[proofplan] We construct an $R$-orientation by pulling back a fixed generator of $H^d(\mathbb{R}^d, \mathbb{R}^d \setminus \{0\}; R)$ through each local trivialization. The key step is to show that when two trivializations overlap, the transition map — which has positive determinant by hypothesis — acts as the identity on fiber cohomology. This follows because $\mathrm{GL}_d^+(\mathbb{R})$ is path-connected, so any positive-determinant linear map is homotopic to the identity and therefore induces the identity on cohomology. [/proofplan] [step:Fix a generator of the fiber cohomology] The pair $(\mathbb{R}^d, \mathbb{R}^d \setminus \{0\})$ has cohomology \begin{align*} H^i(\mathbb{R}^d, \mathbb{R}^d \setminus \{0\}; R) \cong \begin{cases} R & i = d, \\ 0 & \text{otherwise.} \end{cases} \end{align*} Fix a generator $u \in H^d(\mathbb{R}^d, \mathbb{R}^d \setminus \{0\}; R)$. This generator corresponds to a choice of orientation for $\mathbb{R}^d$ over $R$. [/step] [step:Define the local orientation at each fiber using trivializations] Let $\{U_\alpha\}_{\alpha \in I}$ be the given trivializing cover with trivializations $\varphi_\alpha: E|_{U_\alpha} \xrightarrow{\sim} U_\alpha \times \mathbb{R}^d$. For each $\alpha \in I$ and $x \in U_\alpha$, the trivialization restricts to a linear isomorphism on fibers: \begin{align*} (\varphi_\alpha)_x: E_x \xrightarrow{\sim} \{x\} \times \mathbb{R}^d \cong \mathbb{R}^d. \end{align*} Define $\varepsilon_x^\alpha \in H^d(E_x, E_x^\#; R)$ to be the pullback of $u$ along $(\varphi_\alpha)_x$: \begin{align*} \varepsilon_x^\alpha := ((\varphi_\alpha)_x)^*(u). \end{align*} Since $(\varphi_\alpha)_x$ is a linear isomorphism, it induces an isomorphism on relative cohomology, so $\varepsilon_x^\alpha$ is a generator of $H^d(E_x, E_x^\#; R) \cong R$. [guided] We want to assign to each fiber $E_x$ a generator $\varepsilon_x$ of the relative cohomology group $H^d(E_x, E_x^\#; R) \cong R$. Each trivialization $\varphi_\alpha$ provides a canonical way to do this: identify $E_x$ with $\mathbb{R}^d$ and pull back the fixed generator $u$. The question is whether different trivializations produce the same generator — this is where the positive-determinant condition will enter. [/guided] [/step] [step:Show the construction is independent of the trivialization on overlaps] Let $x \in U_\alpha \cap U_\beta$. The transition homeomorphism $g_{\alpha\beta}(x) := (\varphi_\alpha)_x \circ (\varphi_\beta)_x^{-1}: \mathbb{R}^d \to \mathbb{R}^d$ is a linear isomorphism with $\det(g_{\alpha\beta}(x)) > 0$ by hypothesis. The two candidate generators at $x$ are related by \begin{align*} \varepsilon_x^\beta = ((\varphi_\beta)_x)^*(u) = ((\varphi_\alpha)_x^{-1} \circ g_{\alpha\beta}(x))^*(u) = ((\varphi_\alpha)_x)^* \circ g_{\alpha\beta}(x)^*(u). \end{align*} It therefore suffices to show that $g_{\alpha\beta}(x)^*(u) = u$. [guided] The generators $\varepsilon_x^\alpha$ and $\varepsilon_x^\beta$ differ by the action of the transition map $g_{\alpha\beta}(x)$ on the cohomology $H^d(\mathbb{R}^d, \mathbb{R}^d \setminus \{0\}; R)$. We need to show this action is trivial — i.e., $g_{\alpha\beta}(x)$ does not change the generator. This is where we use the positive-determinant hypothesis. [/guided] [/step] [step:Use path-connectedness of $\mathrm{GL}_d^+(\mathbb{R})$ to conclude $g_{\alpha\beta}(x)^* = \operatorname{id}$] Since $\det(g_{\alpha\beta}(x)) > 0$, the linear map $g_{\alpha\beta}(x)$ lies in $\mathrm{GL}_d^+(\mathbb{R})$. The group $\mathrm{GL}_d^+(\mathbb{R})$ is path-connected: for any $A \in \mathrm{GL}_d^+(\mathbb{R})$, there exists a continuous path \begin{align*} \gamma: [0, 1] &\to \mathrm{GL}_d^+(\mathbb{R}), \quad \gamma(0) = I, \quad \gamma(1) = A. \end{align*} Composing with the map $A \mapsto A^*: H^d(\mathbb{R}^d, \mathbb{R}^d \setminus \{0\}; R) \to H^d(\mathbb{R}^d, \mathbb{R}^d \setminus \{0\}; R)$ gives a continuous family of $R$-module homomorphisms from $R$ to $R$, connecting $\operatorname{id}^* = \operatorname{id}$ to $g_{\alpha\beta}(x)^*$. Since homotopic maps of pairs induce equal maps on cohomology, we conclude \begin{align*} g_{\alpha\beta}(x)^*(u) = \operatorname{id}^*(u) = u. \end{align*} Therefore $\varepsilon_x^\alpha = \varepsilon_x^\beta$ for all $x \in U_\alpha \cap U_\beta$. Writing $\varepsilon_x := \varepsilon_x^\alpha$ for any $\alpha$ with $x \in U_\alpha$, we obtain a well-defined family $\{\varepsilon_x\}_{x \in X}$. [guided] Why is $\mathrm{GL}_d^+(\mathbb{R})$ path-connected? Every matrix $A \in \mathrm{GL}_d^+(\mathbb{R})$ can be connected to $I$ by first using polar decomposition to connect $A$ to a rotation (the path $t \mapsto (1-t)A + t R$ stays in $\mathrm{GL}_d^+(\mathbb{R})$ when done carefully via the positive-definite factor), and then using the fact that $\mathrm{SO}(d)$ is connected to reach $I$. The crucial point is that $\mathrm{GL}_d^-(\mathbb{R})$ (negative determinant) is the other connected component of $\mathrm{GL}_d(\mathbb{R})$. A map $A$ with $\det A < 0$ would reverse the generator ($A^*(u) = -u$), which is precisely why negative-determinant transition functions obstruct orientability over $\mathbb{Z}$ (and over any ring where $1 \neq -1$). The positive-determinant hypothesis ensures we stay in the identity component, so the induced map on cohomology is trivially the identity. [/guided] [/step] [step:Verify local compatibility and conclude $R$-orientability] It remains to check the local compatibility condition. For each $\alpha$, the relative Künneth theorem for the trivial bundle $U_\alpha \times \mathbb{R}^d$ provides a class \begin{align*} \varepsilon_{U_\alpha} \in H^d(E|_{U_\alpha}, E|_{U_\alpha}^\#; R) \end{align*} that restricts to $\varepsilon_x = \varepsilon_x^\alpha$ at each fiber $x \in U_\alpha$. Since the family $\{\varepsilon_x\}_{x \in X}$ is well-defined (independent of the trivialization by the previous step) and locally compatible over each $U_\alpha$, the bundle $E$ is $R$-orientable for every commutative ring $R$. [/step]