Proving that the orientation bundle of a non orientable manifold is isomorphic to every other oriented 2-coverings of such manifold I've got some problem proving this statement, recalling that for me, an orientable manifold is a manifold which admits an atlas such that the transition functions have always local degree $1$ (we are working with topological manifolds). 
I cannot image a (reasonable) map between the two coverings, because maybe I do not see which special property should have an oriented 2-covering of a non oriented manifold.
Thanks in advance
 A: Let $\hat M \to M$ is a two sheeted covering where $M$ is non-orientable and $\hat M$ is oriented (in particular it has to be connected). Then there is a single covering translation $\tau$, i.e. $\mathbb Z/2$ acts on $\hat M$. This translation $\tau$ is orientation reversing, since otherwise $M$ would be orientable. But this is the defining property of the orientation covering (see http://www.map.mpim-bonn.mpg.de/Orientation_covering ).
(If you don't know this fact, just note that you can easily write down an isomorphism explicitely.)
A: An equivalent definition of orientability is that the manifold $M$ admits a function that assigns to each $x$ a generator of the top-dimenasional homology $H^n(M, M\setminus \{x\})\simeq\mathbb{Z}$ and this assignment is "compatible" in some neighborhood $U$ of $x$. That is, each $x$ has a neighborhood $U$ such that there exists a generator of $H^n(M,M\setminus U)\simeq\mathbb{Z}$ that is mapped to the chosen generator of $H^n(M,M\setminus\{y\})$ for each $y\in U$.
If you agree with this, then you can construct a canonical isomorphism as follows. If $C$ is the oriented 2-covering and $\pi: C\to M$ the projection map, then you can assign to each $c\in C$ the pair $(\pi(c),o)$ where $o$ the orientation in $\pi(c)$ induced by $\pi$. That is, $o=\pi_*(o_c)$ where $o_c\in H(C,C\setminus\{c\})$ is the orientation in $c$. It shouldn't be hard to show $c\mapsto (\pi(c), o)$ is injective and surjective.
