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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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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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    @palio: I trivially edited joriki's answer, so now you should be able to turn the accidental downvote into a deserved upvote. joriki: after 5 minutes the vote is "locked in" and you can't change the vote until the post is modified. I simply added a blank space to the end of the second paragraph.2011-07-01