What is an example of an etale morphism which does not locally factor as a composition of an open immersion and a finite etale map? Let $f:X\rightarrow Y$ be an etale morphism of schemes. By $f$ factoring locally as a composition of an open immersion and a finite etale map, I mean that for every $x \in X$ there exists an open $U$ of $x$ and an open $V \subset Y$ containing $f(U)$ such that the restriction $f:U \rightarrow V$ factors as $U \hookrightarrow W \rightarrow V$, where $U \hookrightarrow W$ is an open immersion and $W \rightarrow V$ is finite etale.
According to section 7.5 of the "Berkeley lectures" by Scholze and Weinstein, it is not true that every etale $f$ factors locally as a composition of an open immersion and a finite etale map in the category of schemes (though it is true in the category of analytic adic spaces). What is an example of this failure for schemes?
Thanks!
 A: Ariyan Javanpeykar's answer to a related question gives an example that works. Since you're asking for something a priori slightly stronger than in that question (no local factorization into open immersion + finite étale, not just no global factorization), I'll elaborate. The example is as follows: Let $X = \mathbb{A}^1 \setminus \{0, 2/3\}$, and define a morphism $f\colon X \to \mathbb{A}^1$ by $f(x) = x^2 (x - 1)$. Note that $f$ is étale.
Let $U \subseteq X$ be any open subset containing $1 \in X$, and let $V \subseteq \mathbb{A}^1$ be any open subset such that $f(U) \subseteq V$. Note that $0 = f(1) \in V$. Suppose we have a scheme $W$, an open immersion $j\colon U \to W$, and a finite morphism $h\colon W \to V$ such that $f\rvert_U = h \circ j$. Let $Z \subseteq W$ be the irreducible component containing $j(U)$ (with the reduced induced closed subscheme structure). Irreducible components are closed, and closed immersions are finite, so $g := h\rvert_Z\colon Z \to V$ is finite. Finite morphisms are affine, so $Z$ is affine, irreducible, reduced, 1-dimensional, and birational to $\mathbb{A}^1$, which means $Z$ is an open subscheme of $\mathbb{A}^1$. So $g$ is given by a rational function, but since $g$ agrees with $f$ on a dense open subset, this rational function must be $x^2 (x - 1)$.
But since $g$ is finite, all the fibers have the same degree, so $g^{-1}(0)$ consists of three points, counting multiplicity. Since $x^2 (x - 1)$ is ramified above $0$, this means $g$ (and hence $h$) cannot be étale.
Actually, now that I've written this up, I'm a little confused why this doesn't also give a counterexample for analytic adic spaces. I guess it has to do with the more flexible nature of open immersions in the analytic setting?
