Let $k$ be a complete non-archimedean field and let $\varphi \colon X \to Y$ be a morphism of rigid analytic spaces over $k$, where $\newcommand{\Sp}{\operatorname{Sp}}Y = \Sp(B)$ is affinoid. Consider the following condition:


*$(\dagger)$ The morphism $\varphi$ is separated and there exist two finite admissable affinoid coverings $\mathfrak{U} = (U\_i)\_{i \in I}$, $\mathfrak{V} = (V\_i)\_{i \in I}$ of $X$ such that $V\_i \Subset\_Y U\_i$ (i.e. $V\_i$ lies relatively compact in $U\_i$ w.r.t. $Y$) for all $i \in I$.*


A morphism $\varphi \colon X \to Y$ (with $Y$ not necessarily affinoid) is called *proper* if it is separated and there is an admissable affinoid covering $(W\_j)\_{j \in J}$ of $Y$ such that for each $j \in J$ the morphism $\varphi^{-1}(W\_j) \to W\_j$ satisfies condition $(\dagger)$.


If $Y$ is affinoid, then obviously a morphism satisfying condition $(\dagger)$ is proper, whereas the converse is not clear. In fact, Bosch in his book *Lectures of Formal and Rigid Geometry* writes at the beginning of section 6.4 that $(\dagger)$ is slightly stronger than properness.



> 
> Is there an example of a proper morphism with affinoid base which does not satisfy condition $(\dagger)$?
> 
> 
> 



An example of a morphism of rigid analytic spaces with affinoid base which is proper but does not satisfy $(\dagger)$
These two notions are actually equivalent (at least if $k$ is the fraction field of a dvr $R$), but I do not know any direct way to see this. 


The proof I know heavily uses the theory of formal schemes. The three main results we need are Lemma 2.5, Lemma 2.6 and the statement after Corollary 3.2 from Lutkebohmert's paper ``Formal-algebraic and rigid-analytic geometry''.



> 
> **Lemma 2.5**: Let $\mathfrak W \to \mathfrak V=\text{Spf}(A)$ be a morphism of admissible formal schemes whose generic fibers are affinoid and let $\mathfrak U \subset \mathfrak W$ be an open formal subscheme. Then
>  $\mathfrak U\_K$ is relatively compact in $\mathfrak W\_K$ over $\mathfrak V\_K$ if and only if the
>  Zariski-closure $\overline{\mathfrak U}\_0$ of $\mathfrak U\_0$ in $\mathfrak W\_0$ is proper over $\mathfrak V\_0$.
> 
> 
> **Lemma 2.6**: Let $f\colon \mathfrak X \to \mathfrak Y$ be a morphism of admissible formal schemes and let $f\_K\colon \mathfrak X\_K \to \mathfrak Y\_K$ be its generic fiber. If the rigid map $f\_K$ is proper, the formal
>  map $f$ is proper.
> 
> 
> **Claim on page 350**: Let $f\colon \mathfrak X \to \mathfrak Y=\text{Spf} B$ be a proper morphism of admissible formal schemes. Let $\mathfrak U$ be a formal open subscheme of $\mathfrak X$ which is affine. Then
>  there exists an admissible blowing-up $\mathfrak X \to \mathfrak X'$ with a center contained in the complement of $\mathfrak U\_0$ such that there exists an open subscheme $\mathfrak U'$ of $\mathfrak X'$ whose associated
>  rigid subspace $\mathfrak U'\_K$ is affinoid and such that the Zariski-closure $\overline{\mathfrak U\_0}$ of $\mathfrak U\_0$ in $\mathfrak X'\_0$ is contained in $\mathfrak U'\_0$. 
> 
> 
> 


These results easily imply the desired claim. Indeed, suppose that $g\colon X \to \text{Sp}(A)$ is a proper map. Choose some topologically finitely generated ring of definition $A\_0 \subset A$. Then using the standard machinery on formal models, we can find a morphism of admissible formal schemes $f\colon \mathfrak X \to \text{Spf } A\_0$ such that the generic fiber $f\_K$ is equal to $g$. Then Lemma 2.6 reads that $f$ is proper as a map of formal schemes. 


Now comes the key step: we use the last cited claim to find a good covering of $X$. Namely, we choose any covering of $\mathfrak X$ by open affine formal schemes $\mathfrak U\_i$. We apply that claim to find an admissible blow-up $\mathfrak X'\_i \to \mathfrak X$ with an open formal subscheme $\mathfrak U'\_i\subset X'\_i$ such that $(\mathfrak U'\_i)\_K$ is affinoid and $\overline{\mathfrak U\_0}$ (that is proper as a closed subscheme of $\mathfrak X'\_0$) is contained in $(\mathfrak U'\_i)\_0$. Lemma 2.5 implies that $(\mathfrak U\_i)\_K$ is relatively compact in $(\mathfrak U'\_i)\_K$. The last thing to observe is that $(\mathfrak U\_i)\_K$ form a finite covering of the rigid space $X$. 


P.S. Lutkebohmert's paper is written entirely in the noetherian setup. So if one wants to give a proof along these lines without noetherianness assumption on $\mathcal O\_K$, then one needs to reprove these results in the set-up of arbitrary (complete) rank-$1$ valuation rings. My recollection is that it should not be that difficult given the recent finiteness results of Fujiwara and Kato (Fujiwara, Kato ``Foundations of Rigid Geometry I'')


