# Topological puzzle: configuration space

I came across a problem which looks like an exercise from Topology by Munkres, yet I can’t figure it out.

Let $X$ be a connected topological space. The product $X\times X$ is also a topological space, and is connected as well. (Why? Because each “horizontal” fiber $X\times \{x\}$ and each “vertical” fiber $\{x\}\times X$ must lie within some connected component, but since they intersect, there is just one component.) However, if we remove the diagonal $D=\{(x,x)\colon x\in X\}$ from the product, the connectedness may be lost. For example, $(X\times X)\setminus D$ is disconnected when $X$ is a line.

So far so easy. Now let’s take a quotient of $(X\times X)\setminus D$, identifying each pair $(x_1,x_2)$ with the reordered pair $(x_2,x_1)$. In other words, the symmetric group $S_2$ acts on $(X\times X)\setminus D$ by permutation of coordinates, and we take the space of orbits. This new space is denoted $C_2(X)$ and is called the configuration space of $X$.

Is $C_2(X)$ connected?

It is in all examples that I can think of. Of course, this is not much evidence (for one thing, my thinking does not go beyond Hausdorff spaces). I’m happy to assume that $X$ is metrizable, in which case $C_2(X)$ carries the Hausdorff metric $d(\{a,b\},\{a',b'\}) = \min(\max(|a-a'|,|b-b'|),\max(|a-b'|,|b-a'|))$. Here I’m using the $|\cdot -\cdot |$ notation for the metric to reduce the clutter.