# Lifting points via étale morphism of adic spaces

This question was suggested to me during the reading of Huber's book about Etale Cohomology of Adic Spaces. I formulate this question here in the context of adic spaces, but I think, since a morphism of (analytic) adic spaces (suppose affinoid) is étale if and only if the induced morphism between the first components of the Huber pairs defining the spaces is algebraically étale, that one can even prove it in purely algebraic setting getting again the result.

The question is the following. Consider an étale morphism $f:X\rightarrow Y$ of locally of finite type Tate adic spaces over $(\mathbb{Q}_{p},\mathbb{Z}_{p})$. Let $u\in Y(K, K^{+})$ be a point of $Y$ with values in an affinoid pair $(K,K^{+})$, where $K$ is a nonarchimedean extension of $\mathbb{Q}_{p}$, and $K^{+}$ is a valuation ring of $K$. Suppose that this point has a lifting to a $(K,K^{+})$-point of $X$. Then the same is true on a small enough neighborhood of $u$, i.e. there exists a neighborhood of $u$, say $U$, such that every $(K,K^{+})$-valued point of $U$ has a lift to a $(K,K^{+})$-point of $X$.

My idea is the following. First, the whole problem is local both in the target and in the domain, hence we can assume that $X=\text{Spa}(B,B^{+})$, and that $Y=\text{Spa}(A,A^{+})$ for $A,B$ complete affinoid. Moreover, since the map is étale, by a result of Huber, we can assume that $B=A\langle X_{1},\ldots ,X_{n}\rangle/(f_{1},\ldots ,f_{n})$, with $f_{i}\in A[X_{1},\ldots ,X_{n}]$ for all $i$ such that $\text{det}\left(\frac{\partial f_{i}}{\partial X_{j}}\right)_{i,j=1,\ldots ,n}$ is a unit in $B$. Now, we have a morphism $u:(A,A^{+})\rightarrow (K,K^{+})$, which is the $(K,K^{+})$-point of $Y$, lifting to $x:(B,B^{+})\rightarrow (K,K^{+})$. But now I was not able to go on, how to construct the right neighborhood of $u$? Thanks for any suggestion!

• mathoverflow.net/q/242430/93715 – Simone Jun 16 '16 at 22:28
• It seems to be generally accepted that one should pick a site for a question and stick with it. This one is probably better for MO. The last one probably wasn't. – Hoot Jun 17 '16 at 0:10