Proving that $[0,1] \times X \cong [0,1] \times Y$, where $X$ is Möbius strip, $Y$ is curved surface of cylinder, and $\cong$ denotes homeomorphic

Question

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.

Answer 1

From the comments above.

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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, $$ [0,1] \times X \not\cong [0,1] \times Y. $$