TITLE: Are $\mathbb{C}^2$ and $\mathbb{C}^2/(x,y)\sim(y,x)$ homeomorphic?
QUESTION [5 upvotes]: Let $A$ be the set of monic quadratics over $\mathbb C$ and let $B$ be the set of unordered pairs over $\mathbb C$ where possibly the two elements of the pair may be the same. Then the map which takes a quadratic to it roots seems to be a homeomorphism between $A$ and $B$. But I can't reconcile this with the fact that $B$ is $\mathbb C^2$ after it has been "folded in half" and so the points $\{x,x\}\in B$ ought to lie on a boundary near which $B$ is not homeomorphic to $\mathbb C^2$.  Where am I going wrong?

REPLY [2 votes]: Viète's mapping $$V:\mathbb C^2\to A=\mathbb C^2:(x,y)\mapsto (s=x+y,p=xy)$$ yields a bijective continuous map $$v :B=\frac {\mathbb C^2}{\sim}\to A=\mathbb C^2:\widetilde {(x,y)}=\widetilde {(y,x)}\mapsto (s=x+y,p=xy)$$ The mapping $V$ is proper, hence closed and thus the descended map $v$ is closed too, which finishes the proof that $v$ is a homeomorphism.   
Edit
The map $$v^{-1}: A=\mathbb C^2\to B=\frac {\mathbb C^2}{\sim}:(s,p)\mapsto \widetilde {(x,y)}$$ is the one described by Oscar:
It sends the ordered pair $(s,p)$ identified to the monic polynomial $T^2-sT+p$ to the unordered pair $\widetilde {(x,y)}$ consisting of its two roots, given by the good old formula $$x,y=\frac {s\pm\sqrt {s^2-4p}}{2}$$.