2

It is know that $\partial M\approx S^1$, where $\partial M$ is the boundary of the mobius band. But how do we prove that? I am trying to define the homeomorphism explicitly, but it is messy.

The Mobius band $M$ is defined by the quotient $q:I\times I\rightarrow M$ with the equivalence relation $(0,x)\sim(1,1-x)$.

$\partial M=q(I\times\{0,1\})$.

  • 1
    Consider the map $\partial M to \mathbb{C}: (x,y) \mapsto \exp(i(x+y)\pi)$. – user3482749 Dec 14 '20 at 05:17
  • 2
    This might be a silly question. $\partial M$ is not a subset of $\mathbb{R}^2$, right? Then, what is $(x,y)$? –  Dec 14 '20 at 05:20
  • $\partial M$ is a subset of $M$, which is a quotient of $I \times I$. $(x,y)$ are the coordinates induced by that direct sum decomposition. – user3482749 Dec 14 '20 at 23:32

1 Answers1

4

So first of all you need to know that $\partial M=(I\times\{0,1\})/\sim$. Then consider

$$f:I\times\{0,1\}\to S^1$$ $$f(x,y)=\exp(\pi i (x+y))$$

This function maps $I\times\{0\}$ onto the first half circle and $I\times\{1\}$ onto the second half circle. It is not a homeomorphism, because it is not injective. However it induces a well defined continuous bijection

$$F:\partial M\to S^1$$ $$F([x,y])=f(x,y)$$

on the quotient and thus a homeomorphism since $\partial M$ is compact.

freakish
  • 42,851