Is the subset of $\mathbb C^n$ that is conjugate invarivant and permutation invariant simply connected in the quotient space?

Suppose $$\sim$$ is a relation on $$\mathbb C^n$$ such that $$x \sim y$$ if and only if there is some permutation $$\sigma \in S_n$$ with $$x = (x_1, \dots, x_n) = (y_{\sigma(1)}, \dots, y_{\sigma(n)}) = \sigma y.$$ Let $$\mathbb C^n/\sim$$ be the quotient space and $$\mathbb C^n_{\text{conj}}/\sim$$ be a subset in $$\mathbb C^n/\sim$$ that contains all $$v \in \mathbb C^n/\sim$$ such that $$\bar{v} = v$$. In other words, elements in $$v$$ (concerned as unordered list) is invariant under complex conjugation.

My question is whether this set $$\mathbb C^n_{\text{conj}}/\sim$$ is simply connected?

• I suggest you write the answer you found as … an answer! Then you accept it, possibly after some enforced delay to allow others to answer it (I don't know if such a delay is required or not). This way, the question will not remain apparently unanswered forever. An alternative, if you don't wish to keep the question around, you may opt to delete it. (But then you'll lose the reputation the question has earned you, of course.) – Harald Hanche-Olsen Feb 7 at 18:05
• Your definition of $\mathbb C^n_{\text{conj}}/\sim$ should be clarified. It is the subset of all $v \in \mathbb {C}^n/\sim$ such that $\overline{v} = v$. – Paul Frost Feb 7 at 18:15
• @PaulFrost: Thanks. This is more clear. – user1101010 Feb 7 at 18:53

I just realized the answer is quite obvious. Vieta's map gives a homeomorphism between $$\mathbb C^n$$ and $$\mathbb C^n/\sim$$. If we restrict the map on $$\mathbb R^n$$, then the image is exactly $$\mathbb C^n_{\text{conj}}/\sim$$. So it must be simply connected and indeed contractible.