[proofplan] We prove the identity pointwise. At a point $p \in M$, both sides are alternating $(k+\ell)$-covectors on $T_pM$, so it suffices to evaluate them on arbitrary tangent vectors $v_1,\dots,v_{k+\ell} \in T_pM$. The definition of pullback sends each input vector through the differential $d\varphi_p$, and the definition of the wedge product is the same antisymmetrised sum on both sides. [/proofplan] [step:Evaluate the pullback of the wedge product at an arbitrary point] Fix a point $p \in M$ and tangent vectors $v_1,\dots,v_{k+\ell} \in T_pM$. Let \begin{align*} d\varphi_p: T_pM \to T_{\varphi(p)}N \end{align*} be the differential of $\varphi$ at $p$. Let $\operatorname{Sh}(k,\ell)$ denote the set of $(k,\ell)$-shuffles, that is, the set of permutations $\sigma$ of $\{1,\dots,k+\ell\}$ such that \begin{align*} \sigma(1) < \cdots < \sigma(k), \qquad \sigma(k+1) < \cdots < \sigma(k+\ell). \end{align*} For $\sigma \in \operatorname{Sh}(k,\ell)$, let $\operatorname{sgn}(\sigma) \in \{-1,1\}$ denote its sign. Using the definition of pullback for the $(k+\ell)$-form $\alpha \wedge \beta$, we obtain \begin{align*} \bigl(\varphi^*(\alpha \wedge \beta)\bigr)_p(v_1,\dots,v_{k+\ell}) &= (\alpha \wedge \beta)_{\varphi(p)} \bigl(d\varphi_p(v_1),\dots,d\varphi_p(v_{k+\ell})\bigr). \end{align*} Applying the shuffle definition of the wedge product gives \begin{align*} \bigl(\varphi^*(\alpha \wedge \beta)\bigr)_p(v_1,\dots,v_{k+\ell}) &= \sum_{\sigma \in \operatorname{Sh}(k,\ell)} \operatorname{sgn}(\sigma)\, \alpha_{\varphi(p)} \bigl(d\varphi_p(v_{\sigma(1)}),\dots,d\varphi_p(v_{\sigma(k)})\bigr) \\ &\qquad\qquad\cdot \beta_{\varphi(p)} \bigl(d\varphi_p(v_{\sigma(k+1)}),\dots,d\varphi_p(v_{\sigma(k+\ell)})\bigr). \end{align*} [guided] We want to compare two differential forms on $M$. Since differential forms are defined pointwise as alternating covectors, we fix a point $p \in M$ and arbitrary tangent vectors $v_1,\dots,v_{k+\ell} \in T_pM$. The pullback of a differential form is defined by applying the differential of the map to each input vector. For the smooth map $\varphi: M \to N$, the relevant [linear map](/page/Linear%20Map) at $p$ is \begin{align*} d\varphi_p: T_pM \to T_{\varphi(p)}N. \end{align*} Therefore \begin{align*} \bigl(\varphi^*(\alpha \wedge \beta)\bigr)_p(v_1,\dots,v_{k+\ell}) &= (\alpha \wedge \beta)_{\varphi(p)} \bigl(d\varphi_p(v_1),\dots,d\varphi_p(v_{k+\ell})\bigr). \end{align*} Now expand the wedge product. Let $\operatorname{Sh}(k,\ell)$ be the set of permutations $\sigma$ of $\{1,\dots,k+\ell\}$ preserving the order of the first $k$ selected inputs and the last $\ell$ selected inputs: \begin{align*} \sigma(1) < \cdots < \sigma(k), \qquad \sigma(k+1) < \cdots < \sigma(k+\ell). \end{align*} For each such permutation, $\operatorname{sgn}(\sigma)$ denotes its sign. The [shuffle formula for the wedge product](/theorems/3558) gives \begin{align*} \bigl(\varphi^*(\alpha \wedge \beta)\bigr)_p(v_1,\dots,v_{k+\ell}) &= \sum_{\sigma \in \operatorname{Sh}(k,\ell)} \operatorname{sgn}(\sigma)\, \alpha_{\varphi(p)} \bigl(d\varphi_p(v_{\sigma(1)}),\dots,d\varphi_p(v_{\sigma(k)})\bigr) \\ &\qquad\qquad\cdot \beta_{\varphi(p)} \bigl(d\varphi_p(v_{\sigma(k+1)}),\dots,d\varphi_p(v_{\sigma(k+\ell)})\bigr). \end{align*} This is the expression we will compare with the wedge product of the two pulled-back forms. [/guided] [/step] [step:Expand the wedge product of the pulled-back forms] Using the shuffle definition of the wedge product on $T_pM$, we have \begin{align*} \bigl(\varphi^*\alpha \wedge \varphi^*\beta\bigr)_p(v_1,\dots,v_{k+\ell}) &= \sum_{\sigma \in \operatorname{Sh}(k,\ell)} \operatorname{sgn}(\sigma)\, (\varphi^*\alpha)_p(v_{\sigma(1)},\dots,v_{\sigma(k)}) \\ &\qquad\qquad\cdot (\varphi^*\beta)_p(v_{\sigma(k+1)},\dots,v_{\sigma(k+\ell)}). \end{align*} By the definition of pullback for the $k$-form $\alpha$ and the $\ell$-form $\beta$, \begin{align*} (\varphi^*\alpha)_p(v_{\sigma(1)},\dots,v_{\sigma(k)}) &= \alpha_{\varphi(p)} \bigl(d\varphi_p(v_{\sigma(1)}),\dots,d\varphi_p(v_{\sigma(k)})\bigr), \\ (\varphi^*\beta)_p(v_{\sigma(k+1)},\dots,v_{\sigma(k+\ell)}) &= \beta_{\varphi(p)} \bigl(d\varphi_p(v_{\sigma(k+1)}),\dots,d\varphi_p(v_{\sigma(k+\ell)})\bigr). \end{align*} Substituting these identities into the preceding sum gives \begin{align*} \bigl(\varphi^*\alpha \wedge \varphi^*\beta\bigr)_p(v_1,\dots,v_{k+\ell}) &= \sum_{\sigma \in \operatorname{Sh}(k,\ell)} \operatorname{sgn}(\sigma)\, \alpha_{\varphi(p)} \bigl(d\varphi_p(v_{\sigma(1)}),\dots,d\varphi_p(v_{\sigma(k)})\bigr) \\ &\qquad\qquad\cdot \beta_{\varphi(p)} \bigl(d\varphi_p(v_{\sigma(k+1)}),\dots,d\varphi_p(v_{\sigma(k+\ell)})\bigr). \end{align*} [guided] Now we compute the right-hand side. The wedge product $\varphi^*\alpha \wedge \varphi^*\beta$ is formed after pulling back each form to $M$, so its value at $p$ is computed using tangent vectors in $T_pM$: \begin{align*} \bigl(\varphi^*\alpha \wedge \varphi^*\beta\bigr)_p(v_1,\dots,v_{k+\ell}) &= \sum_{\sigma \in \operatorname{Sh}(k,\ell)} \operatorname{sgn}(\sigma)\, (\varphi^*\alpha)_p(v_{\sigma(1)},\dots,v_{\sigma(k)}) \\ &\qquad\qquad\cdot (\varphi^*\beta)_p(v_{\sigma(k+1)},\dots,v_{\sigma(k+\ell)}). \end{align*} Each pulled-back factor is evaluated by applying $d\varphi_p$ to each input vector before evaluating the original form on $N$. Thus \begin{align*} (\varphi^*\alpha)_p(v_{\sigma(1)},\dots,v_{\sigma(k)}) &= \alpha_{\varphi(p)} \bigl(d\varphi_p(v_{\sigma(1)}),\dots,d\varphi_p(v_{\sigma(k)})\bigr), \\ (\varphi^*\beta)_p(v_{\sigma(k+1)},\dots,v_{\sigma(k+\ell)}) &= \beta_{\varphi(p)} \bigl(d\varphi_p(v_{\sigma(k+1)}),\dots,d\varphi_p(v_{\sigma(k+\ell)})\bigr). \end{align*} Substitution gives \begin{align*} \bigl(\varphi^*\alpha \wedge \varphi^*\beta\bigr)_p(v_1,\dots,v_{k+\ell}) &= \sum_{\sigma \in \operatorname{Sh}(k,\ell)} \operatorname{sgn}(\sigma)\, \alpha_{\varphi(p)} \bigl(d\varphi_p(v_{\sigma(1)}),\dots,d\varphi_p(v_{\sigma(k)})\bigr) \\ &\qquad\qquad\cdot \beta_{\varphi(p)} \bigl(d\varphi_p(v_{\sigma(k+1)}),\dots,d\varphi_p(v_{\sigma(k+\ell)})\bigr). \end{align*} This is exactly the same shuffle sum obtained from pulling back $\alpha \wedge \beta$ directly. [/guided] [/step] [step:Conclude equality of the differential forms] The two computations give the same value for every $p \in M$ and every $v_1,\dots,v_{k+\ell} \in T_pM$: \begin{align*} \bigl(\varphi^*(\alpha \wedge \beta)\bigr)_p(v_1,\dots,v_{k+\ell}) = \bigl(\varphi^*\alpha \wedge \varphi^*\beta\bigr)_p(v_1,\dots,v_{k+\ell}). \end{align*} Therefore the two alternating $(k+\ell)$-covectors agree at each point $p \in M$. Hence \begin{align*} \varphi^*(\alpha \wedge \beta) = \varphi^*\alpha \wedge \varphi^*\beta \end{align*} as elements of $\Omega^{k+\ell}(M)$. [/step]