How to show analytification functor commutes with forgetful functor? Let $k$ be a field complete with respect to a non-trivial non-archimedean
absolute value (so that rigid $k$-space makes sense). Denote $K$ a finite field extension of $k$.
Denote $X\rightsquigarrow X^{\mathrm{an}/k}$ the analytification functor from the category of locally of finite type $k$-schemes to the category of rigid $k$-spaces.
Similarly there is an analytification functor $X\rightsquigarrow X^{\mathrm{an}/K}$ over $K$.
There is a well-defind forgetful functor $S:X\rightsquigarrow X$ from $K$-schemes to $k$-schemes ($S$ represents schemes) and a forgetful functor $R:Y\rightsquigarrow Y$ from rigid $K$-spaces to rigid $k$-spaces ($R$ represents rigid).
Let $X$ be a locally of finite type $K$-scheme. I believe that $S(X)^{\mathrm{an}/k}\cong R(X^{\mathrm{an}/K})$ as rigid $k$-spaces. The universal property induces a canonical map $R(X^{\mathrm{an}/K})\to S(X)^{\mathrm{an}/k}$ but I cannot show it is an isomorphism. A proof or reference would be nice.
p.s. the idea comes from proving absolute/relative Frobenius morphism commutes with analytification, but I first need to make sure the maps have the same source.
Also in ME, but I just got advised not to post the same problem on both MS and ME. So I deleted the one on ME, and will possibly undelete it after a week or two.
 A: I believe I got an answer. Denote $X^{\mathrm{an}/K},X^{\mathrm{an}/k}$ the analytifications of $X$ over $K$ and $k$ respectively. Denote $K$-maps (resp. $k$-maps) the maps of locally $G$-ringed $K$-spaces (resp. $k$-spaces).
Lemma: Let $X$ be a locally of finite type $K$-schemes. Then $X^{\mathrm{an}/K}\cong X^{\mathrm{an}/k}$ both as rigid $k$-spaces and rigid $K$-spaces.
Plan: Assume $X$ is affine. Find maps $X^{\mathrm{an}/K}\to X^{\mathrm{an}/k}$ and $X^{\mathrm{an}/k}\to X^{\mathrm{an}/K}$ and prove both of them are both $K$-maps and $k$-maps. Then the result follows from universal properties of $X^{\mathrm{an}/k}$ and $X^{\mathrm{an}/K}$. Then show it generalizes to the general case.
Proof: Assume $X$ is affine. Clearly the canonical $K$-map $X^{\mathrm{an}/K}\to X$ is also $k$-map, hence uniquely factors through the canonical $k$-map $X^{\mathrm{an}/k}\to X$. So we have a $k$-map $X^{\mathrm{an}/K}\to X^{\mathrm{an}/k}$.
To find a $K$-map from $X^{\mathrm{an}/k}$ to $X^{\mathrm{an}/K}$, it suffices to show that the canonical $k$-map $X^{\mathrm{an}/k}\to X$ is actually a $K$-map.
We know $X$ is affine, say $X\cong\mathop{\mathrm{Spec}}\frac{k[x_1,...,x_n]}{I}$. Pick $c\in k$ s.t. $|c|>1$. Denote $T_n^{(i)}=k\langle c^{-i}x_1,...,c^{-i}x_n\rangle$. We know $X^{\mathrm{an}/k}=\bigcup_{i\geq0}\mathop{\mathrm{Sp}}\frac{T_n^{(i)}}{(I)}$, and $X^{\mathrm{an}/k}\to X$ is determined by the compatible maps $\frac{k[x_1,...,x_n]}{I}\to \frac{T_n^{(i)}}{(I)}$. Let $\frac{T_n^{(i)}}{(I)}$ equips with a $K$-algebra structure via $K\to \frac{k[x_1,...,x_n]}{I}\to \frac{T_n^{(i)}}{(I)}$ (note that any ring map from a field to a ring is injective). Then $\frac{k[x_1,...,x_n]}{I}\to \frac{T_n^{(i)}}{(I)}$ is a map of $K$-algebras. So $X^{\mathrm{an}/k}$ is a rigid $K$-space and $X^{\mathrm{an}/k}\to X$ is actually a $K$-map and it induces a $K$-map $X^{\mathrm{an}/k}\to X^{\mathrm{an}/K}$ which is also a $k$-map.
By universal property of $X^{\mathrm{an}/k}$, we have an equality of $k$-maps
$$(X^{\mathrm{an}/k}\to X^{\mathrm{an}/K}\to X^{\mathrm{an}/k})=(X^{\mathrm{an}/k}\stackrel{\mathrm{id}}{\to} X^{\mathrm{an}/k})$$
Next we show that the $k$-map $X^{\mathrm{an}/K}\to X^{\mathrm{an}/k}$ is also a $K$-map so we can prove that the $K$-map $X^{\mathrm{an}/K}\to X^{\mathrm{an}/k}\to X^{\mathrm{an}/K}$ is the identity map hence an identity $k$-map, it follows that $X^{\mathrm{an}/k}\cong X^{\mathrm{an}/K}$ both as rigid $K$-spaces and rigid $k$-spaces.
Just like $X^{\mathrm{an}/k}=\bigcup_{i\geq0}\mathop{\mathrm{Sp}}\frac{T_n^{(i)}}{(I)}$, we have $X^{\mathrm{an}/K}=\bigcup_{j\geq0}\mathop{\mathrm{Sp}}A_j$ for some $K$-algebras $A_j$. It suffices to show that $\mathop{\mathrm{Sp}}A_j\to X^{\mathrm{an}/k}$ is a $K$-map for all $j$. Fix $j\geq 0$, then there exists $\alpha(j)\geq 0$ s.t. $\mathop{\mathrm{Sp}}A_j\to X^{\mathrm{an}/k}\to X$ induces
$$\frac{k[x_1,...,x_n]}{I}\to \frac{T_n^{(\alpha(j))}}{(I)}\to A_j$$
where the composition is a map of $K$-algebras, hence all intermediate maps are of $K$-algebras, and $X^{\mathrm{an}/K}\to X^{\mathrm{an}/k}$ is a $K$-map and we finished the affine case.
The patching part can be annoying. For a general scheme $X$, pick an open affine cover $\{X_i\}_{i\in I}$. For each pair $X_{ij}=X_i \cap X_j$, we know it can be covered by simultaneous distinguished open affines $\{D_{ijk}\}_{k\in S_{ij}}$. We know that $X^{\mathrm{an}/k}$ is obtained by gluing $\{X_i^{\mathrm{an}/k}\}_{i\in I}$ and $X^{\mathrm{an}/K}$ is obtained by gluing $\{X_i^{\mathrm{an}/K}\}_{i\in I}$. In particular, the gluing data of $X^{\mathrm{an}/k}$ over $k$ also has a structure of gluing data over $K$ (i.e. the spaces and the maps are over $K$, if we want to be precise, we will need to use $D_{ijk}$), so $X^{\mathrm{an}/k}$ is a rigid $K$-space. With this we can show
$$(X^{\mathrm{an}/k}\to X^{\mathrm{an}/K}\to X^{\mathrm{an}/k})=(X^{\mathrm{an}/k}\stackrel{\mathrm{id}}{\to} X^{\mathrm{an}/k})$$
Now we need the following pasting lemmas for $K$-maps.

Lemma: Let $X$ and $Y$ be rigid $K$-spaces admitting admissible coverings $\{X_i\}_{i\in I}$ and $\{Y_i\}_{i\in I}$, respectively. Let
$\psi_i :X_i \to Y_i,i\in I$ be $K$-maps such that for all $i,j\in I$,
we have $\psi_i|_{X_i\cap X_j}=\psi_j|_{X_i\cap X_j}:X_i\cap X_j \to
Y_i \cap Y_j$. Then there exists a unique $K$-map $\psi:X\to Y$
extending all $\psi_i,i\in I$. For proof see [Non-Archimedean analysis] by Bosch, 9.3.3
Prop 1.

The $k$-map $\psi:X^{\mathrm{an}/K}\to X^{\mathrm{an}/k}$ descends to a gluing data of $k$-maps: $$\psi_i:X_i^{\mathrm{an}/K}\to X_i^{\mathrm{an}/k},\psi_{ij}:X_{ij}^{\mathrm{an}/K}\to X_{ij}^{\mathrm{an}/k}$$
In particular, $\psi_i$ is already a $K$-map by the proof for affine case. Thus $\psi_{ij}$ is a $K$-map as a restriction map. So we have a gluing data of $K$-maps, then we obtain a $K$-map $\psi_K:X^{\mathrm{an}/K}\to X^{\mathrm{an}/k}$ s.t. $\psi_K=\psi$ as $k$-maps.
The rest of the proof is the same as the affine case, the result follows from the universal properties.$\square$
Bosch, S.; Güntzer, Ulrich; Remmert, Reinhold, Non-Archimedean analysis. A systematic approach to rigid analytic geometry, Grundlehren der Mathematischen Wissenschaften, 261. Berlin etc.: Springer Verlag. XII, 436 p. DM 168.00 (1984). ZBL0539.14017.
