A smooth local diffeomorphism lifts canonically to the orientation double cover

The formula defines maps and the covering and commutation identities. For $(x,o_x)\in \tilde{M}$ the differential $D{f}_x$ is an isomorphism by [F2], so $D{f}_x(o_x)$ is a ray in $\det T_{f(x)}M$ and $\tilde{f}(x,o_x):=(f(x),D{f}_x(o_x))$ is a point of $\tilde{M}$; the inverse linear map sends the opposite ray to the opposite ray, so $D{f}_x(-o_x)=-D{f}_x(o_x)$ and hence $\tau \tilde{f}=\tilde{f}\tau$ and $\tau {\tilde{f}}^{\mathrm{\prime }}={\tilde{f}}^{\mathrm{\prime }}\tau$ from ${\tau }^2=\mathrm{i}\mathrm{d}$. The identities $\pi \circ \tilde{f}=\pi \circ {\tilde{f}}^{\mathrm{\prime }}=f\circ \pi$ are the definitions, using [F1].

Smoothness in the sheet charts. Let $(U,\phi )$ be a chart of $M$ and $(V,\psi )$ a chart of $M$ with $f(U)\subseteq V$; write $\widehat{F}:=\psi \circ f\circ {\phi }^{-1}$ on $\phi (U)$, so $\det D{\widehat{F}}_u\mathrm{\ne }0$ for all $u$ by [F2] and $u\mapsto \det D{\widehat{F}}_u$ is continuous with locally constant sign. In the sheet charts of [F1] the point $(x,s_{U,\phi }^{+}(x))$ is carried by $\tilde{f}$ to the point whose ray is $D{f}_x(s_{U,\phi }^{+}(x))=d{\psi }_{f(x)}^{-1}(D{\widehat{F}}_{\phi (x)}(\text{standard ray}))$, which equals $\mathrm{sign}\det D{\widehat{F}}_{\phi (x)}\cdot s_{V,\psi }^{+}(f(x))$; hence the coordinate expression of $\tilde{f}$ is $(u,ϵ)\mapsto (\widehat{F}(u),\mathrm{sign}\det D{\widehat{F}}_u\cdot ϵ)$, smooth because $\widehat{F}$ is smooth and the sign is locally constant on $\phi (U)$. The expression for $\tau \tilde{f}$ differs only by the locally constant factor $-1$ on the second coordinate, so ${\tilde{f}}^{\mathrm{\prime }}$ is smooth too.

Uniqueness of the two lifts. Suppose $M$ is nonorientable, so that $\tilde{M}$ is connected by [F1]; then $\pi$ is a two-sheeted covering with connected total space and deck group $\{\mathrm{i}\mathrm{d},\tau \}$. Let $g:\tilde{M}\to \tilde{M}$ satisfy $\pi \circ g=f\circ \pi$. Then $g$ and $\tilde{f}$ both lift the map $f\circ \pi$ through $\pi$, so their difference is measured by a deck transformation: at each point, $g(p)=\tilde{f}(p)$ or $g(p)=\tau \tilde{f}(p)$, and continuity on the connected $\tilde{M}$ makes the choice constant; hence $g=\tilde{f}$ or $g=\tau \tilde{f}={\tilde{f}}^{\mathrm{\prime }}$. The two are distinct because $\tau$ has no fixed point on $\tilde{M}$, while ${\tilde{f}}^{\mathrm{\prime }}=\tilde{f}$ would force $\tau$ to fix every point of the nonempty set $\tilde{f}(\tilde{M})$.

Diffeomorphisms lift to diffeomorphisms. If $f$ is a diffeomorphism with inverse $f^{-1}$, form $\widetilde{f^{-1}}$ by the same construction, using the invertible differentials $D(f^{-1}{)}_{f(x)}=(D{f}_x{)}^{-1}$ that follow from the chain rule for $f^{-1}\circ f={\mathrm{i}\mathrm{d}}_M$ (The chain rule for differentials of smooth maps); then $\widetilde{f^{-1}}\circ \tilde{f}(x,o_x)=(x,(D{f}_x{)}^{-1}(D{f}_x(o_x)))=(x,o_x)$ and likewise in the other order, so $\tilde{f}$ is a bijection with smooth inverse $\widetilde{f^{-1}}$ by step 2.1, hence a diffeomorphism; so is ${\tilde{f}}^{\mathrm{\prime }}=\tau \circ \tilde{f}$. The construction uses only the given map, its differential and the cover, so it is canonical.