The proof is almost exactly the same as for Lemma \ref{lemma-branches}. Let $I \subset A$ be the ideal of nilpotents. We have $(A/I)^{sh} = A^{sh}/IA^{sh}$ by (Algebra, Lemma \ref{algebra-lemma-quotient-henselization}). The spectra of $A$, $A^{sh}$, $A'$, and $(A')^h$ are the same as the spectra of $A/I$, $A^{sh}/IA^{sh}$, $A'$, and $(A')^{sh} = A' \otimes_{A/I} A^{sh}/IA^{sh}$. Thus we may replace $A$ by $A_{red} = A/I$ and assume $A$ is reduced. Then $A \subset A'$ which we will use below without further mention. \medskip\noindent Proof of (1). The field extension $\kappa'/\kappa$ is algebraic because $A'$ is integral over $A$. Since $A'$ is integral over $A$, we see that $(A')^{sh}$ is integral over $A^{sh}$. By going up (Algebra, Lemma \ref{algebra-lemma-integral-going-up}) every maximal ideal of $A'$, resp.\ $(A')^{sh}$ lies over the maximal ideal $\mathfrak m$, resp.\ $\mathfrak m^{sh}$ of $A$, resp.\ $A^h$. We have $$ (A')^{sh} \otimes_{A^{sh}} \kappa^{sep} = A' \otimes_A A^h \otimes_{A^h} \kappa^{sep} = (A' \otimes_A \kappa) \otimes_{\kappa} \kappa^{sep} $$ because the residue field of $A^{sh}$ is $\kappa^{sep}$. Thus the fibre of $c$ over $\mathfrak m'$ is the spectrum of $\kappa' \otimes_\kappa \kappa^{sep}$. We conclude (1) is true because there is a bijection $$ \Hom_\kappa(\kappa', \overline{\kappa}) \to \Spec(\kappa' \otimes_\kappa \kappa^{sep}),\quad \sigma \mapsto \Ker( \sigma \otimes 1 : \kappa' \otimes_\kappa \kappa^{sep} \to \overline{\kappa} ) $$ We will use below that the displayed ring is integral over a field hence spectrum of this ring is a profinite space, see Algebra, Lemmas \ref{algebra-lemma-integral-over-field} and \ref{algebra-lemma-ring-with-only-minimal-primes}. \medskip\noindent Proof of (3). The ring $A'$ is a normal ring and in fact a finite product of normal domains, see Algebra, Lemma \ref{algebra-lemma-characterize-reduced-ring-normal}. Since $A^{sh}$ is a filtered colimit of \'etale $A$-algebras, $(A')^{sh}$ is filtered colimit of \'etale $A'$-algebras hence $(A')^{sh}$ is a normal ring by Algebra, Lemmas \ref{algebra-lemma-normal-goes-up} and \ref{algebra-lemma-colimit-normal-ring}. Thus every local ring of $(A')^{sh}$ is a normal domain and we see that every maximal ideal contains a unique minimal prime. By Lemma \ref{lemma-integral-over-henselian-pair} applied to $A^{sh} \to (A')^{sh}$ to see that $((A')^{sh}, \mathfrak m(A')^{sh})$ is a henselian pair. If $\mathfrak q \subset (A')^{sh}$ is a minimal prime (or any prime), then the intersection of $V(\mathfrak q)$ with $V(\mathfrak m (A')^{sh})$ is connected by Lemma \ref{lemma-irreducible-henselian-pair-connected} Since $V(\mathfrak m (A')^{sh}) = \Spec((A')^{sh} \otimes \kappa^{sh})$ is a profinite space by we see there is a unique maximal ideal containing $\mathfrak q$. \medskip\noindent Proof of (2). The minimal primes of $A'$ are exactly the primes lying over a minimal prime of $A$ (by construction). Since $A' \to (A')^{sh}$ is flat by going down (Algebra, Lemma \ref{algebra-lemma-flat-going-down}) every minimal prime of $(A')^{sh}$ lies over a minimal prime of $A'$. Conversely, any prime of $(A')^{sh}$ lying over a minimal prime of $A'$ is minimal because $(A')^{sh}$ is a filtered colimit of \'etale hence quasi-finite algebras over $A'$ (small detail omitted). We conclude that the minimal primes of $(A')^{sh}$ are exactly the primes which lie over a minimal prime of $A$. Similarly, the minimal primes of $A^{sh}$ are exactly the primes lying over minimal primes of $A$. By construction we have $A' \otimes_A Q(A) = Q(A)$ where $Q(A)$ is the total fraction ring of our reduced local ring $A$. Of course $Q(A)$ is the finite product of residue fields of the minimal primes of $A$. It follows that $$ (A')^{sh} \otimes_A Q(A) = A^{sh} \otimes_A A' \otimes_A Q(A) = A^{sh} \otimes_A Q(A) $$ Our discussion above shows the spectrum of the ring on the left is the set of minimal primes of $(A')^{sh}$ and the spectrum of the ring on the right is the is the set of minimal primes of $A^{sh}$. This finishes the proof.