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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.

Mathew
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    If $X=R$ and $n=1$, then $A_n$ is equal to $B_n$, or not? If $X=R^2$ and $n=2$, then I don't think they are equal. Consider two points $(-1, 0)$ and $(1,0)$, rotate them by 180 degrees and you obtain an incontractible loop in $B_n$, or not? The coefficients $\leftrightarrow$ roots map you are suggesting is probably not bicontinuous in the origin – Peter Franek Jul 21 '22 at 19:58
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    Thanks for noticing the error regarding n=1, I should have said n=2. I edited my question. Regarding R^2 when n=2, if you are not convinced by the algebraic proof, consider another representation of unordered pairs of points in the plane, in particular, specify the average of the two points, and the difference (a,d). Notice (a,d) is similar to (a,-d), thus it suffices to show that quotienting by the equivalence relation c ~ -c for c in the complex plane is gives a space homeomorphic to the plane, which is straight forward to see. – Mathew Jul 21 '22 at 20:07
  • Right, interesting. – Peter Franek Jul 21 '22 at 20:41

1 Answers1

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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

  1. (quotient map definition) Given $U \subseteq X^n$, $U$ is open in $X^n$ if and only f $f^{-1}(U)$ is open in $X^n$.
  2. (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.

stowo
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