Show that \(SP_n(I) = \Delta^n\).
Solution
An element of the symmetric product \[ SP_n(I)=I^n/\Sigma_n \] is an unordered \(n\)-tuple of points of \(I=[0,1]\). Every orbit has a unique representative whose coordinates are weakly increasing: \[ 0\le x_1\le x_2\le\cdots\le x_n\le1. \] Therefore the quotient map restricts to a continuous bijection \[ Q=\{(x_1,\ldots,x_n):0\le x_1\le\cdots\le x_n\le1\} \longrightarrow SP_n(I). \] The domain is compact and the target is Hausdorff, so this is a homeomorphism.
Now set \[ y_0=x_1, \qquad y_i=x_{i+1}-x_i\ (1\le i<n), \qquad y_n=1-x_n. \] Then each \(y_i\ge0\) and \[ y_0+\cdots+y_n=1. \] Conversely, \[ x_j=y_0+\cdots+y_{j-1}. \] Thus \(Q\) is affinely homeomorphic to the standard \(n\)-simplex \[ \Delta^n=\{(y_0,\ldots,y_n):y_i\ge0,\ \sum y_i=1\}. \] Hence \[ \boxed{SP_n(I)\cong\Delta^n.} \]