A characteristic feature of [Berkovich spaces](https://en.wikipedia.org/wiki/Berkovich_space) is that they are locally connected (in fact, locally contractible). I'd like to understand the proof. The key ingredient seems to be Corollary 2.2.8 in Berkovich's [book](https://www.google.com/books/edition/Spectral_Theory_and_Analytic_Geometry_Ov/6x5PIl2-DkkC?hl=en). A key step in the proof is to take a connected component $W\_n$ of a certain set $V\_n$, but there is no explanation of why this connected component should be open (as I think it needs to be). I'd like to understand why.


**Question:** Why are Berkovich spaces locally connected?


Ideally, I'd appreciate a clarification of Berkovich's proof, or else a pointer to some other proof in the literature. Also, I'd be happy to understand the case of the Berkovich spectrum of a ring.



Why are Berkovich spaces locally connected?
Let $X = \mathcal{M}(A)$ be an affinoid space. There is a map $\mathcal{M}(A) \to \operatorname{Spec}(A)$ sending a seminorm to its kernel. It is continuous, surjective and induces a bijection between the connected components of $\mathcal{M}(A)$ and $\operatorname{Spec}(A)$. Let me prove the last point.


First assume that $\operatorname{Spec}(A)$ is connected. Then $\mathcal{M}(A)$ is connected too. Indeed, consider a clopen subset $U$ of $\mathcal{M}(A)$. We can construct an analytic function $e$ on $\mathcal{M}(A)$ that is 0 on $U$ and 1 on the complement. But the global sections on $\mathcal{M}(A)$ are nothing but $A$ itself (Tate's theorem), so $e$ defines an element of $A$, which is idempotent. Since $\operatorname{Spec}(A)$ is connected, $A$ has only the trivial idempotents, so $e$ equals 0 or 1, hence $U$ is either the whole $\mathcal{M}(A)$ or the empty set.


For the general case, pick a connected component of $\operatorname{Spec}(A)$. It is a Zariski-closed subset of $\operatorname{Spec}(A)$, hence of the form $\operatorname{Spec}(B)$, with $B$ a quotient of $A$, so in particular an affinoid algebra. The previous argument shows that its preimage $\mathcal{M}(B)$ is connected. This finishes the proof of the bijection.


Finally, recall that the affinoid algebra $A$ is noetherian, so $\operatorname{Spec}(A)$ has finitely many connected components, and similarly for $\mathcal{A}$. As a consequence, those connected components are open.


I think this is what you need to complete the proof of local connectedness.


I am not sure what you mean about the Berkovich spectrum of a ring at the end of your post, so I will not comment on that.


