Restriction of vector bundle to norm 1 is a covering map Given a mainfold $M$, the vector bundle
\begin{equation*}
\pi:\wedge^k T^*M = \sqcup_{p \in M} \wedge^k T_p^*M \rightarrow M
\end{equation*}
has the property that its section are exactly the $k$-forms on $M$.
We assume that $M$ is connected and has dimension $m$.
I have already proven that the vector bundle $\wedge^m T^*M$ has rank 1.
Then we endow every fiber of the total space $\wedge^m T^*M$ with an inner product varying smoothly (i.e. for any two smooth sections, their inner product is a smooth function on $M$). Then $M'$ is the submanifold of $\wedge^m T^*M$ consisting of the vectors with norm 1.
I need to show that
\begin{equation*}
\pi|_{M'}:M' \rightarrow M
\end{equation*}
is a smooth covering map.
I have already proven that this map is both smooth and surjective.
This means that I only need to prove that each point of $M$ has a neighborhood $U$ such that $\pi|_{M'}$ maps each connected component of $\pi|_{M'}^{-1}(U)$ diffeomorphically onto U.
Can anyone tell me how to do this or where to start?
 A: This is general of line bundles, that is, if $p:L\rightarrow M$ is a smooth line bundle endowed with a metric and $E=L_{\|\cdot\|=1}$ denotes the submanifold of unit vectors in $L$, then $L_{\|\cdot\|=1}\rightarrow M$ is a covering map.
The point is that you can make the trivialisation maps of $L$ into fiberwise isometries for the norms. That is, for every $x\in M$, there exists an open neighborhood $U$ of $x$ in $M$ and a diffeomorphism $L_{|U}\rightarrow U\times\mathbb{R}$ that commutes with the projections such that the induced map $p^{-1}(y)\rightarrow\mathbb{R}$ is an isometry for all $y\in U$ with the latter endowed with the standard inner product. This follows from the application of the Gram–Schmidt algorithm to given trivialisations: since it is polynomial in the entries of the matrix, it preserves the smoothness of the trivialisations.
This means that over $U$, $E$ is diffeomorphic to $U\times\{-1,1\}$ (in a way compatible with the projection maps) so that $E\rightarrow M$ is indeed a covering map since it is the product of $U$ and a discrete set; alternatively, up to shrinking $U$, you may assume that $U$ is connected in which case this diffeomorphism implies that $E_{|U}$ has exactly two connected components: (one corresponding to) $U\times\{1\}$ and (one corresponding to) $U\times\{-1\}$, both of which obviously map diffeomorphically to $U$.
