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Consider the Möbius strip as the quotient space obtained from $[0,1]\times[0,1]$ when identifying $(0,t)$ with $(1,1-t)$ for $0\leq t\leq1$. Now consider its center, that is, the image of the set $[0,1]\times\{\frac{1}{2}\}$ under this quotient map. How does one show that this is homeomorphic to the circle $S^1$ (that is, the quotient space obtained from $[0,1]$ by identifying $0$ and $1$)? Intuitively it seems obvious, because I could map $x$ to $\{(x,\frac{1}{2})\}$ for $0<x<1$ and map $x$ to $\{(0,\frac{1}{2}),(1,\frac{1}{2})\}$ for $x=0$ and $x=1$. But I'm not sure how to write down rigorously, the machinery to show that this mapping is a homeomorphism.

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  • $\begingroup$ $(0,1/2)$ gets identified with $(1,1/2)$ and that's all. $\endgroup$
    – Alamos
    Commented Apr 21, 2015 at 12:09
  • $\begingroup$ Yes, I was already quite confident that that was the wanted homeomorphism. But how do you formally prove that this is a homeomorphism, using the definitions of quotient topologies, and things like that? $\endgroup$
    – user233198
    Commented Apr 21, 2015 at 12:10
  • $\begingroup$ You did explicitly write a homeomorphism of $[0,1]$ with $[0,1]\times\{1/2\}$. The subset of the Moebius you want is $C=[0,1]\times\{1/2\}/R$, where $R=\{(0,1/2),(1,1/2)\}$. So you have a bijection from $S^1$ to $C$. You can prove directly that both directions are continuous by describing explicitly all open sets of $C$ (and $S^1$). Open sets of $S^1$ are open sets of $(0,1)$ together with open sets of $[0,1]$ that contain both $0$ and $1$, quotiented by the relation. The open sets of $C$ are the same thing but times $\{1/2\}$. $\endgroup$
    – Alamos
    Commented Apr 21, 2015 at 12:30
  • $\begingroup$ It is clear that the bijection takes the opens on one side to the opens on the other: It is just putting or removing the "times $\{1/2\}$". $\endgroup$
    – Alamos
    Commented Apr 21, 2015 at 12:32

1 Answer 1

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Let $I$ denote the unit interval. Then we have the quotient maps $p : I \to S^1$ onto the circle and $q : I^2 \to M$ onto the Möbius strip.

What you have is a natural map $f : I \to I^2$ given by $f(t) = (t, \frac{1}{2})$, which, when composed with $q$, gives us a natural map from the unit interval into the Möbius strip. Your question is: when does this map $g := q \circ f$ descend to a map $\tilde g : S^1 \to M$?

The answer is exactly what we would expect: you get a continuous map $\tilde g$ when $g$ doesn't send points of $I$ which are mapped to the same point under $p$ to different points of $M$. More precisely, we say that $g$ respects the identifications of $p$. That is, if $x,y \in I$ and $p(x) = p(y)$, then $g(x) = g(y)$.

Moreover, we can say that $\tilde g$ is a homeomorphism if $p,g$ respect each other's identifications, that is, if both $p$ respects the identifications of $g$ and $g$ respects the identifications of $p$. For detailed proofs of these facts, consult John M. Lee's Introduction to Topological Manifolds, 2nd ed. page 72, Theorems 3.73 (Passing to the Quotient) and 3.75 (Uniqueness of Quotient Spaces). These are very handy facts to know, as you will want to use them frequently when dealing with quotient spaces.

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