Oriented Grassmann is a $2$-sheeted covering space of Grassmann Let $G_n(\Bbb R^k)$ denote the Grassmann manifold (consisting of all $n$-planes in $\Bbb R^k$), and let $\tilde{G}_n(\Bbb R^k)$ denote the oriented Grassmann manifold, consisting of all oriented $n$-planes in $\Bbb R^k$. Let $V_n(\Bbb R^k)\subset (\Bbb R^k)^n$ denote the space consisting of all tuples $(v_1,\dots,v_n)$ such that $\{v_1,\dots,v_n\}$ is linearly independent. Then there are natural surjections of $V_n(\Bbb R^k)$ onto $G_n(\Bbb R^k)$ and $\tilde{G}_n(\Bbb R^k)$. We topologize $G_n(\Bbb R^k)$ and $\tilde{G}_n(\Bbb R^k)$ as quotient spaces of $V_n(\Bbb R^k)$. Clearly there is a $2$-$1$ continuous surjection $p:\tilde{G}_n(\Bbb R^k)\to G_n(\Bbb R^k)$ (orientation-ignoring map). How do we know that this map is a covering map?
Edit. I've found a relevant comment in https://pi.math.cornell.edu/~hatcher/VBKT/VB.pdf, p.31 (of the book, not the pdf), but I can't understand what "using for example the local trivializations constructed in Lemma 1.15" means. (I've read the proof of Lemma 1.15)
 A: Write
\begin{align}
\pi &: V^n(\mathbb R^k) \to G_n(\mathbb R^k), \\
\tilde \pi &: V^n(\mathbb R^k) \to \widetilde G_n(\mathbb R^k)
\end{align}
as the quotient maps which defines the topologies of $G_n(\mathbb R^k)$, $\widetilde G_n(\mathbb R^k)$ respectively. Then we have $p \circ \tilde \pi = \pi$, where
$$ p : \widetilde G_n(\mathbb R^k) \to G_n(\mathbb R^k)$$
is the $2-1$ continuous surjection.
Now we show $p$ is a 2-sheeted covering. It more or less follows from definitions. Let $\ell \in G_n(\mathbb R^k)$. Then there is $(v_1, \cdots, v_n) \in V^n(\mathbb R^k)$ so that $\pi (v_1, \cdots, v_n) = \ell$. Note that
$$p^{-1}(\ell) = \{ \ell_+:= \tilde \pi (v_1,v_2, \cdots, v_n),\ell_-:= \tilde \pi (-v_1, v_2, \cdots, v_n)\}.$$
Let $W = \{w_{n+1}, \cdots, w_k\}$ be a fixed set of vectors in $\mathbb R^k$ so that
$$ \{ v_1, \cdots, v_n, w_{n+1}, \cdots, w_k\}$$
forms a basis of $\mathbb R^k$. Let $U_+ \subset V^n(\mathbb R^k)$ be the collection of all $(\bar v_1, \cdots, \bar v_n)$ so that
$$\tag{2} \frac{\det (\bar v_1 , \cdots, \bar v_n , w_{n+1}, \cdots, w_k)}{\det (v_1 , \cdots, v_n , w_{n+1}, \cdots, w_k)} >0.$$
(in particular $\{\bar v_1, \cdots, \bar v_n, w_{n+1}, \cdots, w_n\}$ forms a basis of $\mathbb R^k$) Similarly define $U_-$ by using $-v_1$ instead of $v_1$ in (2).
Clearly $U_\pm$ contains $(\pm v_1,v_2, \cdots, v_n)$, is open in $V_n(\mathbb R^k)$, $U_+\cap U_- = \emptyset$ and
$$ \tilde \pi^{-1} (\tilde \pi(U_\pm)) = U_\pm, \ \ \pi^{-1} (\pi (U_\pm ))= U_- \cup U_+.$$
In particular, $V_\pm = \tilde \pi (U_\pm)$ are open sets in $\widetilde G_n(\mathbb R^k)$, $V = \pi (U_+)$ is open in $G_n(\mathbb R^k)$ and
$$p|_{ V_\pm} : V_\pm \to V$$
is a homeomorphism. Since $\ell$ is arbitrary, we show that $p$ is a 2-sheeted covering.
