I made the following strange observation: For any $n>0$, there is a continuous bijection $\mathbb{C}^n/S_n \rightarrow \mathbb{C}^n$. Now I want to know: are they homeomorphic?
Here, $\mathbb{C}^n/S_n$ denotes the quotient space obtained from $\mathbb{C}^n$ by the action of a permutation group $S_n$; we define $\sigma \cdot (\alpha_1, \cdots \alpha_n) := (\alpha_{\sigma(1)}, \cdots \alpha_{\sigma(n)})$ where $\sigma \in S_n$ is a permutation on $n$ entries.
To see this, define the map $\Phi : \mathbb{C}^n \rightarrow \mathbb{C}^n$ given by $$\Phi(\alpha_1, \cdots \alpha_n) := (\beta_1, \cdots \beta_n) \text{ where } (z-\alpha_1) \cdots (z-\alpha_n) = z^n + \beta_1 z^{n-1} + \cdots + \beta_n z^0$$ By the fundamental theorem of algebra, $\Phi$ is surjective. Also, the fiber of $\Phi$ is the $S_n$-orbit of each $(\alpha_1, \cdots \alpha_n)$. By the universal property of quotients, there is a continuous bijection $\tilde \Phi : \mathbb{C}^n / S_n \rightarrow \mathbb{C}^n$ that factors through $\Phi$.
I have an idea for how to prove that this is a homeomorphism, relying on the fact that a continuous bijection from a compact space to a Hausdorff space is a homeomorphism. For this, it seems to make sense to construct a covering of $\mathbb{C}^n/S_n$ by compact sets, such that their interiors cover $\mathbb{C}^n/S_n$. I didn't spell out this argument fully yet, but I wanted to know if 1. the existence of a continuous bijection is true, and 2. whether the two spaces are homeomorphic indeed.
