A transversal return of a C¹ flow yields a local C¹ Poincaré map, with inverse given by the reversed crossing

By [F1] there are an open interval $J_0\ni t_0$ and an open set $A_0\subseteq W$ with $x_0\in A_0$ with $J_0\times A_0\subseteq \mathrm{\Omega }$ and $\mathrm{\Phi }(J_0\times A_0)\subseteq W$ (shrink $A_0$ if necessary, using that $\mathrm{\Phi }(t_0,x_0)=x_1\in W$ and continuity of $\mathrm{\Phi }$). Define $G:J_0\times A_0\to \mathbb{R}$, $G(t,x):=h(\mathrm{\Phi }(t,x))$; it is $C^1$ by [F1] and [F4], and $G(t_0,x_0)=h(x_1)=0,{\mathrm{\partial }}_{t}G(t_0,x_0)=Dh(x_1)F(x_1)\mathrm{\ne }0.$

Apply [F3] to $G$ at $(t_0,x_0)$: there are open $J_1\subseteq J_0$ containing $t_0$, open $A_1\subseteq A_0$ containing $x_0$ and a unique $C^1$ function $\tau :A_1\to J_1$ with $\tau (x_0)=t_0$ such that for $(t,x)\in J_1\times A_1$ $h(\mathrm{\Phi }(t,x))=0⟺t=\tau (x)\mathrm{.}$ In particular $h(\mathrm{\Phi }(\tau (x),x))=0$ for $x\in A_1$, which is the first part of clause 1.

Reverse-time construction. The field $-F$ is $C^1$ on $U$, and $\mathrm{\Psi }(t,y):=\mathrm{\Phi }(-t,y)$ is its local flow on the reflected domain, because $s\mapsto \mathrm{\Phi }(-s,y)$ solves $y^{\mathrm{\prime }}=-F(y)$ and every solution is a restriction of the corresponding maximal solution [F2]. Apply steps 1.1-2.1 to $-F$ at the initial point $x_1$ and time $t_0$: the reversed reference solution satisfies $\mathrm{\Psi }(t_0,x_1)=\mathrm{\Phi }(-t_0,x_1)=x_0$ by [F2], and its crossing derivative is ${\mathrm{\partial }}_t(h(\mathrm{\Psi }(t,y))){\mid }_{(t_0,x_1)}=Dh(x_0)(-F(x_0))\mathrm{\ne }0$. Hence there are an open interval $J_2\ni t_0$, an open set $B_0\subseteq W$ with $x_1\in B_0$ and a $C^1$ function $\sigma :B_0\to J_2$ with $\sigma (x_1)=t_0$ such that $h(\mathrm{\Phi }(-t,y))=0,(t,y)\in J_2\times B_0⟺t=\sigma (y)\mathrm{.}$ Define $Q:B_0\to \mathrm{\Sigma }$, $Q(y):=\mathrm{\Phi }(-\sigma (y),y)$; then $Q$ is $C^1$ (composition of the $C^1$ map $(t,y)\mapsto \mathrm{\Phi }(t,y)$ with $y\mapsto (-\sigma (y),y)$, whose values lie in $\mathrm{\Omega }$ because $\sigma (y)\in J_2$) and $h(Q(y))=0$ for every $y\in B_0$.

Define $P(x):=\mathrm{\Phi }(\tau (x),x)$ for $x\in A_1$. Then $P$ is $C^1$ as a composition of $C^1$ maps, and $P(x)\in \mathrm{\Sigma }$ for every $x\in A_1$ by step 2.1. Since $P(x_0)=x_1$, continuity of $P$ gives an open set $A_2\subseteq A_1$ with $x_0\in A_2$ and $P(A_2)\subseteq W$.

Shrinking the two neighbourhoods. Choose an open set $A\subseteq A_2$ with $x_0\in A$, $\tau (A)\subseteq J_2$ and $P(A)\subseteq B_0$; this is possible because $\tau (x_0)=t_0\in J_2$, $P(x_0)=x_1\in B_0$ and $\tau ,P$ are continuous. Define $B:=B_0\cap {\sigma }^{-1}(J_1)\cap Q^{-1}(A)\mathrm{.}$ Then $B$ is open (preimages of open sets under the continuous maps $\sigma ,Q$) and contains $x_1$: $\sigma (x_1)=t_0\in J_1$ and $Q(x_1)=\mathrm{\Phi }(-t_0,x_1)=x_0\in A$ by [F2]. By construction $Q(B)\subseteq A$ and $\sigma (B)\subseteq J_1$, so restricting $\tau$ to $A$ and $\sigma$ to $B$ gives the $C^1$ functions of clauses 1 and 2, with $P:A\to \mathrm{\Sigma }$ and $Q:B\to \mathrm{\Sigma }$ both $C^1$ and mapping into $\mathrm{\Sigma }$ (steps 2.1 and 2.2).

First inverse identity. Let $x\in A\cap \mathrm{\Sigma }$ and put $y:=P(x)\in B_0$ and $s:=\tau (x)\in J_2$. Since $h(x)=0$ and $\mathrm{\Phi }(-s,y)=\mathrm{\Phi }(-\tau (x),\mathrm{\Phi }(\tau (x),x))=x$ by [F2], we have $h(\mathrm{\Phi }(-s,y))=0$; the equivalence of step 2.2 with the pair $(s,y)\in J_2\times B_0$ therefore gives $\sigma (y)=s=\tau (x)$. Hence $\sigma (y)=\tau (x)\in J_1$, so $y\in {\sigma }^{-1}(J_1)$, and $Q(y)=\mathrm{\Phi }(-\sigma (y),y)=\mathrm{\Phi }(-\tau (x),y)=x\in A$, so $y\in Q^{-1}(A)$; therefore $y\in B$. Thus $P(A\cap \mathrm{\Sigma })\subseteq B$, and $Q(P(x))=\mathrm{\Phi }(-\sigma (y),y)=x$ for every $x\in A\cap \mathrm{\Sigma }$.

Second inverse identity. Let $y\in B\cap \mathrm{\Sigma }\cap Q^{-1}(A)$ and put $x:=Q(y)\in A\cap \mathrm{\Sigma }$ (the value lies in $\mathrm{\Sigma }$ by step 2.2). By [F2], $\mathrm{\Phi }(\sigma (y),x)=\mathrm{\Phi }(\sigma (y),\mathrm{\Phi }(-\sigma (y),y))=y$, and $h(y)=0$ because $y\in \mathrm{\Sigma }$; the equivalence of step 2.1 with the pair $(\sigma (y),x)\in J_1\times A_1$ therefore gives $\sigma (y)=\tau (x)$. Hence $P(Q(y))=\mathrm{\Phi }(\tau (x),x)=\mathrm{\Phi }(\sigma (y),x)=y$.

Assume the additional exclusion in clause 4. Choose $\epsilon >0$ with $3\epsilon <t_0$ sufficiently small that the reference paths for $0\le s\le \epsilon$ and $t_0-2\epsilon \le s\le t_0+2\epsilon$ lie in $W$, and $Dh(\mathrm{\Phi }(s,x_0))F(\mathrm{\Phi }(s,x_0))$ has a fixed nonzero sign on each of these two intervals. Continuity and compactness give an open neighbourhood $A_3$ of $x_0$ such that all these paths for $x\in A_3$ remain in $W$ with the same respective signs. The compact middle path $K=\{\mathrm{\Phi }(s,x_0):\epsilon \le s\le t_0-\epsilon \}$ lies in the open set $U\setminus {\bar{\mathrm{\Sigma }}}^U$ by the additional hypothesis. By openness of $\mathrm{\Omega }$, continuity of $\mathrm{\Phi }$ and finite product-neighbourhood covers of the compact time intervals, further shrink $A_3$ so that $[0,t_0+2\epsilon ]\times A_3\subseteq \mathrm{\Omega }$ and $\mathrm{\Phi }([\epsilon ,t_0-\epsilon ]\times A_3)\subseteq U\setminus {\bar{\mathrm{\Sigma }}}^U$. Also shrink it so that $\mathrm{\mid }\tau (x)-t_0\mathrm{\mid }<\epsilon$.

Since $Dh(x_i)F(x_i)\mathrm{\ne }0$, one has $Dh(x_i)\mathrm{\ne }0$ at each $x_i$ ($i=0,1$). For $n\ge 2$, after permuting coordinates, [F3] makes $\mathrm{\Sigma }$ near $x_i$ the graph $x_n=f_i(x_1,\dots ,x_{n-1})$ of a $C^1$ function; its graph parametrization and coordinate projection are an embedded section chart. For $n=1$, continuity of $h^{\mathrm{\prime }}$ and $h^{\mathrm{\prime }}(x_i)\mathrm{\ne }0$ give an interval on which $h^{\mathrm{\prime }}$ has a fixed nonzero sign; the mean value theorem [F5] makes $h$ strictly monotone there, so $\mathrm{\Sigma }$ in that interval is the singleton $\{x_i\}$, a zero-dimensional section chart. Shrink $A$ into the chart near $x_0$ and shrink $B_0$ into the target chart near $x_1$; by continuity of $P$ shrink $A$ further so $P(A)\subseteq B_0$, then repeat the construction of $B$ in step 3.2. In dimensions $n\ge 2$, the chart coordinates of $P:A\to \mathrm{\Sigma }$ and $Q:B\to \mathrm{\Sigma }$ are ambient coordinate projections composed with ambient $C^1$ maps, so both are $C^1$; for $n=1$ these maps are locally constant into the singleton charts. Their restrictions to the sections are therefore $C^1$ in the same charts. The inverse identities of steps 3.3--3.4 remain valid after these neighbourhood restrictions.

By steps 3.3 and 3.4, $P(A\cap \mathrm{\Sigma })=B\cap \mathrm{\Sigma }\cap Q^{-1}(A)$: if $y=P(x)$ with $x\in A\cap \mathrm{\Sigma }$, then $y\in B$, $y\in \mathrm{\Sigma }$ and $Q(y)=x\in A$; conversely, if $y\in B\cap \mathrm{\Sigma }\cap Q^{-1}(A)$, then $x:=Q(y)\in A\cap \mathrm{\Sigma }$ and step 3.4 gives $P(x)=y$. Since $B$ and $Q^{-1}(A)$ are open in ${\mathbb{R}}^n$, the set $B\cap \mathrm{\Sigma }\cap Q^{-1}(A)$ is open in the relative topology of $\mathrm{\Sigma }$. On $P(A\cap \mathrm{\Sigma })$ the maps $P$ and $Q$ are inverse to each other by steps 3.3 and 3.4, and both are $C^1$ in the charts of step 4.1, so $P:A\cap \mathrm{\Sigma }\to P(A\cap \mathrm{\Sigma })$ is a $C^1$ diffeomorphism with inverse $Q{\mathrm{\mid }}_{P(A\cap \mathrm{\Sigma })}$. This proves clauses 2 and 3.

Repeat step 3.2 with $A\subseteq A_3$, retaining its inverse identities and charts. For $x\in A\cap \mathrm{\Sigma }$, the scalar function $s\mapsto h(\mathrm{\Phi }(s,x))$ is strictly monotone on $[0,\epsilon ]$ and vanishes at $0$, hence has no zero for $0<s\le \epsilon$. The middle interval has no $\mathrm{\Sigma }$-hits by step 3.5. On $[t_0-2\epsilon ,t_0+2\epsilon ]$ the scalar function is strictly monotone and has its unique zero at $\tau (x)$; thus it has no zero on $[t_0-\epsilon ,\tau (x))$. These intervals cover $(0,\tau (x))$, and $\tau (x)>t_0-\epsilon >0$. This proves the first-return conclusion of clause 4, together with clauses 1-3 already proved.