Orientation and first homology Let $M$ be a differentiable manifold. From the fact that a vector bundle $\pi:V \rightarrow M$ is orientable iff its orientation covering $Or(V) \rightarrow M$ is trivial, one can conclude that if the first integer homology group of $M$ vanishes, then any (differentiable) vector bundle over $M$ is orientable, in particular $M$ is then an orientable manifold.
1) What is the geometric intuition behind this statement? It seems surprising to me that there is a one dimensional phenomenon which is even a stronger property than orientability.
2) Do we have a similar result for topological manifolds?
3) Does the conversion also hold, i.e. if every (differentiable) vector bundle over a (differentiable) manifold $M$ is trivial orientable, do we have $H_1(M;\mathbb{Z})=0$ ?
Edit: Sorry, it seems that i wasn't really awake when i wrote this first. So here is some tidying up. Of course in 3) i meant all bundles to be orientable and not trivial. Also 2) is clearly to be answered with yes since we can define an orientation of a vector bundle in the same manner as one can for smooth ones, i.e. choose orientations of each fiber and make sure they are compatible with a chosen bundle atlas (meaning that if $o_p \in V_p$ is an orienation and $\varphi : \pi^{-1} (U) \rightarrow U \times\mathbb{R}^n$ a bundle chart of this atlas then the iso $\varphi_x:V_x := \pi^{-1}(x)  \cong \mathbb{R}^n$ maps the orientation $o_x$ to the canonical standard orientation of $\mathbb{R}^n$ for all $x \in U$). So we can build the orientation covering $Or(V) = \coprod_\limits{p \in M} Or(V_p) \rightarrow M, (p,o_p) \mapsto p$ and proceed as usual.
 A: So after getting a little more into vector bundle theory, i guess this should work:
Claim Let $M$ be a connected manifold. Then $H^1(M;\mathbb{Z}_2) = 0$ if and only if any vector bundle over $M$ is orientable.
This claim shows that my statement 3) does not hold, e.g. pick the Lens space $L_{3,1} = S^3/\mathbb{Z}_3$ which has $H_1(L_{3,1};\mathbb{Z}) = \mathbb{Z}_3 \neq 0$ but $H^1(L_{3,1};\mathbb{Z}_2) = 0$ by universal coefficients, hence any bundle over $L_{3,1}$ is orientable.
Proof of the claim. Let $H^1(M;\mathbb{Z}_2) = 0$ and let $V \rightarrow M$ be a vector bundle. Consider the orientation covering $\rho : Or(V) \rightarrow M$ which is a double cover and trivial iff $V \rightarrow M$ is orientable. Asumme it is not trivial, so $Or(V)$ is connected, hence path connected (as a locally path connected space). By covering theory we then have an index $[\pi_1(M), \rho_* \pi_1(Or(V)] = 2$, so $\rho_* \pi_1(Or(V)$ is a normal subgroup of $\pi_1(M)$ which leads us to the epimorphism 
$$\pi_1(M) \rightarrow \pi_1(M) / \rho_* \pi_1(Or(V) \cong \mathbb{Z}_2$$ 
But since $Hom(\pi_1(M),\mathbb{Z}_2) \cong Hom(H_1(M;\mathbb{Z}),\mathbb{Z}_2) \cong H^1(M;\mathbb{Z}_2) = 0$ by Hurewicz and universal coefficients this is a contradiction.
On the other hand if $H^1(M;\mathbb{Z}_2) \neq 0$, consider the first Stiefel-Whitney class $w_1$ which is an isomorphism $w_1 : Vect^1(M) \rightarrow H^1(M;\mathbb{Z}_2)$ from isomorphism classes of line bundles on $M$ to the first mod 2 cohomology. We have $w_1(L) = 0$ iff $L \rightarrow M$ is orientable and so there is a non orientable line bundle on $M$.
