The orientation double cover is canonically oriented and preserves closedness

The model. Let $\tilde{M}$ be the set of pairs $(x,o_x)$ with $x\in M$ and $o_x$ a ray in $\det T_{x}M$, with $\pi (x,o_x)=x$ and $\tau (x,o_x)=(x,-o_x)$. For a chart $(U,\phi )$ of $M$ write $s_{U,\phi }^{+}(x):=d{\phi }_x^{-1}(\text{the positive ray of }{\mathbb{R}}^n)$ for the ray in $\det T_{x}M$ pulled back from the standard ray, and $s_{U,\phi }^{-}:=-s_{U,\phi }^{+}$; for open $W\subseteq U$ put $S_{W,\phi }^{\pm }:=\{(x,s_{U,\phi }^{\pm }(x)):x\in W\}$. These sets cover $\tilde{M}$ and are closed under finite intersections: for a second chart $(V,\psi )$ and open $W^{\mathrm{\prime }}\subseteq V$, the intersection $S_{W,\phi }^{ϵ}\cap S_{W^{\mathrm{\prime }},\psi }^{\delta }$ equals $S_{\mathrm{\Omega },\phi }^{ϵ}$ where $\mathrm{\Omega }=\{x\in W\cap W^{\mathrm{\prime }}:ϵs_{U,\phi }^{+}(x)=\delta s_{V,\psi }^{+}(x)\}$, and $\mathrm{\Omega }$ is open in $W\cap W^{\mathrm{\prime }}$ because the comparison of the two chart rays is governed by the sign of the nowhere-zero continuous function $x\mapsto \det d(\psi {\phi }^{-1}{)}_x$, which is locally constant; if no point of $W\cap W^{\mathrm{\prime }}$ satisfies the comparison the intersection is empty. So these sets form a basis of a topology on $\tilde{M}$, and by construction $x\mapsto (x,s_{U,\phi }^{\pm }(x))$ is a homeomorphism of $W$ onto $S_{W,\phi }^{\pm }$.

The basis description gives the covering and the involution. Every point of $\tilde{M}$ lies in some basis element $S_{W,\phi }^{\pm }$, which is homeomorphic to the open set $W\subseteq {\mathbb{R}}^n$, so $\tilde{M}$ is locally Euclidean of dimension $n$. Every chart domain $U$ of $M$ satisfies ${\pi }^{-1}(U)=S_{U,\phi }^{+}\bigsqcup S_{U,\phi }^{-}$, a disjoint union of two open sets on each of which $\pi$ restricts to a homeomorphism onto $U$; since the chart domains cover $M$, $\pi$ is continuous, open, surjective and a two-sheeted covering map. The map $\tau$ is continuous with ${\tau }^2=\mathrm{i}\mathrm{d}$ and $\pi \circ \tau =\pi$, because in the basis it exchanges $S_{W,\phi }^{+}$ with $S_{W,\phi }^{-}$. The space $\tilde{M}$ is Hausdorff: points with distinct images are separated by the inverse images of disjoint open neighbourhoods in the Hausdorff space $M$, and the two points of a fibre lie in the two disjoint sheets over any chart domain containing the image.

Components are covering spaces. Suppose $M$ is nonempty and connected. By [F3] both $M$ and its cover are locally path-connected, their components are open path components, and $M$ is path-connected. Fix $a$ in a component $C$ and let $y\in M$. A path from $\pi (a)$ to $y$ lifts from $a$ by Existence and uniqueness of path lifts through a covering map; its image stays in $C$, so $C$ meets the fibre over every $y$. Hence there are at most two components. If there are two, each meets every two-point fibre exactly once; if there is one, it contains both fibre points. Over a connected evenly covered neighbourhood, each sheet is connected and therefore lies in one component. Thus $\pi {\mathrm{\mid }}_C:C\to M$ is a covering.

Smooth structure. By [F2] each component $C$ of $\tilde{M}$ carries a unique smooth $n$-manifold structure for which $\pi {\mathrm{\mid }}_C$ is a smooth local diffeomorphism; on a nonempty connected base the components are at most two disjoint open sets, so these structures combine into a smooth $n$-manifold structure on $\tilde{M}$ for which $\pi$ is a smooth local diffeomorphism and a two-sheeted covering map. In particular $\tilde{M}$ is a topological $n$-manifold and $\pi$ is a local diffeomorphism, so a chart of $M$ pulls back along $\pi$ on each sheet to a chart of $\tilde{M}$.

The canonical orientation. At a point $p=(x,o_x)$ the differential $d{\pi }_p:T_p\tilde{M}\to T_{x}M$ is an isomorphism, so $d{\pi }_p^{-1}(o_x)$ is a ray in $\det T_p\tilde{M}$. On the sheet $S_{W,\phi }^{+}$ the pulled-back chart $\mathrm{\Phi }:=\phi \circ (\pi {\mathrm{\mid }}_{S_{W,\phi }^{+}})$ has differential $d{\mathrm{\Phi }}_p=d{\phi }_x\circ d{\pi }_p$, so it carries this ray to the ray $d{\phi }_x(o_x)$, which is the standard ray of ${\mathbb{R}}^n$ because $o_x=s_{U,\phi }^{+}(x)$; on the sheet $S_{W,\phi }^{-}$ the same computation gives the opposite standard ray. The ray assignment is therefore constant in these charts, hence a smooth choice of rays, so it is an orientation of $\tilde{M}$ by [F5]. It is canonical: it is defined from $\pi$ and the points $(x,o_x)$ themselves, with no chart or orientation of $M$ chosen.

The deck group. Let $h:\tilde{M}\to \tilde{M}$ be a deck transformation. Since $\pi \circ h=\pi$, $h$ maps each fibre into itself, so $h(x,o_x)$ is $(x,o_x)$ or $(x,-o_x)$, and because $h$ is injective on the two-point fibre it acts by a well-defined sign $h(x,o_x)=(x,ϵ(x)o_x)$ with $ϵ:M\to \{\pm 1\}$. Continuity of $h$ makes $ϵ$ locally constant: over a chart domain $U$ the two sheets $S_{U,\phi }^{\pm }$ are disjoint open sets, and a connected neighbourhood of $x$ maps into one of them, so $ϵ$ is constant near $x$. Hence $ϵ$ is continuous into the discrete group $\{\pm 1\}$ and therefore constant because $M$ is connected; so $h=\mathrm{i}\mathrm{d}$ if $ϵ=+1$ and $h=\tau$ if $ϵ=-1$. Moreover $\tau$ is smooth, being locally the sheet exchange between the charts of $\tilde{M}$, and $\pi \circ \tau =\pi$, so $\tau$ is a deck transformation; thus $\mathrm{Deck}(\pi )=\{\mathrm{i}\mathrm{d},\tau \}\cong \mathbb{Z}\mathrm{/}2$, and it acts freely (only $\mathrm{i}\mathrm{d}$ has a fixed point) and transitively on every two-point fibre.

Orientable case. Suppose $M$ is orientable and let $x\mapsto o_x$ be an orientation of $M$ by [F5]. Then $\mathrm{\Phi }:\tilde{M}\to M\times \mathbb{Z}\mathrm{/}2$, $\mathrm{\Phi }(x,o_x^{\mathrm{\prime }})=(x,ϵ)$ where $o_x^{\mathrm{\prime }}=ϵo_x$, is a bijection over $M$, and in the charts of step 5.1 the map $\mathrm{\Phi }$ and its inverse change only the locally constant sign of the second coordinate, so $\mathrm{\Phi }$ is a diffeomorphism for the product smooth structure on $M\times \mathbb{Z}\mathrm{/}2$ (Products of smooth manifolds have a canonical product smooth structure); it carries $\tau$ to the map exchanging the two components $M\times \{+1\}$ and $M\times \{-1\}$.

Nonorientable case. Suppose $M$ admits no orientation. Then $\tilde{M}$ is connected: if $\tilde{M}=C_1\bigsqcup C_2$ with two components, step 3.1 makes each $C_i$ a covering of $M$ meeting each fibre exactly once, so $\pi {\mathrm{\mid }}_{C_1}$ is a bijective local homeomorphism, i.e. a homeomorphism, and its inverse $s:M\to \tilde{M}$ is a continuous section; writing $s(x)=(x,o_x)$, the assignment $x\mapsto o_x$ is in the pulled-back charts of step 4.1 locally constant, hence a smooth choice of rays and an orientation of $M$ by [F5], a contradiction. So $\tilde{M}$ is connected when $M$ is nonorientable, hence path-connected by [F3] since $M$ is locally path-connected as a manifold and local path connectedness ascends to the cover; the deck group acts transitively on every fibre by step 5.2, so the covering is regular by [F1].

Closedness. If $M$ is closed, i.e. compact and boundaryless, then $\tilde{M}$ is compact by [F4], and it is boundaryless because $\pi$ is a local diffeomorphism onto a boundaryless manifold; hence $\tilde{M}$ is closed. For the empty base $M=\mathrm{\varnothing }$, $\tilde{M}=\mathrm{\varnothing }$, the projection is a covering map vacuously, the deck group is trivial, and the empty ray choice gives its canonical orientation; the two-element deck-group assertion was restricted to a nonempty connected base.