I am trying to prove the following.

Proposition. Let $\alpha:A\to B$ be a fiber bundle and let $b,b^\prime\in B$. If there's a compact connected subspace containing $b,b^\prime$, then their fibers are isomorphic.

My intuition comes from this picture from wikipedia's page on monodromy.

Let $(U_i)$ be a trivializing cover for $\alpha$. Restrict this open cover to a connected subspace containing $b,b^\prime$, and then restrict to a finite subcover $V_1,\dots ,V_n$. By connectedness, for any partition $I\amalg I^c\cong \left\{ 1,\dots ,n \right\}$ we have $\bigcup_IV_i\cap \bigcup_{I^c}V_j\neq\emptyset$, which means $V_i\cap V_j\neq\emptyset$ for some $i\in I,j\in I^c$. From here I want to reason there's a "chain" of members of the covering going from $b$ to $b^\prime$.

My reasoning is below, but it feels sloppy and vague.

  1. Take some member of the cover containing $b$ and relabel it $V_1$. Take the union of the rest of the members. These intersect. Pick some member of the union which intersects and relabel it $V_2$ and take the union $V_1\cup V_2$.
  2. Take some member of the cover intersecting $V_1\cup V_2$ and relabel it $V_3$.
  3. Repeat, repicking members if necessary, until you have a, after relabeling, a collection $V_1,\dots ,V_m$ such that $V_i\cap V_{i+1}\neq\emptyset$.
  4. Finally, apply the fact that if two trivializing opens intersect, the trivializations of the bundles over them have isomorphic fibers.

How can I better justify the existence of the "chain" $V_1,\dots ,V_m$ as in the linked picture?


Hint: The compactness assumption is superfluous, it is only connectedness that matters. Fix a point $b\in B$ and look at the subset $\tilde B\subset B$ consisting of $b'\in B$ for which the fiber is isomorphic to the one over $b$. Then observe directly that both $\tilde B$ and $B\setminus\tilde B$ are open in $B$.

  • $\begingroup$ Sorry, I don't understand how to "directly observe" what you say. I want to use the proposition in my post to prove that path-connected points have isomorphic fibers, so I can't use that fact (sorry if this is not relevant, it's what I understood you're doing). $\endgroup$ – Arrow Jan 4 '17 at 21:45
  • $\begingroup$ The point is that the sets $U_i$ are open in $B$ and since they are trivializing the fibers over any two points lying in the same set $B_i$ are isomorphic. Now think about what this says about $\tilde B$ and its complement and use the definition of being connected. $\endgroup$ – Andreas Cap Jan 5 '17 at 7:52

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