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This MSE question asks for clarification of the local triviality condition imposed in the definition of a fiber bundle.

As mentioned there, the point of local triviality seems to somehow ensure a "continuous variation of fibers". I would like to understand how this can be formalized (as suggested in the comments there).

Let $\pi:E\longrightarrow B$ denote our fiber bundle. The extension space $E$ is partitioned into homeomorphic copies of $F$ by its fibers: $$\left\{ \left\{b\right\}\times F \mid b\in B \right\}$$

We want this collection of sets to be continuously indexed by $B$. In section 2 of his Notes on Compactness, Martín Escardó makes the following definition:

Definition. Let $E$ be a topological space and $\left\{ V_b\mid b\in B \right\}$ be a family of subsets of $E$. This family will be called continuously indexed if the graph $\left\{ (x,b)\in E\times B\mid x\in V_b \right\}$ is open in the product topology in $E\times B$.

At any rate, this seems like exactly the kind of continuity condition we want from the partition corresponding to a fiber bundle, so I was hoping for something along the lines of

Partition by fibers is continuously indexed $(\implies,\iff,\impliedby$ bundle is locally trivial

And/or anything along those lines.

Finally, it would be nice if someone could explain where sheaves are related here, since they are supposed to represent continuously variable sets...

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  • $\begingroup$ Your statement is almost certainly false. Consider $B = \Bbb R$, $V_b = \Bbb Z$ for $b \geq 1$, $\Bbb Z$ for $b \leq 0$, and in between, $\Bbb N \cup (b+-\Bbb N)$. This is not locally trivial but is continuously indexed. $\endgroup$ – user98602 May 23 '16 at 14:52
  • $\begingroup$ @MikeMiller what is $\Bbb N \cup (b+-\Bbb N)$? At any rate, I'm looking for connections and implications more than anything. I just don't really understand why being locally a product projection means continuously varying fibers. For instance, why do fibers vary continuously for trivial fiber bundles? $\endgroup$ – Arrow May 23 '16 at 15:00
  • $\begingroup$ The definition of "continuously indexed" in the question seems inappropriate for the use you want to make of it. It implies that each of the sets $V_b$ has to be open, which is the case in Escardó's situation but not in yours. $\endgroup$ – Andreas Blass May 23 '16 at 18:10
  • $\begingroup$ Sorry, yes, my comment does not apply to the sense of continuously indexed in the question. I just mean that they depend continuously on $b$ in Hausdorff metric which is the most reasonable candidate I can think of for the notion of 'continuously indexed'. $\endgroup$ – user98602 May 23 '16 at 18:39
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Suppose $F$ is homeomorphic to a compact subspace of $\Bbb R^\infty$. (There is some nice classification of which spaces this is true of, but I forget it. Maybe every compact Hausdorff space? At the very least it is true of any compact manifold and of any finite CW complex.)

Consider the space "$B(F)$" (terrible notation, sorry; this is meant to evoke "classifying space of $F$-bundles"), the elements of which are subsets of $\Bbb R^\infty$ which are homeomorphic to $F$ with the subspace topology, and equipped with the Hausdorff metric. This is the quotient of $\text{Emb}(F,\Bbb R^\infty)$ by the free etc action of $\text{Homeo}(F)$; and the first space is contractible, whence $B(F)$ is a $B\text{Homeo}(F)$. There is a tautological $F$-bundle over $B(F)$, given by the subset $E(F) \subset B(F) \times \Bbb R^\infty$, $E(F) = \{(S,x) | x \in S\}$. This is a locally trivial bundle.

This space does literally classify $F$-bundles over (paracompact) spaces $X$. But let's improve this a little bit. Let's say a 'concrete $F$-prebundle' (pre b/c not necessarily locally trivial) is a subset of $E \subset X \times E(F)$ such that the projection $E \to X$ has $F$ as every fiber. This is equivalent to saying that $E$ is the pullback of the tautological bundle over $B(F)$ by some map (not necessarily continuous!) $f: X \to B(F)$.

Then what you're looking for is probably as follows. TFAE. 1: $f$ is continuous; 2: the prebundle $E \to X$ is an honest $F$-bundle; 3: $E$ is a closed subset of $X \times E(F)$.

The proofs of these facts are straightforward and a little tedious; I will not endeavor to write them here, though I encourage someone else to and would gladly upvote such an answer. Note that the above only applies to compact fibers, which is why the example in my comment to the question does not work.

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