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Hi,

Consider a $p$-adic field $K$ (finite extension $\DeclareMathOperator{\bQ}{\mathbb{Q}}$of $\bQ_p$) in Macintyre language $\DeclareMathOperator{\cL}{\mathcal{L}}$ $\cL_{\rm Mac}$. Let $Z$ be a definable (i.e. semialgebraic) analytic subset $Z$ of $K^n$ of $p$-adic dimension $d$ and let $C$ be a definable subset of $Z$ of strictly lower dimension.

My question is:

Is there a an analytic definable isomorphism $f: Z \rightarrow Z \setminus C$ such that $\mid {\rm Jac} f \mid=1$?

I know the answer for $n=1$ is no but I need to know whether it is still no for $n>1$.

Translation with no model theory:by a semialgebraic set we mean a set union of sets of the form $$\lbrace x \in {\bQ}_p^m \mid f(x) =0, g_1(x) \in P_{n_1}, \dots, g_k(x) \in P_{n_k} \rbrace ,$$ where $P_n$ is the set $\lbrace y \in {\bQ}_p^\times \mid \exists x \in {\bQ}_p^\times, y=x^n \rbrace$. A semialgebraic function is a function whose graph is a semialgebraic set.

It was shown that (Scowcroft & van den Dries) and Cluckers (see R. Cluckers: Classification of semi-algebraic p-adic sets up to semi-algebraic bijection, Journal für die reine und angewandte Mathematik, 540, 105 - 114 (2001) math.LO/0311434. ) that dimension is a semialgebraic invariant which means that two semialgebraic sets have a semialgebraic bijection between them if and only if they have the same dimension. (Semialgebraic sets of dimension $n$ have a semialgebraic bijection with an open subset of ${\bQ}_p^n$.) But they did not constrain the semialgebraic bijection to have $p$-adic jacobian 1 which I do now.

Thank you

12 added 40 characters in body; deleted 38 characters in body
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Hi,

Consider a $p$-adic field $K$ (finite extension $\DeclareMathOperator{\bQ}{\mathbb{Q}}$of $\bQ_p$) in Macintyre language $\DeclareMathOperator{\cL}{\mathcal{L}}$ $\cL_{\rm Mac}$. Let $Z$ be a definable (i.e. semialgebraic) subset $Z$ of $K^n$ of $p$-adic dimension $d$ and let $C$ be a definable subset of $Z$ of strictly lower dimension.

My question is:

Is there a definable isomorphism $f: Z \rightarrow Z \setminus C$ such that $\mid {\rm Jac} f \mid=1$?

I know the answer for $n=1$ is no but I need to know whether it is still no for $n>1$.

Translation with no model theory:by a semialgebraic set we mean a set union of sets of the form $$\lbrace x \in {\bQ}_p^m \mid f(x) =0, g_1(x) \in P_{n_1}, \dots, g_k(x) \in P_{n_k} \rbrace ,$$ where $P_n$ is the set $\lbrace y \in {\bQ}_p^\times \mid \exists x \in {\bQ}_p^\times, y=x^n \rbrace$. A semialgebraic function is a function whose graph is a semialgebraic set.

It was shown that (Scowcroft & van den Dries) and Cluckers (see R. Cluckers: Classification of semi-algebraic p-adic sets up to semi-algebraic bijection, Journal für die reine und angewandte Mathematik, 540, 105 - 114 (2001) math.LO/0311434. ) that dimension is a semialgebraic invariant which means that two semialgebraic sets have a semialgebraic bijection between them if and only if they have the same dimension. (Semialgebraic sets of dimension $n$ have a semialgebraic bijection with an open subset of ${\bQ}_p^n$.) But they did not constrain the semialgebraic bijection to have $p$-adic jacobian 1 which I do now.

I add a last detail. In my effort to answer this question, I arrived at a point where it suffices to prove the following claim: given a function $g \colon K \to K$ which is analytic, semialgebraic and a function $f\colon K \to K$ also analytic semialgebraic such that $\mid df /dx \mid \cdot \mid g(x) \mid =1$ then $f$ must be surjective.

Special case: if $\mid g(x) \mid =1$ then the statment of the claim is true.

Thank you