[proofplan] We prove both implications using the pullback identity $T^*(du_1 \wedge \cdots \wedge du_n) = \det(JT)\, du_1 \wedge \cdots \wedge du_n$ for any smooth map $T$ between open subsets of $\mathbb{R}^n$. For the forward direction, we start with an oriented atlas, pull back the standard volume form of $\mathbb{R}^n$ in each chart to obtain local nowhere-vanishing $n$-forms, and glue them with a smooth [partition of unity](/page/Partition%20of%20Unity); positivity of every transition Jacobian forces the convex combination to be nonzero at each point. For the reverse direction, we use a nowhere-vanishing $n$-form $\omega$ to select an oriented atlas: in each connected chart $\omega$ has a coefficient function of constant sign, and we flip the first coordinate where the sign is negative. The pullback identity then forces every transition Jacobian to be positive. [/proofplan]

[step:Record the pullback identity for top-degree forms]We record the algebraic fact used repeatedly below. Let $V, W \subseteq \mathbb{R}^n$ be open and let $T : V \to W$ be a smooth map. Write $u_1, \ldots, u_n$ for the standard coordinate functions on $\mathbb{R}^n$ and let $\eta := du_1 \wedge \cdots \wedge du_n \in \Omega^n(\mathbb{R}^n)$. Then \begin{align*} T^*\eta = \det(JT)\, du_1 \wedge \cdots \wedge du_n, \end{align*} where $JT(v) \in \mathbb{R}^{n\times n}$ is the Jacobian matrix $(JT)_{ij}(v) = \partial_j T_i(v)$. This follows from the definition of pullback and multilinearity: at $v \in V$ with tangent vectors $X_1, \ldots, X_n \in T_v\mathbb{R}^n \cong \mathbb{R}^n$, \begin{align*} (T^*\eta)_v(X_1, \ldots, X_n) = \eta_{T(v)}(JT(v)X_1, \ldots, JT(v)X_n) = \det\bigl[JT(v)X_1 \mid \cdots \mid JT(v)X_n\bigr], \end{align*} and by the multiplicativity of the determinant this equals $\det(JT(v))\cdot \det[X_1\mid\cdots\mid X_n] = \det(JT(v))\,\eta_v(X_1,\ldots,X_n)$.[/step]

[guided]We will repeatedly need to compare $n$-forms expressed in two different coordinate systems. The pullback identity does this in a single line. Let $V, W \subseteq \mathbb{R}^n$ be open, $T: V \to W$ smooth, $u_1, \ldots, u_n$ the standard coordinates, and $\eta := du_1 \wedge \cdots \wedge du_n$. The claim is \begin{align*} T^*\eta = \det(JT)\, du_1 \wedge \cdots \wedge du_n. \end{align*} Why is this true? The pullback is defined pointwise on tangent vectors by $(T^*\eta)_v(X_1,\ldots,X_n) = \eta_{T(v)}(dT_v X_1, \ldots, dT_v X_n)$, and at a point of $\mathbb{R}^n$ the differential $dT_v$ is just the Jacobian matrix $JT(v)$ acting on column vectors. The form $\eta$ at any point of $\mathbb{R}^n$ sends an $n$-tuple of vectors to the determinant of the matrix they form. So \begin{align*} (T^*\eta)_v(X_1, \ldots, X_n) &= \det\bigl[JT(v)X_1 \mid \cdots \mid JT(v)X_n\bigr] \\ &= \det\bigl(JT(v) \cdot [X_1\mid\cdots\mid X_n]\bigr) \\ &= \det(JT(v)) \cdot \det[X_1\mid\cdots\mid X_n] \\ &= \det(JT(v)) \cdot \eta_v(X_1,\ldots,X_n), \end{align*} using the multiplicativity of the determinant. Since the identity holds for every $n$-tuple of tangent vectors at every point, $T^*\eta = \det(JT)\,du_1\wedge\cdots\wedge du_n$ as $n$-forms on $V$. This is the linchpin of the entire proof: positivity of $\det(JT)$ is precisely the positivity of the coefficient that appears when we rewrite a top-degree form in a new coordinate system.[/guided]

[step:Construct a nowhere-vanishing $n$-form from an oriented atlas]Assume $M$ admits an oriented atlas $\mathcal{A} = \{(U_\alpha, \varphi_\alpha)\}_{\alpha \in A}$. For each $\alpha \in A$, define the local $n$-form \begin{align*} \omega_\alpha : U_\alpha &\to \Lambda^n T^*M, \\ p &\mapsto \bigl(\varphi_\alpha^{*}(du_1 \wedge \cdots \wedge du_n)\bigr)_p. \end{align*} Each $\omega_\alpha \in \Omega^n(U_\alpha)$ is smooth and nowhere-vanishing on $U_\alpha$, since it is the pullback of a nowhere-vanishing top form by the diffeomorphism $\varphi_\alpha$. Apply the [Existence of Smooth Partitions of Unity](/theorems/57) to the open cover $\{U_\alpha\}_{\alpha \in A}$ of the smooth (hence paracompact second-countable) manifold $M$, obtaining a smooth [partition of unity](/page/Partition%20of%20Unity) $\{\rho_\alpha\}_{\alpha \in A}$ subordinate to $\{U_\alpha\}$: each $\rho_\alpha \in C^\infty(M, [0,1])$ has $\operatorname{supp}\rho_\alpha \subseteq U_\alpha$, the family $\{\operatorname{supp}\rho_\alpha\}$ is locally finite, and $\sum_\alpha \rho_\alpha \equiv 1$ on $M$. Define \begin{align*} \omega := \sum_{\alpha \in A} \rho_\alpha\, \omega_\alpha \in \Omega^n(M), \end{align*} where each summand $\rho_\alpha \omega_\alpha$ is extended by zero outside $U_\alpha$ (this extension is smooth because $\operatorname{supp}\rho_\alpha \subseteq U_\alpha$). Local finiteness makes the sum locally finite, so $\omega$ is a well-defined smooth $n$-form. Fix $p \in M$ and choose any $\beta \in A$ with $p \in U_\beta$; we show $\omega_p \ne 0$ by exhibiting $\omega$ in the chart $\varphi_\beta$. For $\alpha \in A$ with $p \in U_\alpha \cap U_\beta$, the transition map $\tau_{\alpha\beta} = \varphi_\alpha \circ \varphi_\beta^{-1}$ is defined near $\varphi_\beta(p)$, and \begin{align*} (\varphi_\beta^{-1})^* \omega_\alpha = (\varphi_\beta^{-1})^* \varphi_\alpha^*(du_1 \wedge \cdots \wedge du_n) = \tau_{\alpha\beta}^*(du_1 \wedge \cdots \wedge du_n) = \det(J\tau_{\alpha\beta})\, du_1 \wedge \cdots \wedge du_n \end{align*} by Step 1. For $\alpha = \beta$ the same identity holds with $\tau_{\beta\beta} = \operatorname{id}$, giving $(\varphi_\beta^{-1})^*\omega_\beta = du_1 \wedge \cdots \wedge du_n$. Hence \begin{align*} (\varphi_\beta^{-1})^*\omega = \Bigl(\sum_{\alpha :\, p \in U_\alpha} (\rho_\alpha \circ \varphi_\beta^{-1})\cdot \det(J\tau_{\alpha\beta})\Bigr)\, du_1 \wedge \cdots \wedge du_n \end{align*} on a neighbourhood of $\varphi_\beta(p)$ in $\varphi_\beta(U_\beta)$. Evaluate the coefficient at $\varphi_\beta(p)$. Each $\rho_\alpha(p) \ge 0$, and the oriented-atlas hypothesis gives $\det(J\tau_{\alpha\beta})(\varphi_\beta(p)) > 0$ for every $\alpha$ with $p \in U_\alpha \cap U_\beta$. Since $\sum_\alpha \rho_\alpha(p) = 1$, at least one index $\alpha_0$ has $\rho_{\alpha_0}(p) > 0$, and that index automatically satisfies $p \in U_{\alpha_0}$. The corresponding term $\rho_{\alpha_0}(p)\cdot \det(J\tau_{\alpha_0 \beta})(\varphi_\beta(p))$ is strictly positive, so the entire coefficient is strictly positive. Therefore $\omega_p \ne 0$.[/step]

[guided]We want to manufacture a global nowhere-vanishing $n$-form from the bare data of an oriented atlas $\mathcal{A} = \{(U_\alpha, \varphi_\alpha)\}_{\alpha \in A}$. The natural building blocks are the **local coordinate volume forms** \begin{align*} \omega_\alpha : U_\alpha &\to \Lambda^n T^*M, \\ p &\mapsto \bigl(\varphi_\alpha^{*}(du_1 \wedge \cdots \wedge du_n)\bigr)_p, \end{align*} each of which is smooth and nowhere-vanishing on its chart (pullback of a nowhere-vanishing top form by a diffeomorphism). The challenge is to glue them. A naive sum $\sum_\alpha \omega_\alpha$ has no chance of being well-defined globally; the standard fix is a [partition of unity](/page/Partition%20of%20Unity). The [Existence of Smooth Partitions of Unity](/theorems/57) applies because $M$, being a smooth manifold, is second-countable, Hausdorff, and locally compact — hence paracompact, which is the hypothesis the existence theorem actually needs. It gives a [partition of unity](/page/Partition%20of%20Unity) $\{\rho_\alpha\}_{\alpha \in A}$ with $\operatorname{supp}\rho_\alpha \subseteq U_\alpha$, locally finite supports, and $\sum_\alpha \rho_\alpha \equiv 1$. We define \begin{align*} \omega := \sum_{\alpha \in A} \rho_\alpha\, \omega_\alpha, \end{align*} extending each $\rho_\alpha \omega_\alpha$ by zero outside $U_\alpha$; this extension is smooth because $\operatorname{supp}\rho_\alpha \subseteq U_\alpha$ forces the form to vanish smoothly into the boundary of $U_\alpha$. Local finiteness ensures the sum has only finitely many nonzero terms in a neighbourhood of every point, so $\omega \in \Omega^n(M)$. The crucial question: **why doesn't this sum cancel at some point?** This is where the orientation hypothesis enters. Fix $p \in M$ and pick any chart $(U_\beta, \varphi_\beta)$ containing $p$. We compute $\omega$ in this chart by pulling everything back to $\varphi_\beta(U_\beta) \subseteq \mathbb{R}^n$. For each $\alpha$ with $p \in U_\alpha \cap U_\beta$, set $\tau_{\alpha\beta} := \varphi_\alpha \circ \varphi_\beta^{-1}$ (this is the transition map from chart $\beta$ to chart $\alpha$). By definition $\omega_\alpha = \varphi_\alpha^*(du_1 \wedge \cdots \wedge du_n)$, so \begin{align*} (\varphi_\beta^{-1})^* \omega_\alpha = (\varphi_\beta^{-1})^* \varphi_\alpha^*(du_1\wedge\cdots\wedge du_n) = (\varphi_\alpha \circ \varphi_\beta^{-1})^*(du_1\wedge\cdots\wedge du_n) = \tau_{\alpha\beta}^*(du_1\wedge\cdots\wedge du_n). \end{align*} Applying Step 1's pullback identity, this equals $\det(J\tau_{\alpha\beta})\, du_1\wedge\cdots\wedge du_n$. For $\alpha = \beta$ we get the identity transition map and hence $du_1\wedge\cdots\wedge du_n$ exactly. Summing, \begin{align*} (\varphi_\beta^{-1})^* \omega = \Bigl(\sum_{\alpha :\, p \in U_\alpha} (\rho_\alpha \circ \varphi_\beta^{-1})\cdot \det(J\tau_{\alpha\beta})\Bigr)\, du_1\wedge\cdots\wedge du_n \end{align*} in a neighbourhood of $\varphi_\beta(p)$. Now evaluate the coefficient at the point $\varphi_\beta(p)$: - Every $\rho_\alpha(p)$ is $\ge 0$, since $\rho_\alpha$ takes values in $[0,1]$. - Every $\det(J\tau_{\alpha\beta})(\varphi_\beta(p))$ is strictly positive — **this is exactly the orientation hypothesis**. - Since $\sum_\alpha \rho_\alpha(p) = 1 > 0$, at least one $\alpha_0$ contributes $\rho_{\alpha_0}(p) > 0$, and necessarily $p \in U_{\alpha_0}$. The sum is therefore a non-negative combination with at least one strictly positive term, so the coefficient is strictly positive. We conclude $\omega_p \ne 0$. Since $p$ was arbitrary, $\omega$ is nowhere-vanishing on $M$. **The structural point:** orientation enters only as the sign of the Jacobian determinants. A non-oriented atlas would give some negative determinants, the convex combination could vanish, and the construction collapses. The positivity is exactly what prevents cancellation.[/guided]

[step:From a nowhere-vanishing $n$-form, extract a local coefficient of constant sign in each connected chart] Conversely, assume there exists $\omega \in \Omega^n(M)$ with $\omega_p \ne 0$ for every $p \in M$. Since $M$ is locally Euclidean, the collection of charts $(U, \varphi)$ with $U$ connected and $\varphi(U) \subseteq \mathbb{R}^n$ open forms a smooth atlas on $M$: every point lies in the domain of some chart, and we may shrink any chart to a connected open neighbourhood (e.g., to the preimage of an open ball) without losing smoothness or compatibility. Fix one such chart $(U, \varphi)$ with coordinates $\varphi = (x_1, \ldots, x_n)$. The pullback $(\varphi^{-1})^*\omega \in \Omega^n(\varphi(U))$ is a smooth $n$-form on the [open set](/page/Open%20Set) $\varphi(U) \subseteq \mathbb{R}^n$, so \begin{align*} (\varphi^{-1})^*\omega = f_\varphi\, du_1 \wedge \cdots \wedge du_n \end{align*} for a uniquely determined smooth function $f_\varphi : \varphi(U) \to \mathbb{R}$. We claim $f_\varphi$ has constant sign (strictly positive or strictly negative) on $\varphi(U)$. Indeed, $\varphi^{-1}$ is a diffeomorphism and $\omega$ is nowhere zero, so $(\varphi^{-1})^*\omega$ is a nowhere-vanishing $n$-form on $\varphi(U)$. Hence $f_\varphi$ never vanishes. By continuity on the connected set $\varphi(U)$ (which is connected because $U$ is connected and $\varphi$ is a homeomorphism onto its image), $f_\varphi$ has constant sign. [/step]

[step:Replace charts with negative coefficient by their reflection to obtain a positive atlas]For each chart $(U, \varphi)$ in the connected atlas of Step 3, define a **positive chart** $(U, \tilde\varphi)$ as follows: - If $f_\varphi > 0$ on $\varphi(U)$, set $\tilde\varphi := \varphi$. - If $f_\varphi < 0$ on $\varphi(U)$, set $\tilde\varphi := R \circ \varphi$, where $R : \mathbb{R}^n \to \mathbb{R}^n$ is the reflection \begin{align*} R(u_1, u_2, \ldots, u_n) = (-u_1, u_2, \ldots, u_n). \end{align*} Since $R$ is a diffeomorphism of $\mathbb{R}^n$, $\tilde\varphi$ is a smooth chart on $U$. We compute $f_{\tilde\varphi}$ in the second case. Using $R = R^{-1}$, \begin{align*} (\tilde\varphi^{-1})^*\omega = (\varphi^{-1} \circ R^{-1})^*\omega = (R^{-1})^*\bigl((\varphi^{-1})^*\omega\bigr) = R^*\bigl(f_\varphi\, du_1 \wedge \cdots \wedge du_n\bigr). \end{align*} By Step 1 applied to $R$, $R^*(du_1 \wedge \cdots \wedge du_n) = \det(JR)\, du_1 \wedge \cdots \wedge du_n = -\,du_1 \wedge \cdots \wedge du_n$, so \begin{align*} (\tilde\varphi^{-1})^*\omega = -\,(f_\varphi \circ R)\, du_1 \wedge \cdots \wedge du_n. \end{align*} That is, $f_{\tilde\varphi} = -\,f_\varphi \circ R$ on $\tilde\varphi(U) = R(\varphi(U))$. Since $f_\varphi < 0$ on $\varphi(U)$, $f_\varphi \circ R < 0$ on $R(\varphi(U))$, hence $f_{\tilde\varphi} > 0$. Let $\tilde{\mathcal{A}} := \{(U_\alpha, \tilde\varphi_\alpha)\}$ be the resulting smooth atlas, in which every chart satisfies $f_{\tilde\varphi_\alpha} > 0$.[/step]

[guided]We have built, from $\omega$, a local invariant $f_\varphi$ in every connected chart whose sign is constant on the chart. We now harness this sign to build the oriented atlas. The construction is uniform: in each chart $(U, \varphi)$, leave it alone if $f_\varphi > 0$, and otherwise compose with a reflection to flip the sign. The natural candidate reflection on $\mathbb{R}^n$ is $R(u_1, u_2, \ldots, u_n) = (-u_1, u_2, \ldots, u_n)$, which has Jacobian determinant $-1$. Why does this work? In a chart where $f_\varphi < 0$, set $\tilde\varphi := R \circ \varphi$. The new coefficient is \begin{align*} (\tilde\varphi^{-1})^*\omega &= (\varphi^{-1} \circ R^{-1})^*\omega = (R^{-1})^*\bigl((\varphi^{-1})^*\omega\bigr) \\ &= R^*\bigl(f_\varphi\, du_1\wedge\cdots\wedge du_n\bigr) \quad \text{using } R^{-1} = R \\ &= (f_\varphi \circ R)\cdot R^*(du_1\wedge\cdots\wedge du_n) \\ &= -(f_\varphi \circ R)\, du_1\wedge\cdots\wedge du_n, \end{align*} where in the last step we used Step 1's pullback identity: $R^*(du_1\wedge\cdots\wedge du_n) = \det(JR)\, du_1\wedge\cdots\wedge du_n = -du_1\wedge\cdots\wedge du_n$ because $JR = \operatorname{diag}(-1, 1, \ldots, 1)$ has determinant $-1$. So $f_{\tilde\varphi}(u) = -f_\varphi(Ru)$, which is positive since $f_\varphi$ was negative. The reflection turns a "negatively oriented" chart into a "positively oriented" one. **Why use $R$ rather than $-\operatorname{id}$?** Both work in even dimension, but $-\operatorname{id}$ has determinant $(-1)^n$, which is $+1$ when $n$ is even — useless. The single-coordinate reflection $R$ has determinant $-1$ in every dimension and is the safe choice. The output is an atlas $\tilde{\mathcal{A}} = \{(U_\alpha, \tilde\varphi_\alpha)\}$ on $M$ in which the coefficient of $\omega$ in every chart is strictly positive: $f_{\tilde\varphi_\alpha} > 0$.[/guided]

[step:Verify that the positive atlas is oriented]Let $(U_\alpha, \tilde\varphi_\alpha)$ and $(U_\beta, \tilde\varphi_\beta)$ be two charts of $\tilde{\mathcal{A}}$ with $U_\alpha \cap U_\beta \ne \varnothing$, and let \begin{align*} \tau_{\beta\alpha} := \tilde\varphi_\beta \circ \tilde\varphi_\alpha^{-1} : \tilde\varphi_\alpha(U_\alpha \cap U_\beta) \to \tilde\varphi_\beta(U_\alpha \cap U_\beta) \end{align*} be the transition map. On the overlap, write \begin{align*} (\tilde\varphi_\alpha^{-1})^*\omega = f_\alpha\, du_1 \wedge \cdots \wedge du_n, \qquad (\tilde\varphi_\beta^{-1})^*\omega = f_\beta\, du_1 \wedge \cdots \wedge du_n, \end{align*} with $f_\alpha > 0$ on $\tilde\varphi_\alpha(U_\alpha \cap U_\beta)$ and $f_\beta > 0$ on $\tilde\varphi_\beta(U_\alpha \cap U_\beta)$ by Step 4. Pull the second identity back via $\tau_{\beta\alpha}$: \begin{align*} \tau_{\beta\alpha}^*\bigl((\tilde\varphi_\beta^{-1})^*\omega\bigr) = (\tilde\varphi_\beta^{-1} \circ \tau_{\beta\alpha})^*\omega = (\tilde\varphi_\alpha^{-1})^*\omega = f_\alpha\, du_1 \wedge \cdots \wedge du_n. \end{align*} On the other hand, by Step 1, \begin{align*} \tau_{\beta\alpha}^*\bigl(f_\beta\, du_1 \wedge \cdots \wedge du_n\bigr) = (f_\beta \circ \tau_{\beta\alpha})\cdot \det(J\tau_{\beta\alpha})\, du_1 \wedge \cdots \wedge du_n. \end{align*} Equating coefficients on $\tilde\varphi_\alpha(U_\alpha \cap U_\beta)$, \begin{align*} f_\alpha = (f_\beta \circ \tau_{\beta\alpha})\cdot \det(J\tau_{\beta\alpha}). \end{align*} At every $u \in \tilde\varphi_\alpha(U_\alpha \cap U_\beta)$ we have $f_\alpha(u) > 0$ and $(f_\beta \circ \tau_{\beta\alpha})(u) = f_\beta(\tau_{\beta\alpha}(u)) > 0$, whence \begin{align*} \det(J\tau_{\beta\alpha})(u) = \frac{f_\alpha(u)}{f_\beta(\tau_{\beta\alpha}(u))} > 0. \end{align*} Thus every transition map of $\tilde{\mathcal{A}}$ has positive Jacobian determinant, so $\tilde{\mathcal{A}}$ is an oriented atlas. Therefore $M$ is orientable.[/step]

[guided]We have an atlas $\tilde{\mathcal{A}}$ in which $\omega$ has a strictly positive coefficient in every chart. The remaining task is to verify the orientation condition on overlaps. Pick two overlapping charts $(U_\alpha, \tilde\varphi_\alpha)$ and $(U_\beta, \tilde\varphi_\beta)$ and form the transition map $\tau_{\beta\alpha} := \tilde\varphi_\beta \circ \tilde\varphi_\alpha^{-1}$ from coordinates of chart $\alpha$ to coordinates of chart $\beta$. We must show $\det(J\tau_{\beta\alpha}) > 0$. The strategy is to express the same $n$-form $\omega$ in two different ways and compare. On the overlap, by construction, \begin{align*} (\tilde\varphi_\alpha^{-1})^*\omega = f_\alpha\, du_1\wedge\cdots\wedge du_n, \qquad (\tilde\varphi_\beta^{-1})^*\omega = f_\beta\, du_1\wedge\cdots\wedge du_n, \end{align*} with both $f_\alpha, f_\beta > 0$. The bridge between the two equalities is the pullback by $\tau_{\beta\alpha}$. Specifically, since $\tilde\varphi_\beta^{-1} \circ \tau_{\beta\alpha} = \tilde\varphi_\alpha^{-1}$ (this is just the definition of $\tau_{\beta\alpha}$ rearranged), we have \begin{align*} \tau_{\beta\alpha}^*\bigl((\tilde\varphi_\beta^{-1})^*\omega\bigr) = (\tilde\varphi_\alpha^{-1})^*\omega = f_\alpha\, du_1\wedge\cdots\wedge du_n. \end{align*} Computing the left-hand side a different way using Step 1: \begin{align*} \tau_{\beta\alpha}^*\bigl(f_\beta\, du_1\wedge\cdots\wedge du_n\bigr) = (f_\beta \circ \tau_{\beta\alpha})\cdot \det(J\tau_{\beta\alpha})\, du_1\wedge\cdots\wedge du_n. \end{align*} Comparing the coefficients of $du_1\wedge\cdots\wedge du_n$ (a basis of $\Lambda^n T^*\mathbb{R}^n$, so coefficients are unique) at any point $u \in \tilde\varphi_\alpha(U_\alpha \cap U_\beta)$, \begin{align*} \det(J\tau_{\beta\alpha})(u) = \frac{f_\alpha(u)}{f_\beta(\tau_{\beta\alpha}(u))}. \end{align*} Both numerator and denominator are strictly positive, so $\det(J\tau_{\beta\alpha})(u) > 0$. Since $\alpha, \beta$ and $u$ were arbitrary, every transition map of $\tilde{\mathcal{A}}$ has everywhere-positive Jacobian determinant. By definition, $\tilde{\mathcal{A}}$ is an oriented atlas and $M$ is orientable. **The mechanism in one line:** a nowhere-vanishing top-form is a global object whose coordinate coefficients must compose by Jacobian determinants; matching positive coefficients on both sides forces positive Jacobians, which is the orientation condition. Combining both directions: $M$ is orientable if and only if $\Omega^n(M)$ contains a nowhere-vanishing element. $\square$[/guided]