This question seems like a pretty straightforward generalization of a result from vector bundles to fiber bundles (with fiber an orientable manifold) but its been on MSE for over a week with no answers so I'm reposting https://math.stackexchange.com/questions/4425410/fiber-bundle-orientability-vs-manifold-orientability here:
I read this question about vector bundles
https://math.stackexchange.com/questions/50809/bundle-orientability-vs-manifold-orientability
In the answer to this question the last sentence states the following (I think fairly well known) result about vector bundles
"Let E be a vector bundle over M. Consider the statements (i) M is orientable as a manifold (ii) E is orientable as a manifold (iii) E is orientable as a vector bundle. Any two of the statements being true will imply the third."
I am curious if this statement generalizes to all fiber bundles with orientable fiber.
In other words, is it true that: Let $ F \to E \to M $ be a fiber bundle over M, with $ F $ an orientable manifold. Consider the statements (i) M is orientable as a manifold (ii) E is orientable as a manifold (iii) E is orientable as a fiber bundle. Any two of the statements being true will imply the third.
If not, is a similar 2-out-of-3 theorem true for sphere bundles or some other more restricted class of fiber bundles?
Some stuff that I tried (focused on circle bundles in which case the bundle being orientable is equivalent to it being $ U_1 $ principal see Is every orientable circle bundle principal?)(Note that this is a complete list of all circle bundles over $ S^1,S^2,T^2,K^2,\mathbb{R}P^2 $):
Base $ S^1 $:
All three: $ S^1 \to T^2 \to S^1 $ (in general any trivial bundle $ S^1 \times M $ for any orientable manifold $ M $ has base, bundle and total space all orientable)
Only base orientable: $ S^1 \to K^2 \to S^1 $ where $ K^2 $ is the Klein bottle.
Base $ S^2 $:
- All three: The lens spaces $ S^1 \to L_{n,1} \to S^2 $, where $ n $ is the euler class of the bundle, are all orientable. I believe they are also $ U_1 $ principal (certainly for $ n=0,1,2 $ they are principal since they are $ S^1 \times S^2, S^3\cong SU_2,\mathbb{R}P^3\cong SO_3(\mathbb{R}) $ respectively). $ E^1 \times S^2 $ geometry for $ n=0 $, $ S^3 $ geometry otherwise.
Base $ T^2 $:
All three: The circle bundles $ S^1 \to MT(\begin{bmatrix} 1 & r \\ 0 & 1 \end{bmatrix}) \to T^2 $ where $ MT(\begin{bmatrix} 1 & r \\ 0 & 1 \end{bmatrix}) $ denotes the mapping torus of $ T^2 $ corresponding to the mapping class $ \begin{bmatrix} 1 & r \\ 0 & 1 \end{bmatrix} $, which is the $ r $th power of the Dehn twist $ \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} $. For $ r=0 $ this is $ T^3 $ and admits $ E^3 $ (flat) geometry while for $ r \neq 0
$ these are the nilmanifolds $ N_r $ described in https://math.stackexchange.com/questions/4367670/is-every-nil-manifold-a-nilmanifold and they admit Nil geometry. $ E^3 $ geometry for $ r=0 $ otherwise Nil geometry.
Only base orientable: $ S^1 \rtimes_b T^2 $ two of the four flat compact non orientable three manifolds. For $ b=0 $ this is $ S^1 \times K^2 $ with first homology $ \mathbb{Z}^2 \times C_2 $, for $ b=1 $ this is the mapping torus of the Dehn twist diffeomorphism of $ K^2 $ with first homology $ \mathbb{Z}^2 $. The total space is not orientable. These coincide with the two $ U_1 $ principal bundles over $ K^2 $. $ E^3 $ geometry.
Base $ \mathbb{R}P^2 $:
Only bundle orientable: $ S^1 \to S^1 \times \mathbb{R}P^2 \to \mathbb{R}P^2 $. (in general any trivial bundle $ S^1 \times M $ for any non orientable manifold $ M $ has only the bundle orientable). $ E^1 \times S^2 $ geometry.
Only bundle orientable: $ S^1 \to (S^2 \times S^1)/(-1,-1) \mathbb{R}P^2 $. This is the mapping torus of the antipodal map of $ S^2 $. It is the unique nontrivial $ U_1 $ principal bundle over $ \mathbb{R}P^2 $. $ E^1 \times S^2 $ geometry.
Only total space orientable: $ S^1 \to P_{4n,1} \to \mathbb{R}P^2 $ where $ P_{4n,1} $ is the standard prism manifold with $ 4n $ element dicyclic fundamental group. $ S^3 $ geometry.
Only total space orientable: $ S^1 \to UT(\mathbb{R}P^2) \cong L_{4,1} \to \mathbb{R}P^2 $, the unit tangent bundle of $ \mathbb{R}P^2 $. $ S^3 $ geometry.
Only total space orientable: $ S^1 \to \mathbb{R}P^3 \# \mathbb{R}P^3 \to \mathbb{R}P^2 $. $ E^1 \times S^2 $ geometry.
Base $ K^2 $:
Only the bundle is orientable: The two principal $ U_1 $ bundles over $ K^2 $ coincide with the two non principal $ S^1 $ bundles over $ T^2 $. These are two of the four non orientable compact flat three manifolds they can also be viewed as two of the four mapping tori of $ K^2 $. $ E^3 $ geometry.
none of 3: The other two of the four non orientable compact flat three manifolds, they can also be viewed as the other two of the four mapping tori of $ K^2 $. They are non principal bundles $ S^1 \to S^1 \rtimes_b K^2 \to K^2 $ both with non orientable total space. For $ b=0 $ this is the mapping torus of the Y homoeomorphism of $ K^2 $, it has first homology $ \mathbb{Z}^2 \times C_2 \times C_2 $. For $ b=1 $ this is the mapping torus of $ K^2 $ for the mapping class corresponding to the combination of a Dehn twist and a Y homoemorphism. It has first homology $ \mathbb{Z}^2 \times C_4 $. $ E^3 $ geometry.
Only total space orientable: The circle bundles $ S^1 \to MT(\begin{bmatrix} -1 & -r \\ 0 & -1 \end{bmatrix}) \to K^2 $ where $ MT(\begin{bmatrix} -1 & -r \\ 0 & -1 \end{bmatrix}) $ denotes the mapping torus of $ T^2 $ corresponding to the mapping class $ \begin{bmatrix} -1 & -r \\ 0 & -1 \end{bmatrix} $. These manifolds are double covered by $ MT(\begin{bmatrix} 1 & 2r \\ 0 & 1 \end{bmatrix}) $. For $ r=0 $ this is the unit tangent bundle of the Klein $ UT(K^2) $, which admits $ E^3 $(flat) geometry, while for $ r \neq 0
$ these admit Nil geometry. $ E^3 $ geometry for $ r=0 $, Nil geometry otherwise.
