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Let $(k,v)$ be a Henselian field. Let $O_k$ be the valuation ring. Let $X$ be a scheme separated of finite type over $O_k$. Let $x\in X(O_k)$. Let $f:X\to \mathbf{A}_{O_k}^n$ be an étale morphism of schemes over $O_k$ with $f(x)=0\in \mathbf{A}_{O_k}^n(O_k)$. We use the topology on $X_k(k)$ induced by $v$. Endow $X(O_k)\subset X_k(k)$ with the subspace topology. Is there an open neighborhood $U$ of $x$ in $X(O_k)$ and an open neighborhood $V$ of $0$ in $\mathbf{A}_{O_k}^n(O_k)$, such that $f:X_k(k)\to \mathbf{A}^n(k)=k^n$ restricts to a homeomorphism $U\to V$?

From the implicit function theorem, I learned that the analogous result holds for $k$-points.

Doug
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  • The map on $\mathcal{O}_k$-points, at least if considered as a subspace of the $k$-points which is what you seem to be doing, is induced from the map on $k$-points. – Alex Youcis Jun 04 '25 at 15:16
  • If you use the topology induced by $v$, this would be a topology on $X(k^{\rm sep})$. I don't see what topology it would induce on $X$ (what would the distance between two closed points on the special fibre of $X$ be?). – Damian Rössler Jun 04 '25 at 16:54
  • @DamianRössler I don't know if you are referring to me, but I am not sure what you mean. There is a surjection $X(\mathcal{O}k)\to X(k)$, but it is not bijective. If $X=\mathbb{A}^1{\mathcal{O}_K}$ then $X(\mathcal{O}_K)$ is the unit ball in $X(K)$, with the subspace topology. – Alex Youcis Jun 05 '25 at 20:02
  • @Doug Sorry, my mistake. I didn't pay attention to the fact you weren't assuming smooth. – Alex Youcis Jun 06 '25 at 12:52
  • OK, I am not sure what the relevance of that is. But I see I have been making a mistake in notation, which maybe also affected my comment to @DamianRössler. Most of then a non-archimedean field is $K$ and $k$ is its residue field. I forgot you were using $k$ for the non-archimedean field. My comment, which was in response to my misunderstanding of Rossler in relation to this notational mixup, was just that if $X/\mathcal{O}_k$ is smooth then $X(\mathcal{O}_k)\to X(\text{res. field})$ is surjective by Hensel's lemma. Ignore this. The map, with your notation, $X(\mathcal{O}_k)\to X(k)$ is – Alex Youcis Jun 06 '25 at 13:40
  • essentially never surjective unless $X/\mathcal{O}_k$ is proper, in which case it's a bijection. – Alex Youcis Jun 06 '25 at 13:41

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