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In M Nakahara's Geomertry, Topology and Physics, he states (example 2.4) that the the quotient space(set of equivalence classes) of the real line under the equivalence relation:

$x$ is related to $y$ if $x-y=2n\pi$, $n$ is an integer.

is the circle $S^1$. I can see why that is so. For instance, in the quotient space, $0,2\pi,4\pi$ are all in the same element(in same equivalence class); this is also true for the circle. However, this is also true for the ellipse, or for any closed curve with the perimeter $2\pi$. Why does the quotient space have to be just a circle then?

Mani Jha
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In topology, there is no distinction between a circle and an ellipse. They are both $S^1$. The distinction only appears when you have a metric and start measuring distances and curvature.

Arthur
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  • What about a square loop? Does the same hold for it? – Mani Jha Aug 03 '19 at 08:15
  • @ManiJha Yup. Any polygon with non-intersecting sides. And many others. – Arthur Aug 03 '19 at 08:18
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    Yes. If two spaces can be continuously deformed into each other, then they are homeomorphic (topologically indistinguishable). Imagine a circle made of rubber, then you can easily stretch it into the shape of a square and vice-versa. – Berci Aug 03 '19 at 08:19
  • @Berci Except you normally have to be careful when you say ‘continuously deform’ because a cylinder deformation retracts to a circle but they’re not homeomorphic. – snulty Aug 03 '19 at 08:35
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    @Berci Homeomorphisms aren't really about continuous deformation. That's a homotopy. They are related, but different enough that you have to be careful. For instance, two circles meeting in a point, and two triangles sharing an edge may be continuously deformed to one another, but they aren't homeomorphic. – Arthur Aug 03 '19 at 08:59
  • Yes, yes, right, it's not rigorous at all, I just wanted to visualize it for the OP. – Berci Aug 03 '19 at 09:27