# Are quotients of these two homeomorphic spaces still homeomorphic?

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.

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" ?