First, let me just write down the definitions I will use (see e.g. Lee - Introduction to Smooth Manifolds):
Two (real) vector bundles $(E,B,\pi)$ and $(E',B,\pi')$ (over the same base space $B$) are isomorphic if there exists a homeomorphism $f: E \to E'$ such that $\pi = \pi' \circ f$ and such that the restriction of $f$ to each fiber is a linear map.
A vector bundle $(E,B,\pi)$ is a smooth vector bundle if
- $E$ and $B$ are smooth manifolds with or without boundary,
- $\pi$ is a smooth map,
- local trivializations are diffeomorphisms.
Two smooth vector bundles $(E,B,\pi)$ and $(E',B,\pi')$ are smoothly isomorphic if there exists a diffeomorphism $f: E \to E'$ such that $\pi = \pi' \circ f$ and such that the restriction of $f$ to each fiber is a linear map.
Now, I heard about the following theorem:
Let $(E,B,\pi)$ be a vector bundle. If $B$ is a contractible paracompact space, then $(E,B,\pi)$ is isomorphic to the trivial bundle $(E \times B,B,p)$ (where $p: E \times B \to B$ is the usual projection).
I have two questions:
1- Does this theorem still holds with smooth vector bundles ? I.e.:
Let $(E,B,\pi)$ be a smooth vector bundle. If $B$ is a contractible paracompact space, then $(E,B,\pi)$ is smoothly isomorphic to the trivial bundle $(E \times B,B,p)$.
2- If this is the case, does this theorem still holds with manifolds with corners (i.e. if we replace "$E$ and $B$ are smooth manifolds with or without boundary" by "$E$ and $B$ are smooth manifolds with corners" in the definition of a smooth vector bundle) ?
Thank you for your help.