Thanks! This makes sense. My intuition is that, in the adic setting, you can take $U$ to be a small disc containing 1, and $V$ to be its image via $f$ ($f$ is open since it is etale). Since 1 is not ramified, $f$ is locally 1-1 at 1 (which is not true in the scheme setting, since open sets are too large), so you can take $U$ and $V$ to be so small that already $U\rightarrow V$ is even an isomorphism. I think this picture at least works in the complex analytic setting.

@xlord You have to be careful. There are subtelties between maximal vs. non-maximal points. What you're describing is more accurate on the Berkovich space, but on the adic space you have to contend with the existence of things like Gauss points having non-trivial specialization. In particular, be careful that your intuition almost makes it seem as though for all $x\in X$ and $y=f(x)$ in $Y$ there exists opens $U$ and $V$ neighborhoods around both with $f(U)\subseteq V$ and $f:U\to V$ finite. This is true around maximal points, and in general if $f$ is partially proper, but not in general.

@AlexYoucis thanks, I have not thought of non maximal points. It was more like I was trying to explain what was happening only at the (maximal) evaluation at 1 point. Is this argument incorrect then? Do you know a nice formal way to prove in this case $f$ factors locally as a composition of an open immersion and a finite etale map? (I could in principle check this up in Huber's book, but perhaps it is simpler for this example?)

@xlord Are you asking about Lemma 2.2.8 of Huber's book?

Yes, this is Lemma 2.2.8 in his book.

@xlord What is your exact question then? How to prove the claim when you assume that $x$ (and therefore $y$) are maximal points?

Just for the point x=1. I was under the impression you were saying the sketch in my original comment was incorrect. How would you write it out formally, if not?

@xlord Yo.

hi

Is your x=1 literally a classical point?

yeah

So the like point here is basically this

sorry?

Surprising fact: if x is maximal and f is etale then f:O_{Y,y}--->O_{X,x} is finite.

so if $x$ is classical, $y$ is classical, and if you assume that k is algebraically closed

Second surprising fact: O_{X,x} is Henselian

if k is algebraically closed then it's strictly Henselian

sorry, so this is in what setting?

so finite etale covers which are connected (which O_{Y,y}) is are isomorphisms

these facts

Morphisms of locally Noetherian analytic adic spaces

Moral of the story

is that your intuitive approach is OK for classical points

but it's false even for maximal non-classical poitns

I'm just trying to make sure you didn't have a confusion I once had

which is that if you have an etale morphism that it is locally an isomorphism (i.e. there's an oepn cover of source where it's an isomorphism onto its image)

(PS, OK for classical points in ALGEBRAICALLY CLOSED situation)

thanks, the chat stopped working for me for some reason

very nice, thank you very much

No problem!

Just to give yourself some intuition

right, I was comparing it very freely with the complex analytic situation

the 'Gauss point' is like the avatar

for the unit disk itself

under the squaring map

the Guass point is the only point which maps to the Gauss point

and it very much 'acts like the disk mapping to the disk by squaring'

you can't find a neighborhood of the Guass point where it's an isomorphism

anyways, I'm sure you've heard enough

Best of luck!