[proofplan] At every point of the compact support $K$ we can produce a local flow by the [Existence of Flows](/theorems/1516) theorem. Compactness extracts a finite subcover, yielding a uniform short time $\varepsilon > 0$ such that the flow $\varphi_t$ is defined for all $t \in (-\varepsilon, \varepsilon)$ and on all of $M$ (points outside $K$ are fixed because $X$ vanishes there). We then extend $\varphi$ to arbitrary $t \in \mathbb{R}$ by iterating $\varphi_{\varepsilon/2}$ the appropriate integer number of times and composing with the residual short-time flow. The group law from the local theorem guarantees that the extension is well-defined and smooth, and produces diffeomorphisms with inverse $\varphi_{-t}$. [/proofplan] [step:Produce a uniform short time $\varepsilon > 0$ on a cover of the support] For every $p \in K$, the [Existence of Flows](/theorems/1516) theorem — applied to the smooth vector field $X$ at $p$ — yields $\varepsilon_p > 0$, an open neighbourhood $U_p \subseteq M$ of $p$, and a smooth map $\varphi^{(p)}: (-\varepsilon_p, \varepsilon_p) \times U_p \to M$ satisfying the local flow properties. The family $\{U_p\}_{p \in K}$ is an open cover of $K$. Hypothesis verification for compactness: $K$ is compact by assumption. We apply compactness to extract a finite subcover: there exist $p_1, \dots, p_N \in K$ such that $K \subseteq \bigcup_{i=1}^{N} U_{p_i}$. Set \begin{align*} \varepsilon := \min_{1 \le i \le N} \varepsilon_{p_i} > 0, \end{align*} a strictly positive minimum of finitely many positive numbers. [/step] [step:Assemble local flows on $M \setminus K$ by extension with the identity and glue on overlaps] On $M \setminus K$, the vector field vanishes: $X(x) = 0$ for all $x \in M \setminus K$. By the [Existence of Flows](/theorems/1516) theorem applied pointwise at each $x \in M \setminus K$, the unique local solution of the ODE $\dot\gamma = X \circ \gamma$ with $\gamma(0) = x$ is the constant curve $\gamma(t) \equiv x$. Therefore we may define $\varphi^{\mathrm{out}}: \mathbb{R} \times (M \setminus K) \to M$ by $\varphi^{\mathrm{out}}(t, x) := x$, which is smooth. On each $U_{p_i}$ we have the smooth local flow $\varphi^{(p_i)}: (-\varepsilon_{p_i}, \varepsilon_{p_i}) \times U_{p_i} \to M$. On overlaps $U_{p_i} \cap U_{p_j}$ and for $t$ in the common time interval, uniqueness of solutions of the flow ODE (part of the conclusion of [Existence of Flows](/theorems/1516)) implies \begin{align*} \varphi^{(p_i)}(t, x) = \varphi^{(p_j)}(t, x) \qquad \text{for all } x \in U_{p_i} \cap U_{p_j},\ |t| < \min(\varepsilon_{p_i}, \varepsilon_{p_j}). \end{align*} Similarly, on overlaps $U_{p_i} \cap (M \setminus K)$ — where points $x$ lie outside $\operatorname{supp}(X)$, but within the chart neighbourhood of a support point — the constant curve $\gamma(t) \equiv x$ also satisfies the ODE $\dot\gamma = X(\gamma(t))$ since $X(x) = 0$, so by uniqueness $\varphi^{(p_i)}(t, x) = x = \varphi^{\mathrm{out}}(t, x)$. Define \begin{align*} \Phi: (-\varepsilon, \varepsilon) \times M &\to M \\ (t, x) &\mapsto \begin{cases} \varphi^{(p_i)}(t, x) & x \in U_{p_i} \text{ for some } i, \\ x & x \in M \setminus K. \end{cases} \end{align*} The two cases are consistent on overlaps by the compatibility established above. $\Phi$ is smooth because smoothness is a local property and $\Phi$ restricted to each of the open sets $\{U_{p_i}\}$ and $M \setminus K$ is smooth, and these open sets cover $(-\varepsilon, \varepsilon) \times M$. [/step] [step:Verify the local group law and the identity property of $\Phi$] By construction $\Phi(0, x) = x$ for every $x \in M$ (this holds in each $U_{p_i}$ because $\varphi^{(p_i)}_0 = \mathrm{id}_{U_{p_i}}$, and on $M \setminus K$ because $\Phi_0 = \mathrm{id}$). Thus $\Phi_0 = \mathrm{id}_M$. For the local group law: for $x \in M$ and $s, t \in \mathbb{R}$ with $|s|, |t|, |s + t| < \varepsilon$, we claim \begin{align*} \Phi_{t + s}(x) = \Phi_t(\Phi_s(x)). \end{align*} On $M \setminus K$ both sides equal $x$, so the identity holds. On $U_{p_i}$, both $t \mapsto \Phi_{t+s}(x)$ and $t \mapsto \Phi_t(\Phi_s(x))$ are smooth curves in $M$ satisfying the ODE $\dot\gamma = X(\gamma)$ with the same initial condition $\gamma(0) = \Phi_s(x)$ at $t = 0$: the first by direct computation $\frac{d}{dt}\Phi_{t+s}(x) = X(\Phi_{t+s}(x))$ and $\Phi_{0+s}(x) = \Phi_s(x)$, the second by the defining property of the local flow at the point $\Phi_s(x)$. By uniqueness of ODE solutions (from [Existence of Flows](/theorems/1516)), the two curves agree on their common interval of definition. [/step] [step:Extend $\Phi$ from short times to all of $\mathbb{R}$ via iteration] Given any $t \in \mathbb{R}$, write \begin{align*} t = k \cdot \frac{\varepsilon}{2} + r, \qquad k \in \mathbb{Z},\ r \in [0, \varepsilon/2), \end{align*} using the division algorithm (for $t \ge 0$, take $k = \lfloor 2t/\varepsilon \rfloor$ and $r = t - k\varepsilon/2$; for $t < 0$, adjust signs). Since $|\varepsilon/2| < \varepsilon$ and $|r| < \varepsilon/2 < \varepsilon$, the maps $\Phi_{\varepsilon/2}$ and $\Phi_r$ are defined on all of $M$ and are smooth. Define \begin{align*} \varphi_t := \underbrace{\Phi_{\varepsilon/2} \circ \cdots \circ \Phi_{\varepsilon/2}}_{k \text{ times, if } k \ge 0} \circ \Phi_r \quad \text{or} \quad \varphi_t := \Phi_{-\varepsilon/2}^{|k|} \circ \Phi_r \text{ for } k < 0, \end{align*} where $\Phi_{-\varepsilon/2}$ is the time-$(-\varepsilon/2)$ map from the local flow. Well-definedness: a different representation $t = k'\varepsilon/2 + r'$ with $r' \in [0, \varepsilon/2)$ and $k' \in \mathbb{Z}$ must have $k' = k$ and $r' = r$ by uniqueness of the Euclidean division, so the formula is unambiguous. Globality and smoothness: $\varphi_t$ is a composition of finitely many smooth self-maps $\Phi_{\pm \varepsilon/2}, \Phi_r: M \to M$, hence $\varphi_t: M \to M$ is smooth. Joint smoothness in $(t, x)$ follows because, on any compact time interval $[-T, T] \subseteq \mathbb{R}$, the integer $k(t)$ takes only finitely many values and each local piece $t \mapsto \varphi_t(x)$ is smooth; continuity at the gluing times $t = k\varepsilon/2$ is ensured by the local group law ($\Phi_{\varepsilon/2}^k \circ \Phi_0 = \Phi_{\varepsilon/2}^k$, which agrees with the formula on the adjacent segment at $r \to 0$). [/step] [step:Verify the global group law and the identity $\varphi_0 = \mathrm{id}_M$] $\varphi_0 = \mathrm{id}_M$ because $t = 0$ corresponds to $k = 0$, $r = 0$, yielding the empty composition followed by $\Phi_0 = \mathrm{id}_M$. For the group law: fix $s, t \in \mathbb{R}$. We show $\varphi_{t + s} = \varphi_t \circ \varphi_s$. Write \begin{align*} s = k_1\, \varepsilon/2 + r_1, \qquad t = k_2\, \varepsilon/2 + r_2, \qquad s + t = k_3\, \varepsilon/2 + r_3, \end{align*} with $r_1, r_2, r_3 \in [0, \varepsilon/2)$. By construction, $\varphi_s = \Phi_{\varepsilon/2}^{k_1} \circ \Phi_{r_1}$ and $\varphi_t = \Phi_{\varepsilon/2}^{k_2} \circ \Phi_{r_2}$. Therefore \begin{align*} \varphi_t \circ \varphi_s = \Phi_{\varepsilon/2}^{k_2} \circ \Phi_{r_2} \circ \Phi_{\varepsilon/2}^{k_1} \circ \Phi_{r_1}. \end{align*} We now rearrange using the local group law, which states $\Phi_a \circ \Phi_b = \Phi_{a+b}$ whenever $|a|, |b|, |a+b| < \varepsilon$. Each $\Phi_{\varepsilon/2}$ and $\Phi_{r_j}$ has time in $(-\varepsilon, \varepsilon)$, and pairwise sums of such times that arise in the rearrangement all have magnitude less than $\varepsilon$ (e.g.\ $\Phi_{r_2} \circ \Phi_{\varepsilon/2} = \Phi_{r_2 + \varepsilon/2}$ with $0 \le r_2 + \varepsilon/2 < \varepsilon$). Applying this pairwise and inductively moves all the $\Phi_{\varepsilon/2}$ factors to the left and collects the remainders into a single short-time flow. The final result is \begin{align*} \varphi_t \circ \varphi_s = \Phi_{\varepsilon/2}^{k_1 + k_2 + m} \circ \Phi_{r_1 + r_2 - m\varepsilon/2} \end{align*} where $m \in \{0, 1\}$ accounts for any carry when $r_1 + r_2 \ge \varepsilon/2$. Writing $s + t = (k_1 + k_2 + m) \varepsilon/2 + (r_1 + r_2 - m\varepsilon/2)$ with $r_1 + r_2 - m\varepsilon/2 \in [0, \varepsilon/2)$, we recognise this as the canonical decomposition of $s + t$, so $k_3 = k_1 + k_2 + m$ and $r_3 = r_1 + r_2 - m\varepsilon/2$. Therefore $\varphi_t \circ \varphi_s = \Phi_{\varepsilon/2}^{k_3} \circ \Phi_{r_3} = \varphi_{t + s}$. [/step] [step:Each $\varphi_t$ is a diffeomorphism with inverse $\varphi_{-t}$] From the group law with $s = -t$: \begin{align*} \varphi_t \circ \varphi_{-t} = \varphi_{t + (-t)} = \varphi_0 = \mathrm{id}_M, \qquad \varphi_{-t} \circ \varphi_t = \varphi_{(-t) + t} = \varphi_0 = \mathrm{id}_M. \end{align*} Hence $\varphi_{-t}$ is a two-sided inverse to $\varphi_t$. Both $\varphi_t$ and $\varphi_{-t}$ are smooth (shown in the construction step), so $\varphi_t$ is a smooth map with smooth inverse, i.e.\ a diffeomorphism of $M$. The assignment $t \mapsto \varphi_t$ is a group homomorphism $\mathbb{R} \to \mathrm{Diff}(M)$: the identity $\varphi_{t + s} = \varphi_t \circ \varphi_s$ is precisely the homomorphism property. [/step] [step:Verify uniqueness of the extension] Suppose $\widetilde\varphi: \mathbb{R} \times M \to M$ is another smooth map satisfying (i), (ii), (iii) and whose restriction to $(-\varepsilon, \varepsilon) \times M$ agrees with $\Phi$. Fix $t \in \mathbb{R}$ and write $t = k \varepsilon/2 + r$ as before. By the homomorphism property of $\widetilde\varphi$ and its agreement with $\Phi$ on short times: \begin{align*} \widetilde\varphi_t = \widetilde\varphi_{\varepsilon/2}^k \circ \widetilde\varphi_r = \Phi_{\varepsilon/2}^k \circ \Phi_r = \varphi_t. \end{align*} So $\widetilde\varphi = \varphi$ on all of $\mathbb{R} \times M$. Uniqueness of $\Phi$ on $(-\varepsilon, \varepsilon) \times M$ was established in the gluing step, so the global extension is unique. [/step] [step:Collect the conclusions] We have constructed a unique smooth map $\varphi: \mathbb{R} \times M \to M$ satisfying $\varphi_0 = \mathrm{id}_M$ and the group law $\varphi_{t+s} = \varphi_t \circ \varphi_s$, and shown that each $\varphi_t \in \mathrm{Diff}(M)$ with inverse $\varphi_{-t}$. The map $t \mapsto \varphi_t$ is the global 1-parameter subgroup of $\mathrm{Diff}(M)$ generated by the compactly supported vector field $X$, completing the proof. [/step]