A nonorientable surface $S$ is homeomorphic to the $k$-th connected sum $\mathbb{R}P^2 \sharp \ldots \sharp \mathbb{R}P^2$.

For each nonorientable surface $S$ there exists an oriented $2$-fold covering $\tilde S$, the covering bundle. This is isomorphic to the determinant bundle.

Now the number of $2$-fold coverings on $S$ can be computed with the help of $H^1(S;\mathbb{Z}/2)$, that is $2^k$.

The question is: is there a possibility to identify the orientation bundle $\tilde S$ within the set of all $2$-fold coverings on $S$ uniquely.

It is clear that it has to be oriented and connected.

A nonorientable surface $S$ is homeomorphic to the $k$-th connected sum $\mathbb{R}P^2 \sharp \ldots \sharp \mathbb{R}P^2$.

For each nonorientable surface $S$ there exists an oriented $2$-fold covering $\tilde S$, the covering bundle. This is isomorphic to the determinant bundle.

Now the number of $2$-fold coverings on $S$ can be computed with the help of $H^1(S;\mathbb{Z}/2)$, that is $2^k$.

The question is: is there a possibility to identify the orientation bundle $\tilde S$ within the set of all $2$-fold coverings on $S$ uniquely.

It is clear that it has to be oriented and connected.

Or another question is, if there is a surface $X$ given. Is it possible to find out if it is the orientation bundle $\tilde S$.