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 $z_1\not = \cdots \not = z_k$