[proofplan] We first separate the compact set $\overline V$ from the complement of $U$ by a smooth bump function. Multiplying the local section $s$ by this bump function gives a smooth section over $U$ that vanishes near the part of $M$ where it must be glued to zero. We then define the global section by this product on $U$ and by the zero section outside the support of the bump function, and verify smoothness, agreement near $\overline V$, and the support containment. [/proofplan] [step:Choose a smooth bump function equal to one near $\overline V$ and supported in $U$] Let $K:=\overline V\subset M$. Since $K$ is compact and $K\subset U$, the Smooth Urysohn Lemma for paracompact Hausdorff smooth manifolds gives a smooth function \begin{align*} \chi:M&\to \mathbb R \end{align*} such that there exists an [open set](/page/Open%20Set) $W\subset M$ with $K\subset W\subset U$, $\chi(x)=1$ for every $x\in W$, and $\operatorname{supp}(\chi)\subset U$. [guided] The compactness assumption is used exactly here. We need a smooth function that is identically $1$ on a neighbourhood of the [closed set](/page/Closed%20Set) where we want to preserve $s$, but whose nonzero set stays inside $U$ so that multiplication by $s$ is only performed where $s$ is defined. Let $K:=\overline V$. The hypotheses say that $K$ is compact and $K\subset U$, with $U$ open in the paracompact Hausdorff smooth manifold $M$. By the Smooth Urysohn Lemma for paracompact Hausdorff smooth manifolds, there is a smooth map \begin{align*} \chi:M&\to \mathbb R \end{align*} and an open neighbourhood $W\subset M$ of $K$ such that $W\subset U$, $\chi(x)=1$ for all $x\in W$, and $\operatorname{supp}(\chi)\subset U$. This is the only place where the global topology of $M$ enters the proof. Once $\chi$ is available, the rest is a local vector-bundle construction. [/guided] [/step] [step:Define the global section by multiplying on $U$ and using the zero section elsewhere] Let $0:M\to E$ denote the smooth zero section of the vector bundle $\pi:E\to M$. Define a map \begin{align*} \tilde{s}:M&\to E \end{align*} as follows. For $x\in U$, set $\tilde{s}(x)=\chi(x)s(x)$. For $x\in M\setminus \operatorname{supp}(\chi)$, set $\tilde{s}(x)=0(x)$. This definition is consistent because $M=U\cup (M\setminus \operatorname{supp}(\chi))$, since $\operatorname{supp}(\chi)\subset U$. On the overlap $U\cap (M\setminus \operatorname{supp}(\chi))$, we have $\chi(x)=0$, so $\chi(x)s(x)=0(x)$ in the [vector space](/page/Vector%20Space) $E_x$. [guided] We now turn the local section $s$ into a global object. Let $0:M\to E$ denote the smooth zero section of the vector bundle $\pi:E\to M$. The section $s$ is only defined on $U$, so the product $\chi(x)s(x)$ may only be used for points $x\in U$. This is why the support condition $\operatorname{supp}(\chi)\subset U$ was built into the bump function. Define a map \begin{align*} \tilde{s}:M&\to E \end{align*} by the following two formulae. For $x\in U$, set $\tilde{s}(x)=\chi(x)s(x)$. For $x\in M\setminus \operatorname{supp}(\chi)$, set $\tilde{s}(x)=0(x)$. These two formulae cover all points of $M$ because $\operatorname{supp}(\chi)\subset U$ implies \begin{align*} M=U\cup (M\setminus \operatorname{supp}(\chi)). \end{align*} We also need to check that the two formulae agree where both apply. If $x\in U\cap (M\setminus \operatorname{supp}(\chi))$, then $x$ is outside the support of $\chi$, hence $\chi(x)=0$. Therefore $\chi(x)s(x)=0(x)$ in the fibre vector space $E_x$. Thus the definition of $\tilde{s}$ is unambiguous. [/guided] [/step] [step:Verify that the glued map is a smooth global section] On $U$, the map $x\mapsto \chi(x)s(x)$ is a smooth section of $E|_U\to U$, because scalar multiplication \begin{align*} \mathbb R\times_M E&\to E \end{align*} is smooth for a smooth vector bundle, $\chi|_U:U\to\mathbb R$ is smooth, and $s:U\to E|_U$ is smooth. On the open set $M\setminus \operatorname{supp}(\chi)$, the map $\tilde{s}$ is the smooth zero section. These two smooth formulae agree on the overlap, so they glue to a smooth map $\tilde{s}:M\to E$. For every $x\in M$, the value $\tilde{s}(x)$ lies in the fibre $E_x$, hence $\pi(\tilde{s}(x))=x$. Therefore $\pi\circ \tilde{s}=\operatorname{id}_M$, so $\tilde{s}\in \Gamma(M,E)$. [guided] We verify smoothness using the open cover on which $\tilde{s}$ was defined. On $U$, the formula is $\tilde{s}(x)=\chi(x)s(x)$. This is a smooth section of $E|_U\to U$ because $\chi|_U:U\to\mathbb R$ is smooth, $s:U\to E|_U$ is smooth, and scalar multiplication in a smooth vector bundle is a smooth map \begin{align*} \mathbb R\times_M E&\to E. \end{align*} Here $\mathbb R\times_M E$ denotes the fibre product over $M$, so scalar multiplication is applied fibrewise in $E_x$. On the open set $M\setminus \operatorname{supp}(\chi)$, the formula for $\tilde{s}$ is the zero section $0:M\to E$, which is smooth by the definition of a smooth vector bundle. The two smooth formulae agree on the overlap $U\cap (M\setminus \operatorname{supp}(\chi))$, as checked in the construction step. Therefore the smooth gluing property for maps defined by agreeing smooth formulae on an open cover implies that $\tilde{s}:M\to E$ is smooth. Finally, for every $x\in M$, the value $\tilde{s}(x)$ lies in the fibre $E_x$. Hence \begin{align*} \pi(\tilde{s}(x))=x. \end{align*} Equivalently, $\pi\circ \tilde{s}=\operatorname{id}_M$, so $\tilde{s}$ is a smooth global section, that is $\tilde{s}\in \Gamma(M,E)$. [/guided] [/step] [step:Check agreement with $s$ near $\overline V$ and containment of support] For every $x\in W$, we have $\chi(x)=1$, and since $W\subset U$, \begin{align*} \tilde{s}(x)=\chi(x)s(x)=s(x). \end{align*} Thus $\tilde{s}=s$ on the open neighbourhood $W$ of $\overline V$. If $\tilde{s}(x)\ne 0(x)$, then necessarily $\chi(x)\ne 0$, so \begin{align*} \{x\in M:\tilde{s}(x)\ne 0(x)\}\subset \{x\in M:\chi(x)\ne 0\}. \end{align*} Taking closures in $M$ gives \begin{align*} \operatorname{supp}(\tilde{s})\subset \operatorname{supp}(\chi)\subset U. \end{align*} This proves both required properties of $\tilde{s}$. [guided] It remains to check that the constructed global section has the two properties required in the theorem statement. First, let $x\in W$. By construction of the bump function, $\chi(x)=1$, and since $W\subset U$, the defining formula on $U$ gives \begin{align*} \tilde{s}(x)=\chi(x)s(x)=s(x). \end{align*} Thus $\tilde{s}=s$ on the open neighbourhood $W$ of $\overline V$. Second, we prove the support containment. If $\tilde{s}(x)\ne 0(x)$, then the formula defining $\tilde{s}$ forces $\chi(x)\ne 0$; when $\chi(x)=0$, the product $\chi(x)s(x)$ is the zero vector in $E_x$, and outside $\operatorname{supp}(\chi)$ the section was defined to be the zero section. Therefore \begin{align*} \{x\in M:\tilde{s}(x)\ne 0(x)\}\subset \{x\in M:\chi(x)\ne 0\}. \end{align*} Taking closures in $M$ gives \begin{align*} \operatorname{supp}(\tilde{s})\subset \operatorname{supp}(\chi). \end{align*} The bump function was chosen with $\operatorname{supp}(\chi)\subset U$, so \begin{align*} \operatorname{supp}(\tilde{s})\subset U. \end{align*} This proves that $\tilde{s}$ agrees with $s$ on a neighbourhood of $\overline V$ and has support contained in $U$. [/guided] [/step]