Fact 2.5. Suppose K is a 1-h-minimal valued field. Then there exists a language $\mathcal {L}'\supseteq \mathcal {L}$ , an elementary extension $K'$ of K and an $\aleph _0$ -saturated $\mathcal {L}'$ -structure on $K'$ extending the $\mathcal {L}$ -structure of $K'$ , such that $K'$ is 1-h-minimal as an $\mathcal {L}'$ -structure, and such that $\operatorname {\mathrm {acl}}_{\mathcal {L}'}(A)=\operatorname {\mathrm {dcl}}_{\mathcal {L}'}(A)$ for all $A\subseteq K'$ .

Fact 3.2. Let M be an $\aleph _0$ -saturated structure and $\{\Phi ^l(\bar {D})\}_{l\in I}$ be a family of properties of definable sets $\bar {D}=(D_1,\dots ,D_n)$ in M, indexed by a directed set I. Let b be a tuple in M and S a b-definable set. Assume that

1. (1) For all l, the property $\Phi ^l$ is definable in definable families. That is, if for $i=1,\dots ,n$ , we have $\{D_{i,a}\}_{a\in T}$ b-definable families, then the set $\{a\in T: \Phi ^l({\bar {D}_a})\text { holds}\}$ is b-definable.

2. (2) $\Phi ^l$ implies $\Phi ^{l'}$ for all $l\leq l'$ .

3. (3) For every $a\in S$ , and for $i=1,\dots ,n$ , there are $ba$ -definable sets $D_{i,a}$ , satisfying $\Phi ^{l_a}$ for some $l_a\in I$ .

Then for $i=1,\dots , n$ , there are $\{D_{i,a}\}_{a\in S}$ b-definable families of sets, and a fixed $l\in I$ , such that $\Phi ^l(\bar {D}_{a})$ holds for every $a\in S$ .

Proof. Note that by the previous remark, it suffices to prove the result for $n=1$ , as we shall presently proceed to do. Let $a\in S$ . By hypothesis, there is a b-definable family $\{D_{a'}^a\}_{a'\in S^{0,a}}$ and an element $l^a\in I$ , such that $D_{a}^a$ satisfies $\Phi ^{l^a}$ . Consider $S^a$ to be the set of $a'\in S$ such that $a'\in S^{0,a}$ and such that $\Phi ^{l^a}(D_{a'}^a)$ holds. By hypothesis, this is a b-definable set contained in S and containing a.

We conclude that $S=\bigcup _{a\in S}S^a$ is a cover of S by b-definable sets, and so by compactness and saturation, there is a finite sub-cover, say $S=S^1\cup \dots \cup S^k$ for $S^r=S^{a_r}$ . Indeed, if there was no finite sub-cover, then the partial type expressing $x\in S$ and $x\notin S^a$ for all $a\in S$ is a consistent b-type, and so a realization in M would contradict $S=\bigcup _{a\in S}S^a$ .

Then $D_{a}$ defined as $D_{a}^{a_r}$ if $a\in S^r\setminus \bigcup _{r'<r}S^{r'}$ satisfies that $\{D_{a}\}_{a\in S}$ forms a b-definable family. If we take l such that $l\geq l^{a_1},\dots ,l^{a_k}$ , then we get that $\Phi ^l(D_{a})$ holds for every $a\in S$ , as required.

Notation 3.4. If $D\subset E\times F$ , and $a\in E$ , we often denote $D_a=\{b\in F: (a,b)\in D\}$ . If $b\in F$ , we denote, when no ambiguity can occur, $D_b=\{a\in E: (a,b)\in D\}$ . If $f:D\to C$ is a function, we let $f_a:D_a\to C$ denote the function $f_a(b)=f(a,b)$ , and similarly, $f_b$ for $b\in F$ .

Proof. We proceed by induction on n, the case $n=1$ being [Reference Cluckers, Halupczok, Rideau-Kikuchi and Vermeulen5, Theorem 3.1.2]. Assume the result for n and let $f:U\to K$ be an a-definable function with $U\subset K^n\times K$ open, i a multi-index with $|i|\le r$ . Then for every $x\in K^n$ , there is a finite $ax$ -definable set $C_x\subset K$ and an integer $m_x$ such that

(1) $$ \begin{align} |f_x(y)-\sum_{s<r}\frac{1}{s!}f_x^{(s)}(y_0)(y-y_0)^s| \leq |\frac{1}{r!}f_x^{(r)}(y_0)(y-y_0)^r| \end{align} $$

for every y and $y_0$ in an open ball $m_x$ -away from $C_x$ , and such that $|f_x^{(s)}(y)|$ exists and is constant in any such open ball. By a standard compactness argument (see Fact 3.2), we may assume that the $C_x$ are uniformly definable and that there is some $m\in {\mathbb {N}}$ such that $m_x=m$ for all x. Define $C=\bigcup _x(\{x\}\times C_x)$ . By induction, for each $y\in K$ , we can approximate the functions $g_{s,y}(x)=f_x^{(s)}(y)$ defined on $V_y=\mathrm {Int}(U_y\setminus C_y)$ up to order $r-s$ . By a similar application of Fact 3.2, we obtain a natural number $m'$ and an a-definable family $\{D_y\}_{y\in K^n}$ of subsets $D_y\subseteq V_y$ with $\dim (D_y)<n$ such that $g_{s,y}^{(i)}$ exists and has constant valuation on any ball $m'$ -away from $D_y$ in $V_y$ , for every multi-index i, with $|i|\leq r-s$ . Moreover,

(2) $$ \begin{align} | g_{s,y}(x)-\sum_{\{i:|i|<r-s\}}\frac{1}{i!}g_{s,y}^{(i)}(x_0)(x-x_0)^i | \leq \max_{\{i:|i|=r-s\}} |\frac{1}{i!}g_{s,y}^{(i)}(x_0)(x-x_0)^i|. \end{align} $$

Replacing m and $m'$ by their maximum, we may assume $m=m'$ . Define $D:=\bigcup _y D_y\times \{y\}$ . By additivity of dimension, $\dim (C)\le n$ and $\dim (D)\le n$ . Now take $E=C\cup D\cup \bigcup _y (U_y\setminus V_y)\times \{y\}$ . Similar dimension considerations show that $\dim (E)<n+1$ .

Note that for $(x,y)\in U\setminus E$ , we have that, for i and s such that $|i|+s\leq r$ , $f^{(i,s)}(x,y)$ and $g_{s,y}^{(i)}(x)$ exist and are equal.

Now, for $x\in K^n$ , define $W_x=\mathrm {Int}(U_x\setminus E_x)$ , and for the functions $h_{x,s,i}:y\mapsto f^{(i,s)}(x,y)$ with $s+|i|\leq r$ defined on $W_x$ , we find a finite set $F_x\subset W_x$ such that $\{F_x\}_x$ is an a-definable family, and there is an integer $m'$ , such that in every ball in $W_x$ $m'$ -away from $F_x$ , $h_{x,s,i}$ has constant valuation. We may assume that $m'=m$ as before. Let G be the closure of $E\cup \bigcup _x (\{x\}\times F_x)\cup \bigcup _x (\{x\}\times (U_x\setminus W_x))$ . Note that $\dim (G)<n+1$ .

Take $B_1\times B_2$ a ball in U, m-away from G. Then for every $x\in B_1$ , we get that $B_2$ is m-away from both $C_x$ and $F_x$ and $B_2\subseteq W_x$ . Similarly, for every $y\in B_2$ , $B_1\subseteq V_y$ is m-away from $D_y$ .

We conclude that for every $(x,y)\in B_1\times B_2$ , $f^{(i,s)}(x,y)$ exists and has constant valuation, for every index $(i,s)$ such that $|(i,s)|\leq r$ . Indeed, we have for every $(x,y), (x',y')\in B_1\times B_2$ that

$$\begin{align*}|f^{(i,s)}(x',y')|=|g_{s,y'}^{(i)}(x')|=|g_{s,y'}^{(i)}(x)|=|h_{x,s,i}(y')| =|h_{x,s,i}(y)|=|f^{(i,s)}(x,y)|, \end{align*}$$

as the second equality follows from the condition on $D_{y'}$ , and the fourth from those on $F_x$ and $W_x$ .

Now, if $(x,y)$ and $(x_0,y_0)$ are in $B_1\times B_2$ , then equations 1 and 2 hold, and for the error term of 1, we have $|f_x^{(r)}(y_0)|=|f^{(0,r)}(x,y_0)|= |f^{(0,r)}(x_0,y_0)|$ . Denote

$$\begin{align*}M=\max\left \{\frac{1}{i!s!} |f^{(i,s)}(x_0,y_0)(x-x_0)^i(y-y_0)^s|: |(i,s)|=r\right \}.\end{align*}$$

Then Equation 1 yields $|f(x,y)-\sum _{s<r}\frac {1}{s!}f^{(0,s)}(x,y_0)(y-y_0)^s|\leq M$ . Also, from Equation 2, we have that

$$\begin{align*}\left |\frac{1}{s!}f^{(0,s)}(x,y_0)(y-y_0)^s- \sum_{\{i:|i|<r-s\}}\frac{1}{i!s!}f^{(i,s)}(x_0,y_0)(x-x_0)^i(y-y_0)^s \right | \leq M.\end{align*}$$

Taking the sum over s smaller than r and using the ultrametric inequality, we obtain

$$\begin{align*}\left |\sum_s \frac{1}{s!}f^{(0,s)}(x,y_0)(y-y_0)^s- \sum_{\{(i,s):|i|+s<r\}}\frac{1}{i!s!}f^{(i,s)}(x_0,y_0)(x-x_0)^i(y-y_0)^s\right |\leq M.\end{align*}$$

Summing this with Equation 1 and using the ultrametric inequality once more, we conclude.

Proof. Take a $1$ -dimensional closed $C\subset K\times K$ and m an integer as provided by the Taylor approximation property for errors of order $3$ . We may also assume that $\pi (C)$ and m satisfy the same Taylor approximation property for the function $f\pi $ , where $\pi $ is the coordinate permutation $(x,y)\mapsto (y,x)$ .

Then for $(x,y),(x_0,y_0)\in B_1\times B_2$ in a ball m-away from C, we obtain (see Definition 2.1 for the big-O notation) that

$$ \begin{align*} f(x,y)=& f(x_0,y_0)+ (x-x_0)^2\frac{1}{2}\frac{\partial^2}{\partial x^2}f(x_0,y_0)+ (y-y_0)^2\frac{1}{2}\frac{\partial^2}{\partial y^2}f(x_0,y_0)+ \\ & (x-x_0)(y-y_0)\frac{\partial}{\partial x}\frac{\partial}{\partial y}f(x_0,y_0)+ O((x-x_0,y-y_0)^3). \end{align*} $$

Similarly,

$$ \begin{align*} f\pi(y,x)=f(x,y)=& f(x_0,y_0)+ (x-x_0)^2\frac{1}{2}\frac{\partial^2}{\partial x^2}f(x_0,y_0)+ (y-y_0)^2\frac{1}{2}\frac{\partial^2}{\partial y^2}f(x_0,y_0)+ \\ & (x-x_0)(y-y_0)\frac{\partial}{\partial y}\frac{\partial}{\partial x}f(x_0,y_0)+ O((x-x_0,y-y_0)^3). \end{align*} $$

Taking the difference, we obtain

$$ \begin{align*} (x-x_0)(y-y_0)\frac{\partial}{\partial y}\frac{\partial}{\partial x}f(x_0,y_0)- (x-x_0)(y-y_0)\frac{\partial}{\partial x}\frac{\partial}{\partial y}f(x_0,y_0)= O((x-x_0,y-y_0)^3). \end{align*} $$

Taking $h=(x-x_0)=(y-y_0)$ small, we get $h^2 (\frac {\partial }{\partial y}\frac {\partial }{\partial x}f(x_0,y_0)- \frac {\partial }{\partial x}\frac {\partial }{\partial y}f(x_0,y_0))= O(h^3)$ , so $\frac {\partial }{\partial y}\frac {\partial }{\partial x}f(x_0,y_0)- \frac {\partial }{\partial x}\frac {\partial }{\partial y}f(x_0,y_0)=O(h)$ . This is only possible when the left-hand side is  $0$ , as desired.

The second statement follows from the assumed continuity, as two continuous functions into a Hausdorff space agreeing on a dense subset are equal.

Proof. The problem readily reduces to the case of f a polynomial restricted to some open neighborhood, U, of the origin. That is, we have to show that if $\sum _{i\leq n}b_ix^i=O(x^{n+1})$ in any open $U\subseteq K^r$ , then $b_i=0$ for all i. For $x_0$ fixed, let $x=tx_0$ and consider the single variable polynomial $P(tx_0)=\sum _{i\le n} (b_ix_0^i)t^i=O(t^{n+1})$ . If we knew the result for $r=1$ , this would give that $b_ix_0^i=0$ . Since $x_0\in U$ was arbitrary and U contains a cartesian product of r infinite sets, this implies $b_i=0$ for all i. So we are reduced to proving the result for $r=1$ .

In this case, if i is the smallest with $b_i\neq 0$ , we get $x^i=O(x^{i+1})$ , which is a contradiction.

Proof. For the first statement, we write

$$ \begin{align*} & f(g(a+x))= \\ & f(g(a)+g_1(a)x+...+g_n(a)x^n+O(x^{n+1}))= \\ & f(g(a))+f_1(g(a))h(a,x)+f_2(g(a))h(a,x)^2+...+O(h(a,x)^{n+1})= \\ & fg(a)+b_1(a)x+\cdots+b_n(a)x^n+O(x^{n+1})+O(h(a,x)^{n+1}), \end{align*} $$

where

1. (1) $h(a,x)=g(a+x)-g(a)=g_1(a)x+\cdots +g_n(a)x^n+O(x^{n+1})$

2. (2) The second inequality is the application of the assumption that f is $P_n$ at $g(a)$ .

3. (3) The coefficients $b_i$ arise by expanding the expression

   $$\begin{align*}f_k(a)h(a,x)^k=f_k(a)(g_1(a)x+\cdots+g_n(a)x^n+O(x^{n+1}))^k. \end{align*}$$

To conclude, we note that, as in the proof of Lemma 3.10, $h(a,x)=O(x)$ , and so $O(h(a,x)^{n+1})=O(x^{n+1})$ .

The proof of the second statement is, essentially, similar:

$$ \begin{align*} & f(g(x))=\\ & f(g(x')+g_1(x')(x-x')+...+g_n(x')(x-x')^n+O(x-x')^{n+1})= \\ & f(g(x'))+f_1(g(x'))h(x,x')+f_2(g(x'))h(x,x')^2+...+f_n(g(x'))h(x,x')^n+O(h(x,x')^{n+1})= \\ & f(g(x'))+b_1(x')(x-x')+\cdots+b_n(x')(x-x')^n+O(x-x')^{n+1}, \end{align*} $$

where $h(x,x')=g(x)-g(x')=g_1(x')(x-x')+\cdots +g_n(x')(x-x')^n+O(x-x')^{n+1}= O(x-x')$ , and the coordinates of the coefficients $b_k(x')$ are sums and products of the coordinates of the coefficients of $f_i(g(x'))$ and $g_j(x')$ with $i,j\leq k$ . By what we have just proved, those are $P_{n-i}$ functions. The constant appearing on $h(x,x')=O(x-x')$ does not depend on $x'$ because the $g_i$ are continuous at a. We conclude that the $b_k$ are $P_{n-k}$ at a, as claimed.

Proof. Note that the hypothesis implies $|f(x)-f(x')|=|x-x'|$ . This implies that $f'(a)$ is invertible. Indeed, assume otherwise and take x close to a such that $f'(a)(x-a)=0$ to get $|f(x)-f(a)|<|x-a|$ , a contradiction.

Assume that f is strictly differentiable at a. Take $\epsilon>0$ in $\Gamma $ . Then there is $0<r\in \Gamma $ such that if $|x-a|,|x'-a|<r$ , then $|f(x)-f(x')-f'(a)(x-x')|\le \epsilon |x-x'|$ . If we denote $y=f(x)$ and $y'=f(x')$ , then we have $|y-y'|=|x-x'|$ , so multiplying the above inequality by $f'(a)^{-1}$ , we obtain

$$ \begin{align*} & |f^{-1}(y)-f^{-1}(y')-f'(a)^{-1}(y-y')|= \\ & |f'(a)^{-1}(f'(a)(x-x')-(f(x)-f(x'))|\le \\ & |f'(a)^{-1}||f(x)-f(x')-f'(a)(x-x')|\le \\ & \epsilon |f'(a)^{-1}||x-x'|= \epsilon |f'(a)^{-1}||y-y'|, \end{align*} $$

where, for a linear map, A, represented by the matrix $(a_{ij})_{i,j}$ , we denote $|A|=\max _{i,j}|a_{ij}|$ and use the ultra-metric inequality to get $|Ax|\le |A||x|$ , which we apply to obtain the first inequality in the above computation.

So we conclude that $|f^{-1}(y)-f^{-1}(y')-f'(a)^{-1}(y-y')|\leq \epsilon |f'(a)|^{-1}|y-y'|$ for any $y,y'$ such that $|y-b|=|x-a|<r$ and $|y'-b|=|x'-a|<r$ . We have thus shown that $f^{-1}$ is strictly differentiable at b and $(f^{-1})'(b)=f'(f^{-1}(b))^{-1}$ .

To show that $f^{-1}$ is differentiable if f is, substitute $x'=a$ in the above argument.

Proof. Denote $a=f(b)$ . Note that the hypothesis implies then $|f(x)-f(y)|=|x-y|$ for all distinct $x,y\in U$ , so the inverse map $f^{-1}$ is continuous and in fact satisfies $|f^{-1}(x)-f^{-1}(y)|=|x-y|$ for distinct $x,y\in V$ . In particular, $f^{-1}$ is $T_0$ in V.

Now, assume that f is $P_n$ at b, with $n\ge 1$ . In particular, by Proposition 3.14, it is differentiable and $f'(b)$ is invertible.

Apply the fact that f is $P_n$ (and see also the discussion following the definition) to get

$$\begin{align*}f(y)-f(b)=f'(b)(y-b)+f_2(b)(y-b)^2+\cdots+f_n(b)(y-b)^n+O(y-b)^{n+1}. \end{align*}$$

Rearranging, we get

$$\begin{align*}y-b=f'(b)^{-1}(f(y)-f(b))-f'(b)^{-1}f_2(b)(y-b)^2-\cdots- f'(b)^{-1}f_n(b)(y-b)^n+O(y-b)^{n+1}. \end{align*}$$

Putting $y=f^{-1}(x)$ , and remembering $x-a=O(y-b)$ , we conclude

(◇) $$ \begin{align} f^{-1}(x)-f^{-1}(a)=f'(f^{-1}(a))^{-1}(x-a)+\sum_{2\leq i\leq n}c_i(a)(f^{-1}(x)-f^{-1}(a))^i+O(x-a)^{n+1}, \end{align} $$

for some constants $c_i(a)$ .

Next, we proceed to showing (by induction on $k\le n$ ) that $f^{-1}$ is $P_k$ . As $P_0$ follows from the equality $|f^{-1}(x)-f^{-1}(y)|=|x-y|$ , we assume that $k\geq 1$ . So suppose $f^{-1}$ is $P_{k-1}$ . Using this, we can write $f^{-1}(x)-f^{-1}(a)=\sum _{1\leq j<k}b_j(a)(x-a)^j+O(x-a)^k$ and apply a direct computation to obtain that

$$\begin{align*}c_i(a)(f^{-1}(x)-f^{-1}(a))^i= \sum_{i\leq j\leq k}d_{ij}(a)(x-a)^j+O(x-a)^{k+1} \end{align*}$$

for some constants $d_{ij}$ . Note that as $i\ge 2$ , we obtain the improved error $O(x-a)^{k+1}$ . Substituting this in $(\diamond )$ , it follows that $f^{-1}$ is $P^k$ , as required.

Now suppose f is $T_n$ at b. The proof in this case is similar:

$$\begin{align*}f^{-1}(x)-f^{-1}(x')=f'(f^{-1}(x'))^{-1}(x-x')+\sum_{2\leq i\leq n}\kern-1pt c_i(f^{-1}(x'))(f^{-1}(x)-f^{-1}(x'))^i+O(x-x')^{n+1},\end{align*}$$

so, as above, if $f^{-1}(x)$ is $T_{k-1}$ , we obtain

$$\begin{align*}f^{-1}(x)-f^{-1}(x')=f'(f^{-1}(x'))^{-1}(x-x')+\sum_{2\leq i\leq k}d_i(x')(x-x')^i+O(x-x')^{k+1}.\end{align*}$$

Here, note that $f'$ is $P_{n-1}$ at b and so $f'(f^{-1}(x'))^{-1}$ is $P_{k-1}$ at a. Also, following the above argument, we see that the coordinates of $d_i(x')$ are sums and products of functions of the form $b_{i'}(x')$ with $1\leq i'<i$ , for $b_{i'}(x')$ a $P_{k-1-i'}$ function at a (by the induction hypothesis that $f^{-1}$ is $T_{k-1}$ ), and functions of the form $c_{i'}(f^{-1}(x'))$ for $c_{i'}(y')$ a $P_{k-i'}$ function at b, $i'\leq i$ (by the assumption that f is $T_n$ ). So $d_i(x')$ is $P_{k-i}$ at a.

Proof. By the definition of $T_2$ (with $x=x'$ , $y'=0$ ), we get $f(x,y)=f_0(x)+f_1(x)y+f_2(x)y^2+O(y^3)$ for $f_2$ a $P_0$ function at $0$ , $f_1$ a $P_1$ function at $0$ and $f_0$ a $P_2$ function at $0$ . Fixing x and expanding the Taylor polynomial of $f(x,y)$ , the uniqueness of Taylor coefficients (Lemma 3.10) gives, using our assumption, $axy+f_0(x)+f_1(x)y=0$ .

Expanding $f_0,f_1,f_2$ around $0$ and keeping in mind $f(x,y)= O(x,y)^3$ , we get $f_0(x)= O(x^3)$ and $f_1(x)= O(x^2)$ . Indeed, we have $f_0(x)=b_0+b_1x+b_2x^2+O(x^3)$ , $f_1(x)=c_0+c_1x+O(x^2)$ and $f_2=d_0+O(x)$ , so $f(x,y)=b_0+b_1x+b_2x^2+c_0y+c_1xy+d_0y^2+O(x,y)^3=O(x,y)^3$ . So from the uniqueness of the Taylor coefficients, we get $b_0=b_1=b_2=c_0=c_1=0$ .

Now from $axy=O(x,y)^3$ , we get $a=0$ , by the uniqueness of Taylor coefficients again.

Proof. The proof is similar to spherical completeness in residue characteristic $0$ ; see Lemma 2.7.1 of [Reference Cluckers, Halupczok and Rideau-Kikuchi4]. It is enough consider the $1$ -dimensional case since the higher dimensional case follows by considering the coordinate projections. Note also that our assumption implies that the chain $\{B_i\}$ has no minimal element (as such an element would have valuative radius $0$ ).

The closed case follows from the open case as follows: given a definable chain $\{B_i\}_{i\in I}$ of closed balls, for each i, let $r_i$ be the valuative radius of $B_i$ and let $B_i^{\prime }$ be the unique open ball $B\subseteq B_i$ of valuative radius $r_i$ with the additional property that $B\supseteq B_j$ for all $j<i$ . Obviously, $\bigcap _i B_i=\bigcap _i B_i^{\prime }$ (unless the chain $B_i$ has a minimal element, in which case there is nothing to prove).

Note that, in the above notation, the map $B_i\mapsto r_i$ is injective, so there is no harm assuming that $\{B_i\}$ is indexed by a subset of $\Gamma $ . Thus, our chain $\{B_i\}$ has index set interpretable in $RV$ , so by [Reference Cluckers, Halupczok, Rideau-Kikuchi and Vermeulen5, Proposition 2.3.2], there is a finite set C m-preparing the chain $\{B_i\}$ for some $m\in {\mathbb {N}}$ . We claim that $C\cap B_i\neq \emptyset $ for all $i\in I$ . This would finish the proof since C is finite. Assume, therefore, that this is not the case, and let $i_0\in I$ be such that $B_{i_0}\cap C=\emptyset $ . By assumption, we can find $i<i_0$ such that $r_i<|p^m|r_{i_0}$ . Then $B_i$ is a ball m-away from C, and since our chain has no minimal element, any ball $B\varsubsetneq B_i$ that is an element of our chain is not m-prepared by C, a contradiction.

Proof. Uniqueness is immediate from the hypothesis. For existence, take the family of balls of the form $B(a)_{|f(a)-a|}$ . It is a definable chain of balls indexed by $a\in B_r$ . Indeed, if $a,b\in B_r$ are distinct and the balls are disjoint, then $|f(a)-f(b)|=|a-b|$ , as the distance of points in disjoint balls does not change. Note that in positive residue characteristic, one has the additional hypothesis in Lemma 4.1 because $|f(f(a))-f(a)|\leq |p||f(a)-a|$ , by assumption 2 on f.

By the appropriate version of definable spherical completeness of 1-h-minimal fields (Lemma 2.7.1 of [Reference Cluckers, Halupczok and Rideau-Kikuchi4] for residue characteristic $0$ , Lemma 4.1 otherwise), we obtain a point x in the intersection of all balls. Then x is a fixed point of f. Indeed, if we assume otherwise, then for $y=f(x)$ , we have $|f(y)-y|<|f(x)-x|$ by the hypothesis. However, if a is arbitrary then, as $x\in B(a)_{|f(a)-a|}$ , one has $|f(x)-f(a)|\leq |x-a|\leq |f(a)-a|$ and so $|f(x)-x|\leq |f(a)-a|$ . This is a contradiction and finishes the proof.

Proof. Injectivity of the map follows directly from the hypothesis. The same assumptions also imply that if $x,y\in U$ are distinct, then $|f(x)-f(y)|=|x-y|$ , implying continuity of the inverse.

The main difficulty is showing that $f(U)$ is open. Translating, we may assume $0\in U$ and $f(0)=0$ . We have to find an open neighborhood of $0$ in $f(U)$ . Take $r>0$ such that $0\in B_r\subset U$ . Then $B_{r}\subset f(U)$ . Indeed, if $|a|\leq r$ , then, by the same reasoning as above, the function $g(x)=x+a-f(x)$ satisfies $g(B_r)\subset B_r$ . By the assumptions on f, this implies that g satisfies the hypothesis of Proposition 4.2. So $g(x_0)=x_0$ for some $x_0$ , namely, $a=f(x_0)$ , as claimed.

Differentiability (and strict differentiability) of $f^{-1}$ now follow from 3.14.

Proof. By the definition of strict differentiability, the function $f'(a)^{-1}f$ satisfies the hypothesis of the previous proposition in a small open ball around a. The conclusion follows.

Proof. We proceed by induction on n, the dimension of the domain. We may assume $m=1$ .

First, assume $n=1$ . By the valuative Jacobian property in [Reference Cluckers, Halupczok and Rideau-Kikuchi4, Corollary 3.1.6] and [Reference Cluckers, Halupczok, Rideau-Kikuchi and Vermeulen5, Corollary 3.1.3], there is a finite set $C\subset U$ such that in $U\setminus C$ the function f is locally constant. This implies that the fibers of $f|_{U\setminus C}$ are of dimension 1, and so the image of f is finite. As f is continuous, it is locally constant; the fibers form a finite partition of closed and so open sets on which f is constant.

Now assume the proposition is valid for n, and suppose $U\subset K^n\times K$ . We denote $\pi _1:K^n\times K\to K^n$ the projection onto the first factor and $\pi _2:K^n\times K\to K$ the projection onto the second factor.

Let $V\subset U$ be an open dense subset such that f is differentiable at V with derivative $0$ . Denote $C=U\setminus V$ . Let $T=\{x\in K^n\mid \dim (\pi _1^{-1}(x)\cap C)=1\}$ . Then $\dim (T)<n$ . We conclude that there is an open dense set $W\subset K^{n}$ such that $\dim (\pi _1^{-1}(x)\cap C)=0$ for all $x\in W$ . Similarly, there is an open dense set $P\subset K$ such that $\dim (\pi _2^{-1}(x)\cap C)<n$ for every $x\in P$ . Shrinking V to $V\cap \pi _1^{-1}(W)\cap \pi _2^{-1}(P)$ , we may assume that if $(x,y)\in V$ , then $V_x$ is an open dense subset of $U_x$ and $V_y$ is an open dense subset of $U_y$ . By the induction hypothesis and the $n=1$ case, we conclude that $f_x$ and $f_y$ are locally constant. This implies that the fiber $f^{-1}f(x,y)$ has dimension $n+1$ . Indeed, if B is an open neighborhood of y on which $f_x$ is constant and for each $y'\in B$ we take $R_{y'}\subset U_{x}$ an open neighborhood of x on which $f_{y'}$ is constant, then f is constant on $\bigcup _{y'\in B} R_{y'}\times \{y'\}$ . By dimension considerations, we conclude that the image $f(V)$ is finite. As f is continuous, $f^{-1}f(V)$ is closed in U, and as V is dense in U, we conclude $f(U)=f(V)$ is finite. As f is continuous, we conclude f is locally constant as before.

Proof. The set D of points $x\in U$ such that for every point of $\{x\}\times V$ f is locally of the form $g(x)$ is definable. More precisely, $x\in D$ exactly when for all $y\in V$ , there exists an open ball $B\ni (x,y)$ , such that for all $(x',y'), (x',y")\in B$ , we have $f(x',y')=f(x',y")$ . Thus, the statement that D has dense interior in U is a first order expressible property, so we may assume that $\operatorname {\mathrm {acl}}=\operatorname {\mathrm {dcl}}$ ; see Fact 2.5 and the subsequent remark.

In the course of the proof, we may replace U by a dense open subset a finite number of times. Fix $W\subset U\times V$ , a dense open set where f is differentiable and its derivative with respect to y is $0$ . Shrinking U, we may assume $W_x\subset V$ is dense, open for all $x\in U$ . By Proposition 4.6, we know that $f_x$ is locally constant with finite image for every $x\in U$ (recall that $f_x(y):=f(x,y)$ ). The sets $\mathrm {Im}(f_x)$ form a definable family of finite sets indexed by $x\in U$ , so there is a uniform bound, n, on their cardinalities. Denoting $A_k=\{x\in U: |\mathrm {Im}(f_x)|=k\}$ , we have $U=A_1\cup \cdots \cup A_n$ , so the union of the interiors of the $A_k$ form a dense open subset of U. So we may assume that $|\mathrm {Im}(f_x)|=k$ for all x and some fixed k. Since we assumed that $\operatorname {\mathrm {acl}}=\operatorname {\mathrm {dcl}}$ , there are definable functions $r_1,\dots , r_k:U\to K^m$ such that $\{r_1(x),\dots ,r_k(x)\}=\mathrm {Im}(f_x)$ . By generic continuity of definable functions (Proposition 2.12), we may assume that $r_i$ are all continuous. Then the sets $B_i=\{(x,y): f(x,y)=r_i(x)\}$ form a finite partition of $U\times V$ into closed, and so open subsets.

Proof. After a coordinate permutation in the target, we may assume the principal $k\times k$ minor of $f'(a)$ is invertible. Consider the function $g:U\times K^r\to K^k\times K^r$ defined as $g(x,y)=f(x)+(0,y)$ . Then g is strictly differentiable and has invertible derivative at $(a,0)$ so by the inverse function theorem, Proposition 4.4, we can find a ball B around a and a ball $B_2$ around $0$ , and open set $f(a)\in V$ such that g restrict to a strict diffeomorphism $g:B\times B_2\to V$ . If $i:B\to B\times B_2$ is the inclusion $i(b)=(b,0)$ , then we get that $gi=f$ , so we conclude the statement is valid with $\varphi =g^{-1}$ .

Proof. After applying a coordinate permutation to U, we may assume that the principal $k\times k$ minor of $f'(a)$ is invertible.

Consider the function $g:U\to K^k\times K^r$ defined as $g(x,y)=(f(x,y),y)$ . Then g is strictly differentiable with invertible differential, so by the inverse function theorem, Proposition 4.4, there is an open set $a\in U'\subset U$ such that $g(U')$ is open and $g:U'\to g(U')$ is a strict diffeomorphism.

Shrinking $U'$ , we may assume $g(U')=B\times B_2$ is a product of two balls. Then if $p:B\times B_2\to B$ is the projection $p(b,c)=b$ , we get that $pg=f$ , and so the statement is valid with $\varphi =g^{-1}$ .

Proof. Take a point $(b,c)\in U$ . After a coordinate permutation in U and V, we may assume $f'(b,c)$ has its first $k\times k$ minor invertible. Then by the theorem on submersions, Proposition 4.10, applied to the composition of $f:U\to K^k\times K^s$ with the projection $K^k\times K^s\to K^k$ onto the first factor, we may assume (absorbing the diffeomorphism $\varphi _1$ provided by that proposition) that U is of the form $B_1\times B_2$ and f is of the form ${f(x,y)=(x,g(x,y))}$ . As $f'$ has constant rank equal to k, we conclude that $D_yg=0$ . By Proposition 4.8, we may assume $g(x,y)$ is of the form $g(x,y)=g(x)$ (after passing, if needed, to smaller open sub-balls of $B_1$ and $B_2$ , not necessarily containing $(b,c)$ ). Now the function ${h:B_1\to K^k\times K^s}$ defined by $h(x)=(x,g(x))$ is a definable strictly differentiable immersion, so by the theorem on immersions 4.9, we may, after shrinking $B_1$ and composing with a definable diffeomorphism $\varphi _2$ in the target, assume that h is of the form $h(x)=(x,0)$ . This finishes the proof.

Proof. For the statement about continuity, note that V has the subspace topology (of $V\subset K^r$ ), so we may assume $V=K^r$ . If we denote $U'$ the interior (relative to U) of the set of points of U where f is continuous, then in every ball B where i is a homeomorphism $i:B\cap U\to i(B\cap U)$ , we get that $B\cap U'$ is dense in $B\cap U$ , by generic continuity of definable functions. We conclude that $U'$ is dense as required.

For strict differentiability and $T_k$ , by the above, we may assume that f is continuous. Let $U'$ be the interior of the set of all points where $f:U\to V$ is strictly differentiable and $T_k$ . This is a definable open set. By generic differentiability and generic $T_k$ property for functions defined on open sets, for every point $x\in U$ , there is an open ball $B\ni x$ , such that $B\cap U'$ is dense in $U\cap B$ . Thus, we conclude that $U'$ is dense in U.

Proof. Considering the charts in M, we may assume $M=U\to K^n$ is étale. Now if $(V_i,\tau _i)$ are charts for N, then $f^{-1}\tau _i(V_i)$ cover U, and so the union of their interiors is open dense in U. So we may assume $N=V\to K^m$ is étale. This case is Lemma 5.6.

Proof. If $M=U_1\cup \cdots \cup U_n$ is a covering by open sets and $\varphi _i:U_i\to V_i$ is a homeomorphism onto a set $V_i$ , with an étale map $V_i\to K^n$ , then $\mathrm {dim}(X)=\max _i(\mathrm {dim }(\varphi _i(X\cap U_i)))$ , and the local dimension of X at $x\in X\cap U_i$ is the local dimension of $\varphi _i(X\cap U_i)$ at $\varphi _i(x)$ , so we reduce to the case $M=V$ is étale over $K^n$ . In fact, the result is true whenever $M\subset K^m$ with the subspace topology, as then the local dimension of $X\subset M$ at a point x equals the local dimension of X at x in $K^m$ , and so the result follows from Proposition 2.11(3).

If G is a definable weak topological group and H is a subgroup, then the local dimension of H at any point $h\in H$ is constant independent of h. Indeed, the left translation ${L_h:G\to G}$ is a definable homeomorphism that sends e to h and satisfies $L_h(H)=H$ , so $\mathrm {dim}_e(H)=\mathrm {dim}_h(H)$ .