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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 $\{(a,a) \}_{a \in M}$.

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Interesting question. Certainly looks true for closed manifolds $M$. –  Grumpy Parsnip Sep 19 '11 at 18:55
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If $M$ is a closed manifold we can do the following: Let $U_i$ be a chart in $M$, so we can assume $U_i\cong \mathbb{R}^n$. Now consider the neighborhood $\mathbb{R}^n \times \mathbb{R}^n - \Delta \cong U_i \times U_i - \Delta \subset M \times M - \Delta$. Showing the result for $\mathbb{R}^n \times \mathbb{R}^n - \Delta$ and using a bump function inside an open set of $U_i \times U_i - \Delta$ we can can conclude that $p^{-1}(U_i)\cong \mathbb{R}^n \times (\mathbb{R}^n- \mathbb{R}^{n-1})$. –  Manuel Sep 19 '11 at 19:04
    
Oops, I meant $p^{-1}(U_i)\cong \mathbb{R}^n \times (\mathbb{R}^n-\{0\})$. And of course, we can ignore the $i$ index of the chart $U_i$... –  Manuel Sep 19 '11 at 20:51

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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 $y$).

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