Restricting maps between strict henselisations

Question

$\require{AMScd}$I am currently thinking about (strict) henselisations but I don't know too much literature about the topic. So I am wondering if there is a natural way to restrict maps between strict henselisations to henselisations:

Let $A$, $B$ be local rings with an injective homomorphism $h:A\to B$. If I have a homomorphism $f:A^\text{sh} \to B^\text{sh}$, is there always a homomorphism $g:A^\text h \to B^\text h$ such that the following diagram commutes?

\begin{CD} A^\text{sh} @>f>> B^\text{sh}\\ @AAA @AAA\\ A^\text h @>>g> B^\text h \end{CD}

Equivalently, I am asking if you start with $x \in A^\text h \subset A^\text{sh}$, is $f(x) \in B^\text h$? It feels like this should be related to Galois theory, where Galois extensions fix the base field.

(Note: There might be requirements on $A$ and $B$ like being normal, or $B$ being a field. I'm also interested in answers with more assumptions than stated above.)

Clarifications: The underlying injective homomorphism $h: A \to B$ is not necessarily a local map but $f$ commutes with $h$.

\begin{CD} A^\text{sh} @>f>> B^\text{sh}\\ @AAA @AAA\\ A @>>h> B \end{CD}

The example I have in mind is the following: Pick a ring $R$ and a maximal ideal $\mathfrak{m} \in R$ and a map $\operatorname{Frac}(R) \to \operatorname{Frac}(R)^\text{sep}$ (i.e. a geometric point over the generic point of my curve $\operatorname{Spec}(R)$). Then $A := R_{\mathfrak{m}}$, $B:=\operatorname{Frac}(R)$, the map $h: A \to B$ is not local and $B \to B^\text{sh} = \operatorname{Frac}(R) ^\text{sep}$ is given by above chosen geometric point.

Answer 1

No. Let $A= k[x]_{(x)}$, the localization of the ring of polynomials in one variable at the maximal ideal $(x)$, and $B = k(x)$.

Assume (for simplicity) that the characteristic of $k$ is not $2$.

Then there exists $y \in A^h$ satisfying $y^2 = 1+x$, as that polynomial splits into distinct linear factors modulo $x$.

But there exists no $y \in B^h = B$ satisfying that equation, as writing $y = f/g$ we would have $f^2 = g^2(1+x)$ so $$2 \deg f = \deg f^2= \deg (g^2 (1+x)) = 2\deg g+1$$

So there can be no homomorphism $A^h \to B^h$ sending $x$ to $x$ (as it must to form a commutative diagram with the induced map on strict henselizations).

Answer 2

Edit: This answer is not suitable as the OP does not assume that the map $A\to B$ is local. See clarification by the OP and counter example by Sawin. Sorry.

This is true, without restrictions, and in a slightly more general formulation: star with a local homomorphism $f:A\to B$ and give yourself a morphism $\Omega\to \Omega'$, where $\Omega$ and $\Omega'$ are the residue fields of the strict henselisations of $A$ and $B$ respectively, such that $A\to A^{sh}\to \Omega\to\Omega'$ coincide with $A\to B\to B^{sh}\to \Omega'$. Then, there are unique $f^h:A^h\to B^h$ and $f^{sh}: A^{sh}\to B^{sh}$ making your diagram with $f^h$ (resp. $f^{sh}$) commute. Thus there is such a $g$ and it must be $f^h$.

Reference: [Raynaud, Anneaux locaux henseliens, VIII, 3, Proposition 4].