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I am reading quotient space of topology and I am a little bit confused. I am looking at the relationship between $\mathbb{R},S^1$ and the quotient space $\mathbb{R}/{\sim}$, where the relation $\sim$ corresponds to the partition $\mathbb{R}=\mathbb{Z}\cup(\mathbb{R}-\mathbb{Z})$.

Function $f:\mathbb{R}\to S^1$ give by $f(t)=(\cos(2\pi t),\sin(2\pi t))$ is continuous, onto but not one-to-one. We can also define function $g:\mathbb{R}\to \mathbb{R}/{\sim}$ where $g(t)=[t]$.

The circle $S^1$ and the quotient space $\mathbb{R}/{\sim}$ is not the same mathematical object, I suppose. But intuitively speaking they should be the "same thing". So there should be some relationship (bijective function, I guess) between these two spaces. But I was considering the function $\pi\circ f^{-1}$ and it's not injective. Maybe I should define the equivalence relation in a different way? Like "$x\sim y$ if $x\equiv y\pmod {2\pi}$" Also, suppose we can find such a function, can it be a homeomorphism?

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3x89g2
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  • What is your definition for $S^1$? The set of elements $(x,y)\in\Bbb R^2$ satisfying $x^2+y^2 = 1$? Some people actually define $S^1$ to be $\Bbb R/ \sim$. – Cameron L. Williams May 16 '15 at 03:22
  • Yes, the equivalence relation should be defined as $x \sim y \iff x \equiv y \pmod{2\pi}$. This will give you the desired homeomorphism. Your first equivalence relation gives you some weird, non-Hausdorff version of $\mathbb{Z}$. – shalin May 16 '15 at 03:22
  • @Shalop I see. So would that function be a homeomorphism? When dealing with the continuity, what is the metric of $\mathbb{R}/\sim$? – 3x89g2 May 16 '15 at 03:25
  • Just use the quotient topology induced from the canonical projection $t \mapsto [t]$. – shalin May 16 '15 at 03:26
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    @Misakov I don't think quotients of metric spaces necessarily give rise to metric spaces. The topology you get is the subspace topology of $[0,2\pi]$ wrapped around (which gives rise to the subspace topology on the circle). – Cameron L. Williams May 16 '15 at 03:38
  • @CameronWilliams I see. But we can always make it a topological space, is that correct? – 3x89g2 May 16 '15 at 03:59
  • Yes, of course, by using the quotient topology. If you want an example where a quotient of a metrizable space is not metrizable (as Cameron Williams suggested), look at your first equivalence from above. – shalin May 16 '15 at 04:02
  • To address the question in the title (with the corrected definition of $\mathbb{R}/{\sim}$ as suggested in the comment of @Shalop), they are not "the same thing", but they are "homeomorphic to each other". – Lee Mosher May 16 '15 at 13:54

2 Answers2

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The relation is not given by that partition. As said in the comments, the relation is:

$x \sim y \iff x=y \mod 2\pi$

Now, consider the following map:

$f: \mathbb{R} \rightarrow S^1$; $x \mapsto e^{i x}$

Since it takes equivalents to the same image, the induced map:

$\tilde{f}: \mathbb{R} /{\sim} \rightarrow S^1$

is continuous.

Now, take the map $g: S^1 \rightarrow \mathbb{R}/{\sim}$; $e^{ix} \mapsto [x]$. It is obviously well defined, and easily seen to be continuous. Note that $g$ is the inverse of $f$. Therefore, the spaces are homeomorphic.

Aloizio Macedo
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We could also show that $f\colon\mathbb R\to S^1$ given by $$f(x)=e^{ix}$$ is an open map. To check this, it is enough to check that image of any open interval $(a,b)\subseteq\mathbb R$ is an open set in $S^1$. (If image of every set from a basis is open, then the map is open.)

The we could check continuity of $f$, which can be done in various ways. For example, we can use $f(x)=\cos x+i\sin x$ and the fact that cosine and sine are continuous functions.

It is also clear that $f$ is surjective.

Once we get this, we can use the fact that every continuous surjective open map is a quotient map.

Martin Sleziak
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