# Quotient topology is homeomorphic to nonnegative reals

I'm having some difficult to solve this: Consider the following equivalence relation on $\mathbb R^2$:

$(x_1,x_2)\sim (y_1,y_2)$ if $x_1^2 + y_1^2 = x_2^2 + y_2^2$

Show that the quotient space $\mathbb R^2/\sim$ is homeomorphic to the set $\mathbb R_+ := \{x \in \mathbb{R} \mid x \geq 0\} \subset \mathbb{R}$, provided with the subspace topology.

I tried to find out some continuous bijective function (with inverse continuous) between those two spaces but had no success.

• Hint: For the point $(x_1,y_1)$, what does $x_1^2+y_1^2$ represent? – John Douma Jun 29 '13 at 20:21

## 2 Answers

The intuition was given to you by Brian M. Scott: this quotient contracts the circles centered at $0$ to points. Here is how you can prove that these spaces are homeomorphic, formally.

The mapping $\phi:(x,y)\longmapsto x^2+y^2$ is continuous from $\mathbb{R}^2$ to $\mathbb{R}_+$.

Note that $\phi(x,y)=\phi(x',y')$ whenever $(x,y)\sim (x',y')$. By universal property of a quotient space, the map $\phi$ factors uniquely through the quotient into a continuous map $\widetilde{\phi}:\mathbb{R}^2/\sim\longrightarrow \mathbb{R}_+$.

It is clear that $\widetilde{\phi}$ is bijective. In many similar cases, we can deduce that $\overline{\phi}$ is a homeomorphism by compactness of the domain. But it is not the case here. So we need a little more work.

Note that the map $\theta:r\longmapsto (\sqrt{r},0)$ is continuous from $\mathbb{R}_+$ to $\mathbb{R}^2$. Composing $\theta$ with the quotient map $q:\mathbb{R}^2\longrightarrow \mathbb{R}^2/\sim$ yields $\widetilde{\theta}:\mathbb{R}_+\longrightarrow \mathbb{R}^2/\sim$ continuous. A routine verification shows that $\widetilde{\theta}$ is the inverse of $\widetilde{\phi}$, which is therefore a homeomorphism from $\mathbb{R}^2/\sim$ onto $\mathbb{R}_+$.

HINT: It’s easier to see what’s going on if you use polar coordinates: in polar coordinates the equivalence relation is simply $\langle r_1,\theta_1\rangle\sim\langle r_2,\theta_2\rangle$ if and only if $r_1=r_2$. Thus, the equivalence classes are the circles with centres at the origin. Use the radius coordinate $r$ of the point $\langle r,\theta\rangle$ to find the desired homeomorphism.