[proofplan] We verify each property of the flow map separately. Property (1) follows from the initial condition of the ODE. Property (2) uses the uniqueness of solutions: both $g^{t+s}x_0$ and $g^t(g^s x_0)$ satisfy the same ODE with the same initial data at $t = 0$. Property (3) is an algebraic consequence of (1) and (2). [/proofplan] [step:Verify $g^0 = \operatorname{id}$ from the initial condition] By definition, $g^0(x_0) = x(0)$ where $x(t)$ solves $\dot{x} = V(x)$ with $x(0) = x_0$. Evaluating at $t = 0$ gives $g^0(x_0) = x_0$ for all $x_0$, so $g^0 = \operatorname{id}$. [/step] [step:Establish the group law $g^{t+s} = g^t \circ g^s$ by ODE uniqueness] Fix $x_0 \in \mathbb{R}^m$ and $s \in \mathbb{R}$. Consider two curves in $\mathbb{R}^m$ parametrised by $t$: \begin{align*} \alpha(t) &= g^{t+s}(x_0), \\ \beta(t) &= g^t(g^s(x_0)). \end{align*} Both satisfy the same ODE. For $\alpha$: since $g^{t+s}(x_0)$ is the solution at time $t + s$ starting from $x_0$, differentiating in $t$ gives $\dot{\alpha}(t) = V(\alpha(t))$. For $\beta$: since $g^t(y)$ solves $\dot{x} = V(x)$ with $x(0) = y$ for any $y$, taking $y = g^s(x_0)$ gives $\dot{\beta}(t) = V(\beta(t))$. At $t = 0$, both curves start at the same point: $\alpha(0) = g^s(x_0) = \beta(0)$. By the uniqueness theorem for ODEs (applied to the smooth vector field $V$), a solution to $\dot{x} = V(x)$ with prescribed initial data is unique. Since $\alpha$ and $\beta$ solve the same ODE with the same initial condition, $\alpha(t) = \beta(t)$ for all $t$, giving $g^{t+s}(x_0) = g^t(g^s(x_0))$. [/step] [step:Conclude $(g^t)^{-1} = g^{-t}$ from the group law] Applying property (2) with $s = -t$: \begin{align*} g^t \circ g^{-t} = g^{t + (-t)} = g^0 = \operatorname{id}. \end{align*} Similarly, $g^{-t} \circ g^t = g^0 = \operatorname{id}$. Therefore $g^{-t}$ is the two-sided inverse of $g^t$, i.e., $(g^t)^{-1} = g^{-t}$. Since each $g^t$ is invertible with smooth inverse $g^{-t}$, and both $g^t$ and $g^{-t}$ are smooth (by the smooth dependence on initial conditions guaranteed by the existence theorem), each $g^t$ is a diffeomorphism of $\mathbb{R}^m$. [/step]