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My question is how to prove the affine $n$-space over $p$-adic number $\mathbb{Q}_p$ is simply connected.

To be precise,

Let $X$ be $p$-adically analytic manifold, $f:X\rightarrow \mathbb{A}^n_{\mathbb{Q}_p}$ be a $p$-adically analytic function.
If $f$ is topologically covering map, is $f$ necessarily trivial?

All the termiology are from Lie algebras and Lie groups by Serre.

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    $\begingroup$ What do you mean by a topological covering map? It seems that these spaces are totally disconnected, whereas usually these things are only studied for maps between connected spaces. As an example, you can take any clopen decomposition $\mathbf Q_p = U \amalg V$ and take a 'covering map' given by $U \amalg U \amalg V$ (collapsing the two copies of $U$). Is that a covering map? $\endgroup$
    – R. van Dobben de Bruyn
    Commented Oct 13 at 16:03
  • $\begingroup$ I want a covering map to be analytic. $\endgroup$
    – George
    Commented Oct 14 at 7:05
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    $\begingroup$ If by analytic manifold you mean a rigid analytic space, then can't you take $\mathbf A^n_K \to \mathbf A^n_{\mathbf Q_p}$ for some finite extension $\mathbf Q_p \to K$? I.e. on a small ball given by $\operatorname{sp} K\langle x_1,\ldots,x_n\rangle \to \operatorname{sp} \mathbf Q_p\langle x_1,\ldots,x_n\rangle$. Fundamental groups should only be trivial once your base field is algebraically closed, right? (Over $\mathbf C_p$ you might have a chance, but I'm far from an expert.) $\endgroup$
    – R. van Dobben de Bruyn
    Commented Oct 14 at 14:28
  • $\begingroup$ @R.vanDobbendeBruyn, re, the notion from Serre is much more down-to-earth: an analytic manifold modelled on $\mathbb Q_p$ is a set with an "atlas", i.e., jointly surjective collection of maps from open subsets of $\mathbb Z_p^n$ such that the transition maps are restrictions of analytic functions, i.e., $n$-tuples of convergent power series in $n$ variables, that is then given a topology and notion of sheaf of analytic functions in the usual way. $\endgroup$
    – LSpice
    Commented Oct 15 at 1:39
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    $\begingroup$ @DanielLitt Yes, you are of coure right — $K$ was meant to be algebraically closed of characteristic $0$. Thanks for pointing that out! $\endgroup$
    – Alex Youcis
    Commented Oct 15 at 18:06

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The question is ill-posed since the OP does not define what a covering space is, nor when it is trivial. The usual theory only applies to connected topological spaces (or least assumes the connected components are open), which is not the case here.

At this point, the question becomes a topological question on locally profinite topological spaces. If $f \colon Y \to X$ is a proper (equivalently, quasi-compact) map of locally profinite spaces such that all fibres have cardinality at most $n$, then $f$ is locally constant in the sense that there is a finite clopen decomposition $X = X_0 \amalg \cdots \amalg X_n$ such that $f^{-1}(X_i) \to X_i$ is isomorphic to $X_i \times \{1,\ldots,i\}$. You could also try to study the case of infinite discrete fibres, but there I don't really know what to say.

Conversely, any cover built from a clopen decomposition $X = X_0\amalg \cdots \amalg X_n$ by taking $Y = \coprod_i X_i \times \{0,\ldots,i\}$ is analytic: $Y$ is an open submanifold of $X \times \{0,\ldots,n\}$, and the projection $X \times \{0,\ldots,n\} \to X$ is analytic.

See also "Tournants dangereux" on LG 2.13: all locally constant functions are analytic. This is very far removed from complex analytic geometry, where analytic continuation is unique.

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