hcju/stacksqa huggingface dataset:
By Lemma \ref{lemma-faithful} the functor is faithful. Let $U_1 \to S$ and $U_2 \to S$ be \'etale morphisms and let $a : X \times_S U_1 \to X \times_S U_2$ be a morphism compatible with canonical descent data. We will prove that $a$ is the base change of a morphism $U_1 \to U_2$. \medskip\noindent Let $U'_2 \subset U_2$ be an open subscheme. Consider $W = a^{-1}(X \times_S U'_2)$. This is an open subscheme of $X \times_S U_1$ which is compatible with the canonical descent datum on $V_1 = X \times_S U_1$. This means that the two inverse images of $W$ by the projections $V_1 \times_{U_1} V_1 \to V_1$ agree. Since $V_1 \to U_1$ is surjective (as the base change of $X \to S$) we conclude that $W$ is the inverse image of some subset $U'_1 \subset U_1$. Since $W$ is open, our assumption on $f$ implies that $U'_1 \subset U_1$ is open. \medskip\noindent Let $U_2 = \bigcup U_{2, i}$ be an affine open covering. By the result of the preceding paragraph we obtain an open covering $U_1 = \bigcup U_{1, i}$ such that $X \times_S U_{1, i} = a^{-1}(X \times_S U_{2, i})$. If we can prove there exists a morphism $U_{1, i} \to U_{2, i}$ whose base change is the morphism $a_i : X \times_S U_{1, i} \to X \times_S U_{2, i}$ then we can glue these morphisms to a morphism $U_1 \to U_2$ (using faithfulness). In this way we reduce to the case that $U_2$ is affine. In particular $U_2 \to S$ is separated (Schemes, Lemma \ref{schemes-lemma-compose-after-separated}). \medskip\noindent Assume $U_2 \to S$ is separated. Then the graph $\Gamma_a$ of $a$ is a closed subscheme of $$ V = (X \times_S U_1) \times_X (X \times_S U_2) = X \times_S U_1 \times_S U_2 $$ by Schemes, Lemma \ref{schemes-lemma-semi-diagonal}. On the other hand the graph is open for example because it is a section of an \'etale morphism (Proposition \ref{proposition-properties-sections}). Since $a$ is a morphism of descent data, the two inverse images of $\Gamma_a \subset V$ under the projections $V \times_{U_1 \times_S U_2} V \to V$ are the same. Hence arguing as in the second paragraph of the proof we find an open and closed subscheme $\Gamma \subset U_1 \times_S U_2$ whose base change to $X$ gives $\Gamma_a$. Then $\Gamma \to U_1$ is an \'etale morphism whose base change to $X$ is an isomorphism. This means that $\Gamma \to U_1$ is universally bijective, hence an isomorphism by Theorem \ref{theorem-etale-radicial-open}. Thus $\Gamma$ is the graph of a morphism $U_1 \to U_2$ and the base change of this morphism is $a$ as desired.
This follows from Lemma \ref{lemma-fully-faithful}. For example a closed surjective map of topological spaces is submersive (Topology, Lemma \ref{topology-lemma-closed-morphism-quotient-topology}). Finite, integral, and proper morphisms are universally closed, see Morphisms, Lemmas \ref{morphisms-lemma-integral-universally-closed} and \ref{morphisms-lemma-finite-proper} and Definition \ref{morphisms-definition-proper}. On the other hand an open surjective map of topological spaces is submersive (Topology, Lemma \ref{topology-lemma-open-morphism-quotient-topology}). Flat locally finitely presented, smooth, and \'etale morphisms are universally open, see Morphisms, Lemmas \ref{morphisms-lemma-fppf-open}, \ref{morphisms-lemma-smooth-open}, and \ref{morphisms-lemma-etale-open}. The case of surjective, quasi-compact, flat morphisms follows from Morphisms, Lemma \ref{morphisms-lemma-fpqc-quotient-topology}.
(Recall that pullbacks of descent data are defined in Descent, Definition \ref{descent-definition-pullback-functor}.) Set $X_i = X \times_S S_i$. Denote $(V_i, \varphi_i)$ the pullback of $(V, \varphi)$ to $X_i/S_i$. By assumption (1) we can find an \'etale morphism $U_i \to S_i$ which comes with an isomorphism $X_i \times_{S_i} U_i \to V_i$ compatible with $can$ and $\varphi_i$. By assumption (2) we obtain isomorphisms $\psi_{ij} : U_i \times_{S_i} (S_i \cap S_j) \to U_j \times_{S_j} (S_i \cap S_j)$. By assumption (3) these isomorphisms satisfy the cocycle condition so that $(U_i, \psi_{ij})$ is a descend datum for the Zariski covering $\{S_i \to S\}$. Then Descent, Lemma \ref{descent-lemma-Zariski-refinement-coverings-equivalence} (which is essentially just a reformulation of Schemes, Section \ref{schemes-section-glueing-schemes}) tells us that there exists a morphism of schemes $U \to S$ and isomorphisms $U \times_S S_i \to U_i$ compatible with $\psi_{ij}$. The isomorphisms $U \times_S S_i \to U_i$ determine corresponding isomorphisms $X_i \times_S U \to V_i$ which glue to a morphism $X \times_S U \to V$ compatible with the canonical descent datum and $\varphi$.
By Zariski's main theorem, the scheme $U$ is quasi-affine. In fact, we can find an open immersion $U \to T$ with $T$ affine and $T \to \Spec(A)$ finite, see More on Morphisms, Lemma \ref{more-morphisms-lemma-quasi-finite-separated-pass-through-finite}. Write $Z = \Spec(A/I)$ and denote $U_Z \to T_Z$ the base change. Since $U_Z \to Z$ is finite, we see that $U_Z \to T_Z$ is closed as well as open. Hence by More on Algebra, Lemma \ref{more-algebra-lemma-characterize-henselian-pair} we obtain a unique decomposition $T = T' \amalg T''$ with $T'_Z = U_Z$. Set $U_{fin} = U \cap T'$ and $U_{away} = U \cap T''$. Since $T'_Z \subset U_Z$ we see that all closed points of $T'$ are in $U$ hence $T' \subset U$, hence $U_{fin} = T'$, hence $U_{fin} \to \Spec(A)$ is finite. We omit the proof of uniqueness of the decomposition.
Assume $D$ is a strict normal crossings divisor. Pick $p \in D$ and choose a regular system of parameters $x_1, \ldots, x_d \in \mathfrak m_p$ and $1 \leq r \leq d$ as in Definition \ref{definition-strict-normal-crossings}. Since $\mathcal{O}_{X, p}/(x_i)$ is a regular local ring (and in particular a domain) we see that the irreducible components $D_1, \ldots, D_r$ of $D$ passing through $p$ correspond $1$-to-$1$ to the height one primes $(x_1), \ldots, (x_r)$ of $\mathcal{O}_{X, p}$. By Algebra, Lemma \ref{algebra-lemma-regular-ring-CM} we find that the intersections $D_{i_1} \cap \ldots \cap D_{i_s}$ have codimension $s$ in an open neighbourhood of $p$ and that this intersection has a regular local ring at $p$. Since this holds for all $p \in D$ we conclude that (2) holds. \medskip\noindent Assume (2). Let $p \in D$. Since $\mathcal{O}_{X, p}$ is finite dimensional we see that $p$ can be contained in at most $\dim(\mathcal{O}_{X, p})$ of the components $D_i$. Say $p \in D_1, \ldots, D_r$ for some $r \geq 1$. Let $x_1, \ldots, x_r \in \mathfrak m_p$ be local equations for $D_1, \ldots, D_r$. Then $x_1$ is a nonzerodivisor in $\mathcal{O}_{X, p}$ and $\mathcal{O}_{X, p}/(x_1) = \mathcal{O}_{D_1, p}$ is regular. Hence $\mathcal{O}_{X, p}$ is regular, see Algebra, Lemma \ref{algebra-lemma-regular-mod-x}. Since $D_1 \cap \ldots \cap D_r$ is a regular (hence normal) scheme it is a disjoint union of its irreducible components (Properties, Lemma \ref{properties-lemma-normal-Noetherian}). Let $Z \subset D_1 \cap \ldots \cap D_r$ be the irreducible component containing $p$. Then $\mathcal{O}_{Z, p} = \mathcal{O}_{X, p}/(x_1, \ldots, x_r)$ is regular of codimension $r$ (note that since we already know that $\mathcal{O}_{X, p}$ is regular and hence Cohen-Macaulay, there is no ambiguity about codimension as the ring is catenary, see Algebra, Lemmas \ref{algebra-lemma-regular-ring-CM} and \ref{algebra-lemma-CM-dim-formula}). Hence $\dim(\mathcal{O}_{Z, p}) = \dim(\mathcal{O}_{X, p}) - r$. Choose additional $x_{r + 1}, \ldots, x_n \in \mathfrak m_p$ which map to a minimal system of generators of $\mathfrak m_{Z, p}$. Then $\mathfrak m_p = (x_1, \ldots, x_n)$ by Nakayama's lemma and we see that $D$ is a normal crossings divisor.
As $f$ is flat the pullback is defined by Divisors, Lemma \ref{divisors-lemma-pullback-effective-Cartier-defined} hence the statement makes sense. Let $q \in f^*D$ map to $p \in D$. Choose a regular system of parameters $x_1, \ldots, x_d \in \mathfrak m_p$ and $1 \leq r \leq d$ as in Definition \ref{definition-strict-normal-crossings}. Since $f$ is smooth the local ring homomorphism $\mathcal{O}_{X, p} \to \mathcal{O}_{Y, q}$ is flat and the fibre ring $$ \mathcal{O}_{Y, q}/\mathfrak m_p \mathcal{O}_{Y, q} = \mathcal{O}_{Y_p, q} $$ is a regular local ring (see for example Algebra, Lemma \ref{algebra-lemma-characterize-smooth-over-field}). Pick $y_1, \ldots, y_n \in \mathfrak m_q$ which map to a regular system of parameters in $\mathcal{O}_{Y_p, q}$. Then $x_1, \ldots, x_d, y_1, \ldots, y_n$ generate the maximal ideal $\mathfrak m_q$. Hence $\mathcal{O}_{Y, q}$ is a regular local ring of dimension $d + n$ by Algebra, Lemma \ref{algebra-lemma-dimension-base-fibre-equals-total} and $x_1, \ldots, x_d, y_1, \ldots, y_n$ is a regular system of parameters. Since $f^*D$ is cut out by $x_1 \ldots x_r$ in $\mathcal{O}_{Y, q}$ we conclude that the lemma is true.
As $f$ is flat the pullback is defined by Divisors, Lemma \ref{divisors-lemma-pullback-effective-Cartier-defined} hence the statement makes sense. Let $q \in f^*D$ map to $p \in D$. Choose an \'etale morphism $U \to X$ whose image contains $p$ such that $D \times_X U \subset U$ is a strict normal crossings divisor as in Definition \ref{definition-normal-crossings}. Set $V = Y \times_X U$. Then $V \to Y$ is \'etale as a base change of $U \to X$ (Morphisms, Lemma \ref{morphisms-lemma-base-change-etale}) and the pullback $D \times_X V$ is a strict normal crossings divisor on $V$ by Lemma \ref{lemma-smooth-pullback-strict-normal-crossings}. Thus we have checked the condition of Definition \ref{definition-normal-crossings} for $q \in f^*D$ and we conclude.
The implication (1) $\Rightarrow$ (2) is straightforward and does not need the assumption that $X$ is J-2 or Nagata. Namely, let $p \in D$ and choose an \'etale neighbourhood $(U, u) \to (X, p)$ such that the pullback of $D$ is a strict normal crossings divisor on $U$. Then $\mathcal{O}_{X, p}^{sh} = \mathcal{O}_{U, u}^{sh}$ and we see that the trace of $D$ on $\Spec(\mathcal{O}_{U, u}^{sh})$ is cut out by part of a regular system of parameters as this is already the case in $\mathcal{O}_{U, u}$. \medskip\noindent To prove the implication in the other direction we will use the criterion of Lemma \ref{lemma-characterize-normal-crossings-normalization}. Observe that formation of the normalization $D^\nu \to D$ commutes with strict henselization, see More on Morphisms, Lemma \ref{more-morphisms-lemma-normalization-and-henselization}. If we can show that $D^\nu \to D$ is finite, then we see that $D^\nu \to D$ and the schemes $Z_n$ satisfy all desired properties because these can all be checked on the level of local rings (but the finiteness of the morphism $D^\nu \to D$ is not something we can check on local rings). We omit the detailed verifications. \medskip\noindent If $X$ is Nagata, then $D^\nu \to D$ is finite by Morphisms, Lemma \ref{morphisms-lemma-nagata-normalization}. \medskip\noindent Assume $X$ is J-2. Choose a point $p \in D$. We will show that $D^\nu \to D$ is finite over a neighbourhood of $p$. By assumption there exists a regular system of parameters $f_1, \ldots, f_d$ of $\mathcal{O}_{X, p}^{sh}$ and $1 \leq r \leq d$ such that the trace of $D$ on $\Spec(\mathcal{O}_{X, p}^{sh})$ is cut out by $f_1 \ldots f_r$. Then $$ D^\nu \times_X \Spec(\mathcal{O}_{X, p}^{sh}) =  \coprod\nolimits_{i = 1, \ldots, r} V(f_i) $$ Choose an affine \'etale neighbourhood $(U, u) \to (X, p)$ such that $f_i$ comes from $f_i \in \mathcal{O}_U(U)$. Set $D_i = V(f_i) \subset U$. The strict henselization of $\mathcal{O}_{D_i, u}$ is $\mathcal{O}_{X, p}^{sh}/(f_i)$ which is regular. Hence $\mathcal{O}_{D_i, u}$ is regular (for example by More on Algebra, Lemma \ref{more-algebra-lemma-henselization-regular}). Because $X$ is J-2 the regular locus is open in $D_i$. Thus after replacing $U$ by a Zariski open we may assume that $D_i$ is regular for each $i$. It follows that $$ \coprod\nolimits_{i = 1, \ldots, r} D_i = D^\nu \times_X U \longrightarrow D \times_X U $$ is the normalization morphism and it is clearly finite. In other words, we have found an \'etale neighbourhood $(U, u)$ of $(X, p)$ such that the base change of $D^\nu \to D$ to this neighbourhood is finite. This implies $D^\nu \to D$ is finite by descent (Descent, Lemma \ref{descent-lemma-descending-property-finite}) and the proof is complete.
Fix a base scheme $S$. If $f : X' \to X$ is any $S$-morphism, then the graph $\Gamma_f : X' \to X' \times_S X$ is obtained as the base change of the diagonal $\Delta_{X/S} : X \to X \times_S X$ via the projection $X' \times_S X \to X \times_S X$. If $g : X \to S$ is separated (resp. unramified) then the diagonal is a closed immersion (resp. open immersion) by Schemes, Definition \ref{schemes-definition-separated} (resp.\ Morphisms, Lemma \ref{morphisms-lemma-diagonal-unramified-morphism}). Hence so is the graph as a base change (by Schemes, Lemma \ref{schemes-lemma-base-change-immersion}). In the special case $X' = S$, we obtain (1), resp.\ (2). Part (3) follows on combining (1) and (2).
The maps $f, g : Y \to X$ define maps $f', g' : Y \to X_Y = Y \times_S X$ which are sections of the structure map $X_Y \to Y$. Note that $f = g$ if and only if $f' = g'$. The structure map $X_Y \to Y$ is the base change of $\pi$ and hence unramified and separated also (see Morphisms, Lemmas \ref{morphisms-lemma-base-change-unramified} and Schemes, Lemma \ref{schemes-lemma-separated-permanence}). Thus according to Theorem \ref{theorem-sections-unramified-maps} it suffices to prove that $f'$ and $g'$ pass through the same point of $X_Y$. And this is exactly what the hypotheses (2) and (3) guarantee, namely $f'(y) = g'(y) \in X_Y$.
If $B$ is regular, then $A$ is regular by Algebra, Lemma \ref{algebra-lemma-flat-under-regular}. Assume $A$ is regular. Let $\mathfrak m$ be the maximal ideal of $A$. Then $\dim_{\kappa(\mathfrak m)} \mathfrak m/\mathfrak m^2 = \dim(A) = \dim(B)$ (see Lemma \ref{lemma-etale-dimension}). On the other hand, $\mathfrak mB$ is the maximal ideal of $B$ and hence $\mathfrak m_B/\mathfrak m_B = \mathfrak mB/\mathfrak m^2B$ is generated by at most $\dim(B)$ elements. Thus $B$ is regular. (You can also use the slightly more general Algebra, Lemma \ref{algebra-lemma-flat-over-regular-with-regular-fibre}.)
It is clear from the faithful flatness of $A \to B$ that if $B$ is reduced, so is $A$. See also Algebra, Lemma \ref{algebra-lemma-descent-reduced}. Conversely, assume $A$ is reduced. By assumption $B$ is a localization of a finite type $A$-algebra $B'$ at some prime $\mathfrak q$. After replacing $B'$ by a localization we may assume that $B'$ is \'etale over $A$, see Lemma \ref{lemma-characterize-etale-Noetherian}. Then we see that Algebra, Lemma \ref{algebra-lemma-reduced-goes-up} applies to $A \to B'$ and $B'$ is reduced. Hence $B$ is reduced.
See Algebra, Lemma \ref{algebra-lemma-descent-normal} for descending normality. Conversely, assume $A$ is normal. By assumption $B$ is a localization of a finite type $A$-algebra $B'$ at some prime $\mathfrak q$. After replacing $B'$ by a localization we may assume that $B'$ is \'etale over $A$, see Lemma \ref{lemma-characterize-etale-Noetherian}. Then we see that Algebra, Lemma \ref{algebra-lemma-normal-goes-up} applies to $A \to B'$ and we conclude that $B'$ is normal. Hence $B$ is a normal domain.
We first write the arrow as a composition of three arrows. Namely, set $$ X^{(1)} = X \times_{S_0, F_{S_0}} S_0 $$ and denote $F_{X/S_0} : X \to X^{(1)}$ the relative Frobenius morphism. Denote $\mathcal{E}^{(1)}$ the base change of $\mathcal{E}$ by $\Spec(\sigma)$, in other words the pullback of $\mathcal{E}$ to $\text{Cris}(X^{(1)}/S)$ by the morphism of crystalline topoi associated to the commutative diagram $$ \xymatrix{ X^{(1)} \ar[r] \ar[d] & X \ar[d] \\ S \ar[r]^{\Spec(\sigma)} & S } $$ Then we have the base change map \begin{equation} \label{equation-base-change-sigma} R\Gamma(\text{Cris}(X/S), \mathcal{E}) \otimes^\mathbf{L}_{A, \sigma} A \longrightarrow R\Gamma(\text{Cris}(X^{(1)}/S), \mathcal{E}^{(1)}) \end{equation} see Remark \ref{remark-base-change}. Note that the composition of $F_{X/S_0} : X \to X^{(1)}$ with the projection $X^{(1)} \to X$ is the absolute Frobenius morphism $F_X$. Hence we see that $F_{X/S_0}^*\mathcal{E}^{(1)} = (F_X)_{\text{cris}}^*\mathcal{E}$. Thus pullback by $F_{X/S_0}$ is a map \begin{equation} \label{equation-to-prove} F_{X/S_0}^* : R\Gamma(\text{Cris}(X^{(1)}/S), \mathcal{E}^{(1)}) \longrightarrow R\Gamma(\text{Cris}(X/S), (F_X)^*_{\text{cris}}\mathcal{E}) \end{equation} Finally we can use $F_\mathcal{E}$ to get a map \begin{equation} \label{equation-F-E} R\Gamma(\text{Cris}(X/S), (F_X)^*_{\text{cris}}\mathcal{E}) \longrightarrow R\Gamma(\text{Cris}(X/S), \mathcal{E}) \end{equation} The map of the theorem is the composition of the three maps (\ref{equation-base-change-sigma}), (\ref{equation-to-prove}), and (\ref{equation-F-E}) above. The first is a quasi-isomorphism modulo all powers of $p$ by Remark \ref{remark-base-change-isomorphism}. Hence it is a quasi-isomorphism since the complexes involved are perfect in $D(A)$ see Remark \ref{remark-complete-perfect}. The third map is a quasi-isomorphism after inverting $p$ simply because $F_\mathcal{E}$ has an inverse up to a power of $p$, see Remark \ref{remark-F-crystal-variants}. Finally, the second is an isomorphism after inverting $p$ by Lemma \ref{lemma-pullback-relative-frobenius}.
Denote $\mathcal{C}$ the category of divided power rings $(C, K, \delta)$. Consider the functor $F : \mathcal{C} \longrightarrow \textit{Sets}$ defined by $$ F(C, K, \delta) = \left\{ (\varphi, \psi) \middle| \begin{matrix} \varphi : (A, I, \gamma) \to (C, K, \delta) \text{ homomorphism of divided power rings} \\ \psi : (B, J) \to (C, K)\text{ an } A\text{-algebra homomorphism with }\psi(J) \subset K \end{matrix} \right\} $$ We will show that Divided Power Algebra, Lemma \ref{dpa-lemma-a-version-of-brown} applies to this functor which will prove the lemma. Suppose that $(\varphi, \psi) \in F(C, K, \delta)$. Let $C' \subset C$ be the subring generated by $\varphi(A)$, $\psi(B)$, and $\delta_n(\psi(f))$ for all $f \in J$. Let $K' \subset K \cap C'$ be the ideal of $C'$ generated by $\varphi(I)$ and $\delta_n(\psi(f))$ for $f \in J$. Then $(C', K', \delta|_{K'})$ is a divided power ring and $C'$ has cardinality bounded by the cardinal $\kappa = |A| \otimes |B|^{\aleph_0}$. Moreover, $\varphi$ factors as $A \to C' \to C$ and $\psi$ factors as $B \to C' \to C$. This proves assumption (1) of Divided Power Algebra, Lemma \ref{dpa-lemma-a-version-of-brown} holds. Assumption (2) is clear as limits in the category of divided power rings commute with the forgetful functor $(C, K, \delta) \mapsto (C, K)$, see Divided Power Algebra, Lemma \ref{dpa-lemma-limits} and its proof.
Write $D_{B, \gamma}(J) = (D, \bar J, \bar \gamma)$. The universal property of $D'$ gives us a homomorphism $D' \to D$ of divided power algebras. As $B' \to B$ and $J' \to J$ are surjective, we see that $D' \to D$ is surjective (see remarks above). It is clear that $\bar\gamma_n(k)$ is in the kernel for $n \geq 1$ and $k \in K$, i.e., we obtain a homomorphism $D'/K' \to D$. Conversely, there exists a divided power structure on $\bar J'/K' \subset D'/K'$, see Divided Power Algebra, Lemma \ref{dpa-lemma-kernel}. Hence the universal property of $D$ gives an inverse $D \to D'/K'$ and we win.
In the statement of the lemma we think of $B\langle x_t \rangle$ as a divided power ring with ideal $J' = IB\langle x_t \rangle + B\langle x_t \rangle_{+}$, see Divided Power Algebra, Remark \ref{dpa-remark-divided-power-polynomial-algebra}. The existence of $\Psi$ follows from the universal property of divided power polynomial rings. Surjectivity of $\Psi$ follows from the fact that its image is a divided power subring of $D$, hence equal to $D$ by the universal property of $D$. It is clear that $x_t - f_t$ is in the kernel. Set $$ \mathcal{R} = \{(r_0, r_t) \in I \oplus \bigoplus\nolimits_{t \in T} B \mid \sum r_t f_t = r_0 \text{ in }B\} $$ If $(r_0, r_t) \in \mathcal{R}$ then it is clear that $\sum r_t x_t - r_0$ is in the kernel. As $\Psi$ is a homomorphism of divided power rings and $\sum r_tx_t - r_0 \in J'$ it follows that $\delta_n(\sum r_t x_t - r_0)$ is in the kernel as well. Let $K \subset B\langle x_t \rangle$ be the ideal generated by $x_t - f_t$ and the elements $\delta_n(\sum r_t x_t - r_0)$ for $(r_0, r_t) \in \mathcal{R}$. To show that $K = \Ker(\Psi)$ it suffices to show that $\delta$ extends to $B\langle x_t \rangle/K$. Namely, if so the universal property of $D$ gives a map $D \to B\langle x_t \rangle/K$ inverse to $\Psi$. Hence we have to show that $K \cap J'$ is preserved by $\delta_n$, see Divided Power Algebra, Lemma \ref{dpa-lemma-kernel}. Let $K' \subset B\langle x_t \rangle$ be the ideal generated by the elements \begin{enumerate} \item $\delta_m(\sum r_t x_t - r_0)$ where $m > 0$ and $(r_0, r_t) \in \mathcal{R}$, \item $x_{t'}^{[m]}(x_t - f_t)$ where $m > 0$ and $t', t \in I$. \end{enumerate} We claim that $K' = K \cap J'$. The claim proves that $K \cap J'$ is preserved by $\delta_n$, $n > 0$ by the criterion of Divided Power Algebra, Lemma \ref{dpa-lemma-kernel} (2)(c) and a computation of $\delta_n$ of the elements listed which we leave to the reader. To prove the claim note that $K' \subset K \cap J'$. Conversely, if $h \in K \cap J'$ then, modulo $K'$ we can write $$ h = \sum r_t (x_t - f_t) $$ for some $r_t \in B$. As $h \in K \cap J' \subset J'$ we see that $r_0 = \sum r_t f_t \in I$. Hence $(r_0, r_t) \in \mathcal{R}$ and we see that $$ h = \sum r_t x_t - r_0 $$ is in $K'$ as desired.
Set $D = D_B(J)$ and denote $\bar J \subset D$ its divided power ideal with divided power structure $\bar\gamma$. The universal property of $D$ produces a $B$-algebra map $D \to D_{B'}(JB')$, whence a map as in the lemma. It suffices to show that the divided powers $\bar\gamma$ extend to $D \otimes_B B'$ since then the universal property of $D_{B'}(JB')$ will produce a map $D_{B'}(JB') \to D \otimes_B B'$ inverse to the one in the lemma. \medskip\noindent Choose a surjection $P \to B'$ where $P$ is a polynomial algebra over $B$. In particular $B \to P$ is flat, hence $D \to D \otimes_B P$ is flat by Algebra, Lemma \ref{algebra-lemma-flat-base-change}. Then $\bar\gamma$ extends to $D \otimes_B P$ by Divided Power Algebra, Lemma \ref{dpa-lemma-gamma-extends}; we will denote this extension $\bar\gamma$ also. Set $\mathfrak a = \Ker(P \to B')$ so that we have the short exact sequence $$ 0 \to \mathfrak a \to P \to B' \to 0 $$ Thus $\text{Tor}_1^B(B', B/IB) = 0$ implies that $\mathfrak a \cap IP = I\mathfrak a$. Now we have the following commutative diagram $$ \xymatrix{ B/J \otimes_B \mathfrak a \ar[r]_\beta & B/J \otimes_B P \ar[r] & B/J \otimes_B B' \\ D \otimes_B \mathfrak a \ar[r]^\alpha \ar[u] & D \otimes_B P \ar[r] \ar[u] & D \otimes_B B' \ar[u] \\ \bar J \otimes_B \mathfrak a \ar[r] \ar[u] & \bar J \otimes_B P \ar[r] \ar[u] & \bar J \otimes_B B' \ar[u] } $$ This diagram is exact even with $0$'s added at the top and the right. We have to show the divided powers on the ideal $\bar J \otimes_B P$ preserve the ideal $\Im(\alpha) \cap \bar J \otimes_B P$, see Divided Power Algebra, Lemma \ref{dpa-lemma-kernel}. Consider the exact sequence $$ 0 \to \mathfrak a/I\mathfrak a \to P/IP \to B'/IB' \to 0 $$ (which uses that $\mathfrak a \cap IP = I\mathfrak a$ as seen above). As $B'/IB'$ is flat over $B/IB$ this sequence remains exact after applying $B/J \otimes_{B/IB} -$, see Algebra, Lemma \ref{algebra-lemma-flat-tor-zero}. Hence $$ \Ker(B/J \otimes_{B/IB} \mathfrak a/I\mathfrak a \to B/J \otimes_{B/IB} P/IP) = \Ker(\mathfrak a/J\mathfrak a \to P/JP) $$ is zero. Thus $\beta$ is injective. It follows that $\Im(\alpha) \cap \bar J \otimes_B P$ is the image of $\bar J \otimes \mathfrak a$. Now if $f \in \bar J$ and $a \in \mathfrak a$, then $\bar\gamma_n(f \otimes a) = \bar\gamma_n(f) \otimes a^n$ hence the result is clear.