How to prove that the symmetric product $SP^{n}(\mathbb{C})$ of the complex plane $\mathbb{C}$ is homeomorphic to $\mathbb{C}^{n}$.

Question

I want to prove that the map $f : SP^{n}(\mathbb{C}) \rightarrow \mathbb{C}^{n}$ which to $[z_{1}, z_{2}, \ldots , z_{n}]$ associates the elementary symmetric polynomials is a homeomorphism. I can prove that $f$ is bijective and continuous. I cannot prove that the inverse $f^{-1}$ of $f$ is a homeomorphism.

Answer 1

Let us write $g=f^{-1}$. Note that $g$ can be interpreted as the map which takes the coefficients of a monic polynomial of degree $n$ to its multiset of roots. We also write $p:\mathbb{C}^n\to SP^n(\mathbb{C})$ for the quotient map.

To show that $g$ is continuous, it suffices to show its restriction to every bounded subset of $\mathbb{C}^n$ is continuous. The key fact now is that if $B\subset\mathbb{C}^n$ is bounded, then $p^{-1}(g(B))\subset \mathbb{C}^n$ is also bounded. Concretely, this statement means that if you have a set of monic polynomials of degree $n$ whose coefficients are bounded, then their roots are also bounded. This is obvious: for instance, if all the coefficients of $p(z)=z^n+a_{n-1}z^{n-1}+\dots+a_0$ are bounded by $M$ in absolute value, then $|p(z)|\geq |z|^n-M\sum_{k=0}^{n-1}|z|^{k}$ and so $|p(z)|>0$ as long as $|z|$ is sufficiently large compared to $M$. So, all roots of $p(z)$ can be bounded just in terms of $M$.

So now, let $B\subset\mathbb{C}^n$ be bounded and let $C=\overline{p^{-1}(g(B))}\subset SP^n(\mathbb{C})$. Note that $C$ is compact, since $p^{-1}(g(B))$ is bounded. Thus $p(C)\subset SP^n(\mathbb{C})$ is also compact. But now the restriction of $f$ to $p(C)$ must be a homeomorphism to its image, since it is a continuous bijection from a compact space to a Hausdorff space. This means that $g$ restricted to $f(p(C))$ is continuous. Since $B\subseteq f(p(C))$, this means $g$ is continuous on $B$, as desired.