[proofplan] The argument is the local Moser trick, with one extra adjustment needed to guarantee that the relevant local flow exists for the whole time interval. Near the chosen point $x_0$, we subtract from $\sigma_t$ a closed $1$-form $\alpha_t$ with the same value at $x_0$; this does not change $d\sigma_t$, but it makes the Moser vector field vanish at $x_0$. A compactness and Gronwall argument then gives a neighbourhood $V$ on which the time-dependent flow exists for all $t\in[0,1]$. Finally, Cartan's formula shows that $\psi_t^*\omega_t$ has zero $t$-derivative, hence remains equal to $\omega_0|_V$. [/proofplan]

[step:Choose local closed forms that agree with $\sigma_t$ at $x_0$]Let $m=\dim M$. Choose a coordinate chart $(W,\varphi)$ with $x_0\in W$, $\overline{W}\subset U$ after replacing $W$ by a smaller coordinate neighbourhood if necessary, and write \begin{align*} \varphi=(y_1,\dots,y_m):W\to \varphi(W)\subset \mathbb R^m. \end{align*} For each $i\in\{1,\dots,m\}$, define the smooth function $a_i:[0,1]\to\mathbb R$ by \begin{align*} a_i(t)=\sigma_t(x_0)\bigl((d\varphi_{x_0})^{-1}e_i\bigr), \end{align*} where $(e_1,\dots,e_m)$ is the standard ordered basis of $\mathbb R^m$. Define a smooth family of $1$-forms $\alpha_t\in\Omega^1(W)$ by \begin{align*} \alpha_t=\sum_{i=1}^m a_i(t)\,dy_i. \end{align*} Since the coefficients $a_i(t)$ are constant as functions on $W$, each $\alpha_t$ is closed: \begin{align*} d\alpha_t=0. \end{align*} Moreover, by construction, \begin{align*} (\alpha_t)_{x_0}=(\sigma_t)_{x_0}. \end{align*} Set \begin{align*} \beta_t=\sigma_t|_W-\alpha_t\in\Omega^1(W). \end{align*} Then $\beta_t$ is a smooth family of $1$-forms on $W$ satisfying \begin{align*} d\beta_t=d\sigma_t|_W \end{align*} and \begin{align*} (\beta_t)_{x_0}=0. \end{align*}[/step]

[guided]The point of this preliminary step is to use the freedom in the equation \begin{align*} \frac{d}{dt}\omega_t=d\sigma_t. \end{align*} Only $d\sigma_t$ appears in the hypothesis and in the Moser calculation, so replacing $\sigma_t$ by $\sigma_t$ minus a closed $1$-form changes neither $d\sigma_t$ nor the desired derivative computation. Choose a coordinate chart $(W,\varphi)$ around $x_0$, with \begin{align*} \varphi=(y_1,\dots,y_m):W\to \varphi(W)\subset \mathbb R^m, \end{align*} where $m=\dim M$. We shrink $W$ so that $\overline{W}\subset U$; this gives room for the local flow to remain inside $U$ later. For each coordinate direction $e_i\in\mathbb R^m$, define \begin{align*} a_i(t)=\sigma_t(x_0)\bigl((d\varphi_{x_0})^{-1}e_i\bigr). \end{align*} Thus $a_i(t)$ is the $dy_i$-coordinate of the covector $(\sigma_t)_{x_0}$. Since $(\sigma_t)_{t\in[0,1]}$ is a smooth family of $1$-forms, each function $a_i:[0,1]\to\mathbb R$ is smooth. Now define \begin{align*} \alpha_t=\sum_{i=1}^m a_i(t)\,dy_i. \end{align*} For fixed $t$, the coefficients $a_i(t)$ do not depend on the point of $W$. Therefore \begin{align*} d\alpha_t=\sum_{i=1}^m d(a_i(t))\wedge dy_i+\sum_{i=1}^m a_i(t)\,d(dy_i)=0. \end{align*} The equality $(\alpha_t)_{x_0}=(\sigma_t)_{x_0}$ follows from how the coefficients $a_i(t)$ were chosen in the coordinate coframe at $x_0$. Finally set \begin{align*} \beta_t=\sigma_t|_W-\alpha_t. \end{align*} Then \begin{align*} d\beta_t=d\sigma_t|_W-d\alpha_t=d\sigma_t|_W \end{align*} and \begin{align*} (\beta_t)_{x_0}=(\sigma_t)_{x_0}-(\alpha_t)_{x_0}=0. \end{align*} This vanishing at $x_0$ is the repair that makes the local-in-space flow exist for the full time interval after shrinking the initial neighbourhood.[/guided]

[step:Define the Moser vector field that vanishes at $x_0$] For each $t\in[0,1]$, the form $\omega_t|_W$ is symplectic, hence nondegenerate at every point of $W$. Therefore, for each $x\in W$, the [linear map](/page/Linear%20Map) \begin{align*} T_xW\to T_x^*W,\qquad v\mapsto \iota_v(\omega_t)_x \end{align*} is an isomorphism. Define the time-dependent vector field $X:[0,1]\times W\to TW$ by requiring \begin{align*} \iota_{X_t}\omega_t|_W=-\beta_t \end{align*} for every $t\in[0,1]$, where $X_t\in\mathfrak X(W)$ denotes the vector field $x\mapsto X(t,x)$. The inverse of the nondegenerate bundle map depends smoothly on $(t,x)$ because $(\omega_t)$ and $(\beta_t)$ are smooth. Hence $X$ is a smooth time-dependent vector field on $W$. Since $(\beta_t)_{x_0}=0$, the defining equation gives \begin{align*} X_t(x_0)=0 \end{align*} for every $t\in[0,1]$. [/step]

[step:Shrink the initial neighbourhood so the flow exists for all $t\in[0,1]$] Choose $r>0$ such that the closed Euclidean ball $\overline{B}(\varphi(x_0),r)$ is contained in $\varphi(W)$. Let \begin{align*} Y:[0,1]\times B(\varphi(x_0),r)\to\mathbb R^m \end{align*} be the coordinate expression of $X$, defined by \begin{align*} Y(t,z)=d\varphi_{\varphi^{-1}(z)}\bigl(X_t(\varphi^{-1}(z))\bigr). \end{align*} Since $X_t(x_0)=0$, one has \begin{align*} Y(t,\varphi(x_0))=0 \end{align*} for every $t\in[0,1]$. By smoothness of $Y$ on the compact set $[0,1]\times\overline{B}(\varphi(x_0),r)$, there exists a constant $L>0$ such that \begin{align*} |Y(t,z)-Y(t,z')|\le L|z-z'| \end{align*} for all $t\in[0,1]$ and all $z,z'\in \overline{B}(\varphi(x_0),r)$. Choose $\rho>0$ such that \begin{align*} 0<\rho<r e^{-L}. \end{align*} Define \begin{align*} V=\varphi^{-1}\bigl(B(\varphi(x_0),\rho)\bigr). \end{align*} For $x\in V$, let $z_x:[0,T_x)\to \varphi(W)$ be the maximal solution of the time-dependent [ordinary differential equation](/page/Ordinary%20Differential%20Equation) \begin{align*} \frac{d}{dt}z_x(t)=Y(t,z_x(t)),\qquad z_x(0)=\varphi(x). \end{align*} Local existence and smooth dependence for time-dependent ordinary differential equations gives such a maximal smooth solution. As long as $z_x(t)\in B(\varphi(x_0),r)$, the Lipschitz estimate and $Y(t,\varphi(x_0))=0$ give \begin{align*} \left|\frac{d}{dt}|z_x(t)-\varphi(x_0)|\right|\le L|z_x(t)-\varphi(x_0)| \end{align*} at times where $z_x(t)\ne \varphi(x_0)$, and the same estimate follows by the upper Dini derivative at times where equality holds. Gronwall's inequality gives \begin{align*} |z_x(t)-\varphi(x_0)|\le e^{Lt}|z_x(0)-\varphi(x_0)|<e^L\rho<r \end{align*} for all $t$ for which the solution remains in $B(\varphi(x_0),r)$. Therefore the solution cannot reach the boundary of $B(\varphi(x_0),r)$ before time $1$. The standard [continuation criterion for ordinary differential equations](/theorems/8314) then implies that $z_x(t)$ exists for every $t\in[0,1]$ and remains in $B(\varphi(x_0),r)$. Define \begin{align*} \psi:[0,1]\times V\to U,\qquad \psi(t,x)=\varphi^{-1}(z_x(t)). \end{align*} Then $\psi$ is smooth by smooth dependence on initial conditions, and $\psi_0=\operatorname{id}_V$.[/step]

[guided]We now explain why the vector field has a flow for the full interval $[0,1]$ after shrinking the starting neighbourhood. This is the only delicate local point: a general vector field on $U$ may flow a chosen point out of $U$ before time $1$, and shrinking the initial neighbourhood would not fix the trajectory of that point. Here the repair is that $X_t(x_0)=0$ for every $t$. Choose $r>0$ so that \begin{align*} \overline{B}(\varphi(x_0),r)\subset \varphi(W). \end{align*} Let \begin{align*} Y:[0,1]\times B(\varphi(x_0),r)\to\mathbb R^m \end{align*} be the coordinate representative of $X$, namely \begin{align*} Y(t,z)=d\varphi_{\varphi^{-1}(z)}\bigl(X_t(\varphi^{-1}(z))\bigr). \end{align*} Because $X_t(x_0)=0$, we have \begin{align*} Y(t,\varphi(x_0))=0 \end{align*} for every $t\in[0,1]$. The map $Y$ is smooth, so its first derivatives in the spatial variables are bounded on the compact set $[0,1]\times\overline{B}(\varphi(x_0),r)$. Hence there is a constant $L>0$ such that \begin{align*} |Y(t,z)-Y(t,z')|\le L|z-z'| \end{align*} for every $t\in[0,1]$ and all $z,z'\in\overline{B}(\varphi(x_0),r)$. This is the uniform Lipschitz estimate that controls how fast trajectories can separate from the constant trajectory $z(t)=\varphi(x_0)$. Choose $\rho>0$ with \begin{align*} 0<\rho<r e^{-L} \end{align*} and define \begin{align*} V=\varphi^{-1}\bigl(B(\varphi(x_0),\rho)\bigr). \end{align*} For $x\in V$, let $z_x:[0,T_x)\to\varphi(W)$ be the maximal solution of \begin{align*} \frac{d}{dt}z_x(t)=Y(t,z_x(t)),\qquad z_x(0)=\varphi(x). \end{align*} Local existence and smooth dependence for time-dependent ordinary differential equations gives this solution for at least a short time, and it is smooth in the initial point $x$. As long as the solution lies in $B(\varphi(x_0),r)$, the Lipschitz estimate with $z'=\varphi(x_0)$ gives \begin{align*} |Y(t,z_x(t))|\le L|z_x(t)-\varphi(x_0)|. \end{align*} Thus the distance from $z_x(t)$ to the fixed point $\varphi(x_0)$ satisfies the differential inequality \begin{align*} \left|\frac{d}{dt}|z_x(t)-\varphi(x_0)|\right|\le L|z_x(t)-\varphi(x_0)| \end{align*} at all times where $z_x(t)\ne\varphi(x_0)$. At times where the distance is zero, the same estimate is interpreted through the upper Dini derivative, which is enough for Gronwall's inequality. Therefore \begin{align*} |z_x(t)-\varphi(x_0)|\le e^{Lt}|z_x(0)-\varphi(x_0)|. \end{align*} Since $t\in[0,1]$ and $x\in V$, this yields \begin{align*} |z_x(t)-\varphi(x_0)|<e^L\rho<r. \end{align*} So the trajectory cannot hit the boundary of the coordinate ball before time $1$. The ordinary differential equation continuation criterion says that a maximal solution can fail to extend only by leaving every compact subset of the domain of the vector field. Here the trajectory remains in the compact set $\overline{B}(\varphi(x_0),e^L\rho)$, which is contained in $B(\varphi(x_0),r)\subset\varphi(W)$. Hence the solution exists on all of $[0,1]$. Defining \begin{align*} \psi(t,x)=\varphi^{-1}(z_x(t)) \end{align*} gives a smooth map $\psi:[0,1]\times V\to U$, and the initial condition gives $\psi_0=\operatorname{id}_V$.[/guided]

[step:Differentiate the pulled-back forms along the flow] For each $t\in[0,1]$, write $\psi_t:V\to U$ for $x\mapsto\psi(t,x)$. Since $\psi_t$ is the time-dependent flow of $X_t$, the standard differentiation formula for pullbacks along a time-dependent flow gives \begin{align*} \frac{d}{dt}\bigl(\psi_t^*\omega_t\bigr)=\psi_t^*\left(\frac{d}{dt}\omega_t+\mathcal L_{X_t}\omega_t\right). \end{align*} The family $(\omega_t)$ is symplectic, so each $\omega_t$ is closed: \begin{align*} d\omega_t=0. \end{align*} By Cartan's formula for the Lie derivative of differential forms, \begin{align*} \mathcal L_{X_t}\omega_t=d(\iota_{X_t}\omega_t)+\iota_{X_t}(d\omega_t). \end{align*} Using $d\omega_t=0$ and $\iota_{X_t}\omega_t=-\beta_t$, we obtain \begin{align*} \mathcal L_{X_t}\omega_t=-d\beta_t. \end{align*} Since $d\beta_t=d\sigma_t|_W$ and $\frac{d}{dt}\omega_t=d\sigma_t$, it follows on $W$ that \begin{align*} \frac{d}{dt}\omega_t+\mathcal L_{X_t}\omega_t=d\sigma_t-d\beta_t=0. \end{align*} Therefore \begin{align*} \frac{d}{dt}\bigl(\psi_t^*\omega_t\bigr)=0 \end{align*} as a $2$-form on $V$ for every $t\in[0,1]$. [/step]

[step:Evaluate at $t=0$ to obtain the local symplectic isotopy] Since the derivative of $t\mapsto \psi_t^*\omega_t$ is zero on $[0,1]$, the family $\psi_t^*\omega_t$ is constant in $t$. Evaluating at $t=0$ and using $\psi_0=\operatorname{id}_V$, we get \begin{align*} \psi_t^*\omega_t=\psi_0^*\omega_0=\omega_0|_V \end{align*} for every $t\in[0,1]$. Thus $V\subset U$ and $\psi:[0,1]\times V\to U$ satisfy all required conclusions. [/step]