Describe the identification space $\mathbb{R}^2 / \sim $

Question

Define a relation $\sim$ on $\mathbb{R}^2$ by $(x_1,y_1) \sim (x_2,y_2)$ provided $x_1+y_1^2=x_2+y_2^2$, and I have to describe the identification space $\mathbb{R}^2 / \sim $. \ I take a open set in $\mathbb{R}^2 / \sim $, then $p^{-1} (U)$ is open in $\mathbb{R}^2$, where $$p: \mathbb{R}^2 \to \mathbb{R}^2 / \sim $$ defined as $p(x,y)=[(x,y)]$. But I don't know how to do. Can anybody give me an Idea please

Answer 1

HINT: I’m assuming that you meant to define $\langle x_1,\color{red}{y_1}\rangle\sim\langle x_2,y_2\rangle$.

Let $\langle a,b\rangle\in\Bbb R^2$. Then $[\langle a,b\rangle]=\left\{\langle x,y\rangle\in\Bbb R^2:x+y^2=a+b^2\right\}$. Now $a+b^2$ could be any real number, so the $\sim$-equivalence classes all have the form

$$\left\{\langle x,y\rangle\in\Bbb R^2:x+y^2=c\right\}$$

for some constant $c$. What does such a set look like? Consider the graph of $x=c-y^2$. You can worry about the topology on $\Bbb R^2/\sim$ after you know what the quotient space is.

Added in response to comment:

For any open interval $(a,b)$ let $U(a,b)=\{[\langle x,y\rangle]:x+y^2\in(a,b)\}$. Then

$$p^{-1}[U(a,b)]=\left\{\langle x,y\rangle\in\Bbb R^2:a<x+y^2<b\right\}\,,$$

and it’s not too hard to show that this set, which is the open space between the parabolas $x=a-y^2$ and $x=b-y^2$, is open in $\Bbb R^2$.

Suppose that $\langle x,y\rangle\in p^{-1}[U(a,b)]$, and let $$\epsilon=\min\left\{\left(x+y^2\right)-a,b-\left(x+y^2\right)\right\}\,.$$

Let $\langle u,v\rangle=\langle x+h,y+k\rangle$, and suppose that $|k|<\min\left\{\frac{\epsilon}{6|y|},\sqrt{\frac{\epsilon}3}\right\}$ and $|h|<\frac{\epsilon}3$; then

$$\begin{align*} \left|u+v^2-\left(x+y^2\right)\right|&=\left|x+h+(y+k)^2-x-y^2\right|\\ &=\left|h+2ky+k^2\right|\\ &\le|h|+2|ky|+k^2\\ &<\frac{\epsilon}3+\frac{2|y|\epsilon}{6|y|}+\left(\sqrt{\frac{\epsilon}3}\right)^2\\ &=\epsilon\,, \end{align*}$$

so $a<u+v^2<b$, and $\langle u,v\rangle\in p^{-1}[U(a,b)]$. In other words, if we set $\delta=\min\left\{\frac{\epsilon}3,\min\left\{\frac{\epsilon}{6|y|},\sqrt{\frac{\epsilon}3}\right\}\right\}$, then $(x-\delta,x+\delta)\times(y-\delta,y+\delta)$ is an open nbhd of $\langle x,y\rangle$ contained in $p^{-1}[U(a,b)]$, and $p^{-1}[U(a,b)]$ is therefore open in $\Bbb R^2$.

It follows immediately that if $V$ is any open subset of $\Bbb R$, then

$$p^{-1}\left[\left\{[\langle x,y\rangle]\in\Bbb R^2/\!\sim\,:x+y^2\in V\right\}\right]\tag{1}$$

is open in $\Bbb R^2$ and hence that $\left\{[\langle x,y\rangle]\in\Bbb R^2/\!\sim\,:x+y^2\in V\right\}$ is open in $\Bbb R^2/\!\sim$. To finish the job you need to show that if $V$ is not open in $\Bbb R$, then the set in $(1)$ is not open in $\Bbb R^2/\!\sim$; you can use the same sorts of ideas.

Another way to say this is that the function

$$f:\Bbb R^2/\!\sim\;\to\Bbb R:[\langle x,y\rangle]\mapsto x+y^2$$

is continuous. (In fact $f$ is a homeomorphism.)