# Show that the quotient space is homeomorphic to $\mathbb{R}$

Suppose we have a topological space $X = \mathbb{R}^2$ with standard topology and an equivalence relation on $X$ defined as:

$(a,b) \sim (x,y) \iff a + b^2 = x + y^2$

Show that the quotient space is homeomorphic to topological space $Y$= $\mathbb{R}$ with standard topology (i.e there exists some function $f: X/\sim\to Y$ such that $f$ is continuous, bijective, and the inverse is continuous.

So I know f([(x,y)]) = x + y$^2$, works

Note that f is well defined since any (a,b) $\in$ [(x,y)], a + b$^2$ = x + y$^2$

How do I show that f is continuous and so is its inverse?

Here are some definitions/ theorems I know:

f is continuous iff for every open interval in Y, the preimage of the open interval is open in X\ $\sim$

The topology on X\ $\sim$ is { U $\subseteq$ X\ $\sim$ | p$^{-1}$(U) is open in X= $\mathbb{R}^2$} where p is the quotient map. This is known as the quotient topology.

• Some topological space? I'm not understanding your point. Every quotient space is a topological space in its own right. – dfeuer Nov 15 '13 at 21:47
• I think that the point of the exercise is to identify a more familiar topological space that the quotient is homeomorphic to. – Sammy Black Nov 15 '13 at 21:53
• thats right, sammy black – sarah Nov 15 '13 at 21:55

HINT: Fix $c\in\Bbb R$, and look at the set of points $\langle x,y\rangle\in\Bbb R^2$ such that $x+y^2=c$: that’s one equivalence class. You can even think of it pictorially: it’s the graph of $x=c-y^2$.
• @sarah: Yes, in the case of this particular equivalence relation. Note that there’s one class for each real number $c$; this should suggest a natural map from the quotient space to $\Bbb R$, and with luck it will prove to be the desired homeomorphism. – Brian M. Scott Nov 15 '13 at 22:21
The parabolic contours in this image represent the graphs of the equation $x + y^2 = c$ for various $c$. Each of these curves is an equivalence class that gets identified to a point in the quotient space. What's a good way to identify such a point? How about $c$? (This is essentially the answer that Brian M. Scott gives.)
• So $\mathbb{R}$ is the familiar topological space? – sarah Nov 15 '13 at 22:21
• Yes. And the explicit homeomorphism $X/\!\sim\, \to \Bbb{R}$ is $\{(x, y)\} \mapsto x + y^2$. – Sammy Black Nov 15 '13 at 22:25
• For example how do I show f([(x,y)]) = $\mathbb{R}$ – sarah Nov 15 '13 at 22:50
• To show that the map is surjective, given any $c \in \Bbb{R}$, you have to find some $(x, y) \in \Bbb{R}^2$ with $f(x, y) = c$. The point $(c, 0)$ works, so its equivalence class is a suitable preimage. – Sammy Black Nov 15 '13 at 23:43