $[0,1] \times [0,1]$ but I'm stuck on what to do next.

user85798
  • 1
  • What does the identity on $[0,1]\times [0,1]$ look like under these quotients? – Fimpellizzeri Jan 29 '18 at 19:03
  • 2
    Well, then you are out of luck since the first space is non-orientable and the second one is, which means that they cannot be homeomorphic. – Moishe Kohan Jan 30 '18 at 03:29
  • @MoisheCohen Wait, are you sure? Can't you just rotate? – user85798 Jan 30 '18 at 13:14
  • Rotate what? It is a general fact that the product of a non-orientable manifold with another manifold is again non-orientable. – Moishe Kohan Jan 30 '18 at 13:20
  • @MoisheCohen I mean rotate the map continuously as you go around the strip, so that the normal vectors line up – user85798 Jan 30 '18 at 13:26
  • Is there a proof of that? – user85798 Jan 30 '18 at 13:27
  • In the smooth setting see: https://math.stackexchange.com/questions/550426/product-of-manifolds-orientability. One can also give a purely topological proof but it requires some knowledge of homology (Kunneth formula) on your part. – Moishe Kohan Jan 30 '18 at 13:35
  • By $\simeq$ do you mean homotopy equivalent or homeomorphic? The spaces are homotopy equivalent, but not homeomorphic. – Pedro Jan 30 '18 at 13:38
  • @PedroTamaroff I mean homeomorphic. Is orientability preserved under homomorphisms? – user85798 Jan 30 '18 at 17:10
  • @bwv869: Yes, orientability is preserved by homeomorphisms. Proving this is easy in the setting of smooth manifolds and diffeomorphisms and a bit more complicated in general (you need to know some homology theory). You can also see that the boundary of $[0,1]\times X$ is homeomorphic to the Klein bottle while the boundary of $[0,1]\times Y$ is homeomorphic to the torus, which are not homeomorphic (by comparing their fundamental groups). This is another way to prove that the two spaces are not homeomorphic. – Moishe Kohan Jan 30 '18 at 20:19

1 Answers1

2

From the comments above.

Orientability is preserved under homeomorphisms and it is a general fact that if $X$ is non-orientable and $Y$ and $Z$ are orientable, then $Z \times X$ is non-orientable and $Z \times Y$ is orientable. In your case, we have $Z = [0,1]$, $X = $ Möbius strip and $Y = $ curved surface of cylinder. So, $$ [0,1] \times X \not\cong [0,1] \times Y. $$