Fiber bundle M x M - diagonal

Question

Under what conditions for a space $M$ does the projection map to the first factor $p: M \times M - \Delta \rightarrow M$ has the local triviality condition, i.e. is a fiber bundle? Where $\Delta$ denotes the diagonal $\{(a,a) \}_{a \in M}$.

Answer 1

Jim's comment that it's true for $M$ a manifold is an old result, perhaps first due to Richard Palais (and in greater generality). The result is usually attributed to Fadell and Neuwirth, but their result came later. In their set-up you call $M \times M \setminus \Delta$ to be the configuration space of two points in $M$. In Palais's set-up, $M \times M \setminus \Delta$ is the space of embeddings of a two-point set into $M$. Palais works in the generality of embedding spaces of manifolds, so the domain manifold does not have to be zero-dimensional like in this case. For example, if $Emb(S^2,M)$ denotes the space of embeddings of a 2-sphere in $M$, take any submanifold $X$ of $S^2$, then the restriction map $Emb(S^2,M) \to Emb(X,M)$ is a locally-trivial fiber bundle. These proofs depend pretty heavily on the fact that $M$ is a manifold.

For example, if $M$ were not a manifold, say $M$ is the wedge of a finite collection of intervals (the cone on a finite set). Then your map isn't a fiber bundle, not even a fibration. Because the number of path-components of the fiber changes as you pass over the wedge / cone point.

In particular, if your map is a locally trivial fiber bundle, it means the space $X$ satisfies a weak type of isotopy extension theorem. Because given a path between any two points $x,y \in M$ you can trivialize the bundle $p$ over that path. So if there is a path from $x$ to $y$, $M \setminus \{x\}$ and $M \setminus \{y\}$ are homeomorphic. If you were more ambitious you could turn this line of reasoning into an if and only if statement for $M \times M \setminus \Delta \to M$ to be a fiber bundle. It will amount to saying that the homeomorphisms $M \setminus \{x \} \to M \setminus \{y\}$ can be chosen in a continuous fashion (details suppressed) as you vary $x$ (or $y$).