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Suppose $\def\Sym{\operatorname{Sym}} \def\C{\mathbb{C}} \Sym^n\C$ is the $n$-fold symmetric product space defined by the quotient map $q: \C^n\to \Sym^n\C$. To show $\C^n$ is homeomorphic to $\Sym^n\C$.

My effort:

I want to show: $g: \Sym^n\C \to \C^n$ defined as $$ g \bigl( [r_0, \dots, r_{n-1}] \bigr) = \prod_{i=0\,}^{n-1}(z-r_i) $$ is a homeomorphism, where $z$ is a formal variable. (We write element in $\C^n$ in the form of a $n$-th order polynomial in $z$.)

I have shown that $g$ is a continuous bijective by universal property. I know that one can show the inverse of $g$ is continuous. But the proof is rather long. I would like to know if there is a short proof that $g$ is open or closed.

Sammy Black
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  • Please explain what $z$ is. – Paul Frost Nov 15 '20 at 23:57
  • $g([r_0,\cdots,r_{n-1}])$ is a complex polynomial with roots $r_i$. How do you regard this as an element of $\mathbb C^n$? That was the intention of my previous comment. I am aware what the answer is, but it is an essential ingredient and it should be explicitly mentioned in your question. – Paul Frost Nov 16 '20 at 12:07
  • @PaulFrost, Sorry for the unclarity. I have fixed it. –  Nov 16 '20 at 18:09

1 Answers1

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The question Constructing explicit connected path for coefficients of monic polynomials with roots lying in the open left half plane? has at first glance nothing to do with your present question, but in my answer you can find a proof that $g$ is a closed map.

Paul Frost
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