hcju/stacksqa huggingface dataset:
If $f$ is quasi-finite at $x$ then $\kappa(x)$ is a finite extension of $\kappa(s)$ (by Lemma \ref{lemma-residue-field-quasi-finite}) and $x$ is isolated in $X_s$ (by Lemma \ref{lemma-quasi-finite-at-point-characterize}), hence $\dim_x(X_s) = 0$ by Lemma \ref{lemma-dimension-fibre-at-a-point}. Conversely, if $\dim_x(X_s) = 0$ then by Lemma \ref{lemma-dimension-fibre-at-a-point} we see $\kappa(s) \subset \kappa(x)$ is algebraic and there are no other points of $X_s$ specializing to $x$. Hence $x$ is closed in its fibre by Lemma \ref{lemma-algebraic-residue-field-extension-closed-point-fibre} and by Lemma \ref{lemma-quasi-finite-at-point-characterize} (3) we conclude that $f$ is quasi-finite at $x$.
After shrinking $X$ and $Y$ to open neighborhoods of $x$ and $y$, we can assume that $\dim(X) = \dim_x(X)$ and $\dim(Y) = \dim_y(Y)$, by definition of the dimension of a scheme at a point (Properties, Definition \ref{properties-definition-dimension}). The morphism $f$ is open by Lemmas \ref{lemma-noetherian-finite-type-finite-presentation} and \ref{lemma-fppf-open}. Hence we can shrink $Y$ to arrange that $f$ is surjective. It remains to show that $\dim(X) = \dim(Y) + d$. \medskip\noindent Let $a$ be a point in $X$ with image $b$ in $Y$. By Algebra, Lemma \ref{algebra-lemma-dimension-base-fibre-equals-total}, $$ \dim(\mathcal{O}_{X,a}) = \dim(\mathcal{O}_{Y,b}) + \dim(\mathcal{O}_{X_b, a}). $$ Taking the supremum over all points $a$ in $X$, it follows that $\dim(X) = \dim(Y) + d$, as we want, see Properties, Lemma \ref{properties-lemma-dimension}.
This follows from Lemma \ref{lemma-locally-P} if we show that the property ``$R \to A$ is syntomic'' is local. We check conditions (a), (b) and (c) of Definition \ref{definition-property-local}. By Algebra, Lemma \ref{algebra-lemma-base-change-syntomic} being syntomic is stable under base change and hence we conclude (a) holds. By Algebra, Lemma \ref{algebra-lemma-composition-syntomic} being syntomic is stable under composition and trivially for any ring $R$ the ring map $R \to R_f$ is syntomic. We conclude (b) holds. Finally, property (c) is true according to Algebra, Lemma \ref{algebra-lemma-local-syntomic}.
In the proof of Lemma \ref{lemma-syntomic-characterize} we saw that being syntomic is a local property of ring maps. Hence the first statement of the lemma follows from Lemma \ref{lemma-composition-type-P} combined with the fact that being syntomic is a property of ring maps that is stable under composition, see Algebra, Lemma \ref{algebra-lemma-composition-syntomic}.
In the proof of Lemma \ref{lemma-syntomic-characterize} we saw that being syntomic is a local property of ring maps. Hence the lemma follows from Lemma \ref{lemma-composition-type-P} combined with the fact that being syntomic is a property of ring maps that is stable under base change, see Algebra, Lemma \ref{algebra-lemma-base-change-syntomic}.
Let $s' \in S'$ be a point, and let $s = g(s')$. Then we have $$ X'_{s'} = \Spec(\kappa(s')) \times_{\Spec(\kappa(s))} X_s $$ In other words the fibres of the base change are the base changes of the fibres. Hence the first part is equivalent to Algebra, Lemma \ref{algebra-lemma-lci-field-change-local}. The second part follows from the first because in that case $T$ is the set of points where $f$ is syntomic according to Lemma \ref{lemma-syntomic-locally-standard-syntomic}.
By Algebra, Definition \ref{algebra-definition-relative-global-complete-intersection} $R \to A$ being a relative global complete intersection means all nonzero fibre rings have dimension $n - c$. Thus for a prime $\mathfrak p$ of $R$ the fibre ring $\kappa(\mathfrak p)[x_1, \ldots, x_n]/(\overline{f}_1, \ldots, \overline{f}_c)$ is either zero or a global complete intersection ring of dimension $n - c$. By the discussion following Algebra, Definition \ref{algebra-definition-lci-field} this implies it is equidimensional of dimension $n - c$. Whence the lemma.
By Lemma \ref{lemma-flat-permanence} we see that $q$ is flat. Hence it suffices to show that the fibres of $Y \to S$ are local complete intersections, see Lemma \ref{lemma-syntomic-flat-fibres}. Let $s \in S$. Consider the morphism $X_s \to Y_s$. This is a base change of the morphism $X \to Y$ and hence surjective, and syntomic (Lemma \ref{lemma-base-change-syntomic}). For the same reason $X_s$ is syntomic over $\kappa(s)$. Moreover, $Y_s$ is locally of finite type over $\kappa(s)$ (Lemma \ref{lemma-base-change-finite-type}). In this way we reduce to the case where $S$ is the spectrum of a field. \medskip\noindent Assume $S = \Spec(k)$. Let $y \in Y$. Choose an affine open $\Spec(A) \subset Y$ neighbourhood of $y$. Let $\Spec(B) \subset X$ be an affine open such that $f(\Spec(B)) \subset \Spec(A)$, containing a point $x \in X$ such that $f(x) = y$. Choose a surjection $k[x_1, \ldots, x_n] \to A$ with kernel $I$. Choose a surjection $A[y_1, \ldots, y_m] \to B$, which gives rise in turn to a surjection $k[x_i, y_j] \to B$ with kernel $J$. Let $\mathfrak q \subset k[x_i, y_j]$ be the prime corresponding to $y \in \Spec(B)$ and let $\mathfrak p \subset k[x_i]$ the prime corresponding to $x \in \Spec(A)$. Since $x$ maps to $y$ we have $\mathfrak p = \mathfrak q \cap k[x_i]$. Consider the following commutative diagram of local rings: $$ \xymatrix{ \mathcal{O}_{X, x} \ar@{=}[r] & B_{\mathfrak q} & k[x_1, \ldots, x_n, y_1, \ldots, y_m]_{\mathfrak q} \ar[l] \\ \mathcal{O}_{Y, y} \ar@{=}[r] \ar[u] & A_{\mathfrak p} \ar[u] & k[x_1, \ldots, x_n]_{\mathfrak p} \ar[l] \ar[u] } $$ We claim that the hypotheses of Algebra, Lemma \ref{algebra-lemma-lci-permanence-initial} are satisfied. Conditions (1) and (2) are trivial. Condition (4) follows as $X \to Y$ is flat. Condition (3) follows as the rings $\mathcal{O}_{Y, y}$ and $\mathcal{O}_{X_y, x} = \mathcal{O}_{X, x}/\mathfrak m_y\mathcal{O}_{X, x}$ are complete intersection rings by our assumptions that $f$ and $p$ are syntomic, see Lemma \ref{lemma-syntomic-locally-standard-syntomic}. The output of Algebra, Lemma \ref{algebra-lemma-lci-permanence-initial} is exactly that $\mathcal{O}_{Y, y}$ is a complete intersection ring! Hence by Lemma \ref{lemma-syntomic-locally-standard-syntomic} again we see that $Y$ is syntomic over $k$ at $y$ as desired.
Let $\partial Z' = \overline{Z'} \setminus Z'$ and $\partial Z = \overline{Z} \setminus Z$. These are closed subsets of $X'$ and of $X$. Replacing $X'$ by $X' \setminus \partial Z'$ and $X$ by $X \setminus \big(g^{-1}(\partial Z') \cup \partial Z\big)$ we see that we may assume that $i$ and $i'$ are closed immersions. \medskip\noindent The fact that $g \circ i$ factors through $i'$ implies that $g^*\mathcal{I}'$ maps into $\mathcal{I}$ under the canonical map $g^*\mathcal{I}' \to \mathcal{O}_X$, see Schemes, Lemmas \ref{schemes-lemma-characterize-closed-subspace} and \ref{schemes-lemma-restrict-map-to-closed}. Hence we get an induced map of quasi-coherent sheaves $g^*(\mathcal{I}'/(\mathcal{I}')^2) \to \mathcal{I}/\mathcal{I}^2$. Pulling back by $i$ gives $i^*g^*(\mathcal{I}'/(\mathcal{I}')^2) \to i^*(\mathcal{I}/\mathcal{I}^2)$. Note that $i^*(\mathcal{I}/\mathcal{I}^2) = \mathcal{C}_{Z/X}$. On the other hand, $i^*g^*(\mathcal{I}'/(\mathcal{I}')^2) = f^*(i')^*(\mathcal{I}'/(\mathcal{I}')^2) = f^*\mathcal{C}_{Z'/X'}$. This gives the desired map. \medskip\noindent Checking that the map is locally described as the given map $I'/(I')^2 \to I/I^2$ is a matter of unwinding the definitions and is omitted. Another observation is that given any $x \in i(Z)$ there do exist affine open neighbourhoods $U$, $U'$ with $f(U) \subset U'$ and $Z \cap U$ as well as $U' \cap Z'$ closed such that $x \in U$. Proof omitted. Hence the requirement of the lemma indeed characterizes the map (and could have been used to define it).
Via Lemma \ref{lemma-conormal-functorial} this translates into the following algebra fact. Suppose that $C \to B \to A$ are surjective ring maps. Let $I = \Ker(B \to A)$, $J = \Ker(C \to A)$ and $K = \Ker(C \to B)$. Then there is an exact sequence $$ K/K^2 \otimes_B A \to J/J^2 \to I/I^2 \to 0. $$ This follows immediately from the observation that $I = J/K$.
By Lemma \ref{lemma-differentials-restrict-open} we may replace $X$ and $S$ by $U$ and $V$. Thus we may assume $X = \Spec(A)$ and $S = \Spec(R)$ and we have to show the lemma with $U = X$ and $V = S$. Consider the $A$-module $M = \Gamma(X, \Omega_{X/S})$ together with the $R$-derivation $\text{d}_{X/S} : A \to M$. Let $N$ be another $A$-module and denote $\widetilde{N}$ the quasi-coherent $\mathcal{O}_X$-module associated to $N$, see Schemes, Section \ref{schemes-section-quasi-coherent-affine}. Precomposing by $\text{d}_{X/S} : A \to M$ we get an arrow $$ \alpha : \Hom_A(M, N) \longrightarrow \text{Der}_R(A, N) $$ Using Lemmas \ref{lemma-universal-derivation-universal} and \ref{lemma-affine-case-derivation} we get identifications $$ \Hom_{\mathcal{O}_X}(\Omega_{X/S}, \widetilde{N}) = \text{Der}_S(\mathcal{O}_X, \widetilde{N}) = \text{Der}_R(A, N) $$ Taking global sections determines an arrow $\Hom_{\mathcal{O}_X}(\Omega_{X/S}, \widetilde{N}) \to \Hom_R(M, N)$. Combining this arrow and the identifications above we get an arrow $$ \beta : \text{Der}_R(A, N) \longrightarrow \Hom_R(M, N) $$ Checking what happens on global sections, we find that $\alpha$ and $\beta$ are each others inverse. Hence we see that $\text{d}_{X/S} : A \to M$ satisfies the same universal property as $\text{d} : A \to \Omega_{A/R}$, see Algebra, Lemma \ref{algebra-lemma-universal-omega}. Thus the Yoneda lemma (Categories, Lemma \ref{categories-lemma-yoneda}) implies there is a unique isomorphism of $A$-modules $M \cong \Omega_{A/R}$ compatible with derivations.
This is a special case of Modules, Lemma \ref{modules-lemma-functoriality-differentials-ringed-spaces}. In the case of schemes we can also use the functoriality of the conormal sheaves (see Lemma \ref{lemma-conormal-functorial}) and Lemma \ref{lemma-differentials-diagonal} to define $c_f$. Or we can use the characterization in the last line of the lemma to glue maps defined on affine patches (see Algebra, Equation (\ref{algebra-equation-functorial-omega})).
By Lemma \ref{lemma-base-change-differentials} the composition $p^*\Omega_{X/S} \to \Omega_{X \times_S Y/S} \to \Omega_{X \times_S Y/Y}$ is an isomorphism, and similarly for $q$. Moreover, the cokernel of $p^*\Omega_{X/S} \to \Omega_{X \times_S Y/S}$ is $\Omega_{X \times_S Y/X}$ by Lemma \ref{lemma-triangle-differentials}. The result follows.
This is the sheafified version of Algebra, Lemma \ref{algebra-lemma-differential-seq}. However we should make sure we can define the first arrow globally. Hence we explain the meaning of ``induced by $\text{d}_{X/S}$'' here. Namely, we may assume that $i$ is a closed immersion by shrinking $X$. Let $\mathcal{I} \subset \mathcal{O}_X$ be the sheaf of ideals corresponding to $Z \subset X$. Then $\text{d}_{X/S} : \mathcal{I} \to \Omega_{X/S}$ maps the subsheaf $\mathcal{I}^2 \subset \mathcal{I}$ to $\mathcal{I}\Omega_{X/S}$. Hence it induces a map $\mathcal{I}/\mathcal{I}^2 \to \Omega_{X/S}/\mathcal{I}\Omega_{X/S}$ which is $\mathcal{O}_X/\mathcal{I}$-linear. By Lemma \ref{lemma-i-star-equivalence} this corresponds to a map $\mathcal{C}_{Z/X} \to i^*\Omega_{X/S}$ as desired.
Follows from Algebra, Lemma \ref{algebra-lemma-differential-seq-split}. Clarification: if $g : X \to Z$ is a left inverse of $i$, then $i^*c_g$ is a right inverse of the map $i^*\Omega_{X/S} \to \Omega_{Z/S}$. Also, if $s$ is a section, then it is an immersion $s : Z = S \to X$ over $S$ (see Schemes, Lemma \ref{schemes-lemma-section-immersion}) and in that case $\Omega_{Z/S} = 0$.
Write $\mathcal{F} = \widetilde{M}$ for some $A$-module $M$. Set $N = \Gamma(X, \mathcal{G})$. Let $D : M \to N$ be a differential operator of order $k$. We have to show there exists a unique differential operator $\mathcal{F} \to \mathcal{G}$ of order $k$ which gives rise to $D$ on global sections. Let $U = D(f) \subset X$ be a standard affine open. Then $\mathcal{F}(U) = M_f$ is the localization. By Algebra, Lemma \ref{algebra-lemma-invert-system-differential-operators} the differential operator $D$ extends to a unique differential operator $$ D_f : \mathcal{F}(U) = \widetilde{M}(U) = M_f \to N_f = \widetilde{N}(U) $$ The uniqueness shows that these maps $D_f$ glue to give a map of sheaves $\widetilde{M} \to \widetilde{N}$ on the basis of all standard opens of $X$. Hence we get a unique map of sheaves $\widetilde{D} : \widetilde{M} \to \widetilde{N}$ agreeing with these maps by the material in Sheaves, Section \ref{sheaves-section-bases}. Since $\widetilde{D}$ is given by differential operators of order $k$ on the standard opens, we find that $\widetilde{D}$ is a differential operator of order $k$ (small detail omitted). Finally, we can post-compose with the canonical $\mathcal{O}_X$-module map $c : \widetilde{N} \to \mathcal{G}$ (Schemes, Lemma \ref{schemes-lemma-compare-constructions}) to get $c \circ \widetilde{D} : \mathcal{F} \to \mathcal{G}$ which is a differential operator of order $k$ by Modules, Lemma \ref{modules-lemma-composition-differential-operators}. This proves existence. We omit the proof of uniqueness.
In case $X$, $Y$, and $S$ are affine, this follows, via Lemma \ref{lemma-affine-case-differential-operators}, from the corresponding algebra result, see Algebra, Lemma \ref{algebra-lemma-base-change-differential-operators}. In general, one uses coverings by affines (for example as in Schemes, Lemma \ref{schemes-lemma-affine-covering-fibre-product}) to construct $D'$ globally. Details omitted.
This follows from Lemma \ref{lemma-locally-P} if we show that the property ``$R \to A$ is smooth'' is local. We check conditions (a), (b) and (c) of Definition \ref{definition-property-local}. By Algebra, Lemma \ref{algebra-lemma-base-change-smooth} being smooth is stable under base change and hence we conclude (a) holds. By Algebra, Lemma \ref{algebra-lemma-compose-smooth} being smooth is stable under composition and for any ring $R$ the ring map $R \to R_f$ is (standard) smooth. We conclude (b) holds. Finally, property (c) is true according to Algebra, Lemma \ref{algebra-lemma-locally-smooth}.
The statement is local on $X$ and $S$. By Lemma \ref{lemma-smooth-locally-standard-smooth} above we may assume that $f$ is a standard smooth morphism of affines. In this case the result follows from Algebra, Lemma \ref{algebra-lemma-standard-smooth} (and the definition of a relative global complete intersection, see Algebra, Definition \ref{algebra-definition-relative-global-complete-intersection}).
Note that if $f$ is smooth at $x$, then we see from Lemma \ref{lemma-smooth-locally-standard-smooth} that (5) holds, and (6) is a slightly weakened version of (5). Moreover, $f$ smooth implies that the ring map $\mathcal{O}_{S, s} \to \mathcal{O}_{X, x}$ is flat (see Lemma \ref{lemma-smooth-flat}) and that $\Omega_{X/S}$ is finite locally free of rank equal to $\dim_x(X_s)$ (see Lemma \ref{lemma-smooth-omega-finite-locally-free}). Thus (1) implies (3) and (4). By Lemma \ref{lemma-base-change-smooth} we also see that (1) implies (2). \medskip\noindent By Lemma \ref{lemma-base-change-differentials} the module of differentials $\Omega_{X_s/s}$ of the fibre $X_s$ over $\kappa(s)$ is the pullback of the module of differentials $\Omega_{X/S}$ of $X$ over $S$. Hence the displayed equality in part (4) of the lemma. By Lemma \ref{lemma-finite-type-differentials} these modules are of finite type. Hence the minimal number of generators of the modules $\Omega_{X/S, x}$ and $\Omega_{X_s/s, x}$ is the same and equal to the dimension of this $\kappa(x)$-vector space by Nakayama's Lemma (Algebra, Lemma \ref{algebra-lemma-NAK}). This in particular shows that (3) and (4) are equivalent. \medskip\noindent Algebra, Lemma \ref{algebra-lemma-flat-fibre-smooth} shows that (2) implies (1). Algebra, Lemma \ref{algebra-lemma-characterize-smooth-over-field} shows that (3) and (4) imply (2). Finally, (6) implies (5) see for example Algebra, Example \ref{algebra-example-make-standard-smooth} and (5) implies (1) by Algebra, Lemma \ref{algebra-lemma-standard-smooth}.
Assume first that $f$ locally of finite type. Consider the set $$ T = \{x \in X \mid X_{f(x)}\text{ is smooth over }\kappa(f(x))\text{ at }x\} $$ and the corresponding set $T' \subset X'$ for $f'$. Then we claim $T' = (g')^{-1}(T)$. Namely, let $s' \in S'$ be a point, and let $s = g(s')$. Then we have $$ X'_{s'} = \Spec(\kappa(s')) \times_{\Spec(\kappa(s))} X_s $$ In other words the fibres of the base change are the base changes of the fibres. Hence the claim is equivalent to Algebra, Lemma \ref{algebra-lemma-smooth-field-change-local}. \medskip\noindent Thus case (1) follows because in case (1) $T$ is the (open) set of points where $f$ is smooth by Lemma \ref{lemma-smooth-at-point}. \medskip\noindent In case (2) let $x' \in W'$. Then $g'$ is flat at $x'$ (Lemma \ref{lemma-base-change-module-flat}) and $g \circ f$ is flat at $x'$ (Lemma \ref{lemma-composition-module-flat}). It follows that $f$ is flat at $x = g'(x')$ by Lemma \ref{lemma-flat-permanence}. On the other hand, since $x' \in T'$ (Lemma \ref{lemma-base-change-smooth}) we see that $x \in T$. Hence $f$ is smooth at $x$ by Lemma \ref{lemma-smooth-at-point}.
The algebraic version of this lemma is the following: Given ring maps $A \to B \to C$ with $B \to C$ smooth, then the sequence $$ 0 \to C \otimes_B \Omega_{B/A} \to \Omega_{C/A} \to \Omega_{C/B} \to 0 $$ of Algebra, Lemma \ref{algebra-lemma-exact-sequence-differentials} is exact. This is Algebra, Lemma \ref{algebra-lemma-triangle-differentials-smooth}.
The algebraic version of this lemma is the following: Given ring maps $A \to B \to C$ with $A \to C$ smooth and $B \to C$ surjective with kernel $J$, then the sequence $$ 0 \to J/J^2 \to C \otimes_B \Omega_{B/A} \to \Omega_{C/A} \to 0 $$ of Algebra, Lemma \ref{algebra-lemma-differential-seq} is exact. This is Algebra, Lemma \ref{algebra-lemma-differential-seq-smooth}.
The algebraic version of this lemma is the following: Given ring maps $A \to B \to C$ with $A \to C$ surjective and $A \to B$ smooth, then the sequence $$ 0 \to I/I^2 \to J/J^2 \to C \otimes_B \Omega_{B/A} \to 0 $$ of Algebra, Lemma \ref{algebra-lemma-application-NL} is exact. This is Algebra, Lemma \ref{algebra-lemma-application-NL-smooth}.
By Lemma \ref{lemma-flat-permanence} we see that $q$ is flat. Pick a point $y \in Y$. Pick a point $x \in X$ mapping to $y$. Suppose $f$ has relative dimension $a$ at $x$ and $p$ has relative dimension $b$ at $x$. By Lemma \ref{lemma-smooth-omega-finite-locally-free} this means that $\Omega_{X/S, x}$ is free of rank $b$ and $\Omega_{X/Y, x}$ is free of rank $a$. By the short exact sequence of Lemma \ref{lemma-triangle-differentials-smooth} this means that $(f^*\Omega_{Y/S})_x$ is free of rank $b - a$. By Nakayama's Lemma this implies that $\Omega_{Y/S, y}$ can be generated by $b - a$ elements. Also, by Lemma \ref{lemma-dimension-fibre-at-a-point-additive} we see that $\dim_y(Y_s) = b - a$. Hence we conclude that $Y \to S$ is smooth at $y$ by Lemma \ref{lemma-smooth-at-point} part (2).
After replacing $X$ by an open neighborhood of $x$, there is a natural number $d$ such that all fibers of $X \to Y$ have dimension $d$ at every point, see Lemma \ref{lemma-smooth-omega-finite-locally-free}. Then $f$ is flat (Lemma \ref{lemma-smooth-flat}), locally of finite type (Lemma \ref{lemma-smooth-locally-finite-presentation}), and of relative dimension $d$. Hence the result follows from Lemma \ref{lemma-rel-dimension-dimension}.
The proof of Lemma \ref{lemma-unramified-characterize} shows that being unramified (resp.\ G-unramified) is a local property of ring maps. Hence the first statement of the lemma follows from Lemma \ref{lemma-composition-type-P} combined with the fact that being unramified (resp.\ G-unramified) is a property of ring maps that is stable under composition, see Algebra, Lemma \ref{algebra-lemma-unramified}.
The proof of Lemma \ref{lemma-unramified-characterize} shows that being unramified (resp.\ G-unramified) is a local property of ring maps. Hence the lemma follows from Lemma \ref{lemma-base-change-type-P} combined with the fact that being unramified (resp.\ G-unramified) is a property of ring maps that is stable under base change, see Algebra, Lemma \ref{algebra-lemma-unramified}.
Part (2) follows from part (1) and Lemma \ref{lemma-base-change-unramified}. Let us prove part (1). We first use Algebra, Lemma \ref{algebra-lemma-characterize-unramified}. This lemma implies that if $X$ is a disjoint union of spectra of finite separable field extensions of $k$ then $X \to \Spec(k)$ is unramified. Conversely, suppose that $X \to \Spec(k)$ is unramified. By Algebra, Lemma \ref{algebra-lemma-unramified-at-prime} for every $x \in X$ the residue field extension $k \subset \kappa(x)$ is finite separable. Since $X \to \Spec(k)$ is locally quasi-finite (Lemma \ref{lemma-unramified-quasi-finite}) we see that all points of $X$ are isolated closed points, see Lemma \ref{lemma-quasi-finite-at-point-characterize}. Thus $X$ is a discrete space, in particular the disjoint union of the spectra of its local rings. By Algebra, Lemma \ref{algebra-lemma-unramified-at-prime} again these local rings are fields, and we win.
Note that if $f$ is unramified at $x$, then we see that $\Omega_{X/S} = 0$ in a neighbourhood of $x$ by the definitions and the results on modules of differentials in Section \ref{section-sheaf-differentials}. Hence (1) implies (3) and the vanishing of the right hand vector space in (5). It also implies (2) because by Lemma \ref{lemma-base-change-differentials} the module of differentials $\Omega_{X_s/s}$ of the fibre $X_s$ over $\kappa(s)$ is the pullback of the module of differentials $\Omega_{X/S}$ of $X$ over $S$. This fact on modules of differentials also implies the displayed equality of vector spaces in part (4). By Lemma \ref{lemma-finite-type-differentials} the modules $\Omega_{X/S, x}$ and $\Omega_{X_s/s, x}$ are of finite type. Hence the modules $\Omega_{X/S, x}$ and $\Omega_{X_s/s, x}$ are zero if and only if the corresponding $\kappa(x)$-vector space in (4) is zero by Nakayama's Lemma (Algebra, Lemma \ref{algebra-lemma-NAK}). This in particular shows that (3), (4) and (5) are equivalent. The support of $\Omega_{X/S}$ is closed in $X$, see Modules, Lemma \ref{modules-lemma-support-finite-type-closed}. Assumption (3) implies that $x$ is not in the support. Hence $\Omega_{X/S}$ is zero in a neighbourhood of $x$, which implies (1). The equivalence of (1) and (3) applied to $X_s \to s$ implies the equivalence of (2) and (4). At this point we have seen that (1) -- (5) are equivalent. \medskip\noindent Alternatively you can use Algebra, Lemma \ref{algebra-lemma-unramified} to see the equivalence of (1) -- (5) more directly. \medskip\noindent The equivalence of (1) and (6) follows from Lemma \ref{lemma-unramified-etale-fibres}. It also follows more directly from Algebra, Lemmas \ref{algebra-lemma-unramified-at-prime} and \ref{algebra-lemma-characterize-unramified}.
Let $s' \in S'$ be a point, and let $s = g(s')$. Then we have $$ X'_{s'} = \Spec(\kappa(s')) \times_{\Spec(\kappa(s))} X_s $$ In other words the fibres of the base change are the base changes of the fibres. In particular $$ \Omega_{X_s/s, x} \otimes_{\mathcal{O}_{X_s, x}} \kappa(x') = \Omega_{X'_{s'}/s', x'} \otimes_{\mathcal{O}_{X'_{s'}, x'}} \kappa(x') $$ see Lemma \ref{lemma-base-change-differentials}. Whence $x' \in T'$ if and only if $x \in T$ by Lemma \ref{lemma-unramified-at-point}. The second part follows from the first because in that case $T$ is the (open) set of points where $f$ is G-unramified according to Lemma \ref{lemma-unramified-at-point}.
Assume that $X$ is unramified over $S$. By Lemma \ref{lemma-permanence-finite-type} we see that $f$ is locally of finite type. By assumption we have $\Omega_{X/S} = 0$. Hence $\Omega_{X/Y} = 0$ by Lemma \ref{lemma-triangle-differentials}. Thus $f$ is unramified. If $X$ is G-unramified over $S$ and $Y$ of finite type over $S$, then by Lemma \ref{lemma-finite-presentation-permanence} we see that $f$ is locally of finite presentation and we conclude that $f$ is G-unramified.