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A proper local homeomorphism is a covering map (assuming some mild conditions on the involved spaces). I want to know about the following generalization, which I believe is false but cannot come up with a counterexample to.

Suppose $f : E \to B$ is proper and locally of the form $U\times V \to V$, ie every point $e \in E$ has a neighborhood $W \subset E$ of $e$, some $U = U_e$ and a neighborhood $V \subset B$ of $f(e)$ such that $f(W) \subset V$ and there is a homoeomorphism $W \cong U \times V$ that takes $f$ to the projection. Then $f$ is a fiber bundle.

Here I assume $B$ is locally compact, Hausdorff and locally connected (or locally contractible, or a manifold). $B$ should also be connected if your definition of fiber bundle insists that the fibers are constant (and not just locally constant). The map $f$ being proper means:

  • $f^{-1}(K)$ is compact for $K \subset B$ compact.
  • $f$ is separated: $\{(x, y) \mid f(x) = f(y)\}$ is closed in $E \times E$ (or $E$ is Hausdorff). Without this the line with doubled origin mapping to the line is a counterexample to even the covering space case.

I am also happy to assume some niceness of $f^{-1}(b)$ for $b \in B$ (it is of course compact by properness). As noted above if $f^{-1}(b)$ is discrete then the statement is true.

Even with the strongest variations of the assumptions above (ie $B$ compact manifold, $E$ Hausdorff, fibers of $f$ smooth manifolds), I believe this is false because it would reduce the proof of Ehresmann's theorem (for a proper submersion $f: E \to B$) to:

  1. Use the "submersion theorem" ($\impliedby$ the implicit function theorem) to get that $f$ is a projection locally in $E$.
  2. Use properness to upgrade that to $f$ being a projection locally in $B$, ie a fiber bundle.

This two step process, decoupling the submersion/smoothness and proper/topological parts, is actually how I think of Ehresmann's theorem, but all proofs I have seen leverage differentiability in patching together the different product-like neighborhoods. Except in the covering space case, where the (local) connectivity assumptions are enough to match up nearby fibers. I did find this MO post about Ehresmann's theorem with a comment by Ryan Budney which seems to say that there is a possibly-long-winded proof that uses not much more than the implicit function theorem.

Note that if $f$ is a smooth map of manifolds that is locally (smoothly) a projection then it is automatically a submersion, so the smooth version of the statement is actually equivalent to Ehresmann's theorem.

I am looking for either a proof (possibly with stronger topological assumptions on $B$, $E$ etc) or a counterexample.

ronno
  • 12,914

1 Answers1

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As mentioned by Piotr Achinger in a related MO post, Corollary 6.14 in Siebenmann, L.C. Deformation of homeomorphisms on stratified sets is a statement of this form with some extra assumptions on the fibers $F_b = f^{-1}(b)$ for $b \in B$:

  • $F_b$ is locally connected;
  • the assumption $\mathscr{D}(F_b)$ which roughly says that for compact $K \subset F_b$ and neighborhoods $U \supset K$, maps $j : U \to F_b$ close to the inclusion (in the compact-open topology) can be isotoped to be the inclusion on $K$.

This is in particular satisfied when the fibers are manifolds or locally finite simplicial complexes (Example 1.3 + Theorem 2.3 in the same paper). Again roughly speaking, the isotopies from the assumption $\mathscr{D}(F_b)$ allow matching up $F_b$ with $F_{b'}$ for nearby $b'$.

ronno
  • 12,914