Equivalence between smooth and topological fiber bundles All manifolds in this post are hausdorff and second-countable.
Is it true that two smooth fiber bundles with same fiber, base manifold and structure group (that is a Lie group $G$ of diffeomorphisms of the fiber) are equivalent if and only if are equivalent as continuous fiber bundle (so the equivalence can be only continuous)? If it is true, can you give me a reference?
Thank you.
 A: As long as the Lie groups are finite dimensional, yes, this is true. The key is that you can make finite-dimensional approximations to the classifying space $BG$: that is, there are finite dimensional smooth manifolds $B_iG$ with inclusion maps $B_iG \hookrightarrow B_{i+1}G$ such that, as topological spaces, $BG = \lim B_iG$. I don't have a good reference for this or a sketch of the proof. For your favorite groups, it's obvious: $BO(n) = \text{Gr}(n,\infty) = \lim_k \text{Gr}(n,n+k)$, for instance, or $BU(1) = \lim_k \Bbb{CP}^k$. The general case, then, is the same idea: you find a sequence of spaces $E_nG$ that $G$ acts freely on whose limit is contractible. If I remember a reference or a proof for this I'll edit it in.
Now that we have this: let $M$ be a finite dimensional smooth manifold.
Two $G$-bundles being smoothly equivalent is the same as saying that the smooth maps $M \to BG$ are smoothly homotopic. (To make sense of this, one either makes a Hilbert manifold out of $BG$, in which case the first paragraph wasn't necessary, as we'll see, or you assume $M$ is compact so the image lies in some $B_iG$.) The proof of this is essentially the same as the proof that the same is true continuously of continuous maps.
Now we're in a setting in which we can apply the various smooth map approximation theorems. This is the reason I tried to make things finite dimensional; I don't know how to prove the same approximation theorems in the infinite-dimensional setting, though they're almost certainly true. Anyway, let's get going.
First, any continuous map is homotopic to a smooth map. So for any continuous vector bundle on $M$, this must be isomorphic to a smooth vector bundle. Next, suppose I have smooth vector bundles $f_i: M \to BG$. Suppose they are continuously homotopic: that is, there is a continuous map $f_t: M \times I \to BG$. Now (again, either because we can restrict the codomain to a finite-dimensional manifold or because you're willing to use infinite-dimensional approximation theorems) we know that we can approximate this by a smooth map without changing the values on the boundary, since they're already smooth. So this means that if vector bundles $f_0$, $f_1$ are continuously isomorphic, they are smoothly isomorphic.
This becomes false if you allow the Lie group to become infinite-dimensional (say, $\text{Diff}(M)$.) The simplest examples I know are 4-manifold bundles over the circle which are topologically trivial but not smoothly trivial. But because $M$-bundles over the circle are in bijection with $\pi_0 \text{Homeo}(M)$ topologically or $\pi_0 \text{Diff}(M)$ smoothly, you just need a diffeomorphism that's continuously isotopic to the identity but not smoothly. One is provided, eg, in Ruberman, "An obstruction to smooth isotopy in dimension 4".
