[proofplan] We prove exactness at each term directly from the definitions of restriction and gluing of smooth differential forms. Injectivity says that a form on $M$ is determined by its restrictions to the open cover $U,V$. Exactness in the middle is the local-to-global gluing property for smooth forms: two local forms agreeing on $U \cap V$ assemble uniquely into one global form. Surjectivity of $s_k$ is obtained by multiplying a given form on $U \cap V$ by a smooth [partition of unity](/page/Partition%20of%20Unity) subordinate to $U,V$ and extending the resulting pieces by zero. Finally, commutation with the [exterior derivative](/theorems/1525) follows from the locality of $d$. [/proofplan] [step:Define the maps and verify that their composition is zero] Fix an integer $k \geq 0$. The maps are \begin{align*} r_k: \Omega^k(M) &\to \Omega^k(U) \oplus \Omega^k(V), & \omega &\mapsto (\omega|_U,\omega|_V), \end{align*} and \begin{align*} s_k: \Omega^k(U) \oplus \Omega^k(V) &\to \Omega^k(U \cap V), & (\alpha,\beta) &\mapsto \alpha|_{U \cap V} - \beta|_{U \cap V}. \end{align*} Both maps are linear over $\mathbb{R}$ because restriction of differential forms is linear. For $\omega \in \Omega^k(M)$, we compute \begin{align*} (s_k \circ r_k)(\omega) &= s_k(\omega|_U,\omega|_V) \\ &= (\omega|_U)|_{U \cap V} - (\omega|_V)|_{U \cap V} \\ &= \omega|_{U \cap V} - \omega|_{U \cap V} \\ &= 0. \end{align*} Thus $\operatorname{im} r_k \subseteq \ker s_k$. [/step] [step:Prove that restriction to the open cover is injective] Let $\omega \in \Omega^k(M)$ satisfy $r_k(\omega) = 0$. Then $\omega|_U = 0$ and $\omega|_V = 0$. Since $M = U \cup V$, every point $p \in M$ lies in $U$ or in $V$. Therefore $\omega_p = 0$ at every $p \in M$, so $\omega = 0$ in $\Omega^k(M)$. Hence $r_k$ is injective. [/step] [step:Glue local forms that agree on the overlap] Let $(\alpha,\beta) \in \Omega^k(U) \oplus \Omega^k(V)$ satisfy $s_k(\alpha,\beta)=0$. This means \begin{align*} \alpha|_{U \cap V}=\beta|_{U \cap V}. \end{align*} Define a $k$-form $\omega$ on $M$ pointwise by \begin{align*} \omega_p = \begin{cases} \alpha_p, & p \in U,\\ \beta_p, & p \in V. \end{cases} \end{align*} This definition is independent of the choice of [open set](/page/Open%20Set) containing $p$: if $p \in U \cap V$, then $\alpha_p=\beta_p$ by the displayed equality. The form $\omega$ is smooth because smoothness of a differential form is local on the source manifold. On the open set $U$, its restriction is $\alpha$, and on the open set $V$, its restriction is $\beta$. Hence $\omega \in \Omega^k(M)$ and \begin{align*} r_k(\omega)=(\alpha,\beta). \end{align*} Thus $\ker s_k \subseteq \operatorname{im} r_k$. Together with $\operatorname{im} r_k \subseteq \ker s_k$, this proves \begin{align*} \operatorname{im} r_k=\ker s_k. \end{align*} [guided] Suppose $(\alpha,\beta) \in \Omega^k(U) \oplus \Omega^k(V)$ lies in $\ker s_k$. By definition of $s_k$, this says \begin{align*} 0=s_k(\alpha,\beta)=\alpha|_{U \cap V}-\beta|_{U \cap V}. \end{align*} Equivalently, \begin{align*} \alpha|_{U \cap V}=\beta|_{U \cap V}. \end{align*} We now define the candidate global form. For each point $p \in M$, choose the local expression available at $p$: \begin{align*} \omega_p = \begin{cases} \alpha_p, & p \in U,\\ \beta_p, & p \in V. \end{cases} \end{align*} This is well-defined because $M=U \cup V$, so at least one of the two clauses applies. If both clauses apply, then $p \in U \cap V$, and the equality $\alpha|_{U \cap V}=\beta|_{U \cap V}$ gives $\alpha_p=\beta_p$. It remains to check smoothness. Smoothness of a differential form is local: a form is smooth on $M$ if its restriction to every member of an open cover is smooth. Here $\omega|_U=\alpha$, which is smooth by the assumption $\alpha \in \Omega^k(U)$, and $\omega|_V=\beta$, which is smooth by the assumption $\beta \in \Omega^k(V)$. Since $U,V$ cover $M$, the form $\omega$ is smooth on all of $M$. Therefore $\omega \in \Omega^k(M)$ and \begin{align*} r_k(\omega)=(\omega|_U,\omega|_V)=(\alpha,\beta). \end{align*} So every element of $\ker s_k$ lies in $\operatorname{im} r_k$. Since the reverse inclusion was already proved from $s_k \circ r_k=0$, we obtain \begin{align*} \operatorname{im} r_k=\ker s_k. \end{align*} [/guided] [/step] [step:Use a partition of unity to prove that the difference map is surjective] Let $\gamma \in \Omega^k(U \cap V)$. Choose smooth functions \begin{align*} \rho_U,\rho_V: M \to \mathbb{R} \end{align*} forming a partition of unity subordinate to the open cover $U,V$, so that \begin{align*} \operatorname{supp}\rho_U \subset U,\qquad \operatorname{supp}\rho_V \subset V,\qquad \rho_U+\rho_V=1 \end{align*} on $M$. Define $\alpha \in \Omega^k(U)$ by \begin{align*} \alpha_p = \begin{cases} \rho_V(p)\gamma_p, & p \in U \cap V,\\ 0, & p \in U \setminus \operatorname{supp}\rho_V, \end{cases} \end{align*} and define $\beta \in \Omega^k(V)$ by \begin{align*} \beta_p = \begin{cases} -\rho_U(p)\gamma_p, & p \in U \cap V,\\ 0, & p \in V \setminus \operatorname{supp}\rho_U. \end{cases} \end{align*} These definitions are smooth because $\operatorname{supp}\rho_V \subset V$ and $\operatorname{supp}\rho_U \subset U$; hence the zero extensions occur on neighbourhoods where the multiplying functions vanish. On $U \cap V$, we have \begin{align*} s_k(\alpha,\beta) &= \alpha|_{U \cap V}-\beta|_{U \cap V} \\ &= \rho_V\gamma - (-\rho_U\gamma) \\ &= (\rho_U+\rho_V)\gamma \\ &= \gamma. \end{align*} Thus $s_k$ is surjective. [guided] Let $\gamma \in \Omega^k(U \cap V)$ be arbitrary. To prove surjectivity of $s_k$, we must construct forms $\alpha \in \Omega^k(U)$ and $\beta \in \Omega^k(V)$ such that \begin{align*} \alpha|_{U \cap V}-\beta|_{U \cap V}=\gamma. \end{align*} Choose smooth functions \begin{align*} \rho_U,\rho_V: M \to \mathbb{R} \end{align*} forming a partition of unity subordinate to $U,V$. Thus \begin{align*} \operatorname{supp}\rho_U \subset U,\qquad \operatorname{supp}\rho_V \subset V,\qquad \rho_U+\rho_V=1 \end{align*} on $M$. The reason for multiplying by the opposite cutoff is that $\rho_V\gamma$ can be extended by zero to all of $U$: the support of $\rho_V$ lies inside $V$, so on the part of $U$ outside $V$, the function $\rho_V$ vanishes in a neighbourhood. Define \begin{align*} \alpha_p = \begin{cases} \rho_V(p)\gamma_p, & p \in U \cap V,\\ 0, & p \in U \setminus \operatorname{supp}\rho_V. \end{cases} \end{align*} This gives a smooth $k$-form $\alpha \in \Omega^k(U)$. Indeed, on $U \cap V$ it is the product of the smooth function $\rho_V|_{U \cap V}$ with the smooth form $\gamma$, while near any point of $U \setminus V$ it is identically zero because $\operatorname{supp}\rho_V \subset V$. Similarly, define \begin{align*} \beta_p = \begin{cases} -\rho_U(p)\gamma_p, & p \in U \cap V,\\ 0, & p \in V \setminus \operatorname{supp}\rho_U. \end{cases} \end{align*} The same support argument shows that $\beta \in \Omega^k(V)$ is smooth. Now compute on the overlap $U \cap V$: \begin{align*} s_k(\alpha,\beta) &= \alpha|_{U \cap V}-\beta|_{U \cap V} \\ &= \rho_V\gamma - (-\rho_U\gamma) \\ &= (\rho_U+\rho_V)\gamma \\ &= \gamma. \end{align*} Since $\gamma$ was arbitrary, $s_k$ is surjective. [/guided] [/step] [step:Check compatibility with the exterior derivative] Let $\omega \in \Omega^k(M)$. Since exterior differentiation commutes with restriction to open subsets, \begin{align*} r_{k+1}(d\omega) &=((d\omega)|_U,(d\omega)|_V) \\ &=(d(\omega|_U),d(\omega|_V)) \\ &=(d \oplus d)(r_k(\omega)). \end{align*} Thus $r_\bullet$ is a cochain map. Let $(\alpha,\beta) \in \Omega^k(U) \oplus \Omega^k(V)$. Again using compatibility of $d$ with restriction, \begin{align*} s_{k+1}((d \oplus d)(\alpha,\beta)) &=s_{k+1}(d\alpha,d\beta) \\ &=(d\alpha)|_{U \cap V}-(d\beta)|_{U \cap V} \\ &=d(\alpha|_{U \cap V})-d(\beta|_{U \cap V}) \\ &=d(\alpha|_{U \cap V}-\beta|_{U \cap V}) \\ &=d(s_k(\alpha,\beta)). \end{align*} Therefore $s_\bullet$ is also a cochain map. Combining injectivity of $r_k$, the equality $\operatorname{im} r_k=\ker s_k$, surjectivity of $s_k$, and compatibility with $d$, we obtain a short exact sequence of cochain complexes. [/step]