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Let $X$ be a topological space and $S_3$ the symmetric group acting on $X^3$ by permuting coordinates.

Let $\pi:X^3\rightarrow X^3/S_3$. Denote $[x,y,z]=\pi(x,y,z)$. Let $U_x$ be the neighborhood of $x$ in $X$. Then $(U_x\times U_y\times U_z)/S_3$ is a neighborhood of $[x,y,z]$.

Now consider the point $[x,x,z]\in X^3/S_3$ i.e., $(x=y)$, why a neighborhood of $[x,x,z]$ is $(U_x\times U_x)/S_2 \times U_z$ instead of $(U_x\times U_x\times U_z)/S_3$?

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up vote 3 down vote accepted

It isn't. However, if $U_x$ and $U_z$ are disjoint, then the two are homeomorphic.

If $a,b\in U_x \cap U_z$, then $(U_x\times U_x)/S_2 \times U_z$ contains separate points $([a,b],a)$ and $([a,a],b)$, whereas $(U_x\times U_x\times U_z)/S_3$ and $X^3/S_3$ contain only one point $[a,a,b]$.

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actually i realize that an action of $S_3$ on $U_x\times U_y\times U_z$ need not exist since if we permute $(x,y,z)\in U_x\times U_y\times U_z$ for example to have $(y,z,x)$ and this need not be in $U_x\times U_y\times U_z$.but the question then is why $(U_x\times U_x) /S_2 \times U_z $ is a neighborhood of $[x,x,z]$? and how this relates to $\pi(U_x\times U_x\times U_z)$ which is not necessarely $(U_x\times U_y\times U_z)/S_3$. – palio Jul 1 '11 at 14:17
@palio: As I wrote, I don't think it is. Why do you think it is? – joriki Jul 1 '11 at 14:24
actually i see now that the map $\pi(U_x\times U_x \times U_z)\rightarrow (U_x\times U_x)/S_2\times U_z$ sending $[x_1,x_2,z]\mapsto ([x_1,x_2],z)$ is a homeomorphism – palio Jul 1 '11 at 14:40
Why the downvote? – joriki Jul 1 '11 at 15:24
i'm sorry i did not see it, not yet used to the site!! and also i'm not allowed to vote untill you edit it :( – palio Jul 1 '11 at 15:28

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