Let $k$ be a complete, non-archimedean field, and $X$ a Berkovich space over $k$ (as nice as you like, for arguments sake let's say strictly $k$-analytic, good, and geometrically connected). As discussed in [this article of de Jong](http://www.numdam.org/article/CM_1995__97_1-2_89_0.pdf), covering spaces of $X$ come in two slightly different flavours. One the one hand you can take finite etale covers $Y\rightarrow X$ as you would for schemes, on the other hand you can take a covering space $Y\rightarrow |X|$ of the underlying topological space of $X$, and, roughly speaking, use the Berkovich space structure of $X$ to put one on $Y$. Following de Jong, let us call the first of these 'algebraic' and the second 'topological'. A general covering space is then some kind of mixture of the two.


If $k$ is not separably closed, then one way of producing algebraic covering spaces is via finite separable extensions of $k$: if $L/k$ is such an extension then $X\_L \rightarrow X$ is a finite etale map of Berkovich spaces, where $X\_L$ denotes the base change of $X$ to $L$. My question is then the following:


**Question**: Is it possible that $X\_L \rightarrow X$ is a **topological** covering space, for some non-trivial extension $L/k$?


It's not to hard to see this can't be the case if $X$ has a $k$-rational point (since the fibre of $X\_L\rightarrow X$ over this point will have cardinality 1), but I'm particularly interested in the case when we might have $X(k)=\emptyset$. Concretely, I'm interested in the case when $X$ is (the analyitification of) a smooth projective conic over $k$, without a rational point, and $L/k$ is a quadratic extension over which $X$ does admit a rational point.



Topological and algebraic covering spaces in Berkovich geometry
In your particular case, $X\_L$ has a point, so it is isomorphic to $P^{1,\mathrm{an}}\_L$, hence simply connected. If your covering $X\_L \to X$ were a covering, it would then be a universal covering. But we know that Berkovich curves retract by deformation onto graphs, so the topological fundamental group of $X$ is a free group. In particular, the universal covering of $X$ is either $X$ itself or of infinite degree, and we get a contradiction.


