SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\left\| \tilde {\phi} _ {0} (x) \right\| _ {\mathfrak {g} _ {0}} \leq \left\| \tilde {\phi} _ {0} (x) - \tilde {\phi} _ {0} \left(x ^ {(0)}\right) \right\| _ {\mathfrak {g} _ {0}} + \left\| \tilde {\phi} _ {0} \left(x ^ {(0)}\right) \right\| _ {\mathfrak {g} _ {0}} \leq (1 + 2 4 3 2 / 5 r) \| x - x ^ {(0)} \| _ {\mathfrak {g} _ {0}} <   (1 + 2 4 3 2 / 5 r) r. \tag {48}
$$
Since  $r \leq 1/11$ , we have  $1 + (1 - r)^{-2} \leq 5/2$ , and by combining (46),(47) and (48), we obtain
$$
\| D \tilde {\phi} _ {0} (x) [ w ] \| \leq (5 h / 4) \| D G _ {h, x ^ {(0)}} (x) [ w ] \| _ {\mathfrak {g} _ {0} ^ {- 1}} + 4 8 7 4 (1 - r) ^ {- 7} r \| w \| _ {\mathfrak {g} _ {0}}.
$$
Since  $\| D\tilde{\phi}_0(x)[w]\|_{\mathfrak{g}_0} > \| w\|_{\mathfrak{g}_0} / 2$  and  $(1 - r)^{-7} \leq 2$  with  $r \leq 1 / 11$ , the last inequality becomes
$$
\begin{array}{l} (5 h / 4) \| D G _ {h, x ^ {(0)}} (x) [ w ] \| _ {\mathfrak {g} _ {0} ^ {- 1}} \geq \left\{1 / 2 - 9 7 4 8 r \right\} \| w \| _ {\mathfrak {g} _ {0}}, \\ \geq 1 / 2 - r / \left(4 r _ {\alpha} ^ {*}\right) \geq 1 / 4. \tag {see(33)} \\ \end{array}
$$
In particular,  $\operatorname{Jac}[\mathrm{G}_{h,x^{(0)}}](x)$  is invertible, which concludes the proof.
We prove below that, for any  $z^{(0)} \in \mathrm{T}^{\star}\mathsf{M}$ , the set  $\underline{\mathrm{G}}_h(z^{(0)})$  is reduced to a single point, for  $h$  small enough depending on  $z^{(0)}$ .
Lemma 23. For any  $w\geq 0$  ,any  $r\in (0,1)$  and any  $\alpha \geq 1$  ,we define
$$
h _ {1} (w, r, \alpha) = \min  \left(\frac {1}{1 + (1 - r) ^ {- 2} + 2 (2 r + w)}, \frac {(1 - r) ^ {3}}{3 (r + w)}, \frac {r}{(r + w) (1 + 3 r / 2) + 1 / 2 (r + w) ^ {2}}\right). \tag {49}
$$
We recall that the set-valued map  $\underline{\mathbf{G}}_h$  is defined in (29). Assume A1, A2. Then, for any  $r\in (0,1 / 11]$  any  $z = (x,p)\in \mathrm{T}^{\star}\mathsf{M}$  and any  $h\in (0,h_1(\| p\|_{\mathfrak{g}(x)^{-1}},r,\alpha))$ , there exists a unique  $z'\in \mathsf{B}_{\| \cdot \| _z}(z,r)\subset$ $\mathrm{T}^{\star}\mathsf{M}$  such that  $z'\in \underline{\mathbf{G}}_h(z)$ .
Proof. Assume A1, A2. Let  $r \in (0, 1/11]$ ,  $z = (x,p) \in \mathrm{T}^{\star}\mathsf{M}$  and  $h \in (0,h_1(\| p\|_{\mathfrak{g}(x)^{-1}},r,\alpha))$  where  $h_1$  is defined in (49). We recall that the map  $\underline{\mathrm{g}}_h$  is defined in (30). We define  $\mathsf{B} = \mathsf{B}_{\|\cdot\|_z}(z,r)$  and the map  $\underline{\mathrm{g}}_{h,z}: \mathrm{T}^{\star}\mathsf{M} \to \mathbb{R}^{n} \times \mathbb{R}^{n}$  by  $\underline{\mathrm{g}}_{h,z}(z') = z' - \underline{\mathrm{g}}_h(z,z')$  for any  $z' \in \mathrm{T}^{\star}\mathsf{M}$ . Then,  $\underline{\mathrm{g}}_{h,z}(z') = z'$  if and only if  $z' \in \underline{\mathrm{G}}_h(z)$ . The proof is divided in two steps.
Step 1. We first prove that  $\underline{\mathsf{g}}_{h,z}(\mathsf{B})\subset \mathsf{B}$ . Let  $z^{\prime}\in \mathsf{B}$ , we have by Lemma 9-(a)
$$
\begin{array}{l} \| \underline {{\mathfrak {g}}} _ {h, z} \left(z ^ {\prime}\right) - z \| _ {z} = \frac {h}{2} \| \{\mathfrak {g} (x) ^ {- 1} + \mathfrak {g} \left(x ^ {\prime}\right) ^ {- 1} \} p ^ {\prime} \| _ {\mathfrak {g} (x)} + \frac {h}{4} \| D \mathfrak {g} (x) [ \mathfrak {g} (x) ^ {- 1} p ^ {\prime}, \mathfrak {g} (x) ^ {- 1} p ^ {\prime} ] \| _ {\mathfrak {g} (x) ^ {- 1}} \\ = \frac {h}{2} \| 2 \mathfrak {g} (x) ^ {- 1} p ^ {\prime} + \{\mathfrak {g} (x ^ {\prime}) ^ {- 1} - \mathfrak {g} (x) ^ {- 1} \} p ^ {\prime} \| _ {\mathfrak {g} (x)} + \frac {h}{4} \| D \mathfrak {g} (x) [ \mathfrak {g} (x) ^ {- 1} p ^ {\prime}, \mathfrak {g} (x) ^ {- 1} p ^ {\prime} ] \| _ {\mathfrak {g} (x) ^ {- 1}} \\ \leq \frac {h}{2} (2 \| p ^ {\prime} \| _ {\mathfrak {g} (x) ^ {- 1}} + \| \mathfrak {g} (x) ^ {1 / 2} (\mathfrak {g} (x ^ {\prime}) ^ {- 1} - \mathfrak {g} (x) ^ {- 1}) \mathfrak {g} (x) ^ {1 / 2} \| _ {2} \| p ^ {\prime} \| _ {\mathfrak {g} (x) ^ {- 1}}) + \frac {h}{2} \| p ^ {\prime} \| _ {\mathfrak {g} (x) ^ {- 1}} ^ {2}, \\ \end{array}
$$
where the following inequalities hold
(a)  $\| p' \|_{\mathfrak{g}(x)^{-1}} = \| p' - p + p \|_{\mathfrak{g}(x)^{-1}} \leq r + \| p \|_{\mathfrak{g}(x)^{-1}}$ .
(b)  $\| \mathfrak{g}(x)^{1 / 2}(\mathfrak{g}(x')^{-1} - \mathfrak{g}(x)^{-1})\mathfrak{g}(x)^{1 / 2}\| _2\leq 3\| x - x'\|_{\mathfrak{g}(x)}\leq 3r,$  on the model of (18), using  $r\leq 1 / 11$
$$
\begin{array}{l} \left\| \underline {{\mathrm {g}}} _ {h, z} \left(z ^ {\prime}\right) - z \right\| _ {z} \leq h \left(\left(r + \| p \| _ {\mathfrak {g} (x) ^ {- 1}}\right) + \frac {3}{2} r \left(r + \| p \| _ {\mathfrak {g} (x) ^ {- 1}}\right) + \frac {1}{2} \left(r + \| p \| _ {\mathfrak {g} (x) ^ {- 1}}\right) ^ {2}\right) \\ \leq h \left(\left(r + \| p \| _ {\mathfrak {g} (x) ^ {- 1}}\right) \left(1 + \frac {3}{2} r\right) + \frac {1}{2} \left(r + \| p \| _ {\mathfrak {g} (x) ^ {- 1}}\right) ^ {2}\right) \\ \leq h r / h _ {1} \left(\| p \| _ {\mathfrak {g} (x) ^ {- 1}}, r, \alpha\right) <   r, \\ \end{array}
$$
which proves the statement.
Step 2. We now prove that  $\underline{\mathsf{g}}_{h,z}$  is a contraction on  $\mathsf{B}$ . This proof notably recovers some elements of the proof of Proposition 13 (see Appendix E). Let  $(z_1, z_2) \in \mathsf{B} \times \mathsf{B}$  with  $z_1 = (x_1, p_1)$  and  $z_2 = (x_2, p_2)$ . Remark that  $x_1, x_2 \in \mathsf{W}^0(x, 1) \times \mathsf{W}^0(x, 1)$ . Let us first bound  $\|\underline{\mathsf{g}}_{h,z}^{(1)}(z_1) - \underline{\mathsf{g}}_{h,z}^{(1)}(z_2)\|_{\mathfrak{g}(x)}$ . We have
$$
\begin{array}{l} \underline {{\mathbf {g}}} _ {h, z} ^ {(1)} (z _ {1}) - \underline {{\mathbf {g}}} _ {h, z} ^ {(1)} (z _ {2}) = \frac {h}{2} \left\{\mathfrak {g} (x) ^ {- 1} (p _ {1} - p _ {2}) + \mathfrak {g} (x _ {1}) ^ {- 1} p _ {1} - \mathfrak {g} (x _ {2}) ^ {- 1} p _ {2} \right\} \\ = \frac {h}{2} \left\{\mathfrak {g} (x) ^ {- 1} \left(p _ {1} - p _ {2}\right) + \mathfrak {g} \left(x _ {1}\right) ^ {- 1} \left(p _ {1} - p _ {2}\right) + \left(\mathfrak {g} \left(x _ {1}\right) ^ {- 1} - \mathfrak {g} \left(x _ {2}\right) ^ {- 1}\right) p _ {2} \right\}, \\ \end{array}
$$
and then, since  $r \leq 1/11$ , we have on the model of (18)
$$
\begin{array}{l} \| \underline {{\mathbf {g}}} _ {h, z} ^ {(1)} (z _ {1}) - \underline {{\mathbf {g}}} _ {h, z} ^ {(1)} (z _ {2}) \| _ {\mathfrak {g} (x)} \leq \frac {h}{2} \{1 + \| \mathfrak {g} (x) ^ {1 / 2} \mathfrak {g} (x _ {1}) ^ {- 1} \mathfrak {g} (x) ^ {1 / 2} \| _ {2} \} \| p _ {1} - p _ {2} \| _ {\mathfrak {g} (x) ^ {- 1}} \\ + \frac {h}{2} \| \mathfrak {g} (x) ^ {1 / 2} (\mathfrak {g} (x _ {1}) ^ {- 1} - \mathfrak {g} (x _ {2}) ^ {- 1}) \mathfrak {g} (x) ^ {1 / 2} \| _ {2} \| p _ {2} \| _ {\mathfrak {g} (x) ^ {- 1}} \\ \leq \frac {h}{2} \left\{1 + (1 - r) ^ {- 2} \right\} \| p _ {1} - p _ {2} \| _ {\mathfrak {g} (x) ^ {- 1}} + \frac {3 h}{2} (1 - r) ^ {- 3} \| x _ {1} - x _ {2} \| _ {\mathfrak {g} (x)} (r + \| p \| _ {\mathfrak {g} (x) ^ {- 1}}). \\ \end{array}
$$
Let us now bound  $\| \underline{\mathsf{g}}_{h,z}^{(2)}(z_1) - \underline{\mathsf{g}}_{h,z}^{(2)}(z_2)\|_{\mathfrak{g}(x)^{-1}}$ . We have
$$
\underline {{\mathbf {g}}} _ {h, z} ^ {(2)} (z _ {1}) - \underline {{\mathbf {g}}} _ {h, z} ^ {(2)} (z _ {2}) = \frac {h}{4} \{D \mathfrak {g} (x) [ \mathfrak {g} (x) ^ {- 1} p _ {1}, \mathfrak {g} (x) ^ {- 1} p _ {1} ] - D \mathfrak {g} (x) [ \mathfrak {g} (x) ^ {- 1} p _ {2}, \mathfrak {g} (x) ^ {- 1} p _ {2} ] \},
$$
which gives using Lemma 9-(a)
$$
\begin{array}{l} \left\| \underline {{\mathbf {g}}} _ {h, z} ^ {(2)} \left(z _ {1}\right) - \underline {{\mathbf {g}}} _ {h, z} ^ {(2)} \left(z _ {2}\right) \right\| _ {\mathfrak {g} (x) ^ {- 1}} = \frac {h}{2} \left\| D \mathfrak {g} (x) \left[ \mathfrak {g} (x) ^ {- 1} \left(p _ {1} - p _ {2}\right), \mathfrak {g} (x) ^ {- 1} p _ {2} \right] \right\| _ {\mathfrak {g} (x) ^ {- 1}} \\ + \frac {h}{4} \| D \mathfrak {g} (x) [ \mathfrak {g} (x) ^ {- 1} (p _ {1} - p _ {2}), \mathfrak {g} (x) ^ {- 1} (p _ {1} - p _ {2}) ] \| _ {\mathfrak {g} (x) ^ {- 1}} \\ \leq h \| \mathfrak {g} (x) ^ {- 1} (p _ {1} - p _ {2}) \| _ {\mathfrak {g} (x)} \| \mathfrak {g} (x) ^ {- 1} p _ {2} \| _ {\mathfrak {g} (x)} + \frac {h}{2} \| \mathfrak {g} (x) ^ {- 1} (p _ {1} - p _ {2}) \| _ {\mathfrak {g} (x)} ^ {2} \\ \leq h \left(r + \| p \| _ {\mathfrak {g} (x) ^ {- 1}}\right) \| p _ {1} - p _ {2} \| _ {\mathfrak {g} (x) ^ {- 1}} + h r \| p _ {1} - p _ {2} \| _ {\mathfrak {g} (x) ^ {- 1}}. \\ \end{array}
$$
Finally, it comes that
$$
\begin{array}{l} \left\| \underline {{\mathrm {g}}} _ {h, z} \left(z _ {1}\right) - \underline {{\mathrm {g}}} _ {h, z} \left(z _ {2}\right) \right\| _ {z} \leq \frac {h}{2} \left\{1 + (1 - r) ^ {- 2} + 4 r + 2 \| p \| _ {\mathfrak {g} (x) ^ {- 1}} \right\} \| p _ {1} - p _ {2} \| _ {\mathfrak {g} (x) ^ {- 1}} \\ + \frac {3 h}{2} (1 - r) ^ {- 3} (r + \| p \| _ {\mathfrak {g} (x) ^ {- 1}}) \| x _ {1} - x _ {2} \| _ {\mathfrak {g} (x)} \\ \leq (1 / 2) h \left\{\| x _ {1} - x _ {2} \| _ {\mathfrak {g} (x)} + \| p _ {1} - p _ {2} \| _ {\mathfrak {g} (x) ^ {- 1}} \right\} / h _ {1} (\| p \| _ {\mathfrak {g} (x) ^ {- 1}}, r, \alpha) \\ <   (1 / 2) \| z _ {1} - z _ {2} \| _ {z}, \\ \end{array}
$$
which proves that  $\underline{\mathbf{g}}_{h,z}$  is a contraction on  $\mathsf{B}$ .
Conclusion. We obtain the result of Lemma 23 by applying the fixed point theorem on  $\underline{\mathbf{g}}_{h,z}$  and B.
Elaborating on Lemma 23, we prove in Lemma 24 that the only element of  $\underline{\mathbf{G}}_h(z^{(0)})$  verifies smoothness properties if  $h$  is chosen small enough, depending on  $z^{(0)}$ .
Lemma 24. For any  $z = (x,p)\in \mathrm{T}^{\star}\mathsf{M}$ , any  $r\in (0,1)$  and any  $\alpha \geq 1$ , we define
$$
\bar {h} (z, r, \alpha) = \min  \left(h _ {1} \left(\| p \| _ {\mathfrak {g} (x) ^ {- 1}}, r, \alpha\right), h _ {2} (z), h _ {3} (z, r)\right) \tag {50}
$$
where  $h_1$  is defined in (49), and  $h_2$  and  $h_3$  are defined by
$$
h _ {2} (z) = \frac {1}{3 / 2 + 3 \| p \| _ {\mathfrak {g} (x) ^ {- 1}}}, \quad h _ {3} (z, r) = \frac {1}{1 + (1 - r) ^ {- 1} (r + \| p \| _ {\mathfrak {g} (x) ^ {- 1}})}.
$$
We recall that  $r_{\alpha}^{\star}$  is defined in (33). We also recall that the maps  $\underline{\mathbf{G}}_h$ ,  $\underline{\mathbf{g}}_h$ ,  $\overline{\mathbf{G}}_h$  and  $\mathbf{G}_{h,x^{(0)}}$  are respectively defined in (29), (30), (31) and (32). Assume A1, A2. Then, for any  $z^{(0)} \in \mathrm{T}^{\star}\mathsf{M}$  and any  $h \in (0,\bar{h}(z^{(0)},r_{\alpha}^{\star},\alpha))$ , there exists a unique element  $z_h^{(1/2)} \in \mathrm{T}^{\star}\mathsf{M}$  such that  $z_h^{(1/2)} \in \underline{\mathbf{G}}_h(z^{(0)}) \cap \mathsf{B}_{\|\cdot\|_{z^{(0)}}}(z^{(0)},r_{\alpha}^{\star})$ . Moreover, we have
(a)  $\operatorname{Jac}_{z'}[\underline{\mathbf{g}}_h](z^{(0)}, z_h^{(1/2)})$  and  $\operatorname{Jac}[\overline{\mathbf{G}}_h](z_h^{(1/2)})$  are invertible,
(b)  $\operatorname{Jac}[\mathrm{G}_{h,x^{(0)}}](x_h^{(1/2)})$  is invertible.
Proof. Assume A1, A2. Let  $z^{(0)} \in \mathrm{T}^{\star}\mathsf{M}$  and  $h \in (0,\bar{h}(z^{(0)},r_{\alpha}^{\star},\alpha))$ , where  $\bar{h}$  is defined in (50). Since  $r_{\alpha}^{\star} \leq 1/11$ , Lemma 23 ensures the existence and uniqueness of  $z_h^{(1/2)} = (x_h^{(1/2)},p_h^{(1/2)}) \in \mathrm{T}^{\star}\mathsf{M}$  such that  $z_h^{(1/2)} \in \underline{\mathrm{G}}_h(z^{(0)}) \cap \mathsf{B}_{\|\cdot\|_{z^{(0)}}}(z^{(0)},r_{\alpha}^{\star})$ . Moreover, we have  $x_h^{(1/2)} \in \mathsf{W}^0(x^{(0)},r_{\alpha}^{\star})$  and thus  $\operatorname{Jac}[\mathrm{G}_{h,x^{(0)}}](x_h^{(1/2)})$  is invertible by Lemma 22. Let us prove that  $\operatorname{Jac}_{z'}[\underline{\mathrm{g}}_h](z^{(0)},z_h^{(1/2)})$  and  $\operatorname{Jac}[\overline{\mathrm{G}}_h](z_h^{(1/2)})$  are invertible.
Invertibility of  $\operatorname{Jac}_{z'}[\underline{\mathsf{g}}_h](z^{(0)},z_h^{(1 / 2)})$ . We first remark that  $\operatorname{Jac}_{z'}[\underline{\mathsf{g}}_h](z^{(0)},z_h^{(1 / 2)}) = \operatorname{Jac}[\psi_{0,h}](z_h^{(1 / 2)})$  where  $\psi_{0,h} = \mathrm{Id} + \frac{h}{2}\psi_0$  and  $\psi_0$  is defined on  $\mathbf{T}^{\star}\mathbf{M}$  by
$$
\psi_ {0} (z) = \left(- \{\mathfrak {g} (x ^ {(0)}) ^ {- 1} + \mathfrak {g} (y) ^ {- 1} \} ^ {- 1} p, - \frac {1}{2} D \mathfrak {g} (x ^ {(0)}) [ \mathfrak {g} (x ^ {(0)}) ^ {- 1} p, \mathfrak {g} (x ^ {(0)}) ^ {- 1} p ]\right).
$$
We recall that  $\| z_h^{(1 / 2)} - z^{(0)}\|_{z^{(0)}} < r_\alpha^\star \leq 1 / 11$  and thus define  $r_1 = 1 / 11 - \| z_h^{(1 / 2)} - z^{(0)}\|_{z^{(0)}} > 0$  and  $\mathsf{B}_1 = \mathsf{B}_{\| \cdot \|_{z^{(0)}}}(z_h^{(1 / 2)},r_1)$ . We set  $\tilde{r} = 1 / 11$ . Note that  $\mathsf{B}_1\subset \mathsf{B}_{\| \cdot \|_{z^{(0)}}}(z^{(0)},1 / 11)$ . We show
below that  $\psi_0$  is Lipschitz-continuous on  $\mathsf{B}_1$  with respect to  $\| \cdot \|_{z^{(0)}}$ . Let  $(z,z^{\prime})\in \mathsf{B}_{1}\times \mathsf{B}_{1}$ . Similarly to the proof of Lemma 23, we have
$$
\begin{array}{l} \left\| \psi_ {0} ^ {(1)} (z) - \psi_ {0} ^ {(1)} \left(z ^ {\prime}\right) \right\| _ {\mathfrak {g} \left(x ^ {(0)}\right)} \leq \left\| p - p ^ {\prime} \right\| _ {\mathfrak {g} \left(x ^ {(0)}\right) ^ {- 1}} \\ + \| \mathfrak {g} (x ^ {(0)}) ^ {1 / 2} \mathfrak {g} (x) ^ {- 1} \mathfrak {g} (x ^ {(0)}) ^ {1 / 2} \| _ {2} \| p - p ^ {\prime} \| _ {\mathfrak {g} (x ^ {(0)}) ^ {- 1}} \\ + \| \mathfrak {g} (x ^ {(0)}) ^ {1 / 2} \{\mathfrak {g} (x) ^ {- 1} - \mathfrak {g} (x ^ {\prime}) ^ {- 1} \} \mathfrak {g} (x ^ {(0)}) ^ {1 / 2} \| _ {2} \| p \| _ {\mathfrak {g} (x ^ {(0)}) ^ {- 1}} \\ \leq \left(1 + \left(1 - \tilde {r}\right) ^ {- 2}\right) \| p - p ^ {\prime} \| _ {\mathfrak {g} \left(x ^ {(0)}\right) ^ {- 1}} \\ + 3 (1 - \tilde {r}) ^ {- 3} (\tilde {r} + \| p ^ {(0)} \| _ {\mathfrak {g} (x ^ {(0)}) ^ {- 1}}) \| x - x ^ {\prime} \| _ {\mathfrak {g} (x ^ {(0)})}. \\ \end{array}
$$
In the same manner, we have
$$
\| \psi_ {0} ^ {(2)} (z) - \psi_ {0} ^ {(2)} \left(z ^ {\prime}\right) \| _ {\mathfrak {g} \left(x ^ {(0)}\right) ^ {- 1}} \leq 2 \left(2 \tilde {r} + \| p ^ {(0)} \| _ {\mathfrak {g} \left(x ^ {(0)}\right) ^ {- 1}}\right) \| p - p ^ {\prime} \| _ {\mathfrak {g} \left(x ^ {(0)}\right) ^ {- 1}}.
$$
Therefore, recalling that  $\tilde{r} = 1 / 11$ , we obtain with the previous inequalities
$$
\begin{array}{l} \left\| \psi_ {0} (z) - \psi_ {0} \left(z ^ {\prime}\right) \right\| _ {z ^ {(0)}} \leq \left\{3 + 6 \| p ^ {(0)} \| _ {\mathfrak {g} \left(x ^ {(0)}\right) ^ {- 1}} \right\} \| z - z ^ {\prime} \| _ {z ^ {(0)}} \\ \leq 2 \| z - z ^ {\prime} \| _ {z ^ {(0)}} / h _ {2} \left(z ^ {(0)}\right). \\ \end{array}
$$
Hence,  $\psi_0$  is  $(2 / h_{2}(z^{(0)}))$ -Lipschitz-continuous on  $\mathsf{B}_1$  with respect to  $\| \cdot \|_{z^{(0)}}$ . Then, since  $h < h_2(z^{(0)})$ ,  $\operatorname{Jac}[\psi_{0,h}](z)$  is invertible for any  $z \in \mathsf{B}_1$  by Lemma 21. In particular,  $\operatorname{Jac}_{z'}[\underline{\mathsf{g}}_h](z^{(0)}, z_h^{(1/2)}) = \operatorname{Jac}[\psi_{0,h}](z_h^{(1/2)})$  is invertible.
Invertibility of  $\operatorname{Jac}[\overline{\mathbf{G}}_h](z_h^{(1/2)})$ . Let us remark that  $\overline{\mathbf{G}}_h = \mathrm{Id} + h\zeta$  where  $\zeta$  is defined on  $\mathbf{T}^*\mathbf{M}$  by
$$
\zeta (z) = (0, \frac {1}{4} D \mathfrak {g} (x) [ \mathfrak {g} (x) ^ {- 1} p, \mathfrak {g} (x) ^ {- 1} p ]).
$$
We define  $r_2 = 1/2$  and  $\mathsf{B}_2 = \mathsf{B}_{\|\cdot\|_{z_h^{(1/2)}}}(z_h^{(1/2)}, r_2)$ . We show below that  $\zeta$  is Lipschitz-continuous on  $\mathsf{B}_2$  with respect to  $\|\cdot\|_{z_h^{(1/2)}}$ , with a Lipschitz constant which does not depend on  $z_h^{(1/2)}$ . Let  $(z, z') \in \mathsf{B}_2 \times \mathsf{B}_2$ . Using Lemma 9-(a) with  $r_2 = 1/2$ , it is clear that
$$
\| \zeta (z) - \zeta (z ^ {\prime}) \| _ {z _ {h} ^ {(1 / 2)}} \leq (1 + \| p _ {h} ^ {(1 / 2)} \| _ {\mathfrak {g} (x _ {h} ^ {(1 / 2)}) ^ {- 1}}) \| z - z ^ {\prime} \| _ {z _ {h} ^ {(1 / 2)}}.
$$
Moreover, since  $\| z_h^{(1 / 2)} - z^{(0)}\|_{z^{(0)}} < r_\alpha^\star$ , we have by Lemma 12
$$
\begin{array}{l} \left\| p _ {h} ^ {(1 / 2)} \right\| _ {\mathfrak {g} \left(x _ {h} ^ {(1 / 2)}\right) ^ {- 1}} \leq \left(1 - r _ {\alpha} ^ {\star}\right) ^ {- 1} \left\| p _ {h} ^ {(1 / 2)} \right\| _ {\mathfrak {g} \left(x ^ {(0)}\right) ^ {- 1}} \\ \leq \left(1 - r _ {\alpha} ^ {\star}\right) ^ {- 1} \left(r _ {\alpha} ^ {\star} + \| p ^ {(0)} \| _ {\mathfrak {g} (x ^ {(0)}) ^ {- 1}}\right), \\ \end{array}
$$
$$
\begin{array}{l} \left\| \zeta (z) - \zeta \left(z ^ {\prime}\right) \right\| _ {z _ {h} ^ {(1 / 2)}} \leq \left\{1 + \left(1 - r _ {\alpha} ^ {\star}\right) ^ {- 1} \left(r _ {\alpha} ^ {\star} + \left\| p ^ {(0)} \right\| _ {\mathfrak {g} \left(x ^ {(0)}\right) ^ {- 1}}\right) \right\} \| z - z ^ {\prime} \| _ {z _ {h} ^ {(1 / 2)}} \\ \leq \left\| z - z ^ {\prime} \right\| _ {z _ {h} ^ {(1 / 2)}} / h _ {3} \left(z ^ {(0)}, r _ {\alpha} ^ {\star}\right). \\ \end{array}
$$
Hence,  $\zeta$  is  $(1 / h_{3}(z^{(0)},r_{\alpha}^{\star}))$ -Lipschitz-continuous on  $\mathsf{B}_2$  with respect to  $\| \cdot \|_{z_h^{(1 / 2)}}$ . Since  $h < h_3(z^{(0)},r_\alpha^\star)$ ,  $\operatorname {Jac}[\overline{\mathrm{G}}_h](z)$  is invertible for any  $z\in \mathsf{B}_2$  by Lemma 21. In particular,  $\operatorname {Jac}[\overline{\mathrm{G}}_h](z_h^{(1 / 2)})$  is invertible, which concludes the proof.
The following lemma states the existence of a diffeomorphism between  $z^{(0)} \in \mathrm{T}^{\star}\mathsf{M}$  and  $z^{(1)} \in \mathrm{G}_h(z^{(0)})$  under smoothness conditions verified by  $z^{(1)}$ .
Lemma 25. Let  $(z^{(0)},z^{(1)})\in \mathrm{T}^{\star}\mathsf{M}\times \mathrm{T}^{\star}\mathsf{M}$ . We recall that the maps  $\underline{\mathfrak{g}}_h$ ,  $\overline{\mathrm{G}}_h$  and  $\mathrm{G}_h$  are respectively defined in (30), (31) and (6). Assume that  $\mathfrak{g}\in \mathrm{C}^2 (\mathsf{M},\mathbb{R}^{d\times d})$  and that there exist  $h > 0$  and  $z_{h}^{(1 / 2)}\in \mathrm{T}^{\star}\mathsf{M}$  with the following properties:
(a)  $\underline{\underline{\mathbf{g}}}_h(z^{(0)},z_h^{(1 / 2)}) = 0$  and  $z^{(1)} = \overline{\mathrm{G}}_h(z_h^{(1 / 2)})$
(b)  $\operatorname{Jac}_{z'}[\underline{\mathbf{g}}_h](z^{(0)}, z_h^{(1/2)})$  and  $\operatorname{Jac}[\overline{\mathbf{G}}_h](z_h^{(1/2)})$  are invertible.
Then, there exists a neighbourhood  $\mathsf{U} \subset \mathsf{T}^{\star}\mathsf{M}$  of  $z^{(0)}$  and a  $C^1$ -diffeomorphism  $\xi : \mathsf{U} \to \xi(\mathsf{U}) \subset \mathsf{T}^{\star}\mathsf{M}$  such that (i)  $\xi(z^{(0)}) = z^{(1)}$ , (ii) for any  $z \in \mathsf{U}$ ,  $\xi(z) \in \mathrm{G}_h(z)$  and (iii)  $|\operatorname*{det}\mathrm{Jac}\xi| \equiv 1$ .
Proof. Let  $(z^{(0)}, z^{(1)}) \in \mathrm{T}^{\star}\mathsf{M} \times \mathrm{T}^{\star}\mathsf{M}$ . Assume that  $\mathfrak{g} \in \mathrm{C}(\mathsf{M}, \mathbb{R}^{d \times d})$  and consider  $h > 0$  and  $z_h^{(1/2)} \in \mathrm{T}^{\star}\mathsf{M}$  as described in Lemma 25. Under the assumption on  $\mathfrak{g}$ ,  $\underline{\mathbf{g}}_h$  and  $\overline{\mathbf{G}}_h$  are continuously differentiable respectively on  $\mathrm{T}^{\star}\mathsf{M} \times \mathrm{T}^{\star}\mathsf{M}$  and  $\mathrm{T}^{\star}\mathsf{M}$ . We are going to prove Lemma 25, by first deriving intermediary results on  $\underline{\mathbf{g}}_h$  and  $\overline{\mathbf{G}}_h$ .
Result on  $\underline{\mathbf{g}}_h$ . We recall that  $\underline{\mathbf{g}}_h(z^{(0)},z_h^{(1 / 2)}) = 0$  and  $\mathrm{Jac}_{z'}[\underline{\mathbf{g}}_h](z^{(0)},z_h^{(1 / 2)})$  is invertible. Then, by applying the implicit function theorem on  $\underline{\mathbf{g}}_h$  at  $(z^{(0)},z_h^{(1 / 2)})$ , we obtain the existence of a neighbourhood  $\mathsf{U}_0\subset \mathrm{T}^\star \mathsf{M}$  of  $z^{(0)}$  and a  $\mathrm{C}^1$ -diffeomorphism  $\xi_0:\mathsf{U}_0\to \xi_0(\mathsf{U}_0)\subset \mathrm{T}^\star \mathsf{M}$  such that  $\xi_0(z^{(0)}) = z_h^{(1 / 2)}$  and for any  $z\in \mathsf{U}_0$ ,  $\underline{\mathbf{g}}_h(z,\xi_0(z)) = 0$ , i.e.,  $\xi_0(z)\in \underline{\mathbf{G}}_h(z)$ . Moreover, for any  $z\in \mathsf{U}_0$ , the Jacobian of  $\xi_0$  at  $z$  is given by
$$
\operatorname {J a c} [ \xi_ {0} ] (z) = - \operatorname {J a c} _ {z ^ {\prime}} [ \underline {{\mathbf {g}}} _ {h} ] (z, \xi_ {0} (z)) ^ {- 1} \operatorname {J a c} _ {z} [ \underline {{\mathbf {g}}} _ {h} ] (z, \xi_ {0} (z)).
$$
Result on  $\overline{\mathrm{G}}_h$ . We recall that  $z^{(1)} = \overline{\mathrm{G}}_h(z_h^{(1/2)})$  and  $\operatorname{Jac}[\overline{\mathrm{G}}_h](z_h^{(1/2)})$  is invertible. Then, we apply the inverse function theorem on  $\overline{\mathrm{G}}_h$  at  $z_h^{(1/2)}$  and obtain the existence of a neighbourhood  $\mathsf{U}_1 \subset \mathbb{T}^\star \mathsf{M}$  of  $z_h^{(1/2)}$  such that  $\xi_1 = \overline{\mathrm{G}}_h|_{\mathsf{U}_1}$  is a  $\mathbf{C}^1$ -diffeomorphism on  $\mathsf{U}_1$ .
Final result. We now consider the subset  $\mathsf{U}$  and the map  $\xi$ , respectively defined by
(a)  $\mathsf{U} = \xi_0^{-1}\left(\mathsf{U}_1\cap \xi_0\big(\mathsf{U}_0\big)\right)\subset \mathrm{T}^{\star}\mathsf{M},$  neighbourhood of  $z^{(0)}$
(b)  $\xi = \xi_1|_{\xi_0(\mathsf{U})} \circ \xi_0$ ,  $\mathrm{C}^1$ -diffeomorphism on  $\mathsf{U}$  such that  $\xi(z^{(0)}) = z^{(1)}$ , and for any  $z \in \mathsf{U}$ ,  $\xi(z) \in \mathrm{G}_h(z)$ .
Let us now prove that  $|\operatorname{det}\mathrm{Jac} \xi| \equiv 1$ . Let  $z \in \mathsf{U}$ . We define  $z_0 = \xi_0(z)$ . By the chain rule, we have
$$
\begin{array}{l} \operatorname {J a c} [ \xi ] (z) = \operatorname {J a c} [ \xi_ {1} \circ \xi_ {0} ] (z) \\ = \operatorname {J a c} [ \xi_ {1} ] (\xi_ {0} (z)) \operatorname {J a c} [ \xi_ {0} ] (z) \\ = - \mathrm {J a c} [ \xi_ {1} ] (\xi_ {0} (z)) \mathrm {J a c} _ {z ^ {\prime}} [ \underline {{\mathsf {g}}} _ {h} ] (z, \xi_ {0} (z)) ^ {- 1} \mathrm {J a c} _ {z} [ \underline {{\mathsf {g}}} _ {h} ] (z, \xi_ {0} (z)), \\ \end{array}
$$
(a)  $\operatorname{Jac}[\xi_1](z') = \begin{pmatrix} \mathrm{I}_d & 0 \\ -\frac{h}{2}\partial_{x,x}H_2(z') & \mathrm{I}_d - \frac{h}{2}\partial_{x,p}H_2(z') \end{pmatrix}$ ,
(b)  $\operatorname{Jac}_{z'}[\underline{\mathsf{g}}_h](z,z') = \left( \begin{array}{cc}\mathrm{I}_d - \frac{h}{2}\partial_{x,p}H_2(z') & -\frac{h}{2}\{\partial_{p,p}H_2(x,p') + \partial_{p,p}H_2(x',p')\} \\ 0 & \mathrm{I}_d + \frac{h}{2}\partial_{x,p}H_2(x,p') \end{array} \right),$
(c)  $\operatorname{Jac}_z[\underline{\mathrm{g}}_h](z,z') = \left( \begin{array}{cc} -\mathrm{I}_d - \frac{h}{2}\partial_{x,p}H_2(x,p') & 0\\ \frac{h}{2}\partial_{x,x}H_2(x,p') & -\mathrm{I}_d \end{array} \right)$ .
$$
\begin{array}{l} | \det  \operatorname {J a c} [ \xi ] (z) | \\ = \det (\mathrm {I} _ {d} - \frac {h}{2} \partial_ {x, p} H _ {2} (z _ {0})) \det (\{\mathrm {I} _ {d} - \frac {h}{2} \partial_ {x, p} H _ {2} (z _ {0}) \} \{\mathrm {I} _ {d} + \frac {h}{2} \partial_ {x, p} H _ {2} (x, p _ {0}) \}) ^ {- 1} \det (\mathrm {I} _ {d} + \frac {h}{2} \partial_ {x, p} H _ {2} (x, p _ {0})) = 1 , \\ \end{array}
$$
Lemma 26. Let  $h > 0$ . We recall that the set-valued map  $\mathrm{F}_h$  is defined in Section 3.1. Let  $z \in \mathrm{T}^{\star}\mathsf{M}$ . Assume that there exist  $(z', z'') \in \mathrm{T}^{\star}\mathsf{M} \times \mathrm{T}^{\star}\mathsf{M}$  such that
(a)  $z^{\prime}\in \mathrm{F}_{h}(z)$
(b)  $z'' \in \mathrm{F}_h(z)$ , and
(c)  $x^{\prime} = x^{\prime \prime}$
Then, we have  $p' = p''$ , and thus  $z' = z''$ .
Proof. Let  $h > 0$ . Let  $z \in \mathrm{T}^{\star}\mathsf{M}$ . We consider  $(z', z'') \in \mathrm{F}_h(z) \times \mathrm{F}_h(z)$  such that  $x' = x'' = \tilde{x}$ . By using the last two aligns from (6) with  $z'$  and  $z''$ , we have  $p' = -p^{(1/2)} + \frac{h}{2}\mathfrak{g}(\tilde{x})^{-1}p^{(1/2)} = p''$ , where  $p^{(1/2)} = \frac{2}{h}(\mathfrak{g}(x)^{-1} + \mathfrak{g}(\tilde{x})^{-1})^{-1}(\tilde{x} - x)$ , which concludes the proof.
Under the assumption A2, we define, for any  $z^{(0)} \in \mathrm{T}^{\star}\mathsf{M}$ ,  $h^{\star}(z^{(0)}) = \bar{h}(s(z^{(0)}), r_{\alpha}^{\star}, \alpha)$ , where  $\alpha$  is given by A2, and  $r_{\alpha}^{\star}$  and  $\bar{h}$  are respectively defined in (33) and (50). Note that  $h^{\star}$  appears in Proposition 2, for which we derive the proof below.