Prove that $[0,1]^n / S_n \cong \Delta^n$, ( $S_n$ denotes the permutation group on n symbols, $\Delta^n:=\{(x_0,\dots,x_n)\in\Bbb R^{n+1} : x_0,\dots,x_n \ge 0 , x_0+\dots+x_n=1\}$, and the action is defined by permuting coordinates)
My attempt :
Intuitively I can visualize the orbit spaces to be homeomorphic to the $\Delta^n$ upto $n\le 2$. But facing real difficulty in writing an explicit quotient map from $[0,1]^n \to \Delta^n$ so that I can implement the Universal Property Of Quotient Topology.
Thanks in Advance for help!
Hint: $\Delta^n$ is homeomorphic to the space $(x_1, x_2, \ldots, x_n)$ with $0\leq x_1 \leq x_2\leq \cdots \leq x_n\leq 1$ via the map
$$ (x_0,x_1,\ldots, x_n)\mapsto (x_0,x_0+x_1, x_0+x_1+x_2, \ldots, \sum_{i=0}^{n-1} x_i)$$ with inverse map
$$(x_1,\ldots, x_n)\mapsto (x_1, x_2-1_x, x_3-x_2, \ldots, x_n-x_{n-1}, 1-x_n).$$
Here is how to use the hint. We first note that every orbit contains a unique elements whose coordinates are sorted. For example, in the orbit of $(3/4, 1/4, 1/2)$, there is $(1/4, 1/2, 3/4)$. This gives a bijection between the space of orbits in $[0,1]^n$ and the space of weakly increasing sequences of length $n$, which by the hint is homeomorphic to the $n$-simplex. What remains is to show that this bijection with the orbit space (the quotient by the group action) is actually a homeomorphism. Or, phrased slightly differently, we must show that the topology we inherit as a subspace of $\mathbb R^n$ is the same as the quotient topology.
If a surjective map $f:X\to Y$ is a quotient map, then $U\subset Y$ is open if and only if $f^{-1}(U)$ is open in $X$. We wish to show that the sorting map is a quotient map. However, there is an easier way than by directly using the definition. Instead, we use the following result:
Theorem: If a continuous and surjective is an open map or a closed map, then it is a quotient map.
It isn't too hard to show that the sorting map is continuous, and it is clearly surjective (as it is a map from a larger space onto a subspace that is the identity on the subspace). To see that it is open, consider an open set $U$ in $[0,1]^n$. The image of $U$ is the union of $gU$ for $g\in S_n$ intersected with the set of sorted sequences. Since unions of open sets are open, and since sets are open in the subspace topology if they are intersections of open sets with the subspace, we are done.