Let $D$ denote the open unit disk in $\mathbb C$ and $H$ denote the open left half plane in $\mathbb C$. We know there is a homeomorphism $f : D \to H$. Now let $$g = (g^1, \dots, g^n) := (f, \dots, f): D \times D \times \dots \times D \to H \times \dots H,$$ i.e., $g$ is $n$ copies of $f$. Now $g$ defines a homeomorphism between two subsets in $\mathbb C^n$, namely, $D_n := D \times \dots \times D$ and $H_n := H \times \dots \times H$. Let us define an equivalence $\sim$ relation on $\mathbb C^n$: $x \sim y$ if and only if there is a permutation $\sigma \in \mathbb S_n$ such that $\sigma x = (x_{\sigma(1)}, \dots, x_{\sigma(n)}) = (y_1, \dots, y_n)$.

My question is whether the quotient space $D_n/\sim$ and $H_n/\sim$ are still homeomorphic.


1 Answer 1


For a space $X$, define $Sym^n(X)$ to be $X^n/\sim$ where $\sim$ is defined as in your question.

Now given $f: X\to Y$ a continuous map between spaces, there is an induced map $f^n=(f,...,f) : X^n\to Y^n$ and it passes to the quotient uniquely : there is a unique continuous map $Sym^n(f): Sym^n(X)\to Sym^n(Y)$ such that $Sym^n(f)\circ\pi_X = \pi_Y\circ f^n$, where $\pi_X$ and $\pi_Y$ are the obvious projections.

One can check the two following facts : $Sym^n(f\circ g) = Sym^n(f)\circ Sym^n(g)$ and $Sym^n(id_X) = id_{Sym^n(X)}$.

I recommend you try and prove these two facts using the characterization of $Sym^n(f)$ I just gave (using uniqueness especially).

Can you now prove from these two facts that if $f$ is a homeomorphism, $Sym^n(f)$ is one as well ? Can you now say something even better than "$D_n/\sim$ and $H_n/\sim$ are homeomorphic" ?


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