# Quotient space homeomorphism

I was reading "Introduction to topology, pure and applied" and in the chapter about quotient spaces there was an exercise that asked this:

We have $\mathbb{R}^2-\left\lbrace 0 \right\rbrace$, and the partition it takes in the space are all lines of the form $y=mx$, for all real $m$, including vertical lines. What topological space is that homeomorphic to?

The book actually asks it with a drawing, but I can't paste it here.

So after thinking for a while I've concluded that it's homeomorphic to $S^1$. This is my reasoning that I would like to know if it's right:

First of all, we can identify lines by the angle they have with the horizontal axis, having that angle is $\varphi\in [0,\pi)$. Now, for a union of lines to be an open set, we must have some $\bigcup_{\varphi\in(a,b)} r_\varphi$, being $r_\varphi$ the line with that angle. I identify this with the open sets of the circle, if we name the points of $S^1$ by the same angle, this time $\phi\in[0,2\pi)$, so the points are $p_\phi$ then we can set the function $$r_\varphi \longmapsto p_{2\varphi}$$

This obviously sends open spaces to open spaces. Is my intuition correct? I think it is. How could I formalize this.

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It's quite formal already. You should find prove that your map is actually a homeomorphism. A good way of looking at it is that instead of identifying lines passing through zero, you can consider $S^1$ and identify $x$ with $-x$. Then it's it's easy to see that resulting space is compact, thus you just have to exhibit a continuous bijection, because then the continuity of the inverse you get for free (since maps from compact sets send closed sets to closed sets). –  xyzzyz May 8 '13 at 19:23
Thanks for your comment. I will try that. –  MyUserIsThis May 8 '13 at 19:38
@xyzzyz By the way. Can you link me some proof of this: since maps from compact sets send closed sets to closed sets? I didn't know that. Is that restricted to some kind of maps? –  MyUserIsThis May 8 '13 at 19:43
First of all, i'm assuming all spaces involved are Hausdorff. Then, the image of a compact set is a compact set (this is an easy exercise), a compact subset of Hausdorff space is closed, a closed subset of a compact set is compact. Thus, if $f: X \to Y$ is continuous, $X$ is compact, and $K \subset X$ is closed, then $K$ is compact, so $f(K) \subset Y$ is also compact, thus it's closed. –  xyzzyz May 8 '13 at 21:06
@xyzzyz Ok, thanks for that. –  MyUserIsThis May 8 '13 at 21:23

You're almost there. But you need to think about lines with angle approaching $\pi$. Maybe you want to think about $[0,\pi]$ and a new quotient topology question :)
What I thought about that was that the line $r_\pi$ is the same line as the line $r_0$. And so the only possible discontinuity that could appear, which were at the limits of the interval, disappears because the two points are actually the same. What I didn't know how to formalize was that: How to send open sets to open sets when the set contains the limit angle $r_0$. What exactly do you mean by new quotient topology question? Just taking $\pi$ as another line and defining the new quotient space with a partition containing bot $r_0$ and $r_\pi$? Like gluing the extremes of a line? –  MyUserIsThis May 8 '13 at 19:41
You're right there! You want to see that if you take $[0,\pi]$ and identify $0$ and $\pi$, the resulting topological space is homeomorphic to $S^1$. –  Ted Shifrin May 8 '13 at 19:47
Well, conceptually, you're thinking of the space of lines as being homeomorphic to $[0,\pi]$ with the endpoints identified, and then you compose with the homeomorphism to $S^1$. Your original attempt was flawed because you didn't know how to make lines whose angles were close to $\pi$ be close to lines whose angles were close to $0$. There are other ways around this, but I was suggesting the way I thought best fit the way you were thinking. –  Ted Shifrin May 8 '13 at 22:54