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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?

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The "singular set" $(x,x)$ is codimension 2, not 1, so you should not visualuze this as something with boundary. To convince you "why" this works, let's instead consider $X_n = \Bbb R^n \times \Bbb R^n/(x,y)\sim (y,x)$. By considering the map $(x,y) \mapsto (x+y,x-y)$ we see this is the same as $\Bbb R^n \times \Bbb R^n/(x,y)\sim (x,-y)$. This is the same as $\Bbb R^n \times C(\Bbb{RP}^{n-1})$, the infinite cone on $\Bbb{RP}^{n-1}$. This is a manifold when, and only when, $n=2$ (because $\Bbb{RP}^1$ is a circle!) In the case $n=1$ you get $\Bbb R \times [0,\infty)$, like you expect.

It is probably worth adding further that $\Bbb C^k/S_k$, modding out by the action of the symmetric group, is also homeomorphic to $\Bbb C^k$. Taking symmetric powers of complex curves is a powerful tool.

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  • $\begingroup$ Amazing, thank you! That problem's been bugging me for ages. $\endgroup$ – Oscar Cunningham Jan 25 '16 at 21:17
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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 $$x,y=\frac {s\pm\sqrt {s^2-4p}}{2}$$.

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