Direct limit of strict henselizations Assume we have a map $A \rightarrow A'$ of strictly henselian local rings, such that the induced map between spectra $S'\rightarrow S$ is essentially smooth. Is is true that $S'$ is a direct limit of strict henselizations at geometric points on the closed fiber of an affine scheme smooth over $S$?
 A: We will be using the following definitions which I think are the definitions most often used in the literature.
A ring map $A \to B$ is called essentially smooth if there is a smooth ring map $A \to C$ and a multiplicative subset $S \subset C$ such that $B$ is isomorphic to $S^{-1}C$ as an $A$-algebra. Not everybody agrees with this definition so you should be careful when you read a paper or a book!
A morphism of schemes $f : X \to Y$ is called essentially smooth if there exists an affine open covering $Y = \bigcup V_j$ and for each $j$ an affine open covering $f^{-1}(V_j) = \bigcup U_{ij}$ such that the ring maps $H^0(V_j, \mathcal{O}_{V_j}) \to H^0(U_{ij}, \mathcal{O}_{U_{ij}})$ are essentially smooth. Not everybody agrees with this definition, so you should be careful when you read a paper or a book.
Now if $X$ is a local scheme and $Y$ is affine, then it follows (by a tiny argument) that $f$ is essentially smooth if and only if $H^0(Y, \mathcal{O}_Y) \to H^0(X, \mathcal{O}_X)$ is essentially smooth. (I do not think this is true for general affine schemes.)
OK, so I think an honest translation of your question into a more precise language is: when is a local ring homomorphism $A \to B$ of strictly henselian local rings essentially smooth? I claim this happens if and only if $A = B$.
Aside: I understand from the discussion in the comments this is not at all what you are interested in, but it just occurred to me that it is a fun question with a (kinda) fun answer. Please don't edit the question, just ask a new one if you want to change your question.
Proof of the claim. If $A = B$, then clearly $A \to B$ is essentially smooth. Conversely, assume $A \to B$ is essentially smooth. If $A \to B$ has relative dimension $> 0$, then $R = B/\mathfrak m_A B$ is a local ring which is the localization of a smooth algebra over the field $\kappa = A/\mathfrak m_A$ and is strictly henselian (as a quotient of a strictly henselian local ring).
Fact: if $\dim(R) > 0$, then $R$ is not henselian. For example, let $t \in R$ be an element which is transcendental over $\kappa$ such that $t$ maps to $1$ in $R/\mathfrak m_R$. (We omit the proof that such an $t$ exists; here you have to use that $\dim(R) > 0$.) Choose a prime $\ell$ different from the characteristic of $\kappa$. Then if $R$ is henselian, it contains elements $u_n$ such that $u_n^{\ell^n} = t$ for all $n > 1$. Clearly, the field generated by $t, u_n$ in the fraction field of $R$ is not finitely generated over $\kappa$. Hence the fraction field of $R$ is not finitely generated over $\kappa$. Hence $R$ isn't the localization of any finite type $\kappa$-algebra, in particular $R$ isn't the localization of a smooth $\kappa$-algebra.
Since $\dim(R) = 0$ we see that $A \to B$ is essentially \'etale (because being \'etale is the same thing as being smooth of relative dimension $0$). Since $A$ is strictly henselian we get $A = B$ by one of the many characterizations of strictly henselian local rings.
