[proofplan] The sequence has three exactness conditions to verify: injectivity of $\alpha$, exactness at the middle term, and surjectivity of $\beta$. Injectivity and exactness at the middle are immediate from the definitions of restriction. The crucial content is surjectivity of $\beta$: given a form on $U \cap V$, we must extend it to a pair on $(U, V)$. We produce this extension using a smooth [partition of unity](/page/Partition%20of%20Unity) subordinate to the open cover $\{U, V\}$, whose existence relies critically on the hypothesis that $U$ and $V$ are open. The key mechanism is that multiplying the form on $U \cap V$ by a bump function that vanishes near the boundary of $U \cap V$ inside each of $U$ and $V$ allows a smooth extension by zero. [/proofplan]

[step:Verify that $\alpha$ and $\beta$ are chain maps and the sequence has degree-preserving maps] Both $\alpha$ and $\beta$ are $\mathbb R$-linear and preserve the grading degree since they are defined by restriction, which commutes with the exterior derivative: for any open $W' \subseteq W$ and $\omega \in \Omega^k(W)$, \begin{align*} d(\omega|_{W'}) = (d\omega)|_{W'}. \end{align*} Hence $\alpha \circ d = d \circ \alpha$ and $\beta \circ d = d \circ \beta$, so $\alpha$ and $\beta$ are morphisms of cochain complexes. [/step]

[step:Establish injectivity of $\alpha$ by the open cover property] Suppose $\omega \in \Omega^k(M)$ satisfies $\alpha(\omega) = 0$, i.e. $\omega|_U = 0$ and $\omega|_V = 0$. Then $\omega_p = 0$ for every $p \in U$ and every $p \in V$. Since $M = U \cup V$, every point $p \in M$ lies in $U$ or in $V$, so $\omega_p = 0$ for all $p \in M$. Hence $\omega = 0$ as a section of $\bigwedge^k T^*M$. Therefore $\ker(\alpha) = \{0\}$ and $\alpha$ is injective. [/step]

[step:Verify exactness at $\Omega^*(U) \oplus \Omega^*(V)$] We show $\operatorname{Range}(\alpha) = \ker(\beta)$. *Inclusion $\operatorname{Range}(\alpha) \subseteq \ker(\beta)$.* For $\omega \in \Omega^k(M)$, \begin{align*} \beta(\alpha(\omega)) = \beta(\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*} where we used that restriction is transitive: for $U \cap V \subseteq U \subseteq M$, $(\omega|_U)|_{U \cap V} = \omega|_{U \cap V}$. *Inclusion $\ker(\beta) \subseteq \operatorname{Range}(\alpha)$.* Suppose $(\tilde\alpha, \tilde\beta) \in \Omega^k(U) \oplus \Omega^k(V)$ satisfies $\beta(\tilde\alpha, \tilde\beta) = 0$, i.e. $\tilde\alpha|_{U \cap V} = \tilde\beta|_{U \cap V}$. Define the pointwise assignment \begin{align*} \omega: M &\to \bigwedge\nolimits^k T^*M, \\ p &\mapsto \begin{cases} \tilde\alpha_p & p \in U, \\ \tilde\beta_p & p \in V. \end{cases} \end{align*} This is well-defined: for $p \in U \cap V$, $\tilde\alpha_p = \tilde\beta_p$ by hypothesis. It is smooth: on the open set $U$, $\omega$ agrees with the smooth form $\tilde\alpha$, and on the open set $V$, $\omega$ agrees with the smooth form $\tilde\beta$. Smoothness is a local property, and $\{U, V\}$ is an open cover of $M$, so $\omega \in \Omega^k(M)$. By construction $\omega|_U = \tilde\alpha$ and $\omega|_V = \tilde\beta$, hence $\alpha(\omega) = (\tilde\alpha, \tilde\beta)$. [/step]

[step:Choose a smooth partition of unity subordinate to the open cover $\{U, V\}$]To prove surjectivity of $\beta$, we first produce the geometric data. Since $\{U, V\}$ is an open cover of the smooth manifold $M$, by the [Existence of Partitions of Unity](/theorems/???) there exist smooth functions \begin{align*} \rho_U: M &\to [0,1], & \rho_V: M &\to [0,1] \end{align*} such that \begin{align*} \operatorname{supp}(\rho_U) \subseteq U, \qquad \operatorname{supp}(\rho_V) \subseteq V, \qquad \rho_U + \rho_V = 1 \text{ on } M. \end{align*} The hypothesis that $U$ and $V$ are **open** is essential here: the general existence theorem for partitions of unity requires an open cover, because the defining condition $\operatorname{supp}(\rho_U) \subseteq U$ (a closed-in-$M$ set contained in $U$) is meaningful only when $U$ is open.[/step]

[guided]The engine that drives this proof is a partition of unity, and this step explains why the openness hypothesis is non-negotiable. A partition of unity subordinate to a cover $\{U_i\}$ is a collection of smooth non-negative functions $(\rho_i)$ summing to $1$ with $\operatorname{supp}(\rho_i) \subseteq U_i$ (locally finite). The standard existence theorem requires the cover to be open — the reason is that the construction uses bump functions, and for a bump function $\rho$ to satisfy $\operatorname{supp}(\rho) \subseteq U_i$, the set $U_i$ must be open so that $\rho$ can be non-zero on an open neighbourhood inside $U_i$ while vanishing smoothly towards its boundary. In our setting the cover is $\{U, V\}$, which we are given is open. Invoking the partition-of-unity theorem produces $\rho_U, \rho_V: M \to [0,1]$ smooth with \begin{align*} \operatorname{supp}(\rho_U) \subseteq U, \qquad \operatorname{supp}(\rho_V) \subseteq V, \qquad \rho_U + \rho_V = 1. \end{align*} The key geometric fact we will use in the next step: because $\operatorname{supp}(\rho_V)$ is a closed subset of $M$ contained in $V$, its *relative* behaviour inside $U$ is that $\rho_V$ vanishes on an open neighbourhood of each point of $U \setminus V$ in $U$. This is what lets us extend $\rho_V \cdot \omega$ by zero from $U \cap V$ to all of $U$ smoothly. What if we had only asked that $M = U \cup V$ as a set without openness? Then partitions of unity would generally not exist — for instance, if $U$ were a single point and $V$ its complement, no smooth $\rho_V$ could have support inside $V$ while being continuous and equal to $1$ near that point.[/guided]

[step:Surjectivity of $\beta$ by smooth extension by zero using the partition of unity]Let $\omega \in \Omega^k(U \cap V)$. We construct $(\omega_U, \omega_V) \in \Omega^k(U) \oplus \Omega^k(V)$ with $\beta(\omega_U, \omega_V) = \omega$. *Construction.* Define \begin{align*} \omega_U: U &\to \bigwedge\nolimits^k T^*U, \\ p &\mapsto \begin{cases} \rho_V(p)\, \omega_p & p \in U \cap V, \\ 0 & p \in U \setminus (U \cap V) = U \setminus V. \end{cases} \end{align*} We check smoothness. On the open set $U \cap V$, $\omega_U$ equals the smooth form $\rho_V \cdot \omega$. On the open set $U \setminus \operatorname{supp}(\rho_V)$ (open in $U$ because $\operatorname{supp}(\rho_V)$ is closed in $M$, hence in $U$), the function $\rho_V$ vanishes identically, so the definition $\rho_V \cdot \omega$ agrees with $0$. These two open sets cover $U$: \begin{align*} (U \cap V) \cup (U \setminus \operatorname{supp}(\rho_V)) = U, \end{align*} since any $p \in U$ either lies in $V$ (first set) or lies outside $\operatorname{supp}(\rho_V) \subseteq V$ (second set). On the overlap $(U \cap V) \cap (U \setminus \operatorname{supp}(\rho_V))$, both expressions evaluate to $0$, so the two local definitions agree. Hence $\omega_U \in \Omega^k(U)$. Symmetrically define \begin{align*} \omega_V: V &\to \bigwedge\nolimits^k T^*V, \\ p &\mapsto \begin{cases} -\rho_U(p)\, \omega_p & p \in U \cap V, \\ 0 & p \in V \setminus U, \end{cases} \end{align*} and by the same argument $\omega_V \in \Omega^k(V)$. *Verification.* On $U \cap V$, \begin{align*} \beta(\omega_U, \omega_V) = \omega_U|_{U \cap V} - \omega_V|_{U \cap V} = \rho_V \omega - (-\rho_U \omega) = (\rho_U + \rho_V)\omega = \omega, \end{align*} using $\rho_U + \rho_V = 1$ on $M$ restricted to $U \cap V$. This proves surjectivity of $\beta$.[/step]

[guided]We are given $\omega \in \Omega^k(U \cap V)$ and must produce a preimage under $\beta$ — that is, two forms, one on $U$ and one on $V$, whose difference on the overlap equals $\omega$. The natural attempt is to split $\omega = \omega \cdot 1 = \omega(\rho_U + \rho_V) = \rho_U \omega + \rho_V \omega$ and declare $\omega_U := \rho_V \omega$ on $U \cap V$ and $\omega_V := -\rho_U \omega$ on $U \cap V$. But these are currently defined only on $U \cap V$; we need them on $U$ and $V$ respectively. The trick is that $\rho_V \omega$ can be smoothly extended by zero from $U \cap V$ to all of $U$. Why? Consider the formula \begin{align*} \omega_U(p) = \begin{cases} \rho_V(p)\, \omega_p & p \in U \cap V, \\ 0 & p \in U \setminus V. \end{cases} \end{align*} Smoothness is a local property. At a point $p \in U \cap V$, $\omega_U = \rho_V \omega$ locally, which is smooth. At a point $p \in U \setminus V$, observe that $p \notin V \supseteq \operatorname{supp}(\rho_V)$, so $p \notin \operatorname{supp}(\rho_V)$. Since $\operatorname{supp}(\rho_V)$ is closed in $M$ (by definition of support, the closure of the set where $\rho_V \ne 0$), its complement $M \setminus \operatorname{supp}(\rho_V)$ is open, and contains an open neighbourhood $W$ of $p$ in $U$. On $W$, $\rho_V \equiv 0$, so the first-branch formula $\rho_V \omega$ would also be zero there if $W \subseteq U \cap V$, and the second-branch formula is zero. Both branches agree and equal $0$ on $W$. Hence $\omega_U$ is smooth at $p$. This is precisely the "smooth extension by zero" that would fail without openness: if $V$ were not open, then $U \setminus V$ might not be open, and the argument that $\rho_V \omega$ extends smoothly to zero would break down because we could not find an open neighbourhood on which $\rho_V$ vanishes. Finally we verify the equation $\beta(\omega_U, \omega_V) = \omega$. On $U \cap V$, \begin{align*} (\omega_U - \omega_V)|_{U \cap V} = \rho_V \omega - (-\rho_U \omega) = (\rho_V + \rho_U)\omega. \end{align*} Since $\rho_U + \rho_V = 1$ on $M$ and in particular on $U \cap V$, this equals $\omega$. Hence $\beta(\omega_U, \omega_V) = \omega$, and $\beta$ is surjective.[/guided]

[step:Combine the three exactness conditions] The injectivity of $\alpha$, exactness at $\Omega^*(U) \oplus \Omega^*(V)$, and surjectivity of $\beta$ together establish that \begin{align*} 0 \to \Omega^*(M) \xrightarrow{\alpha} \Omega^*(U) \oplus \Omega^*(V) \xrightarrow{\beta} \Omega^*(U \cap V) \to 0 \end{align*} is a short exact sequence of cochain complexes, since both $\alpha$ and $\beta$ commute with $d$ by the first step. [/step]