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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 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

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The question was posted on math.SE as well, see: – Asaf Karagila Oct 4 '11 at 9:11
I am sure that people working with $p$-adic numbers but not in model-theory would be happy to think about this question if they could understand it. Do you think it would be possible to translate it as a pure question on $p$-adic number (that is by explaining the meaning of definable in this context, and perhaps also of dimension)? – Joël Oct 6 '11 at 17:24
Joel: I will translate this question soon purely in algebraic and analytic terms. – user16974 Oct 6 '11 at 18:41

Observe that $f$ is an analytic isomorphism to conclude that $Z$ and $Z\setminus C$ are both closed in the relative topology (i.e. the topology induced on $Z$ by the topology of the ambiant space). But $Z \setminus C$ cannot be closed in the relative topology since $C$ is not open in the ambiant topology. Hence there is no such isomorphism.

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I added back the answer. Please correct me if I am wrong! – user16974 Oct 23 '11 at 9:57

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