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There's a nice result in algebraic topology saying that given a fiber bundle, its pullbacks along homotopic maps are isomorphic as bundles.
Thinking of a bundle as a comb with the "teeth" as its fibers and the base as the base space, I am looking for some geometric intuition that would make the theorem intuitively obvious, or at least expectable.
What is the geometric intuition behind the homotopy invariance of fiber bundles?
First of all, this does not hold generally but over paracompact spaces.
Secondly it is easier to see and proof the result over compact spaces and many authors omit the proof for the more general, paracompact, spaces.
Thirdly the proof in the compact case basically gives you the intuition you need. What you do is to look at a fiber bundle with base space $B\times [0,1]$ which is given by the homotopy, in particular at $0$ and $1$ you obtain the two "different" bundles. Now you check that on all small enough open intervals in $[0,1]$ the bundle does not change too much, it only deforms slightly in an isomorphic way. Hence, by compactness, it does not change too much over $0$ to $1$ and hence gives isomorphic bundles over those fibers (by fiber here I mean all $B \times t$).
