# Given $X=\mathbb{R}^2$ and the equivalence relation $(x,y)\sim(x',y')$ iff $y-x^2=y'-x'^2$. What space is this homeomorphic to?

Given $$X=\mathbb{R}^2$$ and the equivalence relation $$(x,y)\sim(x',y')$$ iff $$y-x^2=y'-x'^2$$. What space is this homeomorphic to?

There is also a hint that $$g(x,y)=y-x^2$$

So I tried drawing the equivalence relations on $$\mathbb{R}^2$$ and noticed that the equivalence relations are parabolas. The $$[0]$$ is $$y=x^2$$, then $$[1]$$ is $$y=x^2+1$$, etc. So I believe $$[c]$$ is $$y=x^2+c$$

I want to show this space is homeomorphic to $$\mathbb{R}$$ and I have the map $$g:\mathbb{R}^2\to \mathbb{R}$$. which is going to map equivalence classes to that $$c$$ value. I believe I need to find $$f$$ such that $$f\circ p=g$$. where $$p$$ is the quotient map. Then show it is continuous in both directions and bijective.

Can I take $$f:\mathbb{R}^2\setminus \sim\to \mathbb{R}$$ as $$f([(x,y)])=g(x,y)$$?

I know it's surjective since $$g=f\circ p$$ is surjective

Since if $$r\in \mathbb{R}$$ then $$g(0,r)=r$$.

And it's clearly injective, by the equivalence relation.

But I'm not sure how to show it is continuous.

I have this theorem which I believe might be useful:

Let $$X,Z$$ be two topological spaces, $$\sim$$ an equivalence relation on $$X$$ and $$f:X\setminus \sim \to Z$$ a function. Then $$f$$ is continuous if and only if $$f\circ p$$ is continuous.

So I could show $$g$$ is continuous and get that $$f$$ is continuous.

So let $$(a,b)$$ be an open interval in $$\mathbb{R}$$.

then I want to show $$g^{-1}[(a,b)]$$ is open.

I believe it looks like a big open parabola, a union of $$U_c=\{(x,y)\in\mathbb{R}^2: y=x^2+c, a

So I want to show that $$\bigcup U_c$$ is open in $$\mathbb{R}^2$$

Let $$p\in \bigcup U_c$$ then $$p\in U_c$$ for some $$c$$.

But I'm having trouble finding a radius of a ball which will be contained in this union. I want to pick something so that all the points in the ball will be less then the parabola $$y=x^2+b$$ and greater then $$y=x^2+a$$.

• Why not $f([(x,y)])=g(x,y)$? – Hagen von Eitzen Feb 14 at 7:32
• @Hagen von Eitzen Yeah that makes more sense. Since $[c]$ isn't actually what the equivalence classes should look like. – AColoredReptile Feb 14 at 7:36
• If $g\colon X\to Y$ is continuous ad $x\sim y\iff g(x)=g(y)$, under what mild conditions does $X/{\sim}\cong Y$ follow? – Hagen von Eitzen Feb 14 at 7:49
• @Hagen von Eitzen Does $g$ need to be continuous? – AColoredReptile Feb 14 at 8:17

## 1 Answer

It is more or less obvious that $$g$$ is continuous. It is well-known (and easy to verify) that the following is true:

Let $$f, g : X \to \mathbb R$$ be continuous maps defined on a topological space $$X$$ and $$a \in \mathbb R$$. Then $$f + g, f \cdot g$$ and $$a \cdot f$$ are continuous.

The projections $$p_1. p_2 : \mathbb R^2 \to \mathbb R, p_1(x,y) = x, p_2(x,y) = y$$, are clearly continuous. Thus $$g(x,y) = p_2(x,y) - p_1(x,y) \cdot p_1(x,y)$$ is continuous.

Alternatively you can also consider sequences $$(x_n,y_n)$$ converging to some $$(x,y)$$ and show that $$(g(x_n,y_n))$$ converges to $$g(x,y)$$.

Let $$\pi : \mathbb R^2 \to Y = \mathbb R^2/\sim$$ be the quotient map. Define $$j : \mathbb R \to \mathbb R^2, j(t) = (0,t)$$. This is a continuous map, hence $$J = \pi \circ j : \mathbb R \to Y$$ is continuous. We have $$f(J(t)) = f([0,t]) = g(0,t) = t,$$ $$J(f([x,y]) = J(g(x,y) = \pi(0,y - x^2) =[0,y-x^2] = [x,y]$$ because $$(y-x^2) - 0^2 = y - x^2$$. This shows that $$f$$ and $$J$$ are inverse to each other. This means that $$f,J$$ are homeomorphisms such that $$f^{-1} = J$$.

• So if $g$, I get $f$ is continuous. But then I need to show $f^{-1}$ is continuous, right? – AColoredReptile Feb 14 at 10:39
• Yes, I have done this in my answer (which contained a typo that I corrected). We have $f^{-1} = J$. – Paul Frost Feb 14 at 10:50