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I'm just learning about quotient spaces, and I'm having some difficulty visualizing a particular example:

If the group $G =(\mathbb{Z},+)$ acts on the topological space $\mathbb{R}$ (with standard topology) by $n*x=n+x$ ($n\in \mathbb{Z}$ and $x\in\mathbb{R}$), then the quotient space $\mathbb{R}/\mathbb{Z}$ is just the unit circle $S^1$. I can't visualize this though... my understanding is that you glue $x$ to all elements in its equivalence class, so all $n+x$, but I don't see how you get a circle from this. Could someone please explain this clearly?

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  • $\begingroup$ You are taking the quotient by $\mathbb{Z} $, so the equivalence class $[ x ]$ is the set $\{x+n\vert n\in \mathbb{Z}\} $. Now imagine moving through $\mathbb{R} $ and looking at the image in the quotient space. As $ x $ moves along the line, starting at $0$, you get corresponding movement in the quotient space. But when $ x $ reaches $1$, the "shadow" $[x] $ in the quotient space is back to where it started, since $[0]=[1] $. $\endgroup$ – Unwisdom Feb 9 '14 at 20:08
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Your equivalence classes are all the numbers with the same digits after the decimal point (i.e. 1.475, 3.475, 295.475, are all identified). Therefore every unit interval maps back to $[0,1)$, and going off the "right" edge loops back to the beginning (since 1 is identified with 0). That "looping" means for instance the open set $(0.9,1.1)$ would be identified with $[0,0.1)\cup(0.9,1)$ which gives you the circle.

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  • $\begingroup$ That helped a lot. Thank you! $\endgroup$ – Adam Feb 9 '14 at 20:14

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