[proofplan] We prove the two implications by translating orientation data into positive generators of the determinant line. Starting from an oriented trivializing atlas, the local top exterior products agree by positive transition factors; a smooth [partition of unity](/page/Partition%20of%20Unity) lets us average these local positive generators without cancellation. Conversely, a nowhere-vanishing determinant section decides which frames are positive, and local nonvanishing of its coefficient forces the resulting transition determinants to be positive. The final observation identifies sections giving the same orientation exactly up to multiplication by a positive smooth function. [/proofplan]

[step:Construct local positive generators from an oriented atlas] Assume $E$ is orientable. Choose an oriented trivializing atlas $\{(U_i,e_i)\}_{i\in I}$ for $E$, where each \begin{align*} e_i=(e_{i,1},\dots,e_{i,r}) \end{align*} is a smooth local frame of $E$ over $U_i$, and where every transition matrix between these frames has positive determinant. For each $i\in I$, define the smooth local section $\omega_i:U_i\to \det E$ by \begin{align*} \omega_i(x)=e_{i,1}(x)\wedge \cdots \wedge e_{i,r}(x). \end{align*} If $U_i\cap U_j\neq \varnothing$, let \begin{align*} A_{ij}:U_i\cap U_j&\to GL(r,\mathbb{R}) \end{align*} be the smooth transition map determined by \begin{align*} (e_{i,1},\dots,e_{i,r})=(e_{j,1},\dots,e_{j,r})A_{ij}. \end{align*} By multilinearity and alternation of the exterior product, \begin{align*} \omega_i(x)=\det(A_{ij}(x))\,\omega_j(x) \end{align*} for every $x\in U_i\cap U_j$. Since the atlas is oriented, $\det(A_{ij}(x))>0$ on $U_i\cap U_j$. [/step]

[step:Average the local generators without cancellation]Because $M$ is paracompact, choose a locally finite smooth partition of unity $\{\rho_i:M\to [0,1]\}_{i\in I}$ subordinate to the cover $\{U_i\}_{i\in I}$, with $\operatorname{supp}\rho_i\subset U_i$ after refinement if necessary. This uses the standard existence theorem for smooth partitions of unity subordinate to locally finite smooth covers (citing a result not yet in the wiki: smooth partition of unity). For each $i\in I$, define the smooth global section \begin{align*} \widetilde{\omega}_i:M&\to \det E \end{align*} by For $x\in U_i$, set $\widetilde{\omega}_i(x)=\rho_i(x)\omega_i(x)$, and for $x\notin \operatorname{supp}\rho_i$, set $\widetilde{\omega}_i(x)=0$. This is smooth because $\rho_i$ vanishes on a neighbourhood of $M\setminus U_i$. Since the family is locally finite, the pointwise sum defines a map $\sigma:M\to \det E$ by \begin{align*} \sigma(x)=\sum_{i\in I}\widetilde{\omega}_i(x). \end{align*} This is a well-defined smooth section. Fix $x\in M$. Choose $j\in I$ with $\rho_j(x)>0$, which exists because $\sum_i\rho_i(x)=1$. For every $i$ with $\rho_i(x)>0$, we have $x\in U_i\cap U_j$, hence \begin{align*} \omega_i(x)=\det(A_{ij}(x))\,\omega_j(x) \end{align*} with $\det(A_{ij}(x))>0$. Therefore \begin{align*} \sigma(x)=\left(\sum_{i\in I}\rho_i(x)\det(A_{ij}(x))\right)\omega_j(x), \end{align*} where the sum is finite, every summand is nonnegative, and the $j$-summand is positive. Thus the coefficient of $\omega_j(x)$ is strictly positive, so $\sigma(x)\neq 0$. Hence $\det E$ admits a nowhere-vanishing smooth section.[/step]

[guided]The goal is to turn many local positive generators of the line bundle $\det E$ into one global generator. The only possible obstruction is cancellation: if two local generators pointed in opposite directions in the same one-dimensional fibre, their sum could vanish. The oriented atlas prevents this because all local generators lie in the same positive ray on overlaps. Since $M$ is paracompact, we choose a locally finite smooth partition of unity \begin{align*} \{\rho_i:M\to [0,1]\}_{i\in I} \end{align*} subordinate to the oriented trivializing cover $\{U_i\}_{i\in I}$, with $\operatorname{supp}\rho_i\subset U_i$ after refinement. This invokes the standard [smooth partition of unity theorem](/theorems/3917) for paracompact smooth manifolds (citing a result not yet in the wiki: smooth partition of unity). The local finiteness is essential because it makes the sum of sections locally finite and therefore smooth. For each index $i\in I$, define a global section \begin{align*} \widetilde{\omega}_i:M&\to \det E \end{align*} by multiplying the local positive generator $\omega_i$ by the cutoff function $\rho_i$ on $U_i$ and extending by zero outside the support: For $x\in U_i$, set $\widetilde{\omega}_i(x)=\rho_i(x)\omega_i(x)$, and for $x\notin \operatorname{supp}\rho_i$, set $\widetilde{\omega}_i(x)=0$. This extension is smooth because $\rho_i$ vanishes on a neighbourhood of $M\setminus U_i$. Now define $\sigma:M\to \det E$ by \begin{align*} \sigma(x)=\sum_{i\in I}\widetilde{\omega}_i(x). \end{align*} The sum is locally finite, so it defines a smooth section of $\det E$. It remains to prove that $\sigma$ never vanishes. Fix $x\in M$. Since the partition of unity sums to $1$, there is an index $j\in I$ such that $\rho_j(x)>0$. For every $i\in I$ with $\rho_i(x)>0$, the point $x$ lies in $U_i\cap U_j$. On this overlap the oriented transition matrix $A_{ij}(x)\in GL(r,\mathbb{R})$ satisfies $\det(A_{ij}(x))>0$, and the corresponding determinant generators satisfy \begin{align*} \omega_i(x)=\det(A_{ij}(x))\,\omega_j(x). \end{align*} Therefore the value of $\sigma$ at $x$ can be written in the single basis vector $\omega_j(x)$ of the one-dimensional [vector space](/page/Vector%20Space) $\det E_x$: \begin{align*} \sigma(x)=\left(\sum_{i\in I}\rho_i(x)\det(A_{ij}(x))\right)\omega_j(x). \end{align*} Only finitely many terms occur. Each coefficient $\rho_i(x)\det(A_{ij}(x))$ is nonnegative, and the term with $i=j$ is positive because $\rho_j(x)>0$ and $\det(A_{jj}(x))=1$. Hence the scalar coefficient is strictly positive. Since $\omega_j(x)\neq 0$ in $\det E_x$, we get $\sigma(x)\neq 0$. Thus the determinant line bundle has a nowhere-vanishing smooth section.[/guided]

[step:Use a nonvanishing determinant section to choose positive frames] Conversely, suppose $\sigma:M\to \det E$ is a nowhere-vanishing smooth section. Let $(U,e)$ be any smooth local trivialization of $E$, with frame \begin{align*} e=(e_1,\dots,e_r). \end{align*} Define $\omega_e:U\to \det E$ by \begin{align*} \omega_e(x)=e_1(x)\wedge\cdots\wedge e_r(x). \end{align*} Since $\omega_e$ is a local frame of the line bundle $\det E$, there is a unique smooth function \begin{align*} a_e:U&\to \mathbb{R} \end{align*} such that \begin{align*} \sigma(x)=a_e(x)\omega_e(x) \end{align*} for every $x\in U$. Since $\sigma$ is nowhere zero and $\omega_e(x)\neq 0$, we have $a_e(x)\neq 0$ for every $x\in U$. Refine the trivializing cover so that each [open set](/page/Open%20Set) is connected. On each connected member, the smooth function $a_e:U\to \mathbb{R}\setminus\{0\}$ has constant sign. If $a_e>0$, keep the frame $e$. If $a_e<0$, replace it by \begin{align*} e'=(-e_1,e_2,\dots,e_r). \end{align*} Then \begin{align*} \omega_{e'}=-\omega_e \end{align*} and therefore \begin{align*} \sigma=(-a_e)\omega_{e'} \end{align*} with $-a_e>0$. Thus, after this refinement and possible sign change in the first frame vector, we obtain local frames whose determinant generators are positive multiples of $\sigma$. [/step]

[step:Verify that the selected frames have positive transition determinants] Let $e=(e_1,\dots,e_r)$ over $U$ and $f=(f_1,\dots,f_r)$ over $V$ be two of the local frames constructed above. Define \begin{align*} \omega_e:U&\to \det E, & \omega_f:V&\to \det E \end{align*} by taking the top exterior product of each frame. By construction, there are smooth functions \begin{align*} a:U&\to \mathbb{R}_{+}, & b:V&\to \mathbb{R}_{+} \end{align*} such that \begin{align*} \sigma=a\omega_e \quad \text{on } U, \qquad \sigma=b\omega_f \quad \text{on } V. \end{align*} On $U\cap V$, let \begin{align*} A:U\cap V&\to GL(r,\mathbb{R}) \end{align*} be the transition map determined by \begin{align*} (e_1,\dots,e_r)=(f_1,\dots,f_r)A. \end{align*} Then \begin{align*} \omega_e=\det(A)\omega_f. \end{align*} Using the two expressions for $\sigma$ on $U\cap V$ gives \begin{align*} a\,\det(A)\omega_f=b\omega_f. \end{align*} Since $\omega_f$ is nowhere zero and $a,b>0$, we obtain \begin{align*} \det(A)=\frac{b}{a}>0. \end{align*} Thus all transition determinants between the selected frames are positive. These frames form an oriented atlas, so $E$ is orientable. [/step]

[step:Identify when two determinant sections define the same orientation] Let $\sigma_1,\sigma_2:M\to \det E$ be nowhere-vanishing smooth sections. If $\sigma_2=f\sigma_1$ for a positive smooth function \begin{align*} f:M&\to \mathbb{R}_{+}, \end{align*} then a frame has top exterior product a positive multiple of $\sigma_1$ exactly when it has top exterior product a positive multiple of $\sigma_2$, so the two sections define the same orientation. Conversely, suppose $\sigma_1$ and $\sigma_2$ define the same orientation. Since $\det E$ is a line bundle and $\sigma_1$ is nowhere zero, there is a unique smooth function \begin{align*} f:M&\to \mathbb{R}\setminus\{0\} \end{align*} such that $\sigma_2=f\sigma_1$. In any positively oriented local frame $e=(e_1,\dots,e_r)$, both sections are positive multiples of \begin{align*} \omega_e=e_1\wedge\cdots\wedge e_r. \end{align*} Hence $f>0$ locally, and therefore $f:M\to\mathbb{R}_{+}$ globally. This proves the stated equivalence modulo positive smooth functions and completes the proof. [/step]