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 Post subject: Homeomorphism
PostPosted: Tue Nov 10, 2015 12:37 pm 
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
Consider the set \(\displaystyle{X=S^{1}-\left\{\left(\cos\,\phi,\sin\,\phi\right)\right\}\subseteq \mathbb{R}^2}\) , where \(\displaystyle{\phi\in\left(0,2\,\pi\right)}\).

Prove that there exists \(\displaystyle{k\in\mathbb{Z}}\) such that the sets \(\displaystyle{X}\) and \(\displaystyle{\left(\phi+2\,k\,\pi,\phi+2\,(k+1)\,\pi\right)\subseteq \mathbb{R}}\)

are homeomorphic.


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 Post subject: Re: Homeomorphism
PostPosted: Thu Nov 26, 2015 12:49 am 

Joined: Thu Nov 19, 2015 7:27 pm
Posts: 6
Since the homeomorphism relation is an equivalence relation, suffices to show both spaces are homeomorphic to the open interval $(0,1)$.

The latter space certainly is.

For the first one, wlog $\phi = 0$ and parametrise $X$ as $ \lbrace (\cos x , \sin x) \mid 0<x< 2 \pi \rbrace $ . Define $f : X \rightarrow (0,1)$ by

$$ f((\cos x, \sin x)) =\frac{1} {2 \pi} \cdot x $$ .

This is a homeomorphism.

May I just say that any $k$ will do and in fact, any interval will do.


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 Post subject: Re: Homeomorphism
PostPosted: Thu Nov 26, 2015 1:53 pm 
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Joined: Mon Nov 09, 2015 1:52 pm
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Hi Nikos. Thank you for your solution.

I would like to make a geometrical comment.

My opinion is that \(\displaystyle{k\in\mathbb{Z}}\) is the rotation index of the circle \(\displaystyle{S^1}\)

having the paramatrization \(\displaystyle{S^1=\left\{\left(\cos\,x,\sin\,x\right)\in\mathbb{R}^2: x\in\mathbb{R}\right\}}\).

For example, if we want to paramatrize the circle from \(\displaystyle{\left(\cos\,\phi,\sin\,\phi\right)}\) (\(\displaystyle{k=0}\)

then, \(\displaystyle{X\simeq Y=\left(\phi,\phi+2\,\pi\right)}\) by defining \(\displaystyle{f(x)=(\cos\,x,\sin\,x)\,,x\in Y}\) .

If we want to "run" the circle one time (\(\displaystyle{k=1}\)), then the above function with domain

\(\displaystyle{\left(\phi+2\,\pi,\phi+4\,\pi\right)}\), is appropriate.

What's your opinion ?


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