# determining a homeomorphic copy of a quotient of a subspace

let $X$ and $Y$ be two topological spaces. let $S_n$ be the symmetric group and let $Z=X\times Y\sqcup Y\times X$ and take the quotient of $Z$ by identifying $(x,y)$ with $(y,x)$ what is this quotient and is it homeomorphic to some simpler space? i thought it is $(X \times Y)/S_2$ but there is no action from $S_2$ on $X\times Y$ unless $X=Y$ but what if $X\not = Y$? i asked this question because i'm trying to describe the space $K$ of unordered couples $[x,*]\in (X\times X)/S_2$ such that $x\not = *$ here $*$ is the base point of $X$ so i wrote that $K$ is the space $(X-\{*\})\times \{*\} \sqcup \{*\}\times (X-\{*\})$ modulo the identification $(x,*)\sim (*,x)$. I know that $K=(X\vee X)-\{(*,*)\}/S_2$ but i want a more compact formula that can be generalized to the general case of the space $A_n$ of unordered $n$-tuples $[*,x_1,\cdots,x_{n-1}]\in X^n/S_n$ such that $x_1\not = \cdots \not =x_{n-1}\not = *$

Added I think this space $A_n$ has the form $F(X-*,n-1)/S_{n-1}$ where $F(Z,k)\subset Z^k$ is the subspace of all distinct entries, i.e; $(z_1,\cdots, z_k)\in Z^k$ such that $z_1\not = \cdots \not = z_k$

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I might be missing something, but to me it seems that the quotient space you get is just $X\times Y$. To clarify a little: Let $Z=X\times Y\times\{1\}\sqcup X\times Y\times\{2\}$. You are identifying $(x,y,1)$ with $(y,x,2)$, right? – Martin Sleziak Dec 16 '11 at 12:54
No i'm just identifying tuples with same elements but in different order.. i'm just ignoring order of elements so $(x,y)$ is the same as $(y,x)$ and i think what i added in my post is correct – palio Dec 16 '11 at 13:56
Making $(x,y)$ the same as $(y,x)$ is exactly what Martin said you were doing, and you do get just $X\times Y$, even when $Y=X$. I suspect that what you really want is $X^2$ modulo the equivalence relation that identifies $(x,y)$ and $(y,x)$, which is not the same as the quotient of $X^2\sqcup X^2$ that identifies $(x,y)$ in the first $X^2$ with $(y,x)$ in the second $X^2$. – Brian M. Scott Dec 16 '11 at 17:51

For $n=2$ there is a very simple description of $A_n$, assuming that you’ve correctly described the space that you want.

Let $Y=X\setminus\{*\}$. Your $A_2$, as I understand it, can be constructed as follows. Start with $X\times X$. Let $S=\big(\{*\}\times Y\big)\cup\big(Y\times\{*\}\big)$. Then $A_2$ is the quotient of $S$ under the identification of $\langle*,y\rangle$ with $\langle y,*\rangle$ for each $y\in Y$. This $A_2$ is clearly homeomorphic to $Y$ via the map $$h:A_2\to Y:\big[\langle*,y\rangle\big]\mapsto y\;,$$ where $\big[\langle*,y\rangle\big]=\big\{\langle*,y\rangle,\langle y,*\rangle\big\}$ is the equivalence class of $\langle*,y\rangle$ under the identification.

The general case requires a little more work, and you may or may not find the resulting description satisfactorily simple.

Let $[Y]^n=\{F\subseteq Y:|F|=n\}$, and let $\tau$ be the topology on $Y$. For $\mathscr{V}=\langle V_1,\dots,V_n\rangle \in \tau^n$ let $\mathscr{V}^\#=\{F\in[Y]^n:F\cap V_k\ne \varnothing\text{ for }k=1,\dots,n\}$. Let $Y_n$ be the space with underlying set $[Y]^n$ and topology $\tau_n$ generated by the base $\{\mathscr{V}^\#:\mathscr{V}\in\tau^n\}$.

In other words, a point of $Y_n$ is an $n$-element subset $F=\{y_1,\dots,y_n\}$ of $Y$. A basic open nbhd of $F\,$ in $Y_n$ is a set of the form

$$\big\{\{z_1,\dots,z_n\}\in[Y]^n:z_k\in V_k\text{ for }k=1,\dots,n\big\}\;,$$

where $V_k$ is an open nbhd of $y_k$ in $Y\,$ for $k=1,\dots,n$.

Then $Y_n$ is homeomorphic to the subspace of $X^n/S_n$ generated by $n$-tuples of distinct points of $Y$, which in turn is homeomorphic to your $A_{n+1}$. Including the point $*$ in every tuple is akin to taking a Cartesian product with a one-point space and has no effect on the homeomorphism type.

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