[proofplan] We construct $\Phi$ as in the statement and an inverse $\Psi: \mathrm{MOA}(M) \to \mathrm{Or}(M)$ sending a maximal oriented atlas $\mathcal{A}$ to the pointwise orientation in which $(\partial_{x_1}|_p, \dots, \partial_{x_n}|_p)$ is declared positive for every chart $(U, \varphi) \in \mathcal{A}$ and every $p \in U$. The key linear-algebraic input, used in both directions, is that on the overlap of two charts the two coordinate frames are related by the Jacobian of the transition map, so they represent the same orientation on each tangent space exactly when the transition Jacobian has positive determinant. The covering property of $\Phi(\omega)$ is obtained by taking a smooth local frame for $\omega$ and, if necessary, flipping the first coordinate of a chart to invert the orientation of its coordinate frame; the smoothness of $\Psi(\mathcal{A})$ is witnessed by the coordinate frame of any chart in $\mathcal{A}$. Finally $\Phi$ and $\Psi$ are mutual inverses by direct check. [/proofplan] [step:Reduce orientation comparison on a chart overlap to the sign of the transition Jacobian] Let $(U_\alpha, \varphi_\alpha), (U_\beta, \varphi_\beta) \in \mathcal{S}$ with $U_\alpha \cap U_\beta \neq \varnothing$. Write $\varphi_\alpha = (x_1, \dots, x_n)$ and $\varphi_\beta = (y_1, \dots, y_n)$ as $\mathbb{R}$-valued coordinate functions on $U_\alpha \cap U_\beta$, and let \begin{align*} T_{\beta\alpha} : \varphi_\alpha(U_\alpha \cap U_\beta) &\to \varphi_\beta(U_\alpha \cap U_\beta) \\ u &\mapsto \varphi_\beta(\varphi_\alpha^{-1}(u)) \end{align*} be the transition map, with components $T_{\beta\alpha} = (T_1, \dots, T_n)$ where $T_j = y_j \circ \varphi_\alpha^{-1}$. For each $p \in U_\alpha \cap U_\beta$, the chain rule applied to $x_i \mapsto y_j$ gives \begin{align*} \frac{\partial}{\partial x_i}\bigg|_p = \sum_{j=1}^n \frac{\partial T_j}{\partial u_i}(\varphi_\alpha(p)) \cdot \frac{\partial}{\partial y_j}\bigg|_p = \sum_{j=1}^n (JT_{\beta\alpha}(\varphi_\alpha(p)))_{ji} \cdot \frac{\partial}{\partial y_j}\bigg|_p. \end{align*} Thus the matrix expressing the ordered basis $(\partial_{x_1}|_p, \dots, \partial_{x_n}|_p)$ in the ordered basis $(\partial_{y_1}|_p, \dots, \partial_{y_n}|_p)$ is exactly $JT_{\beta\alpha}(\varphi_\alpha(p))$. By the definition of orientation on a [vector space](/page/Vector%20Space), the two ordered bases represent the same orientation of $T_pM$ if and only if this change-of-basis matrix has positive determinant, i.e. \begin{align*} \det\big(JT_{\beta\alpha}(\varphi_\alpha(p))\big) > 0. \end{align*} [guided] The whole proof rests on translating "two coordinate frames at $p$ define the same orientation of $T_pM$" into "the transition map has positive Jacobian determinant at $\varphi_\alpha(p)$." We make this translation once, here, and reuse it throughout. Let $(U_\alpha, \varphi_\alpha)$ and $(U_\beta, \varphi_\beta)$ be two charts overlapping at $p$, with coordinates $(x_1, \dots, x_n)$ and $(y_1, \dots, y_n)$. The transition map is the smooth map between open subsets of $\mathbb{R}^n$ \begin{align*} T_{\beta\alpha} : \varphi_\alpha(U_\alpha \cap U_\beta) &\to \varphi_\beta(U_\alpha \cap U_\beta) \\ u = (u_1, \dots, u_n) &\mapsto T_{\beta\alpha}(u) = (T_1(u), \dots, T_n(u)) \end{align*} where $T_j = y_j \circ \varphi_\alpha^{-1}$ expresses each $y$-coordinate as a smooth function of the $u$'s (the values of the $x$-coordinates). The chain rule for the smooth function $y_j$ written in $x$-coordinates yields, at $p$, \begin{align*} \frac{\partial}{\partial x_i}\bigg|_p = \sum_{j=1}^n \frac{\partial T_j}{\partial u_i}(\varphi_\alpha(p)) \cdot \frac{\partial}{\partial y_j}\bigg|_p. \end{align*} The numbers $\frac{\partial T_j}{\partial u_i}(\varphi_\alpha(p))$ are precisely the entries $(JT_{\beta\alpha}(\varphi_\alpha(p)))_{ji}$ of the Jacobian matrix. So the matrix whose $i$-th column lists the coefficients of $\partial_{x_i}|_p$ in the basis $(\partial_{y_j}|_p)$ is exactly $JT_{\beta\alpha}(\varphi_\alpha(p))$. By definition, two ordered bases of a real [vector space](/page/Vector%20Space) define the same orientation if and only if the change-of-basis matrix between them has positive determinant. Applying this to $T_pM$: \begin{align*} (\partial_{x_1}|_p, \dots, \partial_{x_n}|_p) \text{ and } (\partial_{y_1}|_p, \dots, \partial_{y_n}|_p) \text{ represent the same orientation of } T_pM \end{align*} if and only if $\det(JT_{\beta\alpha}(\varphi_\alpha(p))) > 0$. This is the dictionary we will use repeatedly. [/guided] [/step] [step:Show that $\Phi(\omega)$ covers $M$] Fix an orientation $\omega$ on $M$. Define \begin{align*} \Phi(\omega) := \{(U, \varphi) \in \mathcal{S} : (\partial_{x_1}|_p, \dots, \partial_{x_n}|_p) \text{ represents } \omega_p \text{ for every } p \in U\}. \end{align*} Fix $p \in M$. By the local consistency in the definition of orientation, there exist an open neighborhood $V$ of $p$ and a smooth local frame $(E_1, \dots, E_n) \in \Gamma(TM|_V)^n$ with $(E_1(q), \dots, E_n(q))$ representing $\omega_q$ for every $q \in V$. Pick any chart $(W, \varphi) \in \mathcal{S}$ with $p \in W$; shrinking $W$ if necessary, we may assume $W \subseteq V$ and $W$ is connected. Write $\varphi = (x_1, \dots, x_n)$ and let $A : W \to \mathbb{R}^{n \times n}$ be the matrix-valued function whose entries $A_{ji}(q)$ are defined by \begin{align*} \frac{\partial}{\partial x_i}\bigg|_q = \sum_{j=1}^n A_{ji}(q)\, E_j(q). \end{align*} The entries $A_{ji}$ are smooth because both frames are smooth and $A(q)$ is the matrix of one smooth frame in terms of the other. The basis $(\partial_{x_1}|_q, \dots, \partial_{x_n}|_q)$ is a basis of $T_qM$ for every $q \in W$ (charts give rise to coordinate frames), so $\det A(q) \neq 0$ on $W$. Hence the smooth function $q \mapsto \det A(q)$ is continuous and nowhere zero on the connected set $W$, so it has constant sign. If $\det A > 0$ on $W$, then $(\partial_{x_i}|_q)$ and $(E_j(q))$ represent the same orientation of $T_qM$ for every $q \in W$, namely $\omega_q$; hence $(W, \varphi) \in \Phi(\omega)$. If $\det A < 0$ on $W$, define \begin{align*} \sigma : \mathbb{R}^n &\to \mathbb{R}^n \\ (u_1, u_2, \dots, u_n) &\mapsto (-u_1, u_2, \dots, u_n), \end{align*} and set $\tilde\varphi := \sigma \circ \varphi : W \to \mathbb{R}^n$. Since $\sigma$ is a smooth diffeomorphism, $(W, \tilde\varphi)$ is a chart in $\mathcal{S}$. Writing $\tilde\varphi = (\tilde x_1, \dots, \tilde x_n) = (-x_1, x_2, \dots, x_n)$, we have $\partial_{\tilde x_1}|_q = -\partial_{x_1}|_q$ and $\partial_{\tilde x_i}|_q = \partial_{x_i}|_q$ for $i \geq 2$. The change-of-basis matrix from $(\partial_{x_i}|_q)$ to $(\partial_{\tilde x_i}|_q)$ is $\operatorname{diag}(-1, 1, \dots, 1)$ with determinant $-1$, so the new coordinate frame represents the orientation opposite to that of $(\partial_{x_i}|_q)$, hence equal to $\omega_q$. Thus $(W, \tilde\varphi) \in \Phi(\omega)$ and $p \in W$. In either case, $p$ lies in a chart of $\Phi(\omega)$, so $\Phi(\omega)$ covers $M$. [guided] We want to show every point of $M$ lies in some chart whose coordinate frame represents $\omega$ pointwise. The strategy: start with any chart at $p$; its coordinate frame either already represents $\omega$ on a small connected neighborhood, or it represents $-\omega$ there — in which case we flip a single coordinate to flip the orientation. Fix $p \in M$. By the orientation's local consistency, choose a neighborhood $V$ of $p$ and a smooth local frame $(E_1, \dots, E_n)$ on $V$ pointwise representing $\omega$. Then pick any smooth chart $(W, \varphi) \in \mathcal{S}$ with $p \in W \subseteq V$ and $W$ connected (charts always admit such shrinkings, since $\varphi(W)$ is an open subset of $\mathbb{R}^n$ and we may restrict to a small connected open ball around $\varphi(p)$). The question is: does the coordinate frame $(\partial_{x_i})$ represent $\omega$ everywhere on $W$? Write each $\partial_{x_i}|_q$ in the basis $(E_j(q))$, getting a matrix $A(q)$ with $\partial_{x_i}|_q = \sum_j A_{ji}(q)\, E_j(q)$. The orientations represented by these two bases at $q$ agree iff $\det A(q) > 0$. The crucial observation is that $\det A$ is continuous (in fact smooth) on $W$ and never zero (because both are bases), so $\det A$ has constant sign on the connected set $W$. Two cases: (i) If $\det A > 0$ on $W$: every $\partial_{x_i}|_q$ represents $\omega_q$, so $(W, \varphi)$ is in $\Phi(\omega)$. (ii) If $\det A < 0$ on $W$: the coordinate frame represents $-\omega$ throughout $W$. Apply the smooth diffeomorphism $\sigma : \mathbb{R}^n \to \mathbb{R}^n$ that flips the first coordinate, giving a new chart $(W, \sigma \circ \varphi)$ in $\mathcal{S}$. Its coordinate frame differs from $(\partial_{x_i})$ only in the sign of the first vector, so the change-of-basis matrix is $\operatorname{diag}(-1, 1, \dots, 1)$ with determinant $-1$. Negating one basis vector flips the orientation. So the new coordinate frame represents $\omega_q$ at every $q \in W$, placing $(W, \sigma \circ \varphi)$ in $\Phi(\omega)$. In either case we have found a chart of $\Phi(\omega)$ containing $p$. Since $p \in M$ was arbitrary, $\Phi(\omega)$ covers $M$. [/guided] [/step] [step:Verify that $\Phi(\omega)$ is an oriented atlas and is maximal] Let $(U_\alpha, \varphi_\alpha), (U_\beta, \varphi_\beta) \in \Phi(\omega)$ with $U_\alpha \cap U_\beta \neq \varnothing$. For any $p \in U_\alpha \cap U_\beta$, both coordinate frames $(\partial_{x_1}|_p, \dots, \partial_{x_n}|_p)$ and $(\partial_{y_1}|_p, \dots, \partial_{y_n}|_p)$ represent $\omega_p$ by definition of $\Phi(\omega)$. By Step 1, this is equivalent to \begin{align*} \det\big(JT_{\beta\alpha}(\varphi_\alpha(p))\big) > 0. \end{align*} As $p \in U_\alpha \cap U_\beta$ was arbitrary, this positivity holds throughout $\varphi_\alpha(U_\alpha \cap U_\beta)$. Combined with the covering property from Step 2, $\Phi(\omega)$ is an oriented atlas. For maximality, suppose $\mathcal{A}'$ is an oriented atlas with $\Phi(\omega) \subseteq \mathcal{A}'$, and let $(U, \varphi) \in \mathcal{A}'$ with coordinate functions $(z_1, \dots, z_n)$. For each $p \in U$, by Step 2 there is a chart $(U_\alpha, \varphi_\alpha) \in \Phi(\omega) \subseteq \mathcal{A}'$ with $p \in U_\alpha$. Since $\mathcal{A}'$ is an oriented atlas, the transition from $(U, \varphi)$ to $(U_\alpha, \varphi_\alpha)$ has positive Jacobian determinant at $\varphi(p)$, hence by Step 1 the coordinate frame $(\partial_{z_1}|_p, \dots, \partial_{z_n}|_p)$ represents the same orientation of $T_pM$ as $(\partial_{x_1}|_p, \dots, \partial_{x_n}|_p)$, which is $\omega_p$. As $p$ was arbitrary, $(U, \varphi) \in \Phi(\omega)$. Thus $\mathcal{A}' = \Phi(\omega)$, so $\Phi(\omega)$ is maximal. [/step] [step:Construct the inverse $\Psi$ and verify it produces an orientation] Let $\mathcal{A} \in \mathrm{MOA}(M)$. For each $p \in M$, choose any chart $(U, \varphi) \in \mathcal{A}$ with $p \in U$ (such a chart exists because $\mathcal{A}$ is in particular an atlas, hence covers $M$), and define $\omega_p^{\mathcal{A}}$ to be the orientation of $T_pM$ represented by $(\partial_{x_1}|_p, \dots, \partial_{x_n}|_p)$, where $(x_1, \dots, x_n) = \varphi$. **Well-definedness.** Suppose $(U', \varphi') \in \mathcal{A}$ also contains $p$, with $\varphi' = (y_1, \dots, y_n)$. Since $\mathcal{A}$ is an oriented atlas, the transition between $\varphi$ and $\varphi'$ has positive Jacobian determinant at $\varphi(p)$. By Step 1, the two coordinate frames at $p$ represent the same orientation of $T_pM$, so $\omega_p^{\mathcal{A}}$ does not depend on the choice of chart. **Local consistency.** Fix $p \in M$ and a chart $(U, \varphi) \in \mathcal{A}$ with $p \in U$. The coordinate vector fields $\partial_{x_i} \in \Gamma(TM|_U)$ are smooth, and $(\partial_{x_1}|_q, \dots, \partial_{x_n}|_q)$ represents $\omega_q^{\mathcal{A}}$ for every $q \in U$ by construction. Thus $(\partial_{x_i})$ is a smooth local frame on $U$ pointwise representing $\omega^{\mathcal{A}}$, verifying the local consistency condition. Define $\Psi(\mathcal{A}) := \omega^{\mathcal{A}}$. By the above, $\Psi(\mathcal{A}) \in \mathrm{Or}(M)$. [/step] [step:Verify that $\Phi$ and $\Psi$ are mutual inverses] **$\Psi \circ \Phi = \mathrm{id}_{\mathrm{Or}(M)}$.** Let $\omega \in \mathrm{Or}(M)$ and set $\omega' := \Psi(\Phi(\omega))$. For any $p \in M$, by Step 2 there is a chart $(U, \varphi) \in \Phi(\omega)$ containing $p$. By definition of $\Psi$, $\omega'_p$ is the orientation of $T_pM$ represented by $(\partial_{x_1}|_p, \dots, \partial_{x_n}|_p)$, which equals $\omega_p$ by definition of $\Phi(\omega)$. Hence $\omega' = \omega$. **$\Phi \circ \Psi = \mathrm{id}_{\mathrm{MOA}(M)}$.** Let $\mathcal{A} \in \mathrm{MOA}(M)$ and set $\mathcal{A}' := \Phi(\Psi(\mathcal{A}))$. For the inclusion $\mathcal{A} \subseteq \mathcal{A}'$: let $(U, \varphi) \in \mathcal{A}$ with $\varphi = (x_1, \dots, x_n)$. For each $p \in U$, the orientation $\Psi(\mathcal{A})_p$ is, by Step 4, represented by $(\partial_{x_1}|_p, \dots, \partial_{x_n}|_p)$, so $(U, \varphi) \in \Phi(\Psi(\mathcal{A})) = \mathcal{A}'$. For the reverse inclusion $\mathcal{A}' \subseteq \mathcal{A}$: by Step 3 applied to $\omega := \Psi(\mathcal{A})$, the set $\mathcal{A}' = \Phi(\Psi(\mathcal{A}))$ is a maximal oriented atlas. Since $\mathcal{A} \subseteq \mathcal{A}'$ and $\mathcal{A}$ is itself a maximal oriented atlas (by hypothesis), the inclusion forces $\mathcal{A} = \mathcal{A}'$. This completes the proof that $\Phi : \mathrm{Or}(M) \to \mathrm{MOA}(M)$ is a bijection with inverse $\Psi$. [guided] We verify the two compositions one at a time. **Round trip starting from an orientation.** Given $\omega \in \mathrm{Or}(M)$, the orientation $\omega' := \Psi(\Phi(\omega))$ is defined pointwise: at each $p \in M$, $\omega'_p$ is the orientation represented by the coordinate frame of any chart in $\Phi(\omega)$ at $p$. Step 2 guarantees that some such chart $(U, \varphi)$ exists, and the very definition of $\Phi(\omega)$ requires the coordinate frame of $(U, \varphi)$ to represent $\omega_p$. So $\omega'_p = \omega_p$ at every $p$, i.e. $\omega' = \omega$. **Round trip starting from a maximal oriented atlas.** Given $\mathcal{A} \in \mathrm{MOA}(M)$, set $\mathcal{A}' := \Phi(\Psi(\mathcal{A}))$. We must show $\mathcal{A} = \mathcal{A}'$. (i) For any chart $(U, \varphi) \in \mathcal{A}$ and any $p \in U$, the orientation $\Psi(\mathcal{A})_p$ is defined to be the one represented by $(\partial_{x_i}|_p)$ (Step 4). So the coordinate frame of $(U, \varphi)$ represents $\Psi(\mathcal{A})$ pointwise on $U$, which is exactly the condition for $(U, \varphi)$ to belong to $\Phi(\Psi(\mathcal{A})) = \mathcal{A}'$. Hence $\mathcal{A} \subseteq \mathcal{A}'$. (ii) By Step 3 applied with $\omega := \Psi(\mathcal{A})$, the atlas $\mathcal{A}' = \Phi(\Psi(\mathcal{A}))$ is a maximal oriented atlas. We now have two maximal oriented atlases with $\mathcal{A} \subseteq \mathcal{A}'$, and by maximality of $\mathcal{A}$ this inclusion is an equality: any proper extension of $\mathcal{A}$ as an oriented atlas is forbidden by definition. So $\mathcal{A} = \mathcal{A}'$. Therefore $\Phi$ and $\Psi$ are mutually inverse, and we have established the desired bijection between $\mathrm{Or}(M)$ and $\mathrm{MOA}(M)$. [/guided] [/step]