[proofplan] The exterior algebra $\Lambda^*(\mathbb{R}^n)^* = \bigoplus_{k=0}^n \Lambda^k(\mathbb{R}^n)^*$ is, fibrewise, an associative graded-commutative $\mathbb{R}$-algebra with unit $1 \in \Lambda^0(\mathbb{R}^n)^* = \mathbb{R}$. Since the wedge product on $\Omega^*(U)$ is defined fibrewise, associativity, the unit law, and graded commutativity descend immediately from the pointwise statements once we verify that the wedge product of two smooth forms is again smooth. We do this by expanding both factors in the standard multi-index basis $\{dx_I\}_{|I|=k}$: the coefficients of $\alpha \wedge \beta$ in this basis are finite $\mathbb{R}$-linear combinations of products of the smooth coefficients of $\alpha$ and $\beta$, hence smooth. The $C^\infty(U)$-module structure and the compatibility identity follow from the bilinearity of the pointwise wedge and the commutativity of multiplication in $\mathbb{R}$. [/proofplan]

[step:Fix the pointwise model and the standard multi-index basis]Let $(e_1, \dots, e_n)$ be the standard basis of $\mathbb{R}^n$ and let $(dx_1, \dots, dx_n)$ denote its [dual basis](/theorems/414) in $(\mathbb{R}^n)^*$, characterised by $dx_i(e_j) = \delta_{ij}$. For each integer $k$ with $0 \le k \le n$, write \begin{align*} \mathcal{I}_k &:= \{\, I = (i_1, \dots, i_k) \in \mathbb{N}^k : 1 \le i_1 < i_2 < \dots < i_k \le n \,\} \end{align*} for the set of strictly increasing multi-indices of length $k$, and set $dx_I := dx_{i_1} \wedge \dots \wedge dx_{i_k} \in \Lambda^k(\mathbb{R}^n)^*$ for $I \in \mathcal{I}_k$; by convention $dx_\varnothing := 1 \in \Lambda^0(\mathbb{R}^n)^* = \mathbb{R}$. The family $\{dx_I\}_{I \in \mathcal{I}_k}$ is a basis of $\Lambda^k(\mathbb{R}^n)^*$, so any $\alpha \in \Omega^k(U)$ admits a unique pointwise decomposition \begin{align*} \alpha(x) &= \sum_{I \in \mathcal{I}_k} a_I(x)\, dx_I, \qquad x \in U, \end{align*} with scalar coefficients $a_I : U \to \mathbb{R}$. By definition, $\alpha$ is smooth as a map $U \to \Lambda^k(\mathbb{R}^n)^*$ if and only if each coefficient $a_I$ is smooth, i.e. $a_I \in C^\infty(U)$.[/step]

[guided]We first pin down precisely what an element of $\Omega^k(U)$ looks like, because the entire proof rests on writing forms in the coordinate basis and reading off coefficients. The exterior algebra $\Lambda^k(\mathbb{R}^n)^*$ is a finite-dimensional real [vector space](/page/Vector%20Space) of dimension $\binom{n}{k}$. A canonical basis is obtained by wedging the coordinate covectors $dx_i$ in strictly increasing order: for each strictly increasing multi-index $I = (i_1 < \dots < i_k)$ in $\mathcal{I}_k$, the element $dx_I = dx_{i_1} \wedge \dots \wedge dx_{i_k}$ lies in $\Lambda^k(\mathbb{R}^n)^*$, and the family $\{dx_I\}_{I \in \mathcal{I}_k}$ is a basis. A smooth $k$-form on $U$ is by definition a smooth map $\alpha: U \to \Lambda^k(\mathbb{R}^n)^*$. Because the codomain is a finite-dimensional [vector space](/page/Vector%20Space) with a fixed basis, smoothness of $\alpha$ is equivalent to smoothness of each coordinate of $\alpha$ in that basis: writing \begin{align*} \alpha(x) &= \sum_{I \in \mathcal{I}_k} a_I(x)\, dx_I, \qquad x \in U, \end{align*} we have $\alpha \in \Omega^k(U)$ if and only if $a_I \in C^\infty(U)$ for every $I \in \mathcal{I}_k$. This is the working description we use throughout.[/guided]

[step:Establish the pointwise exterior algebra facts to be inherited] For each $x \in U$, the space $\Lambda^*(\mathbb{R}^n)^* = \bigoplus_{k=0}^n \Lambda^k(\mathbb{R}^n)^*$ is the exterior algebra of $(\mathbb{R}^n)^*$. The wedge product \begin{align*} \wedge_x : \Lambda^p(\mathbb{R}^n)^* \times \Lambda^q(\mathbb{R}^n)^* &\to \Lambda^{p+q}(\mathbb{R}^n)^* \end{align*} is $\mathbb{R}$-bilinear and satisfies, for all $\xi, \xi' \in \Lambda^p(\mathbb{R}^n)^*$, $\eta \in \Lambda^q(\mathbb{R}^n)^*$, $\zeta \in \Lambda^r(\mathbb{R}^n)^*$, and $c \in \mathbb{R} = \Lambda^0(\mathbb{R}^n)^*$: \begin{align*} (\xi \wedge \eta) \wedge \zeta &= \xi \wedge (\eta \wedge \zeta) && \text{(associativity)}, \\ 1 \wedge \xi = \xi \wedge 1 &= \xi && \text{(unit)}, \\ \xi \wedge \eta &= (-1)^{pq}\, \eta \wedge \xi && \text{(graded commutativity)}, \\ c\,(\xi \wedge \eta) = (c\xi) \wedge \eta &= \xi \wedge (c\eta) && \text{($\mathbb{R}$-bilinearity in the scalar slot)}. \end{align*} These are the standard properties of the exterior algebra of a finite-dimensional real [vector space](/page/Vector%20Space). [/step]

[step:Show that the wedge product of two smooth forms is smooth]Let $\alpha \in \Omega^p(U)$ and $\beta \in \Omega^q(U)$, and write \begin{align*} \alpha(x) = \sum_{I \in \mathcal{I}_p} a_I(x)\, dx_I, \qquad \beta(x) = \sum_{J \in \mathcal{I}_q} b_J(x)\, dx_J, \end{align*} with $a_I, b_J \in C^\infty(U)$. Define $(\alpha \wedge \beta)(x) := \alpha(x) \wedge_x \beta(x)$. Using bilinearity of $\wedge_x$ and pulling scalars out, \begin{align*} (\alpha \wedge \beta)(x) &= \sum_{I \in \mathcal{I}_p}\sum_{J \in \mathcal{I}_q} a_I(x)\, b_J(x)\, \bigl( dx_I \wedge dx_J \bigr). \end{align*} The wedge $dx_I \wedge dx_J \in \Lambda^{p+q}(\mathbb{R}^n)^*$ is a fixed element (independent of $x$): it equals $0$ if $I \cap J \neq \varnothing$, and otherwise equals $\operatorname{sgn}(\sigma_{I,J})\, dx_K$ where $K \in \mathcal{I}_{p+q}$ is the strictly increasing rearrangement of $I \sqcup J$ and $\sigma_{I,J}$ is the permutation sorting $(I, J)$ into $K$. Re-indexing the double sum by $K \in \mathcal{I}_{p+q}$ gives \begin{align*} (\alpha \wedge \beta)(x) &= \sum_{K \in \mathcal{I}_{p+q}} c_K(x)\, dx_K, \qquad c_K(x) := \sum_{\substack{I \in \mathcal{I}_p,\, J \in \mathcal{I}_q \\ I \sqcup J = K \text{ (as sets)}}} \operatorname{sgn}(\sigma_{I,J})\, a_I(x)\, b_J(x). \end{align*} Each $c_K$ is a finite $\mathbb{R}$-linear combination of products of smooth functions on $U$; since $C^\infty(U)$ is closed under finite sums and products (it is a commutative $\mathbb{R}$-algebra under pointwise operations), $c_K \in C^\infty(U)$ for every $K \in \mathcal{I}_{p+q}$. By the criterion from the first step, $\alpha \wedge \beta \in \Omega^{p+q}(U)$. Hence \begin{align*} \wedge : \Omega^p(U) \times \Omega^q(U) &\to \Omega^{p+q}(U) \end{align*} is a well-defined operation, and extending it $\mathbb{R}$-bilinearly to the direct sum $\Omega^*(U) = \bigoplus_{k=0}^n \Omega^k(U)$ yields a graded product on $\Omega^*(U)$.[/step]

[guided]We want to show two things in this step: that the pointwise wedge $(\alpha \wedge \beta)(x) := \alpha(x) \wedge_x \beta(x)$ lands back in $\Omega^{p+q}(U)$ (i.e. the result is smooth), and that this gives a well-defined $\mathbb{R}$-bilinear graded product on the direct sum. The strategy is to expand both factors in the multi-index basis from the first step and read off the coefficients of the product. Write \begin{align*} \alpha(x) = \sum_{I \in \mathcal{I}_p} a_I(x)\, dx_I, \qquad \beta(x) = \sum_{J \in \mathcal{I}_q} b_J(x)\, dx_J, \end{align*} with $a_I, b_J \in C^\infty(U)$ (by the [basis criterion](/theorems/3308)). The pointwise wedge $\wedge_x$ is $\mathbb{R}$-bilinear, so distributing the sums and pulling out the scalar factors $a_I(x), b_J(x)$: \begin{align*} (\alpha \wedge \beta)(x) &= \sum_{I, J} a_I(x)\, b_J(x)\, \bigl(dx_I \wedge dx_J\bigr). \end{align*} The crucial observation is that $dx_I \wedge dx_J$ is a constant element of $\Lambda^{p+q}(\mathbb{R}^n)^*$ — it does not depend on $x$. We compute it: if $I$ and $J$ share an index, the wedge is $0$ by graded commutativity ($dx_i \wedge dx_i = 0$). Otherwise, $I \sqcup J$ is a set of $p+q$ distinct indices in $\{1, \dots, n\}$; reordering the wedge into strictly increasing order $K = (k_1 < \dots < k_{p+q})$ introduces a sign $\operatorname{sgn}(\sigma_{I,J})$ from the permutation that sorts $(I, J)$ to $K$, yielding $\operatorname{sgn}(\sigma_{I,J})\, dx_K$. Re-indexing the sum by the resulting $K$ gives \begin{align*} (\alpha \wedge \beta)(x) &= \sum_{K \in \mathcal{I}_{p+q}} c_K(x)\, dx_K, \end{align*} where \begin{align*} c_K(x) &= \sum_{\substack{I \in \mathcal{I}_p,\, J \in \mathcal{I}_q \\ I \sqcup J = K}} \operatorname{sgn}(\sigma_{I,J})\, a_I(x)\, b_J(x). \end{align*} Now we ask: is $c_K$ smooth? It is a finite sum (the multi-indices range over a finite set) of constant-signed products of two smooth functions on $U$. The space $C^\infty(U)$ is a commutative $\mathbb{R}$-algebra under pointwise operations: products and sums of smooth real-valued functions are smooth (this is the multivariable chain/product rule applied componentwise). Therefore each $c_K \in C^\infty(U)$, and by the [basis criterion](/theorems/3308) from the first step, $\alpha \wedge \beta \in \Omega^{p+q}(U)$. This makes $\wedge$ a well-defined map $\Omega^p(U) \times \Omega^q(U) \to \Omega^{p+q}(U)$. Extending it $\mathbb{R}$-bilinearly to $\Omega^*(U) = \bigoplus_k \Omega^k(U)$ — that is, defining the wedge of $\alpha = \sum_p \alpha_p$ and $\beta = \sum_q \beta_q$ as $\sum_{p,q} \alpha_p \wedge \beta_q$ — produces a graded product on the direct sum, which is the structure we want.[/guided]

[step:Inherit associativity, the unit law, and graded commutativity from the pointwise algebra]Let $\alpha \in \Omega^p(U)$, $\beta \in \Omega^q(U)$, $\gamma \in \Omega^r(U)$. By the definition of the wedge product on $\Omega^*(U)$, for every $x \in U$, \begin{align*} \bigl((\alpha \wedge \beta) \wedge \gamma\bigr)(x) &= \bigl(\alpha(x) \wedge_x \beta(x)\bigr) \wedge_x \gamma(x) \\ &= \alpha(x) \wedge_x \bigl(\beta(x) \wedge_x \gamma(x)\bigr) \\ &= \bigl(\alpha \wedge (\beta \wedge \gamma)\bigr)(x), \end{align*} where the middle equality uses associativity of $\wedge_x$ (the second step). Since this holds for every $x \in U$, the equality of smooth forms $(\alpha \wedge \beta) \wedge \gamma = \alpha \wedge (\beta \wedge \gamma)$ in $\Omega^{p+q+r}(U)$ follows. For the unit, observe that the constant function $1 \in C^\infty(U) = \Omega^0(U)$ satisfies $1(x) = 1 \in \mathbb{R} = \Lambda^0(\mathbb{R}^n)^*$ for every $x \in U$. The pointwise unit law gives, for every $\alpha \in \Omega^k(U)$ and every $x \in U$, \begin{align*} (1 \wedge \alpha)(x) = 1(x) \wedge_x \alpha(x) = 1 \wedge_x \alpha(x) = \alpha(x) = \alpha(x) \wedge_x 1 = (\alpha \wedge 1)(x), \end{align*} so $1 \wedge \alpha = \alpha = \alpha \wedge 1$ in $\Omega^k(U)$. For graded commutativity, let $\alpha \in \Omega^p(U)$ and $\beta \in \Omega^q(U)$. For every $x \in U$, the pointwise identity yields \begin{align*} (\alpha \wedge \beta)(x) &= \alpha(x) \wedge_x \beta(x) = (-1)^{pq}\, \beta(x) \wedge_x \alpha(x) = (-1)^{pq}\, (\beta \wedge \alpha)(x), \end{align*} and again equality at every $x$ gives the equality $\alpha \wedge \beta = (-1)^{pq}\, \beta \wedge \alpha$ in $\Omega^{p+q}(U)$. Combining the three identities with the well-definedness and $\mathbb{R}$-bilinearity established in the previous step, $(\Omega^*(U), +, \wedge)$ is an associative graded $\mathbb{R}$-algebra with unit $1 \in \Omega^0(U)$, and the graded commutativity relation holds. This proves (i) and (ii).[/step]

[guided]With the wedge product already known to be well-defined on $\Omega^*(U)$ and $\mathbb{R}$-bilinear, the algebraic axioms (associativity, unit, graded commutativity) reduce to pointwise checks. The slogan is: an equality of smooth maps $U \to \Lambda^{p+q}(\mathbb{R}^n)^*$ holds if and only if it holds at every point of $U$. For associativity, fix any $x \in U$ and unwind both sides at $x$: \begin{align*} \bigl((\alpha \wedge \beta) \wedge \gamma\bigr)(x) &= \bigl(\alpha(x) \wedge_x \beta(x)\bigr) \wedge_x \gamma(x), \\ \bigl(\alpha \wedge (\beta \wedge \gamma)\bigr)(x) &= \alpha(x) \wedge_x \bigl(\beta(x) \wedge_x \gamma(x)\bigr). \end{align*} The right-hand sides are equal by associativity of the pointwise exterior algebra $\Lambda^*(\mathbb{R}^n)^*$ (the second step). Equality at every $x$ gives the equality of smooth forms. For the unit, the constant function $1 \in C^\infty(U)$ is identified with the constant section $x \mapsto 1 \in \mathbb{R} = \Lambda^0(\mathbb{R}^n)^*$. The pointwise unit law $1 \wedge_x \xi = \xi \wedge_x 1 = \xi$ then immediately gives $1 \wedge \alpha = \alpha \wedge 1 = \alpha$ in $\Omega^*(U)$. For graded commutativity, the same pointwise principle applies: at each $x$, $\alpha(x) \wedge_x \beta(x) = (-1)^{pq}\, \beta(x) \wedge_x \alpha(x)$, and the sign $(-1)^{pq}$ is a constant independent of $x$, so it commutes with the pointwise evaluation. Equality at every $x$ promotes this to an equality in $\Omega^{p+q}(U)$. Note that the degree of each form is well-defined and is what determines the sign — that is precisely why we work in the graded decomposition $\Omega^*(U) = \bigoplus_k \Omega^k(U)$ rather than treating $\Omega^*(U)$ as a single ungraded [vector space](/page/Vector%20Space).[/guided]

[step:Establish the $C^\infty(U)$-module structure and the bilinearity identity]For $f \in C^\infty(U)$ and $\alpha \in \Omega^k(U)$, define $f\alpha : U \to \Lambda^k(\mathbb{R}^n)^*$ by $(f\alpha)(x) := f(x)\, \alpha(x)$, where the scalar multiplication on the right is the $\mathbb{R}$-action on the finite-dimensional [vector space](/page/Vector%20Space) $\Lambda^k(\mathbb{R}^n)^*$. Writing $\alpha = \sum_I a_I\, dx_I$ in the basis from the first step, \begin{align*} (f\alpha)(x) &= \sum_{I \in \mathcal{I}_k} \bigl(f(x)\, a_I(x)\bigr)\, dx_I, \end{align*} and each coefficient $f \cdot a_I$ is smooth on $U$ (product of two smooth functions in the $\mathbb{R}$-algebra $C^\infty(U)$). Hence $f\alpha \in \Omega^k(U)$. The module axioms — $1 \cdot \alpha = \alpha$, $(f+g)\alpha = f\alpha + g\alpha$, $f(\alpha + \alpha') = f\alpha + f\alpha'$, $(fg)\alpha = f(g\alpha)$ — all follow from the corresponding pointwise statements in the $\mathbb{R}$-[vector space](/page/Vector%20Space) $\Lambda^k(\mathbb{R}^n)^*$, verified at each $x \in U$. Thus $\Omega^k(U)$ is a $C^\infty(U)$-module. For the bilinearity identity, fix $f \in C^\infty(U)$, $\alpha \in \Omega^p(U)$, $\beta \in \Omega^q(U)$. By the third pointwise property of step two ($\mathbb{R}$-bilinearity in the scalar slot of $\wedge_x$), for every $x \in U$, \begin{align*} f(x)\, \bigl(\alpha(x) \wedge_x \beta(x)\bigr) &= \bigl(f(x)\, \alpha(x)\bigr) \wedge_x \beta(x) = \alpha(x) \wedge_x \bigl(f(x)\, \beta(x)\bigr). \end{align*} Reading off both sides as smooth forms on $U$ via the definitions of $f(\alpha \wedge \beta)$, $(f\alpha) \wedge \beta$, and $\alpha \wedge (f\beta)$ gives \begin{align*} f\, (\alpha \wedge \beta) = (f\alpha) \wedge \beta = \alpha \wedge (f\beta) \quad \text{in } \Omega^{p+q}(U). \end{align*} This proves (iii) and completes the proof.[/step]

[guided]The $C^\infty(U)$-action and the bilinearity identity are again pointwise consequences, with one extra ingredient: we need to check that multiplying a smooth form by a smooth function produces a smooth form. Define $(f\alpha)(x) := f(x)\,\alpha(x)$, using the scalar multiplication on the finite-dimensional [vector space](/page/Vector%20Space) $\Lambda^k(\mathbb{R}^n)^*$. Why is $f\alpha$ smooth? Expand $\alpha$ in the multi-index basis: \begin{align*} (f\alpha)(x) &= \sum_{I \in \mathcal{I}_k} \bigl(f(x)\, a_I(x)\bigr)\, dx_I. \end{align*} The new coefficients are pointwise products $f \cdot a_I$ of smooth functions on $U$. Since $C^\infty(U)$ is a commutative $\mathbb{R}$-algebra under pointwise operations, $f \cdot a_I \in C^\infty(U)$, so by the [basis criterion](/theorems/3308) $f\alpha \in \Omega^k(U)$. The module axioms ($1 \cdot \alpha = \alpha$; distributivity in both arguments; $(fg)\alpha = f(g\alpha)$) are straightforward translations of the corresponding statements in the $\mathbb{R}$-[vector space](/page/Vector%20Space) $\Lambda^k(\mathbb{R}^n)^*$, checked at each $x \in U$. For the compatibility identity $f(\alpha \wedge \beta) = (f\alpha) \wedge \beta = \alpha \wedge (f\beta)$, fix $x \in U$. The pointwise wedge $\wedge_x$ is $\mathbb{R}$-bilinear; in particular, scalars (elements of $\mathbb{R} = \Lambda^0(\mathbb{R}^n)^*$) can be pulled into either factor: \begin{align*} f(x)\, \bigl(\alpha(x) \wedge_x \beta(x)\bigr) &= \bigl(f(x)\, \alpha(x)\bigr) \wedge_x \beta(x) = \alpha(x) \wedge_x \bigl(f(x)\, \beta(x)\bigr). \end{align*} By the definitions of $f(\alpha \wedge \beta)$, $(f\alpha) \wedge \beta$, and $\alpha \wedge (f\beta)$, the three displayed expressions are the pointwise evaluations at $x$ of these three smooth forms. Equality at every $x \in U$ promotes to equality in $\Omega^{p+q}(U)$, which is (iii). A subtle point worth flagging: we used that scalars can be pulled across $\wedge_x$ from either factor. This is special to scalar factors, i.e. degree-$0$ elements. For higher-degree forms, moving a factor across the wedge incurs the sign $(-1)^{pq}$ from graded commutativity; but a degree-$0$ factor gives sign $(-1)^{0 \cdot q} = 1$, so no sign appears. This is what makes $C^\infty(U)$ behave as a true (ungraded) scalar ring acting on each $\Omega^k(U)$. Combining (i), (ii), and (iii), $(\Omega^*(U), +, \wedge)$ is an associative graded-commutative $\mathbb{R}$-algebra with unit, each graded piece is a $C^\infty(U)$-module, and the wedge product is $C^\infty(U)$-bilinear. This completes the proof.[/guided]