# Homeomorphism Between Quotient Space and $\Bbb{R}$

Define an equivalence relation on the plane $X =\mathbb{R^2}$ as follows:

$$(x_0, y_0) \sim (x_1,y_1) \mbox{ if } x_0 + y_0^2 = x_1 +y_1^2$$

Let $X^*$ be the corresponding quotient space. It is homeomorphic to a familiar space: what is it?

Let $g : \mathbb{R}^2 \to \mathbb{R}$ be defined by $g(x,y) := x + y^2$. Then certainly it is continuous; moreoever, $\mathbb{R} \ni x = g(x,0)$ which shows that $g$ is surjective. Letting $[(x,y)]$ denotes the equivalence classes associated to the equivalence relation already defined, given $a \in \mathbb{R}$, $a = w + z^2$ for some $w,z \in \mathbb{R}$ we see that $g^{-1}(\{w+z^2\}) = \{(x,y) \in \mathbb{R}^2 ~|~ g(x,y) = w + z^2 \} = \{(x,y) ~|~ x + y^2 = w + z^2 \} = [(w,z)]$, which proves that $X^* = \{g^{-1}(\{a\}) ~|~ a \in \mathbb{R} \}$. Therefore, there exists a continuous bijection $f : X^* \to \mathbb{R}$ such that $f$ is a homeomorphism if and only if $g$ is a quotient map, which we now prove.

It suffices to show that $g$ is an open map. Given $(a,b) \times (c,d)$, I will prove $g((a,b) \times (c,d)) = \bigcup_{y \in (c,d)} [(a,b) + y^2]$. If $z \in g((a,b) \times (c,d))$, the $z = g(x,y) = x + y^2$ for some $x \in (a,b)$ and $y \in (c,d)$, which means by definition $z \in (a,b) + y^2$. Now, if $z \in \bigcup_{y \in (c,d)} [(a,b) + y^2]$, then $z \in (a,b) + y^2$ for some $y \in (c,d)$; furthermore, there is some $x \in (a,b)$ such that $z = x + y^2 = g(x,y)$ and hence $x \in g((a,b) \times (c,d))$. Since $(a,b)$ is open and addition is a homeomorphism, each $(a,b) + y^2$ is open and hence $g((a,b) \times (c,d))$ is a union of open sets.

How does this sound? I couldn't really find any full solutions online, so I am looking for a critique of my proof.

EDIT:

Here is an alternative proof, based on Hurkyl's suggestion. For convenience, let $[x,y]$ denote the equivalence classes instead of $[(x,y)]$. First note that $x+y^2 + 0 = x + y^2$ which implies $[x+y^2,0] = [x,y]$. One thing I left out above is the fact that $f \circ p = g$, where $p : X \to X^*$ defined by $p(x,y) = [x,y]$ is the continuous surjection that projects points to the equivalence class to which they belong; this means that $f$ is given explicitly by $f[x,y] = x + y^2$. Now define $k : \Bbb{R} \to X^*$ by $k(x)=[x,0]$. Then $f(k(x)) = f([x,0])=x + 0^2 = x$ and $k(f[x,y]) = k(x+y^2)=[x+y^2,0]=[x,y]$, so that $k =f^{-1}$. But clearly $k$ is equal to the composition $p \circ h$ of continuous functions. Hence $f$ is a homeomorphism.

• Don't overlook the direct way to show something is a homeomorphism: construct its inverse! Especially if you can do so in a way that makes it obvious the inverse is continuous! E.g. in this problem, you might use the continuous map $h : \mathbb{R} \to \mathbb{R}^2 : t \mapsto (t,0)$ in some fashion. – Hurkyl Oct 16 '17 at 18:13
• @Hurkyl Okay. I added a second proof based on your suggestion. Perhaps you could have a look at it when you have a chance. – user193319 Oct 17 '17 at 11:41
• You might add the precise reference to the problem you are solving. I recall seeing it in Munkres at some point. – Stephen Oct 17 '17 at 20:19