For which topological spaces $X$ is $X^n$ homeomorphic to the space of unordered n tuples of points in $X$ Consider a topological space $X$. Consider the spaces $A_n=X^n$ and $B_n=X^n/q_n$, the space of where $q_n$ is the equivalence relation where two points in $X^n$ are equivalent if one can be constructed from another via a permutation of the points.
For which topological spaces $X$ are $A_n$ and $B_n$ homeomorphic for all natural numbers $n$? For example, when $X=\mathbb{R}$, $A_n$ and $B_n$ are not homeomorphic in general, consider $n=2$ in particular.
Another example, when $X=\mathbb{R}^2$, $A_n,B_n$ are homeomorphic for each natural $n$. This can be proven by thinking of $X$ as the complex plane, $A_n$ as the space of monic polynomials of degree $n+1$ over $\mathbb{C}$, and thinking of $B_n$ as the space of unordered $n$ tuples of roots. It is a theorem that the map which takes the coefficients of a polynomial to its roots is bijective and bicontinuous, thus these spaces are homeomorphic.
Some more details on this proof:
1: Each list of coefficients gives a unique set of roots and vice versa
2: the coefficient list as a function of the roots is continuous because this function is a polynomial
3: the roots as a function of the coefficients is continuous in the quotient space $B_n$ (for a proof, see Bhatia matrix analysis, beginning of chapter 6).
I'd like to know for which other spaces this works.
 A: Via this post, we have the following characterization:

Let $X$ be a topological space. Then, $X^n$ is homeomorphic to $X^n / q_n$ if and only if there exists a quotient map $f : X^n \to X^n$ such that $f$ induces the equivalence relation $q_n$ on $X^n$.

So, this gives a (presumably difficult) way to test if $X$ satisfies the property you like.

It suffices to find a surjection $f : X^n \to X^n$ such that

*

*(quotient map definition) Given $U \subseteq X^n$, $U$ is open in $X^n$ if and only f $f^{-1}(U)$ is open in $X^n$.

*(induced equivalence relation) Given $x = (x_1,\ldots,x_n),y=(y_1,\ldots,y_n) \in X^n$, $x$ is equivalent to $y$ under $q_n$ if and only if $f(x)=f(y)$. In other words, the coordinates $y_1,\ldots,y_n$ are a permutation of the coordinates $x_1,\ldots,x_n$ if and only if $f(x)=f(y)$.


You can replace 1. by any of the other equivalent ways of defining quotient maps, as outlined again here.
This is perhaps not a satisfying answer, but at least it gives a basic characterization. Maybe others will be able to develop this thought process further. Feel free to comment ideas below. I was thinking maybe 1. and 2. can be shown to prove that $f$ must satisfy a stronger property, since we know what $q_n$ is, but I couldn't work that out.
