Thom classes and orientability. Assume that $\xi$ is the vector bundle $E \stackrel{p} \to B$. Then I know that an orientation of this bundle, gives a unique Thom class in $\tilde{H}^r(\mathrm{Th}(E))$. However I was wondering whether the opposite is also true. That is, if we assume the existence of a Thom class for $\xi$, is it true that dually this class gives rise to unique orientation for $\xi$? In other words, is it true that exists an one-to-one correspondence between Thom classes and orientations for an arbitrary vector bundle? 
Also, what if instead of the ordinary singular cohomology theory we replace an extraordinary cohomology theory, does the stage change dramatically?
 A: It is true that the existence of a Thom class yields an orientation for the bundle.  For $b \in B$, there is a subspace $E_b^+ \subset \operatorname{Th}(\xi)$ corresponding to the one-point compactification of the fiber of $E$ above $b$.  By definition, restricting the Thom class $u_\xi$ to each fiber $E_b^+$ gives a generator of $\tilde{H}^r(E_b^+) \cong \tilde{H}^r(S^r) \cong \mathbb{Z}$, which is an orientation of the fiber $E_b$.  These local orientations are clearly compatible because they all come from $\operatorname{Th}(\xi)$.  
In the situation of a generalized cohomology theory $E$, we typically define $E$-orientability as the existence of a Thom class for the theory $E$.  For example, we just showed that $H\mathbb{Z}$-orientability is just the usual notion of orientability.  On the other hand, $H\mathbb{F}_2$-orientability is no condition at all.  A more exotic example, due to Atiyah-Bott-Shapiro is that $K$-orientability (resp. $KO$-orientability) for a stable vector bundle corresponds to having a $\operatorname{Spin}^\mathbb{C}$ (resp. $\operatorname{Spin}$) structure.  
