[proofplan]
We verify the three properties of the flow $g^t$ of a smooth vector field $V$ directly from the existence and uniqueness theorem for ODEs. Property (1), $g^0 = \operatorname{id}$, is immediate from the initial condition. Property (2), the group law $g^{t+s} = g^t \circ g^s$, follows by showing that both sides satisfy the same ODE with the same initial condition and invoking uniqueness. Property (3), that $g^t$ is a diffeomorphism with inverse $g^{-t}$, is a consequence of the group law.
[/proofplan]
[step:Verify that $g^0$ is the identity map]
Let $x_0 \in \mathbb{R}^n$. By definition, $g^t(x_0)$ is the value at time $t$ of the unique solution $x: I \to \mathbb{R}^n$ to the initial value problem
\begin{align*}
\dot{x} = V(x), \qquad x(0) = x_0.
\end{align*}
Evaluating at $t = 0$ gives $g^0(x_0) = x(0) = x_0$. Since $x_0$ was arbitrary, $g^0 = \operatorname{id}$.
[/step]
[step:Establish the group law $g^{t+s} = g^t \circ g^s$ by uniqueness of ODE solutions]
Fix $s \in \mathbb{R}$ and $x_0 \in \mathbb{R}^n$, and set $y_0 := g^s(x_0)$. We compare two curves in $\mathbb{R}^n$, both regarded as functions of the variable $t$.
**Curve 1.** Define $\alpha(t) := g^{t+s}(x_0)$. Since $\tau \mapsto g^\tau(x_0)$ solves $\dot{x} = V(x)$, the time-shifted curve $\alpha$ satisfies
\begin{align*}
\dot{\alpha}(t) = \frac{d}{dt} g^{t+s}(x_0) = V(g^{t+s}(x_0)) = V(\alpha(t)),
\end{align*}
with initial condition $\alpha(0) = g^s(x_0) = y_0$.
**Curve 2.** Define $\beta(t) := g^t(y_0) = g^t(g^s(x_0))$. By definition of the flow, $\beta$ satisfies
\begin{align*}
\dot{\beta}(t) = V(g^t(y_0)) = V(\beta(t)),
\end{align*}
with initial condition $\beta(0) = g^0(y_0) = y_0$.
Both $\alpha$ and $\beta$ solve the same ODE $\dot{x} = V(x)$ with the same initial condition $y_0$. Since $V$ is smooth (and in particular locally Lipschitz), the Picard-Lindelof theorem guarantees uniqueness, so $\alpha(t) = \beta(t)$ for all $t$ in the common interval of existence. That is,
\begin{align*}
g^{t+s}(x_0) = g^t(g^s(x_0))
\end{align*}
for all $t, s$ and all $x_0$.
[guided]
The strategy is to reduce the group law to a uniqueness statement. Both $g^{t+s}(x_0)$ and $g^t(g^s(x_0))$ are obtained by following the vector field $V$, but starting at different moments: the first follows the trajectory of $x_0$ from time $0$ to time $t + s$, while the second first flows $x_0$ to time $s$ (reaching $y_0 = g^s(x_0)$) and then restarts the flow from $y_0$ for an additional time $t$.
The key observation is that the ODE $\dot{x} = V(x)$ is autonomous -- the right-hand side does not depend on time. Therefore, shifting the time origin does not change the equation. The curve $t \mapsto g^{t+s}(x_0)$ is simply the original solution with its time parameter shifted by $s$, and it still satisfies $\dot{x} = V(x)$. Since both curves solve the same autonomous ODE with the same initial value $y_0$ at $t = 0$, uniqueness (guaranteed by the smoothness of $V$, which implies local Lipschitz continuity) forces them to agree.
This is where the autonomous nature of the vector field is essential. For a time-dependent vector field $V(x, t)$, the flow would satisfy a two-parameter group law $g^{t,s} \circ g^{s,r} = g^{t,r}$ instead.
[/guided]
[/step]
[step:Conclude that $g^t$ is a diffeomorphism with inverse $g^{-t}$]
Applying the group law from the previous step with $s$ replaced by $-t$, we obtain
\begin{align*}
g^t \circ g^{-t} = g^{t + (-t)} = g^0 = \operatorname{id}.
\end{align*}
Similarly, replacing $t$ by $-t$ and $s$ by $t$ gives
\begin{align*}
g^{-t} \circ g^t = g^{-t + t} = g^0 = \operatorname{id}.
\end{align*}
Therefore $g^t$ is invertible with $(g^t)^{-1} = g^{-t}$. Since the solution of a smooth ODE depends smoothly on initial data (by the smooth dependence theorem for ODEs), both $g^t$ and $g^{-t}$ are smooth maps $\mathbb{R}^n \to \mathbb{R}^n$. A smooth bijection with a smooth inverse is a diffeomorphism, so $g^t$ is a diffeomorphism for each $t$.
[/step]
