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 $U_c=\{(x,y)\in\mathbb{R}^2: y=x^2+c, a<c<b\}$

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 $y=x^2+a$.

  • $\begingroup$ Why not $f([(x,y)])=g(x,y)$? $\endgroup$ – Hagen von Eitzen Feb 14 at 7:32
  • $\begingroup$ @Hagen von Eitzen Yeah that makes more sense. Since $[c]$ isn't actually what the equivalence classes should look like. $\endgroup$ – AColoredReptile Feb 14 at 7:36
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    $\begingroup$ 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? $\endgroup$ – Hagen von Eitzen Feb 14 at 7:49
  • $\begingroup$ @Hagen von Eitzen Does $g$ need to be continuous? $\endgroup$ – AColoredReptile Feb 14 at 8:17

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 $f^{-1} = J$.

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

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