The *orientation sheaf* of an $n$-manifold $M$ is $\mathcal{O}\_n=Sheaf(U\mapsto H\_n(M,M-U;\mathbb{Z}))$, with stalks given by $(\mathcal{O}\_n)\_x = lim H\_n(M,M-U)=H\_n(M,M-x)=\mathbb{Z}$ (the limit is over neighborhoods $U$ containing $x\in M$).


Suppose $M$ is orientable and closed, so that $H\_n(M)=\mathbb{Z}$. Then $\mathcal{O}\_n=M\times\mathbb{Z}$ is a constant sheaf.


Suppose now that $M$ is nonorientable. Then we can take a neighborhood $N$ of each point $x\in M$ so that $\mathcal{O}\_n|\_N$ is constant. Thus $\mathcal{O}\_n$ is a locally constant sheaf.


The *orientable double cover* of $M$ is $\tilde{M}=\lbrace(x,\mu\_x)\rbrace$ where $\mu\_x$ is a local orientation at $x\in M$ (a choice of generator of the local homology). If $M$ is nonorientable then $\tilde{M}$ is connected, and if $M$ is orientable then $\tilde{M}\approx M\times\mathbb{Z}\_2$. In either case, there is an embedding $\tilde{M}\hookrightarrow \mathcal{O}\_n$. 


**Is there anything else I can say about this enlargement from $\tilde{M}$ to $\mathcal{O}\_n$ ? Does one encode more information than the other?**


If we instead use $\mathbb{R}$-coefficients, I believe (but can easily be wrong) that $\tilde{M}$ is a deformation retract of $\mathcal{O}\_n$ for nonorientable $M$, and is a deformation retract of $\mathcal{O}\_n-\Gamma^0$ for orientable $M$ (where $\Gamma^0$ is the zero section of this real line bundle). In other words, $\tilde{M}$ and $\mathcal{O}\_n$ are the same up to homotopy (over $\mathbb{R}$-coefficients), and one does not encode more information than the other.



Orientation Sheaf and Double Cover
An elaboration.


$\mathcal O\_n$ is a bundle over $M$ with fiber $\mathbb Z$. There is an action of the integers on it, because the integers act on homology. Earlier I said this was a principal bundle, I was too tired! The action is of course not free. In particular, this bundle $\mathcal O\_n \to M$ has a section. 


Given a manifold $M$, its orientation cover is a $\mathbb Z\_2$-principal bundle. The action of $\mathbb Z\_2$ is given by reversal of orientations. 


So the product $\tilde M \times \mathbb Z$ has an action of $\mathbb Z\_2$ given by 


$$t.(x,n) = (t.x, (-1)^tn)$$


where $t.x$ is the action of $\mathbb Z\_2$ on $\tilde M$. 


$\tilde M \times\_{\mathbb Z\_2} \mathbb Z$ is the quotient of $\tilde M \times \mathbb Z$ by the action of $\mathbb Z\_2$. By design, it is a $\mathbb Z$-principal bundle -- you can project onto $\tilde M / \mathbb Z\_2 \equiv M$ -- and by design it is isomorphic to $\mathcal O\_n$. The map $\tilde M \times\_{\mathbb Z\_2} \mathbb Z \to \mathcal O\_n$ is induced by your inclusion $\tilde M \to \mathcal O\_n$. 


