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Suppose that $E$ is a vector bundle over a compact, hausdorf space $X$. Then $E^n$ is a vector bundle over $X^n$. If $D(E)$ is the disk bundle, there is a map on fibers $D(E^n)_x \rightarrow D(E)^n_x$. Does this induce a homotopy equivalence $D(E^n)\simeq D(E)^n$?

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Does it work when $E$ is trivial? If it does, surely the homotopy equivalence can be done more or less canonically, so it should glue to a homotopy equivalence for arbitrary locally trivial bundles. –  Mariano Suárez-Alvarez Jun 6 '12 at 16:48

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Yes. For any vector space $V$, there is a canonical (basis-independent) deformation retraction of $D(V)^n$ onto $D(V^n)$, obtained by moving directly inwards along a line through the origin. Performing this on each fiber, we obtain a deformation retraction of $D(E)^n$ onto $D(E^n)$.

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