[proofplan] We prove both directions using the equivalence between orientations and nowhere-vanishing top forms. For the forward direction, we start from a nowhere-vanishing volume form $\omega \in \Omega^n(M)$ and select the sub-atlas on which $\omega$ evaluates positively on the coordinate frame; we verify this sub-atlas covers $M$ (by post-composing with a reflection if necessary) and that its transition maps preserve orientation. For the reverse direction, we start from an atlas with positive transition determinants and glue the local top forms $\varphi_\alpha^*(dy_1 \wedge \cdots \wedge dy_n)$ using a subordinate partition of unity, checking that the positivity of the Jacobians prevents cancellation in the sum and produces a nowhere-vanishing global form. [/proofplan]

[step:Reformulate orientability in terms of a nowhere-vanishing top form]By the standard equivalence, an $n$-manifold $M$ is orientable if and only if there exists a nowhere-vanishing [differential $n$-form](/page/Differential%20Form) $\omega \in \Omega^n(M)$, called a [volume form](/page/Volume%20Form). We use this characterisation throughout the proof; the task reduces to converting between volume forms and atlases with positive transition maps.[/step]

[guided]The intrinsic definition of orientability is phrased in terms of compatible bases on tangent spaces, but for computational purposes the equivalent statement "$M$ admits a nowhere-vanishing top form" is far more tractable because a top form is an object we can explicitly manipulate in local coordinates. We will use this equivalence implicitly: instead of proving "$M$ is orientable iff atlas exists", we prove "$M$ admits a volume form iff atlas exists". Both reformulations use the fact that on an $n$-manifold, $\Lambda^n T^*_p M$ is one-dimensional, so a non-zero element picks out a preferred orientation of the tangent space at each point.[/guided]

[step:Forward direction: select charts where $\omega$ is positive on the coordinate frame] Assume $M$ is orientable and fix a nowhere-vanishing $\omega \in \Omega^n(M)$. Define \begin{align*} \mathcal{A} := \{(U_\alpha, \varphi_\alpha) : U_\alpha \text{ connected}, \ \omega_p(\partial_{x_1}\big|_p, \dots, \partial_{x_n}\big|_p) > 0 \text{ for all } p \in U_\alpha\}, \end{align*} where $\partial_{x_i}\big|_p$ are the coordinate vector fields associated to $\varphi_\alpha$. The set $\mathcal{A}$ is an atlas because every point admits some connected chart $(V, \psi)$, and on $V$ the smooth function $p \mapsto \omega_p(\partial_{y_1}\big|_p, \dots, \partial_{y_n}\big|_p)$ is nowhere zero (since $\omega$ is nowhere zero and the coordinate frame is a basis), hence of constant sign on the connected set $V$. If the sign is positive, $(V, \psi) \in \mathcal{A}$; if negative, define \begin{align*} \tilde{\psi} : V &\to \mathbb{R}^n \\ p &\mapsto (-y_1(p), y_2(p), \dots, y_n(p)), \end{align*} a smooth homeomorphism onto its image. The coordinate frame of $\tilde\psi$ is $(-\partial_{y_1}, \partial_{y_2}, \dots, \partial_{y_n})$, so $\omega$ evaluated on this frame has the opposite sign, which is now positive. Thus $(V, \tilde\psi) \in \mathcal{A}$, and $\mathcal{A}$ covers $M$. [/step]

[step:Show that transition maps in $\mathcal{A}$ have positive Jacobian determinant]Let $(U_\alpha, \varphi_\alpha), (U_\beta, \varphi_\beta) \in \mathcal{A}$ with $U_\alpha \cap U_\beta \ne \varnothing$, and set $T := \varphi_\beta \circ \varphi_\alpha^{-1}$ on $\varphi_\alpha(U_\alpha \cap U_\beta)$. Write coordinates $x$ for $\varphi_\alpha$ and $y$ for $\varphi_\beta$. On the overlap, the coordinate frames are related by \begin{align*} \partial_{x_i} = \sum_{j=1}^{n} \frac{\partial y_j}{\partial x_i}\,\partial_{y_j}. \end{align*} Since $\omega$ is a top form, it is multilinear and alternating, so \begin{align*} \omega(\partial_{x_1}, \dots, \partial_{x_n}) = \det\!\left(\frac{\partial y_j}{\partial x_i}\right) \omega(\partial_{y_1}, \dots, \partial_{y_n}) = \det(JT) \cdot \omega(\partial_{y_1}, \dots, \partial_{y_n}), \end{align*} where $JT$ is the Jacobian matrix of $T$. By construction of $\mathcal{A}$, both sides with $\omega(\partial_{x_1}, \dots, \partial_{x_n})$ and $\omega(\partial_{y_1}, \dots, \partial_{y_n})$ are strictly positive. Therefore $\det(JT) > 0$ at every point of $\varphi_\alpha(U_\alpha \cap U_\beta)$.[/step]

[guided]The computation relies on the multilinear-alternating property of $\omega$: applied to a linear combination of arguments, $\omega$ produces a determinant. Concretely, write $v_i := \partial_{x_i} = \sum_j M_{ji} \partial_{y_j}$ with $M_{ji} = \partial y_j / \partial x_i$; then $\omega(v_1, \dots, v_n) = (\det M) \omega(\partial_{y_1}, \dots, \partial_{y_n})$ is the fundamental identity for a top form under a change of basis in the tangent space. The matrix $M$ is precisely the Jacobian matrix $JT$ of the transition $T = \varphi_\beta \circ \varphi_\alpha^{-1}$ evaluated on $\varphi_\alpha(U_\alpha \cap U_\beta)$ via the chain rule. Both $\omega$-evaluations are positive by the definition of $\mathcal{A}$, and their ratio is $\det(JT)$; this ratio of positives is positive.[/guided]

[step:Reverse direction: glue pullbacks of the standard top form via a partition of unity] Conversely, assume we have an atlas $\{(U_\alpha, \varphi_\alpha)\}$ with all transition maps satisfying $\det(J(\varphi_\beta \circ \varphi_\alpha^{-1})) > 0$ on overlaps. By paracompactness of $M$, there exists a smooth [partition of unity](/page/Partition%20of%20Unity) $\{\rho_\alpha\}$ subordinate to $\{U_\alpha\}$ with $\rho_\alpha \ge 0$, $\operatorname{supp}(\rho_\alpha) \subset U_\alpha$, the collection $\{\operatorname{supp}(\rho_\alpha)\}$ locally finite, and $\sum_\alpha \rho_\alpha \equiv 1$. For each $\alpha$, define \begin{align*} \omega_\alpha := \varphi_\alpha^*(dy_1 \wedge \cdots \wedge dy_n) \in \Omega^n(U_\alpha), \end{align*} and extend $\rho_\alpha \omega_\alpha$ by zero to all of $M$ (valid because $\operatorname{supp}(\rho_\alpha) \subset U_\alpha$). Set \begin{align*} \omega := \sum_\alpha \rho_\alpha \omega_\alpha \in \Omega^n(M). \end{align*} The sum is locally finite, so $\omega$ is smooth. [/step]

[step:Verify $\omega$ is nowhere vanishing using positivity of transition determinants]Fix $p \in M$ and pick any chart $(U_\gamma, \varphi_\gamma)$ with $p \in U_\gamma$. Evaluate $\omega$ on the coordinate frame of $\varphi_\gamma$ at $p$: \begin{align*} \omega_p(\partial_{x_1^\gamma}, \dots, \partial_{x_n^\gamma}) = \sum_{\alpha : p \in U_\alpha} \rho_\alpha(p) \cdot (\omega_\alpha)_p(\partial_{x_1^\gamma}, \dots, \partial_{x_n^\gamma}). \end{align*} For each $\alpha$ with $p \in U_\alpha$, the multilinear-alternating computation from the forward direction gives \begin{align*} (\omega_\alpha)_p(\partial_{x_1^\gamma}, \dots, \partial_{x_n^\gamma}) = \det(J(\varphi_\alpha \circ \varphi_\gamma^{-1})(\varphi_\gamma(p))), \end{align*} which is strictly positive by the hypothesis on the atlas. Since $\rho_\alpha(p) \ge 0$ and at least one $\rho_\alpha(p) > 0$ (because $\sum_\alpha \rho_\alpha(p) = 1$), the sum is a sum of non-negative terms with at least one strictly positive term, hence strictly positive. Therefore $\omega_p \ne 0$.[/step]

[guided]The central subtlety is preventing cancellation in the sum defining $\omega$. If the transition maps were orientation-reversing between some pair of charts, the local forms $\omega_\alpha$ and $\omega_\beta$ would disagree in sign on overlaps, and the partition-of-unity sum could vanish at a point where exactly two positive weights $\rho_\alpha, \rho_\beta$ are active. The assumption $\det(J(\varphi_\beta \circ \varphi_\alpha^{-1})) > 0$ is precisely what prevents this: it forces all $\omega_\alpha$ to agree in orientation on every overlap, so the partition-of-unity convex combination is a positive linear combination of co-oriented top forms and cannot sum to zero. The normalisation $\sum_\alpha \rho_\alpha = 1$ supplies at least one strictly positive weight at every point, completing the argument.[/guided]

[step:Conclude the equivalence] The forward direction constructed an atlas in $\mathcal{A}$ with positive transition determinants from a volume form. The reverse direction constructed a nowhere-vanishing top form $\omega$ from such an atlas. Using the equivalence between orientability and the existence of a nowhere-vanishing top form established in Step 1, we conclude that $M$ is orientable if and only if it admits an atlas with everywhere-positive Jacobian determinants of its transition maps. [/step]