TITLE: Is the configuration space a manifold? a CW complex?
QUESTION [8 upvotes]: The ordered configuration space of $n$ points in a topological space $X$ is defined as $F(X,n)=\{(x_1,\ldots,x_n)\in X^{n} | x_i\neq x_j \mbox{ for } i\neq j\}$ and the unordered configuration space is the orbit space $C(X,n)$ under the action of symmetric group $S_n$. My question is:
Is $F(X,n)$ a manifold when $X$ is a manifold? If not, is it a (finite) CW complex? What about $C(X,n)$?
Why $F(X,n)$ is called "ordered" configuration space?

REPLY [10 votes]: When $X$ is a manifold, $X^n$ is also a manifold. Since $F(X,n)$ is an open subset of $X^n$, it is a manifold as well. Next, we have a finite group acting on $F(X,n)$ freely, that is, no group element other than identity has any fixed points. Therefore, the quotient is also a manifold. 
The word unordered fits the space $C$ because taking quotient by permutations means forgetting the order of the elements in $n$-tuples. Simply to emphasize the difference between $F$ and $C$, the former space is called ordered.