Proof. The proof is divided into two parts. First we use the properness of $f$ to lift ${\it\gamma}$ to an arc on $X$ and then we conclude thanks to Puiseux theorem.

Consider the following diagram

Let $X_{1}=\tilde{f}^{-1}((0,{\it\varepsilon}))$ . Since $f$ is proper, $\overline{X}_{1}\setminus X_{1}\subset \tilde{X}$ . Let $x_{0}\in \overline{X}_{1}\setminus X_{1}$ , then by the curve selection lemma ([Reference Bochnak, Coste and Roy8, Proposition 8.1.13] for the semialgebraic case) there exists ${\it\gamma}_{1}:[0,{\it\eta})\rightarrow \tilde{X}$ analytic (resp. Nash) such that ${\it\gamma}_{1}(0)=x_{0}$ and ${\it\gamma}_{1}((0,{\it\eta}))\subset X_{1}$ . We have the following diagram

Then, $h(0)=0$ and $h((0,{\it\eta}))\subset (0,{\it\varepsilon})$ . Hence there exists ${\it\alpha}\in (0,{\it\eta})$ such that $h:[0,{\it\alpha})\rightarrow [0,{\it\beta})$ is a subanalytic (resp. semialgebraic) homeomorphism.

By Puiseux theorem ([Reference Bochnak, Coste and Roy8, Proposition 8.1.12] for the semialgebraic case; see also [Reference Pawłucki42]), there exist $m\in \mathbb{N}_{{>}0}$ and ${\it\delta}\leqslant {\it\beta}^{\frac{1}{m}}$ such that $h^{-1}(t^{m})$ is analytic (resp. Nash) for $t\in [0,{\it\delta})$ .

Finally, $\tilde{{\it\gamma}}:[0,{\it\delta})\rightarrow X$ defined by $\tilde{{\it\gamma}}(t)=\text{pr}_{X}{\it\gamma}_{1}h^{-1}(t^{m})$ satisfies $f\circ \tilde{{\it\gamma}}(t)={\it\gamma}(t^{m})$ .◻

Proof. Let $S$ be as in Definition 2.22. By the Jacobian criterion and the Nash implicit function theorem we may assume that $S$ is locally a Nash subset of $\mathbb{R}^{d}$ . Taking the Zariski closure we may moreover assume that $S$ is an algebraic subset of $\mathbb{R}^{d}$ since it does not change the dimension. Let ${\it\gamma}:(-{\it\varepsilon},{\it\varepsilon})\rightarrow \mathbb{R}^{d}$ be an analytic arc entirely included in $S$ .

As in [Reference Kurdyka30, Corollaire 2.7], by Puiseux theorem, we may assume that

$$\begin{eqnarray}\displaystyle f({\it\gamma}(t)) & = & \displaystyle \displaystyle \mathop{\sum }_{i\geqslant 0}b_{i}t^{i/p},\quad t\geqslant 0\nonumber\\ \displaystyle f({\it\gamma}(t)) & = & \displaystyle \displaystyle \mathop{\sum }_{i\geqslant 0}c_{i}(-t)^{i/r},\quad t\leqslant 0.\nonumber\end{eqnarray}$$

By [Reference Kurdyka30, Corollaire 2.8 and Corollaire 2.9], two phenomena may prevent $f({\it\gamma}(t))$ from being analytic: either one of these expansions has a noninteger exponent or these expansions do not coincide.

To handle the first case, we assume that one of these expansions, for instance for $t\geqslant 0$ , has a noninteger exponent, that is,

$$\begin{eqnarray}f({\it\gamma}(t))=\mathop{\sum }_{i=0}^{m}b_{i}t^{i}+bt^{\frac{p}{q}}+\cdots \,,\quad b\neq 0,\;m<\frac{p}{q}<m+1,\;t\geqslant 0.\end{eqnarray}$$

It follows from Remark 2.20 there exists $N\in \mathbb{N}$ such that for every analytic arc ${\it\delta}$ we have $f({\it\gamma}(t)+t^{N}{\it\delta}(t))\equiv f({\it\gamma}(t))\operatorname{mod}t^{m+1}$ . We are going to prove that for ${\it\eta}\in \mathbb{R}^{d}$ generic, the arc $\tilde{{\it\gamma}}(t)={\it\gamma}(t)+t^{N}{\it\eta}$ is not entirely included in $S$ in order to get a contradiction since $f(\tilde{{\it\gamma}}(t))\equiv f({\it\gamma}(t))\operatorname{mod}t^{m+1}$ .

Let $t_{0}\in (-{\it\varepsilon},{\it\varepsilon})\setminus \{0\}$ . Since $\operatorname{dim}S<d$ , there is $\tilde{{\it\eta}}\in \mathbb{R}^{d}\setminus C_{{\it\gamma}(t_{0})}S$ where $C_{{\it\gamma}(t_{0})}S$ is the tangent cone of $S$ at ${\it\gamma}(t_{0})$ . Thus there exists $F\in I(S)$ with $F({\it\gamma}(t_{0})+x)=F_{{\it\mu}}(x)+\cdots +F_{{\it\mu}+r}(x)$ where $\operatorname{deg}F_{i}=i$ and such that $F_{{\it\mu}}(\tilde{{\it\eta}})\neq 0$ . Then $F\left({\it\gamma}(t_{0})+st_{0}^{N}\tilde{{\it\eta}}\right)=\left(st_{0}^{N}\right)^{{\it\mu}}F_{{\it\mu}}(\tilde{{\it\eta}})+\left(st_{0}^{N}\right)^{{\it\mu}+1}G(s,t)$ and hence for $s$ small enough the arc ${\it\gamma}(t)+t^{N}s\tilde{{\it\eta}}$ is not entirely included in $S$ .

Then we prove that the expansions coincide in a similar way. Assume that the expansions are different, that is,

$$\begin{eqnarray}\displaystyle f({\it\gamma}(t)) & = & \displaystyle \displaystyle \mathop{\sum }_{i=0}^{m-1}a_{i}t^{i}+bt^{m}+\cdots \,,\quad t\geqslant 0\nonumber\\ \displaystyle f({\it\gamma}(t)) & = & \displaystyle \displaystyle \mathop{\sum }_{i=0}^{m-1}a_{i}t^{i}+ct^{m}+\cdots \,,\quad t\leqslant 0\nonumber\end{eqnarray}$$

with $b\neq c$ . As in the previous case, we may construct an arc $\tilde{{\it\gamma}}$ not entirely included in $S$ such that $f{\it\gamma}(t)$ and $f\tilde{{\it\gamma}}(t)$ coincide up to order $m+1$ . That leads to a contradiction.◻

Proof. Assume that $f$ is generically arc-analytic. Let ${\it\rho}:U\rightarrow X$ be a resolution of singularities given by a sequence of blowings-up with nonsingular centers, then $f\circ {\it\rho}:U\rightarrow Y$ is semialgebraic and generically arc-analytic with $U$ nonsingular. Thus $f\circ {\it\rho}$ is arc-analytic by Lemma 2.23. By Theorem 2.15, there exists a sequence of blowings-up with nonsingular centers ${\it\eta}:M\rightarrow U$ such that $f\circ {\it\rho}\circ {\it\eta}$ is Nash. Finally $f\circ {\it\sigma}$ is Nash where ${\it\sigma}={\it\rho}\circ {\it\eta}:M\rightarrow X$ is a sequence of blowings-up with nonsingular centers.

Assume that $f$ is blow-Nash. Then there is ${\it\sigma}:M\rightarrow X$ a sequence of blowings-up with nonsingular centers such that $f\circ {\it\sigma}:M\rightarrow Y$ is Nash. Let ${\it\gamma}$ be an arc on $X$ not entirely included in the singular locus of $X$ and the center of ${\it\sigma}$ , then there is $\tilde{{\it\gamma}}$ an arc on $M$ such that ${\it\gamma}={\it\sigma}(\tilde{{\it\gamma}})$ . Thus $f({\it\gamma}(t))=f\circ {\it\sigma}(\tilde{{\it\gamma}}(t))$ is analytic.◻

Proof. The following proof comes from [Reference Lam33, Lemma 6.14]. After an affine change of coordinates, we may assume that $f(a,b_{1})<0<f(a,b_{2})$ with $a=(a_{1},\ldots ,a_{N-1})$ . Let $g\in I(V(f))$ and assume that $f\nmid g$ in $\mathbb{R}[x_{1},\ldots ,x_{N}]$ . In the PID (and hence UFD) $\mathbb{R}(x_{1},\ldots ,x_{N-1})[x_{N}]$ , $f$ is also irreducible and $f\nmid g$ too. In this PID, we may find ${\it\varphi}$ and ${\it\gamma}$ such that ${\it\varphi}f+{\it\gamma}g=1$ . Let ${\it\varphi}={\it\varphi}_{0}/h$ and ${\it\gamma}={\it\gamma}_{0}/h$ with $0\neq h\in \mathbb{R}[x_{1},\ldots ,x_{N-1}]$ and ${\it\varphi}_{0},{\it\gamma}_{0}\in \mathbb{R}[x_{1},\ldots ,x_{N-1}][x_{N}]$ . Then ${\it\varphi}_{0}f+{\it\gamma}_{0}g=h$ . Let $V$ be a neighborhood of $a$ in $\mathbb{R}^{N-1}$ such that $\text{for all }v\in V,f(v,b_{1})<0<f(v,b_{2})$ . By the IVT, for all $v\in V$ , there is $b_{1}\leqslant b_{v}\leqslant b_{2}$ such that $f(v,b_{v})=0$ , and so $g(v,b_{v})=0$ . Then $\text{for all }v\in V,h(v)=0$ and hence $h\equiv 0$ which is a contradiction.◻

Proof. (iii) $\Rightarrow$ (ii) is obvious using Hensel’s lemma and Artin approximation theorem [Reference Artin2].

(ii) $\Rightarrow$ (i) is obvious since ${\it\pi}_{n}={\it\pi}_{n}^{n+1}\circ {\it\pi}_{n+1}$ .

(i) $\Rightarrow$ (iii): Assume that $0$ is a singular point of $X$ . We can find ${\it\gamma}={\it\alpha}t\in {\mathcal{L}}_{1}(X)$ which does not lie in the tangent cone of $X$ at $0$ , that is, such that $f({\it\alpha}t)\not \equiv 0\operatorname{mod}t^{m+1}$ for some $f\in I(X)$ of order $m$ . Such a $1$ -jet cannot be lifted to ${\mathcal{L}}_{m}(X)$ .◻

Proof. We first notice that (i) is a direct consequence of (iii).

(ii) $({\it\pi}_{0}^{n})^{-1}(X\setminus X_{\text{sing}})$ is of dimension $(n+1)d$ since the fiber of ${\it\pi}_{0}^{n}$ over a nonsingular point is of dimension $nd$ .

(iii) We may assume that $m=n+1$ . Let ${\it\gamma}\in {\it\pi}_{n}({\mathcal{L}}(X))$ . We may assume that ${\it\gamma}\in (\mathbb{R}_{n}[t])^{N}$ . We consider the following diagram

with $p_{1}(x,t)={\it\gamma}(t)+t^{n+1}x$ and $p_{2}(x,t)=t$ . Let $\mathfrak{X}=\overline{p_{1}^{-1}(X)\cap \{t\neq 0\}}^{\,Zar}$ . For $c\neq 0$ , $\mathfrak{X}\cap p_{2}^{-1}(c)\simeq X$ and $\operatorname{dim}\mathfrak{X}\cap p_{2}^{-1}(c)=\operatorname{dim}\mathfrak{X}-1$ . Hence $\operatorname{dim}\mathfrak{X}\cap p_{2}^{-1}(0)\leqslant \operatorname{dim}\mathfrak{X}-1=\operatorname{dim}X$ .

We are looking for objects of the form ${\it\pi}_{n+1}({\it\gamma}(t)+t^{n+1}{\it\alpha}(t))$ with ${\it\gamma}(t)+t^{n+1}{\it\alpha}(t)\in {\mathcal{L}}(X)$ . Such an ${\it\alpha}$ is equivalent to a section of ${p_{2}}_{|\mathfrak{X}}$ , that is, $\begin{array}{@{}rcl@{}}\mathbb{R}\ & \rightarrow \ & \mathfrak{X}\\ ~\\ ~\\ t\ & \mapsto \ & ({\it\alpha}(t),t)\end{array}\!\!$ . Since we want an arc modulo $t^{n+2}$ , we are looking for the constant term of ${\it\alpha}$ , therefore $(\tilde{{\it\pi}}_{n}^{n+1})^{-1}({\it\gamma})\subset \mathfrak{X}\cap p_{2}^{-1}(0)$ .

(iv) Let ${\it\gamma}\in {\mathcal{L}}_{n}(X)$ . Let ${\it\eta}\in \mathbb{R}^{N}$ . Assume that $I(X)=(f_{1},\ldots ,f_{r})$ . By Taylor expansion we get

$$\begin{eqnarray}f_{i}({\it\gamma}+t^{n+1}{\it\eta})\equiv f_{i}({\it\gamma}(t))+t^{n+1}\left({\rm\nabla}_{{\it\gamma}(t)}f_{i}\right)({\it\eta})~\mod ~t^{n+2}\end{eqnarray}$$

Assume that $f_{i}({\it\gamma}(t))\equiv t^{n+1}{\it\alpha}_{i}\operatorname{mod}t^{n+2}$ . Since

$$\begin{eqnarray}t^{n+1}({\rm\nabla}_{{\it\gamma}(t)}f_{i})({\it\eta})\equiv t^{n+1}({\rm\nabla}_{{\it\gamma}(0)}f_{i})({\it\eta})\operatorname{mod}t^{n+2},\end{eqnarray}$$

we have

$$\begin{eqnarray}f_{i}({\it\gamma}+t^{n+1}{\it\eta})\equiv t^{n+1}\left({\it\alpha}_{i}+\left({\rm\nabla}_{{\it\gamma}(0)}f_{i}\right)({\it\eta})\right)~\mod ~t^{n+2}\end{eqnarray}$$

Hence, ${\it\gamma}(t)+t^{n+1}{\it\eta}$ is in the fiber $({\it\pi}_{n}^{n+1})^{-1}({\it\gamma})$ if and only if ${\it\alpha}_{i}+({\rm\nabla}_{{\it\gamma}(0)}f_{i})({\it\eta})=0,i=1,\ldots ,r$ .◻

Proof. Let $x\notin V(H)$ then there exist $f_{1},\ldots ,f_{N-d}\in I(X)$ , ${\it\delta}$ a $(N-d)$ -minor of $\left(\frac{\partial f_{i}}{\partial x_{j}}\right)_{{i=1,\ldots ,N-d\atop j=1,\ldots ,N}}$ and $h\in \mathbb{R}[x_{1},\ldots ,x_{N}]$ with $hI(X)\subset (f_{1},\ldots ,f_{N-d})$ and $h{\it\delta}(x)\neq 0$ . Since ${\it\delta}(x)\neq 0$ , $x$ is a nonsingular point of $V(f_{1},\ldots ,f_{N-d})$ . Furthermore we have $X=V(I(X))\subset V(f_{1},\ldots ,f_{N-d})\subset V(hI(X))$ and, since $h(x)\neq 0$ , in an open neighborhood $U$ of $x$ in $\mathbb{R}^{N}$ we have $V(hI(X))\cap U=X\cap U$ . Hence $V(f_{1},\ldots ,f_{N-d})\cap U=X\cap U$ . So $x$ is a nonsingular point of $X$ by [Reference Bochnak, Coste and Roy8, Proposition 3.3.10]. We proved that $X_{\text{sing}}\subset V(H)$ .

Now, assume that $x\in X\setminus X_{\text{sing}}$ . With the notation of [Reference Bochnak, Coste and Roy8, Section 3], the local ring ${\mathcal{R}}_{X,x}={\mathcal{R}}_{\mathbb{R}^{N},x}/I(X){\mathcal{R}}_{\mathbb{R}^{N},x}$ is regular, so we may find a regular system of parameters $(f_{1},\ldots ,f_{N})$ of ${\mathcal{R}}_{X,x}$ such that $I(X){\mathcal{R}}_{\mathbb{R}^{N},x}=(f_{1},\ldots ,f_{N-d}){\mathcal{R}}_{\mathbb{R}^{N},x}$ by [Reference Kunz25, VI.1.8 and VI.1.10]Footnote ⁹ (see also [Reference Bochnak, Coste and Roy8, Proposition 3.3.7]). Moreover, we may assume that the $f_{1},\ldots ,f_{N-d}$ are polynomials. We may use the following classical argument. ${\it\theta}:\mathbb{R}[x_{1},\ldots ,x_{N}]\rightarrow \mathbb{R}^{N}$ defined by $f\mapsto f(x)$ induces an isomorphism ${\it\theta}^{\prime }:\mathfrak{m}_{x}/\mathfrak{m}_{x}^{2}\rightarrow \mathbb{R}^{N}$ . Then $\operatorname{rk}\left(\frac{\partial f_{i}}{\partial x_{j}}(x)\right)=\operatorname{dim}{\it\theta}((f_{1},\ldots ,f_{N-d}))$ which is, by ${\it\theta}^{\prime }$ , the dimension of $((f_{1},\ldots ,f_{N-d})+\mathfrak{m}_{x}^{2})/\mathfrak{m}_{x}^{2}$ as a subspace of $\mathfrak{m}_{x}/\mathfrak{m}_{x}^{2}$ . If we denote by $\mathfrak{m}$ the maximal ideal of ${\mathcal{R}}_{X,x}=\left(\mathbb{R}[x_{1},\ldots ,x_{N}]/(f_{1},\ldots ,f_{N-d})\right)_{\mathfrak{m}_{x}}$ , we have $\mathfrak{m}/\mathfrak{m}^{2}\simeq \mathfrak{m}_{x}/((f_{1},\ldots ,f_{N-d})+\mathfrak{m}_{x}^{2})$ . So we have $\operatorname{dim}\left(\mathfrak{m}/\mathfrak{m}^{2}\right)+\operatorname{rk}\left(\frac{\partial f_{i}}{\partial x_{j}}(x)\right)=N$ .

Furthermore, since ${\mathcal{R}}_{X,x}$ is a $d$ -dimensional regular local ring, $\operatorname{dim}\left(\mathfrak{m}/\mathfrak{m}^{2}\right)=d$ . Hence $\left(\frac{\partial f_{i}}{\partial x_{j}}(x)\right)_{{i=1,\ldots ,N-d\atop j=1,\ldots ,N}}$ is of rank $N-d$ and so there exists ${\it\delta}$ a $(N-d)$ -minor of $\left(\frac{\partial f_{i}}{\partial x_{j}}\right)_{{i=1,\ldots ,N-d\atop j=1,\ldots ,N}}$ such that ${\it\delta}(x)\neq 0$ . Assume that $I(X)=(g_{1},\ldots ,g_{r})$ in $\mathbb{R}[x_{1},\ldots ,x_{N}]$ . Then $g_{i}=\sum \frac{f_{j}}{q_{j}}$ with $q_{j}(x)\neq 0$ , so $g_{i}h_{i}\subset (f_{1},\ldots ,f_{N-d})$ with $h_{i}=\prod q_{j}$ . Then $h=\prod h_{i}$ satisfies $h(x)\neq 0$ and $hI(X)\subset (f_{1},\ldots ,f_{N-d})$ . So $x\notin V(H)$ . Hence $V(H)\subset X_{\text{sing}}\cup (\mathbb{R}^{N}\setminus X)$ .

To complete the proof, it remains to prove that $V(H)\subset X$ . Let $x\notin X$ . There exist $f_{1},\ldots ,f_{N-d}\in I(X)$ such that $f_{i}(x)\neq 0$ . We construct by induction $N-d$ polynomials of the form $g_{i}=a_{i}f_{i}$ with $g_{i}(x)\neq 0$ and $(\text{d}g_{1}\wedge \cdots \wedge \text{d}g_{N-d})_{x}\neq 0$ . Suppose that $g_{1},\ldots ,g_{j-1}$ are constructed, if $(\text{d}g_{1}\wedge \cdots \wedge \text{d}g_{j-1}\wedge \text{d}f_{j})_{x}\neq 0$ , we can take $a_{j}=1$ , so we may assume that $(\text{d}g_{1}\wedge \cdots \wedge \text{d}g_{j-1}\wedge \text{d}f_{j})_{x}=0$ . Then we just have to take some $a_{j}$ satisfying $(\text{d}g_{1}\wedge \cdots \wedge \text{d}g_{j-1}\wedge \text{d}a_{j})_{x}\neq 0$ and $a_{j}(x)\neq 0$ since $(\text{d}g_{1}\wedge \cdots \wedge \text{d}g_{j-1}\wedge \text{d}(a_{j}f_{j}))_{x}=f_{j}(x)(\text{d}g_{1}\wedge \cdots \wedge \text{d}g_{j-1}\wedge \text{d}a_{j})_{x}$ . Then we have $g_{1},\ldots ,g_{N-d}\in I(X)$ whose a $(N-d)$ -minor ${\it\delta}$ satisfies ${\it\delta}(x)\neq 0$ . Moreover we have $g_{i}(x)\neq 0$ and $g_{i}I\subset (g_{1},\ldots ,g_{N-d})$ . So $x\notin V(H)$ .◻

Proof. By Noetherianity, we may assume that $H=(h_{1}{\it\delta}_{1},\ldots ,h_{r}{\it\delta}_{r})$ with $h_{i},{\it\delta}_{i}$ as desired. Therefore, ${\mathcal{L}}^{(e)}(X)\subset \cup A_{h_{i},{\it\delta}_{i}}$ .

Finally,

$$\begin{eqnarray}\displaystyle {\mathcal{L}}(X)\cap A_{h,{\it\delta}} & = & \displaystyle \left\{{\it\gamma}\in {\mathcal{L}}\left(\mathbb{R}^{N}\right),\forall f\in I(X),f({\it\gamma})=0,h{\it\delta}({\it\gamma})\not \equiv 0\operatorname{mod}t^{e+1}\right\}\nonumber\\ \displaystyle & = & \displaystyle \left\{{\it\gamma}\in {\mathcal{L}}\left(\mathbb{R}^{N}\right),f_{1}({\it\gamma})=\cdots =f_{N-d}({\it\gamma})=0,h{\it\delta}({\it\gamma})\not \equiv 0\operatorname{mod}t^{e+1}\right\}.\nonumber\end{eqnarray}$$

Indeed, for the second equality, if $f\in I(X)$ then $hf\in (f_{1},\ldots ,f_{N-d})$ , hence if ${\it\gamma}$ vanishes the $f_{i}$ , then $hf({\it\gamma})=0$ , and so $f({\it\gamma})=0$ since $h({\it\gamma})\neq 0$ .◻

Proof of Lemma 4.5.

We first notice that 4.5 (iii) is a consequence of 4.5 (ii): $\text{for all }{\it\pi}_{n}({\it\gamma})\in {\rm\Delta}_{e,e^{\prime },n}$ we have

$$\begin{eqnarray}\displaystyle & & \displaystyle {\it\pi}_{n}({\it\gamma})\in {\it\sigma}_{\ast n}^{-1}({\it\sigma}_{\ast n}({\it\pi}_{n}({\it\gamma})))\nonumber\\ \displaystyle & & \displaystyle \quad =\left\{{\it\pi}_{n}({\it\eta}),{\it\eta}\in {\mathcal{L}}(M),{\it\sigma}({\it\eta})\equiv {\it\sigma}({\it\gamma})\operatorname{mod}t^{n+1}\right\}\!\,\text{using that }{\mathcal{L}}(M)\rightarrow {\mathcal{L}}_{n}(M)\nonumber\\ \displaystyle & & \displaystyle \qquad \text{is surjective since }M\text{ is smooth and that }{\it\pi}_{n}\circ {\it\sigma}_{\ast }={\it\sigma}_{\ast n}\circ {\it\pi}_{n}.\nonumber\\ \displaystyle & & \displaystyle \quad \subset \left\{{\it\eta}\in {\rm\Delta}_{e,e^{\prime },n},{\it\gamma}\equiv {\it\eta}\operatorname{mod}t^{n-e+1}\right\}\subset {\rm\Delta}_{e,e^{\prime },n}\text{ by 4.5(ii)}.\nonumber\end{eqnarray}$$

Next 4.5 (ii) is a direct consequence of 4.5 (i). We apply 4.5 (i) to ${\it\gamma}$ with ${\it\delta}={\it\sigma}_{\ast }({\it\eta})$ , hence there exists a unique $\tilde{{\it\eta}}$ such that $\tilde{{\it\eta}}\equiv {\it\gamma}\operatorname{mod}t^{n-e+1}$ and ${\it\sigma}_{\ast }(\tilde{{\it\eta}})={\it\sigma}_{\ast }({\it\eta})$ . By the assumptions on ${\it\sigma}$ and the definition of ${\rm\Delta}_{e,e^{\prime }}$ , for ${\it\varphi}_{1}\in {\mathcal{L}}(M)$ and ${\it\varphi}_{2}\in {\rm\Delta}_{e,e^{\prime }}$ with ${\it\varphi}_{1}\neq {\it\varphi}_{2}$ we have ${\it\sigma}({\it\varphi}_{1})\neq {\it\sigma}({\it\varphi}_{2})$ . Hence ${\it\eta}=\tilde{{\it\eta}}$ and ${\it\eta}\equiv {\it\gamma}\operatorname{mod}t^{n-e+1}$ . Since ${\it\sigma}({\it\gamma})\equiv {\it\sigma}({\it\eta})\operatorname{mod}t^{n+1}$ and $n\geqslant e^{\prime }$ , ${\it\sigma}({\it\eta})\in {\mathcal{L}}^{(e^{\prime })}(X)$ . We may write ${\it\eta}(t)={\it\gamma}(t)+t^{n+1-e}u(t)$ and applying Taylor expansion to $\operatorname{Jac}_{{\it\sigma}}({\it\gamma}(t)+t^{n+1-e}u(t))$ we get that $\operatorname{Jac}_{{\it\sigma}}({\it\eta}(t))\equiv \operatorname{Jac}_{{\it\sigma}}({\it\gamma}(t))\operatorname{mod}t^{e+1}$ since $n+1-e\geqslant e+1$ . So ${\it\eta}\in {\rm\Delta}_{e,e^{\prime }}$ .

So we just have to prove 4.5 (i) and 4.5 (iv).

We begin to refine the cover of Lemma 4.7: for $e^{\prime \prime }\leqslant e^{\prime }$ , we set

$$\begin{eqnarray}\displaystyle A_{h,{\it\delta},e^{\prime \prime }} & = & \displaystyle \left\{{\it\gamma}\in A_{h,{\it\delta}},\operatorname{ord}_{t}{\it\delta}({\it\gamma})=e^{\prime \prime }\text{and }\operatorname{ord}_{t}{\it\delta}^{\prime }({\it\gamma})\geqslant e^{\prime \prime }\right.\nonumber\\ \displaystyle & & \displaystyle \quad \left.\text{for all }(N-d)\text{-minor }{\it\delta}^{\prime }\text{of }\left(\frac{\partial f_{i}}{\partial x_{j}}\right)_{{i=1,\ldots ,N-d\atop j=1,\ldots ,N}}\right\}.\nonumber\end{eqnarray}$$

Fix some $A=A_{h,{\it\delta},e^{\prime \prime }}$ , then it suffices to prove the lemma for ${\rm\Delta}_{e,e^{\prime }}\cap {\it\sigma}^{-1}(A)$ .

Up to renumbering the coordinates, we may also assume that ${\it\delta}$ is the determinant of the first $N-d$ columns of ${\rm\Delta}=\left(\frac{\partial f_{i}}{\partial x_{j}}\right)_{{i=1,\ldots ,N-d\atop j=1,\ldots ,N}}$ .

We choose a local coordinate system of $M$ at ${\it\gamma}(0)$ in order to define $\operatorname{Jac}_{{\it\sigma}}$ and express arcs of $M$ as elements of $\mathbb{R}\{t\}^{d}$ .

Now, a crucial observation is that the first $N-d$ rows of $\operatorname{Jac}_{{\it\sigma}}({\it\gamma})$ are $\mathbb{R}\{t\}$ -linear combinations of the last $d$ rows: the application

$$\begin{eqnarray}\begin{array}{@{}ccccc@{}}M & \longrightarrow & X & \longrightarrow & \mathbb{R}^{N-d}\\ y & \longmapsto & {\it\sigma}(y) & \longmapsto & \left(f_{i}({\it\sigma}(y))\right)_{i=1,\ldots ,N-d}\end{array}\end{eqnarray}$$

is identically zero, so its Jacobian matrix is identically zero too and thus ${\rm\Delta}({\it\sigma}({\it\gamma}))\operatorname{Jac}_{{\it\sigma}}({\it\gamma})=0$ . Let $P$ be the transpose of the comatrix of the submatrix of ${\rm\Delta}$ given by the first $N-d$ columns of ${\rm\Delta}$ , then $P{\rm\Delta}=({\it\delta}I_{N-d},W)$ . Moreover, we have $W({\it\sigma}({\it\gamma}))\equiv 0\operatorname{mod}t^{e^{\prime \prime }}$ . Indeed, if we denote ${\rm\Delta}_{1},\ldots ,{\rm\Delta}_{N-d}$ the $N-d$ first columns of ${\rm\Delta}$ and $W_{1},\ldots ,W_{d}$ the columns of $W$ , then $W_{j}({\it\sigma}({\it\gamma}))$ is solution of $({\rm\Delta}_{1}({\it\sigma}({\it\gamma})),\ldots ,{\rm\Delta}_{N-d}({\it\sigma}({\it\gamma})))X={\it\delta}({\it\sigma}({\it\gamma})){\rm\Delta}_{N-d+j}({\it\sigma}({\it\gamma}))$ since

$$\begin{eqnarray}\displaystyle {\it\delta}({\it\sigma}({\it\gamma})){\rm\Delta}({\it\sigma}({\it\gamma})) & = & \displaystyle \left({\rm\Delta}_{1}({\it\sigma}({\it\gamma})),\ldots ,{\rm\Delta}_{N-d}({\it\sigma}({\it\gamma}))\right)P({\it\sigma}({\it\gamma})){\rm\Delta}({\it\sigma}({\it\gamma}))\nonumber\\ \displaystyle & = & \displaystyle \left({\rm\Delta}_{1}({\it\sigma}({\it\gamma})),\ldots ,{\rm\Delta}_{N-d}({\it\sigma}({\it\gamma}))\right)\left({\it\delta}({\it\sigma}({\it\gamma}))I_{N-d},W({\it\sigma}({\it\gamma}))\right).\nonumber\end{eqnarray}$$

So, by Cramer’s rule,

$$\begin{eqnarray}\displaystyle \left(W_{j}({\it\sigma}({\it\gamma}))\right)_{i} & = & \displaystyle \operatorname{det}\left({\rm\Delta}_{1}({\it\sigma}({\it\gamma})),\ldots ,{\rm\Delta}_{i-1}({\it\sigma}({\it\gamma})),{\rm\Delta}_{N-d+j}({\it\sigma}({\it\gamma})),\right.\nonumber\\ \displaystyle & & \displaystyle \left.{\rm\Delta}_{i+1}({\it\sigma}({\it\gamma})),\ldots ,{\rm\Delta}_{N-d}({\it\sigma}({\it\gamma}))\right).\nonumber\end{eqnarray}$$

Finally, the congruence arises because the minor formed by the $N-d$ first columns is of minimal order by definition of $A$ .

Now the columns of $\operatorname{Jac}_{{\it\sigma}}({\it\gamma})$ are solutions of

(2) $$\begin{eqnarray}\left(t^{-e^{\prime \prime }}\cdot P({\it\sigma}({\it\gamma}))\cdot {\rm\Delta}({\it\sigma}({\it\gamma}))\right)X=0\end{eqnarray}$$

but since $t^{-e^{\prime \prime }}\cdot P({\it\sigma}({\it\gamma}))\cdot {\rm\Delta}({\it\sigma}({\it\gamma}))=(t^{-e^{\prime \prime }}{\it\delta}({\it\sigma}({\it\gamma}))I_{N-d},t^{-e^{\prime \prime }}W({\it\sigma}({\it\gamma})))$ we may express the first $N-d$ coordinates of each solution in terms of the last $d$ coordinates. This completes the proof of the observation.

For 4.5 (i), it suffices to prove that for all $v\in \mathbb{R}\{t\}^{N}$ satisfying ${\it\sigma}({\it\gamma})+t^{n+1}v\in {\mathcal{L}}(X)$ there exists a unique $u\in \mathbb{R}\{t\}^{d}$ such that

(3) $$\begin{eqnarray}{\it\sigma}({\it\gamma}+t^{n+1-e}u)={\it\sigma}({\it\gamma})+t^{n+1}v.\end{eqnarray}$$

By Taylor expansion, we have

(4) $$\begin{eqnarray}{\it\sigma}({\it\gamma}(t)+t^{n+1-e}u)={\it\sigma}({\it\gamma}(t))+t^{n+1-e}\operatorname{Jac}_{{\it\sigma}}({\it\gamma}(t))u+t^{2(n+1-e)}R({\it\gamma}(t),u)\end{eqnarray}$$

with $R({\it\gamma}(t),u)$ analytic in $t$ and $u$ . By (4), (3) is equivalent to

(5) $$\begin{eqnarray}t^{-e}\operatorname{Jac}_{{\it\sigma}}({\it\gamma}(t))u+t^{n+1-2e}R({\it\gamma}(t),u)=v\end{eqnarray}$$

with $n+1-2e\geqslant 1$ by hypothesis.

Since ${\it\sigma}({\it\gamma}(t))+t^{n+1}v\in {\mathcal{L}}(X)$ and using Taylor expansion, we get

$$\begin{eqnarray}0=f_{i}({\it\sigma}({\it\gamma}(t))+t^{n+1}v)=t^{n+1}{\rm\Delta}({\it\sigma}({\it\gamma}(t)))v+t^{2(n+1)}S({\it\gamma}(t),v)\end{eqnarray}$$

with $S({\it\gamma}(t),v)$ analytic in $t$ and $v$ . So $v$ is a solution of (2) and hence the first $N-d$ coefficients of $v$ are $\mathbb{R}\{t\}$ -linear combinations of the last $d$ coefficients with the same relations that for $\operatorname{Jac}_{{\it\sigma}}({\it\gamma})$ . This allows us to reduce (5) to

(6) $$\begin{eqnarray}\displaystyle t^{-e}\operatorname{Jac}_{p\circ {\it\sigma}}({\it\gamma}(t))u+t^{n+1-2e}p\left(R({\it\gamma}(t),u)\right)=p(v) & & \displaystyle\end{eqnarray}$$

where $p:\mathbb{R}^{N}\rightarrow \mathbb{R}^{d}$ is the projection on the last $d$ coordinates. The observation ensures that $\operatorname{ord}_{t}\operatorname{Jac}_{p\circ {\it\sigma}}({\it\gamma}(t))=\operatorname{ord}_{t}\operatorname{Jac}_{{\it\sigma}}({\it\gamma}(t))=e$ and thus (6) is equivalent to

(7) $$\begin{eqnarray}\displaystyle u=\left(t^{-e}\operatorname{Jac}_{p\circ {\it\sigma}}({\it\gamma}(t))\right)^{-1}p(v)-t^{n+1-2e}\left(t^{-e}\operatorname{Jac}_{p\circ {\it\sigma}}({\it\gamma}(t))\right)^{-1}p\left(R({\it\gamma}(t),u)\right). & & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

Applying the implicit function theorem to $u(t,v)$ ensures that given an analytic arc $v(t)$ there exists a solution $u_{v}(t)=u(t,v(t))$ . Using the same argument as in the proof of 4.5 (ii), the solution $u_{v}(t)$ is unique. This proves 4.5 (i).

Let us prove 4.5 (iv). Let ${\it\gamma}\in {\rm\Delta}_{e,e^{\prime }}\cap {\it\sigma}^{-1}(A)$ then

$$\begin{eqnarray}\displaystyle & & \displaystyle {\it\sigma}_{\ast n}^{-1}({\it\pi}_{n}({\it\sigma}_{\ast }({\it\gamma})))\nonumber\\ \displaystyle & & \displaystyle \quad =\left\{{\it\eta}\in {\mathcal{L}}_{n}(M),{\it\sigma}_{\ast n}({\it\eta})={\it\pi}_{n}({\it\sigma}_{\ast }({\it\gamma}))\right\}\nonumber\\ \displaystyle & & \displaystyle \quad =\left\{{\it\pi}_{n}({\it\eta}),{\it\eta}\in {\mathcal{L}}(M),{\it\sigma}({\it\eta})\equiv {\it\sigma}({\it\gamma})\operatorname{mod}t^{n+1}\right\}\,\!\text{using that }{\mathcal{L}}(M)\rightarrow {\mathcal{L}}_{n}(M)\nonumber\\ \displaystyle & & \displaystyle \qquad \;\text{is surjective since }M\text{ is smooth and that }{\it\pi}_{n}\circ {\it\sigma}_{\ast }={\it\sigma}_{\ast n}\circ {\it\pi}_{n}.\nonumber\\ \displaystyle & & \displaystyle \quad =\left\{{\it\gamma}(t)+t^{n+1-e}u(t)\operatorname{mod}t^{n+1},u\in \mathbb{R}\{t\}^{d},\operatorname{Jac}_{p\circ {\it\sigma}}({\it\gamma}(t))u(t)\equiv 0\operatorname{mod}t^{e}\right\}\nonumber\\ \displaystyle & & \displaystyle \qquad \;\text{by 4.5(ii) and (6)}.\nonumber\end{eqnarray}$$

Thus, the fiber is an affine subspace of $\mathbb{R}^{de}$ . There are invertible matrices $A$ and $B$ with coordinates in $\mathbb{R}\{t\}$ such that $A\operatorname{Jac}_{p\circ {\it\sigma}}({\it\gamma}(t))B$ is diagonal with entries $t^{e_{1}},\ldots ,t^{e_{d}}$ such that $e=e_{1}+\cdots +e_{d}$ . Therefore the fiber is of dimension $e$ .

Since ${\it\sigma}$ is not assumed to be birational, we cannot use the section argument of [Reference Denef and Loeser9, 3.4] or [Reference Koike and Parusiński23, 4.2], instead we use a topological Noetherianity argument to prove that ${\it\sigma}_{\ast n|{\rm\Delta}_{e,e^{\prime },n}}$ is a piecewise trivial fibration.

We may assume that $M$ is semialgebraically connected, then by Artin-Mazur theorem [Reference Bochnak, Coste and Roy8, 8.4.4], there exist $Y\subset \mathbb{R}^{p+q}$ a nonsingular irreducible algebraic set of dimension $\operatorname{dim}M$ , $M^{\prime }\subset Y$ an open semialgebraic subset of $Y$ , $s:M\rightarrow M^{\prime }$ a Nash-diffeomorphism and $g:Y\rightarrow \mathbb{R}^{N}$ a polynomial map such that the following diagram commutes

Thus, we have

$$\begin{eqnarray}\displaystyle & & \displaystyle {\it\sigma}_{\ast n}^{-1}({\it\pi}_{n}({\it\sigma}_{\ast }({\it\gamma})))=\left\{\vphantom{u\in \mathbb{R}\{t\}^{d},\operatorname{Jac}_{g\circ s}({\it\gamma}(t))u(t)\equiv 0\operatorname{mod}t^{e}}{\it\gamma}(t)+t^{n+1-e}u(t)\operatorname{mod}t^{n+1},\right.\nonumber\\ \displaystyle & & \displaystyle \qquad \qquad \qquad \qquad \left.u\in \mathbb{R}\{t\}^{d},\operatorname{Jac}_{g\circ s}({\it\gamma}(t))u(t)\equiv 0\operatorname{mod}t^{e}\right\}.\nonumber\end{eqnarray}$$

So ${\rm\Delta}_{e,e^{\prime },n}$ is constructible and we may assume that ${\it\sigma}_{\ast n}:{\rm\Delta}_{e,e^{\prime },n}\rightarrow {\it\sigma}_{\ast n}({\rm\Delta}_{e,e^{\prime },n})$ is polynomial up to working with arcs over $M^{\prime }$ via $s$ . The fibers (i.e., $\mathbb{R}^{e}$ ) have odd Euler characteristic with compact support, so by Theorem 2.8 the image ${\it\sigma}_{\ast n}({\rm\Delta}_{e,e^{\prime },n})$ is constructible.

Let $V=\{u_{0}+u_{1}t+\cdots +u_{n}t^{n},u_{i}\in \mathbb{R}^{d}\}$ and fix ${\rm\Lambda}_{0}:V\rightarrow V_{0}$ a linear projection on a subspace of dimension $e$ . The set ${\rm\Omega}_{0}=\{{\it\pi}_{n}({\it\gamma}(t))\in {\rm\Delta}_{e,e^{\prime },n},\operatorname{dim}{\rm\Lambda}_{0}({\it\sigma}_{\ast n}^{-1}({\it\pi}_{n}({\it\sigma}_{\ast }({\it\gamma}))))<e\}$ is closed, constructible and union of fibers of ${\it\sigma}_{\ast n}$ . Therefore $({\it\sigma}_{\ast n},{\rm\Lambda}_{0}):{\rm\Delta}_{e,e^{\prime },n}\setminus {\rm\Omega}_{0}\rightarrow {\it\sigma}_{\ast n}({\rm\Delta}_{e,e^{\prime },n}\setminus {\rm\Omega}_{0})\times V_{0}$ is a constructible isomorphism. We now repeat the argument to the closed constructible subset ${\it\sigma}_{\ast n}({\rm\Omega}_{0})$ and so on. Indeed, assume that ${\rm\Delta}_{e,e^{\prime },n}\supsetneq {\rm\Omega}_{0}\supsetneq {\rm\Omega}_{1}\supsetneq \cdots \supsetneq {\rm\Omega}_{i-1}$ are constructed as previously and that ${\rm\Omega}_{i-1}\neq \varnothing$ , then we may choose ${\rm\Lambda}_{i}$ such that ${\rm\Omega}_{i}\subsetneq {\rm\Omega}_{i-1}$ . So on the one hand the process continues until one ${\rm\Omega}_{i}$ is empty, on the other hand it must stop because of the Noetherianity of the ${\mathcal{A}}{\mathcal{S}}$ -topology. Therefore after a finite number of steps, one ${\rm\Omega}_{i}$ is necessarily empty.◻

Proof. Let ${\it\gamma}\in {\mathcal{B}}_{\mathbf{j}}$ and choose a local coordinate system of $M$ at ${\it\gamma}(0)$ such that the critical locus of ${\it\sigma}$ is locally described by the equation $\prod _{i\in J}x_{i}^{{\it\nu}_{i}}=0$ and $E_{i}$ by the equation $x_{i}=0$ . Since $\operatorname{ord}_{{\it\gamma}}E_{i}=j_{i}$ , we have ${\it\gamma}_{i}(t)=c_{j_{i}}t^{j_{i}}+\cdots \,$ and $c_{j_{i}}\neq 0$ . Then $\prod _{i\in J}{\it\gamma}_{i}^{{\it\nu}_{i}}=ct^{e(\mathbf{j})}+\cdots \,$ with $c\neq 0$ .

So we have $\operatorname{ord}_{t}\left(\operatorname{Jac}_{{\it\sigma}}({\it\gamma}(t))\right)=e(\mathbf{j})$ .

In the same way, $\operatorname{ord}_{{\it\gamma}}{\it\sigma}^{-1}(H)=e^{\prime }(\mathbf{j})$ thus $\operatorname{ord}_{{\it\sigma}({\it\gamma})}(H)=e^{\prime }(\mathbf{j})$ .◻

Proof. Consider $\mathbf{j}$ such that $E_{J}^{\bullet }\neq \varnothing$ and $\text{for all }i\in I,0\leqslant j_{i}\leqslant n$ . The fiber of ${\mathcal{B}}_{\mathbf{j},n}\rightarrow E_{J}^{\bullet }$ is

$$\begin{eqnarray}\mathop{\prod }_{i\in J}(\mathbb{R}^{\ast }\times \mathbb{R}^{n-j_{i}})\times (\mathbb{R}^{n})^{d-|J|}\simeq (\mathbb{R}^{\ast })^{|J|}\times \mathbb{R}^{dn-s_{\mathbf{j}}}\end{eqnarray}$$

since truncating the coordinates of ${\it\gamma}\in {\mathcal{B}}_{\mathbf{j}}$ to degree $n$ produces $d-|J|$ polynomials of degree $n$ with fixed constant terms and for $i\in J$ a polynomial of the form $c_{j_{i}}t^{j_{i}}+c_{j_{i}+1}t^{j_{i}+1}+\cdots +c_{n}t^{n}$ with $c_{j_{i}}\in \mathbb{R}^{\ast }$ and other $c_{k}\in \mathbb{R}$ . We conclude that $\operatorname{dim}{\mathcal{B}}_{\mathbf{j},n}=d(n+1)-s_{\mathbf{j}}$ .

We first assume that $\mathbf{j}\in A_{n}({\it\sigma})$ . By Lemma 4.9, ${\mathcal{B}}_{\mathbf{j}}\subset {\rm\Delta}_{e(\mathbf{j}),e^{\prime }(\mathbf{j})}({\it\sigma})$ . Hence by 4.5 (iv), $X_{\mathbf{j},n}({\it\sigma})$ is constructible since it is the image of the constructible set ${\mathcal{B}}_{\mathbf{j},n}$ by the map ${\it\sigma}_{\ast n|{\rm\Delta}_{e(\mathbf{j}),e^{\prime }(\mathbf{j}),n}}$ with fibers of odd Euler characteristic with compact support. Let ${\it\gamma}_{1}\in {\mathcal{B}}_{\mathbf{j},n}$ and ${\it\gamma}_{2}\in {\rm\Delta}_{e(\mathbf{j}),e^{\prime }(\mathbf{j}),n}$ with ${\it\sigma}_{\ast n}({\it\gamma}_{1})={\it\sigma}_{\ast n}({\it\gamma}_{2})$ , then, by 4.5 (ii), ${\it\gamma}_{1}\equiv {\it\gamma}_{2}\operatorname{mod}t^{n-e(\mathbf{j})+1}$ with $n-e(\mathbf{j})\geqslant e(\mathbf{j})$ and hence ${\it\gamma}_{2}\in {\mathcal{B}}_{\mathbf{j},n}$ . Thus by 4.5 (iv) the map ${\mathcal{B}}_{\mathbf{j},n}\rightarrow X_{\mathbf{j},n}({\it\sigma})$ is a piecewise trivial fibration with fiber $\mathbb{R}^{e(\mathbf{j})}$ . So we have $\operatorname{dim}X_{\mathbf{j},n}({\it\sigma})=d(n+1)-s_{\mathbf{j}}-e(\mathbf{j})$ as claimed.

Otherwise $\mathbf{j}\notin A_{n}({\it\sigma})$ and then $\operatorname{dim}X_{\mathbf{j},n}\leqslant \operatorname{dim}{\mathcal{B}}_{\mathbf{j},n}=d(n+1)-s_{\mathbf{j}}<d(n+1)-\frac{n}{c}$ (since $\frac{n}{2}<e(\mathbf{j})\leqslant {\it\nu}_{\operatorname{max}}s_{\mathbf{j}}$ or $n<e^{\prime }(\mathbf{j})\leqslant {\it\lambda}_{\operatorname{max}}s_{\mathbf{j}}$ ).◻

Notation 4.13. We set

$$\begin{eqnarray}\displaystyle & \displaystyle \widetilde{{\it\pi}_{n}({\mathcal{L}}(X))}:=\overline{Z_{n}({\it\sigma})\sqcup ({\it\pi}_{n}({\mathcal{L}}(X))\setminus \operatorname{Im}{\it\sigma}_{\ast n})}^{\,{\mathcal{A}}S}\sqcup \mathop{\sqcup }_{\mathbf{j}\in A_{n}({\it\sigma})}X_{\mathbf{j},n}({\it\sigma}) & \displaystyle \nonumber\\ \displaystyle & \displaystyle \left(\text{resp. }\widetilde{\operatorname{Im}\tilde{{\it\sigma}}_{\ast n}}:=\overline{Z_{n}(\tilde{{\it\sigma}})}^{\,{\mathcal{A}}S}\sqcup \mathop{\sqcup }_{\mathbf{j}\in A_{n}(\tilde{{\it\sigma}})}X_{\mathbf{j},n}(\tilde{{\it\sigma}})\right) & \displaystyle \nonumber\end{eqnarray}$$

where the closure is taken in the complement of $\sqcup _{\mathbf{j}\in A_{n}({\it\sigma})}X_{\mathbf{j},n}({\it\sigma})(\text{resp. in }\widetilde{{\it\pi}_{n}({\mathcal{L}}(X))}\setminus \sqcup _{\mathbf{j}\in A_{n}(\tilde{{\it\sigma}})}X_{\mathbf{j},n}(\tilde{{\it\sigma}}))$ . Hence we still have the inclusion $\widetilde{\operatorname{Im}\tilde{{\it\sigma}}_{\ast n}}\subset \widetilde{{\it\pi}_{n}({\mathcal{L}}(X))}$ , the unions are still disjoint and the dimensions remain the same.

Proof. We have

$$\begin{eqnarray}\displaystyle {\it\beta}\left(X_{\mathbf{j},n}({\it\sigma})\right) & = & \displaystyle {\it\beta}\left({\mathcal{B}}_{\mathbf{j},n}\right)u^{-\sum {\it\nu}_{i}j_{i}}~~\text{ by Lemmas 4.5 and 4.9}\nonumber\\ \displaystyle & = & \displaystyle {\it\beta}\left(E_{J}^{\bullet }\times (\mathbb{R}^{\ast })^{|J|}\times \mathbb{R}^{dn-s_{\mathbf{j}}}\right)u^{-\sum {\it\nu}_{i}j_{i}}\nonumber\\ \displaystyle & & \displaystyle \quad \text{by the beginning of the proof of Lemma 4.10}\nonumber\\ \displaystyle & = & \displaystyle {\it\beta}\left(E_{J}^{\bullet }\right)(u-1)^{|J|}u^{nd-s_{\mathbf{j}}-\sum {\it\nu}_{i}j_{i}}.\nonumber\end{eqnarray}$$

The same argument works for $\tilde{{\it\sigma}}$ too.◻

Proof. Applying the virtual Poincaré polynomial to the partitions of Notation 4.13, we get

$$\begin{eqnarray}\displaystyle & & \displaystyle {\it\beta}\left(\widetilde{{\it\pi}_{n}({\mathcal{L}}(X))}\right)-{\it\beta}\left(\widetilde{\operatorname{Im}\tilde{{\it\sigma}}_{\ast n}}\right)-\mathop{\sum }_{\mathbf{j}\in A_{n}({\it\sigma})\cap A_{n}(\tilde{{\it\sigma}})}\left({\it\beta}(X_{\mathbf{j},n}({\it\sigma}))-{\it\beta}(X_{\mathbf{j},n}(\tilde{{\it\sigma}}))\right)\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\sum }_{\mathbf{j}\in A_{n}({\it\sigma})\setminus A_{n}(\tilde{{\it\sigma}})}{\it\beta}(X_{\mathbf{j},n}({\it\sigma}))-\mathop{\sum }_{\mathbf{j}\in A_{n}(\tilde{{\it\sigma}})\setminus A_{n}({\it\sigma})}{\it\beta}(X_{\mathbf{j},n}(\tilde{{\it\sigma}}))\nonumber\\ \displaystyle & & \displaystyle \qquad +\,{\it\beta}\left(\overline{Z_{n}({\it\sigma})\sqcup ({\it\pi}_{n}({\mathcal{L}}(X))\setminus \operatorname{Im}{\it\sigma}_{\ast n})}^{\,{\mathcal{A}}{\mathcal{S}}}\right)-{\it\beta}\left(\overline{Z_{n}(\tilde{{\it\sigma}})}^{\,{\mathcal{A}}{\mathcal{S}}}\right).\nonumber\end{eqnarray}$$

We set

$$\begin{eqnarray}\displaystyle & \displaystyle P_{n}={\it\beta}\left(\widetilde{{\it\pi}_{n}({\mathcal{L}}(X))}\right)-{\it\beta}\left(\widetilde{\operatorname{Im}\tilde{{\it\sigma}}_{\ast n}}\right), & \displaystyle \nonumber\\ \displaystyle & \displaystyle Q_{n}=-\mathop{\sum }_{\mathbf{j}\in A_{n}({\it\sigma})\cap A_{n}(\tilde{{\it\sigma}})}\left({\it\beta}(X_{\mathbf{j},n}({\it\sigma}))-{\it\beta}(X_{\mathbf{j},n}(\tilde{{\it\sigma}}))\right), & \displaystyle \nonumber\\ \displaystyle & \displaystyle R_{n}=\mathop{\sum }_{\mathbf{j}\in A_{n}({\it\sigma})\setminus A_{n}(\tilde{{\it\sigma}})}{\it\beta}(X_{\mathbf{j},n}({\it\sigma})),\quad \displaystyle S_{n}=-\mathop{\sum }_{\mathbf{j}\in A_{n}(\tilde{{\it\sigma}})\setminus A_{n}({\it\sigma})}{\it\beta}(X_{\mathbf{j},n}(\tilde{{\it\sigma}})), & \displaystyle \nonumber\\ \displaystyle & \displaystyle T_{n}={\it\beta}\left(\overline{Z_{n}({\it\sigma})\sqcup ({\it\pi}_{n}({\mathcal{L}}(X))\setminus \operatorname{Im}{\it\sigma}_{\ast n})}^{\,{\mathcal{A}}{\mathcal{S}}}\right),\quad \displaystyle U_{n}=-{\it\beta}\left(\overline{Z_{n}(\tilde{{\it\sigma}})}^{\,{\mathcal{A}}{\mathcal{S}}}\right). & \displaystyle \nonumber\end{eqnarray}$$

Assume there exists $i_{0}\in I$ such that ${\it\nu}_{i_{0}}>\tilde{{\it\nu}}_{i_{0}}$ .

Then for $n$ big enough,

$$\begin{eqnarray}K_{n}=\left\{s_{\mathbf{j}}+\mathop{\sum }_{i\in I}\tilde{{\it\nu}}_{i}j_{i},~\mathbf{j}\in A_{n}({\it\sigma})\cap A_{n}(\tilde{{\it\sigma}}),\mathop{\sum }_{i\in I}({\it\nu}_{i}-\tilde{{\it\nu}}_{i})j_{i}>0\right\}\end{eqnarray}$$

is not empty. The minimum $k_{n}=\min K_{n}$ stabilizes for $n$ greater than some rank $n_{0}$ . Let $k=k_{n_{0}}$ . Then, for $n\geqslant n_{0}$ , the degree of $Q_{n}$ is $\operatorname{max}\left\{d(n+1)-s_{\mathbf{j}}-\sum _{i\in I}\tilde{{\it\nu}}_{i}j_{i}\right\}=d(n+1)-k$ using the computation at the beginning of the proof of Lemma 4.10.

The leading coefficients of $P_{n}$ is positive since $P_{n}={\it\beta}\left(\widetilde{{\it\pi}_{n}({\mathcal{L}}(X))}\setminus \widetilde{\operatorname{Im}\tilde{{\it\sigma}}_{\ast n}}\right)$ . The leading coefficient of $Q_{n}$ is also positive. Hence the degree of the LHS is at least $d(n+1)-k$ .

Moreover, we have $\operatorname{deg}R_{n}<d(n+1)-\frac{n}{\tilde{c}}$ , $\operatorname{deg}S_{n}<d(n+1)-\frac{n}{c}$ , $\operatorname{deg}T_{n}<d(n+1)-\frac{n}{\operatorname{max}(c,1)}$ and $\operatorname{deg}U_{n}<d(n+1)-\frac{n}{\tilde{c}}$ . Indeed, for $T_{n}$ , ${\it\pi}_{n}({\mathcal{L}}(X))\setminus \!\operatorname{Im}{\it\sigma}_{\ast n}\subset {\it\pi}_{n}({\mathcal{L}}(X_{\text{sing}}))$ and $\operatorname{dim}\left({\it\pi}_{n}({\mathcal{L}}(X_{\text{sing}}))\right)\leqslant (n+1)(d-1)<d(n+1)-n$ by 2.33 (i). So the degree of the RHS is less than $d(n+1)-\frac{n}{\operatorname{max}(c,\tilde{c},1)}$ .

We get a contradiction for $n$ big enough.◻

Proof. Let ${\it\gamma}$ be an analytic arc on $X$ not entirely in $S$ and not entirely in the singular locus of $X$ .

Assume that ${\it\gamma}\notin \operatorname{Im}\tilde{{\it\sigma}}_{\ast }$ . Then, by Proposition 2.21, we have

$$\begin{eqnarray}\tilde{{\it\sigma}}^{-1}({\it\gamma}(t))=\mathop{\sum }_{i=0}^{m}b_{i}t^{i}+bt^{\frac{p}{q}}+\cdots \,,\quad b\neq 0,\;m<\frac{p}{q}<m+1,\;t\geqslant 0.\end{eqnarray}$$

Since $\tilde{{\it\sigma}}^{-1}$ is locally Hölder by Remark 2.20, there is $N\in \mathbb{N}$ such that for every analytic arc ${\it\eta}$ on $X$ with ${\it\gamma}\equiv {\it\eta}\operatorname{mod}t^{N}$ we have $\tilde{{\it\sigma}}^{-1}({\it\eta}(t))\equiv \tilde{{\it\sigma}}^{-1}({\it\gamma}(t))\operatorname{mod}t^{m+1}$ . Hence such an analytic arc ${\it\eta}$ is not in the image of $\tilde{{\it\sigma}}_{\ast }$ and for $n\geqslant N$ , ${\it\pi}_{n}({\it\eta})$ is not in the image of $\tilde{{\it\sigma}}_{\ast n}:{\mathcal{L}}_{n}(M)\rightarrow {\it\pi}_{n}({\mathcal{L}}(X))$ . Hence $({{\it\pi}_{N}^{n}}_{|{\it\pi}_{n}({\mathcal{L}}(X))})^{-1}({\it\pi}_{N}({\it\gamma}))\subset {\it\pi}_{n}({\mathcal{L}}(X))\setminus \operatorname{Im}(\tilde{{\it\sigma}}_{\ast n})$ .

The first step consists in computing the dimension of the fiber $({{\it\pi}_{N}^{n}}_{|{\it\pi}_{n}({\mathcal{L}}(X))})^{-1}({\it\pi}_{N}({\it\gamma}))$ where $n\geqslant N$ . For that, we will work with a resolution ${\it\rho}:\tilde{X}\rightarrow X$ (for instance ${\it\sigma}$ ) instead of $\tilde{{\it\sigma}}$ since every analytic arc on $X$ not entirely included in $X_{\text{sing}}$ may be lifted to $\tilde{X}$ by ${\it\rho}$ . Let ${\it\theta}$ be the unique analytic arc on $\tilde{X}$ such that ${\it\rho}({\it\theta})={\it\gamma}$ . Let $e=\operatorname{ord}_{t}(\operatorname{Jac}_{{\it\rho}}({\it\theta}(t)))$ and $e^{\prime }$ be such that ${\it\gamma}\in {\mathcal{L}}^{(e^{\prime })}(X)$ . We may assume that $N\geqslant \operatorname{max}(2e,e^{\prime })$ in order to apply Lemma 4.5 to ${\it\rho}$ for ${\it\gamma}$ .

We consider the following diagram

Since the fibers of ${{\it\rho}_{\ast n}}_{|{\rm\Delta}_{e,e^{\prime },n}}$ and ${{\it\rho}_{\ast N}}_{|{\rm\Delta}_{e,e^{\prime },N}}$ are of dimension $e$ , and since the fibers of ${\it\pi}_{N}^{n}:{\mathcal{L}}_{n}(\tilde{X})\rightarrow {\mathcal{L}}_{N}(\tilde{X})$ are of dimension $(n-N)d$ , we have $\operatorname{dim}\big(\big({{\it\pi}_{N}^{n}}_{|{\it\pi}_{n}({\mathcal{L}}(X))}\big)^{-1}({\it\pi}_{N}({\it\gamma}))\big)=(n-N)d$ .

Hence $\operatorname{dim}\left({\it\pi}_{n}({\mathcal{L}}(X))\setminus \operatorname{Im}(\tilde{{\it\sigma}}_{\ast n})\right)\geqslant (n-N)d$ . And so, with the notation of Lemma 4.15, we have

$$\begin{eqnarray}P_{n}+0=R_{n}+S_{n}+T_{n}+U_{n}\end{eqnarray}$$

with $\operatorname{deg}P_{n}\geqslant (n-N)d=(n+1)d-(N+1)d$ and $\operatorname{deg}(R_{n}+S_{n}+T_{n}+U_{n})<(n+1)d-\frac{n}{\operatorname{max}(c,\tilde{c},1)}$ .

We get a contradiction for $n$ big enough.◻

End of the proof of Theorem 3.5.

Let ${\it\gamma}$ be an analytic arc on $X$ not entirely included in $S\cup X_{\text{sing}}$ . By Lemma 4.17 and since ${\it\gamma}$ is not entirely included in $S\cup X_{\text{sing}}$ , $\tilde{{\it\sigma}}^{-1}({\it\gamma}(t))$ is well defined and analytic. Hence $f^{-1}({\it\gamma}(t))={\it\sigma}(\tilde{{\it\sigma}}^{-1}({\it\gamma}(t)))$ is real analytic. Finally $f^{-1}$ is generically arc-analytic in dimension $d=\operatorname{dim}X$ .

So $f^{-1}$ is blow-Nash by Proposition 2.27 and $\text{for all }i\in I,{\it\nu}_{i}=\tilde{{\it\nu}}_{i}$ by Lemma 4.15. Then, arguing as in Lemma 3.1, $f^{-1}$ satisfies the Jacobian hypothesis too.◻