Finding well-known topological space that is homeomorphic to given quotient space.

Question

A quotient space is induced by the equivalence relation on $\mathbb{R}^2$ as $(a_0, b_0)\sim (a_1, b_1)$ iff $a_0+b_0^2=a_1+b_1^2$.

I have identified that the equivalence classes are level curves of the function $f(x,y)=x+y^2$. What is the familiar topological space that is homeomorphic with this space?

Answer 1

You're almost done. The function $f:\Bbb R^2 \to \Bbb R$ is continuous, surjective, and $(a_0,b_0)\sim (a_1,b_1)$ if and only if $f(a_0,b_0)= f(a_1,b_1)$. This means that $f$ survives on the quotient as a continuous bijection $\widetilde{f}:\Bbb R^2/_{\sim}\to \Bbb R$, given by $\widetilde{f}[x,y] = x+y^2$. To check that $\widetilde{f}$ is a homeomorphism, it remains to check that the inverse function of $\widetilde{f}$ is continuous. It is $\widetilde{f}^{-1}:\Bbb R \to \Bbb R^2/_{\sim}$ given by $\widetilde{f}^{-1}(c)= [c,0]$. It is continuous as the composition of two continuous maps ($c\mapsto (c,0)$ followed by the quotient projection), and it fits the bill, for:

(1) $\widetilde{f}[c,0] = c+0^2 = c$;

(2) $[\widetilde{f}[x,y],0]=[x+y^2,0] = [x,y]$.

Geometrically: the plane is being foliated by horizontal parabolas, the $x$-axis intersects (transversely) each of them exactly once, and $\widetilde{f}^{-1}$ maps each parabola to its intersection with the $x$-axis in a continuous fashion.

Bottom line: $\Bbb R^2/_{\sim}$ is $\Bbb R$ in disguise.

Answer 2

Your function $f$ is a continuous function $f:\mathbb{R}^2\to\mathbb{R}$ with the same fibers as your given quotient map. If you show that $f$ is a quotient map, then the universal property of quotient maps will give a homeomorphism between these two quotient spaces of $\mathbb{R}^2$.