hcju/stacksqa huggingface dataset:
By assumption (1) there exists a scheme $T$ and an object $\xi$ of $\mathcal{Y}$ over $T$ such that the corresponding $1$-morphism $\xi : (\Sch/T)_{fppf} \to \mathcal{Y}$ is smooth an surjective. Then $\mathcal{U} = (\Sch/T)_{fppf} \times_{\xi, \mathcal{Y}} \mathcal{X}$ is an algebraic stack by assumption (2). Choose a scheme $U$ and a surjective smooth $1$-morphism $(\Sch/U)_{fppf} \to \mathcal{U}$. The projection $\mathcal{U} \longrightarrow \mathcal{X}$ is, as the base change of the morphism $\xi : (\Sch/T)_{fppf} \to \mathcal{Y}$, surjective and smooth, see Algebraic Stacks, Lemma \ref{algebraic-lemma-base-change-representable-transformations-property}. Then the composition $(\Sch/U)_{fppf} \to \mathcal{U} \to \mathcal{X}$ is surjective and smooth as a composition of surjective and smooth morphisms, see Algebraic Stacks, Lemma \ref{algebraic-lemma-composition-representable-transformations-property}. Hence $\mathcal{X}$ is an algebraic stack by Algebraic Stacks, Lemma \ref{algebraic-lemma-smooth-surjective-morphism-implies-algebraic}.
Choose a representable stack in groupoids $\mathcal{U}$ and a surjective smooth $1$-morphism $\mathcal{U} \to \mathcal{X}$. Let $T$ be a scheme and let $\xi$ be an object of $\mathcal{Y}$ over $T$. The morphism of $2$-fibre products $$ (\Sch/T)_{fppf} \times_{\xi, \mathcal{Y}} \mathcal{U} \longrightarrow (\Sch/T)_{fppf} \times_{\xi, \mathcal{Y}} \mathcal{X} $$ is representable by algebraic spaces, surjective, and smooth as a base change of $\mathcal{U} \to \mathcal{X}$, see Algebraic Stacks, Lemmas \ref{algebraic-lemma-base-change-representable-by-spaces} and \ref{algebraic-lemma-base-change-representable-transformations-property}. By our condition on the diagonal of $\mathcal{Y}$ we see that the source of this morphism is representable by an algebraic space, see Algebraic Stacks, Lemma \ref{algebraic-lemma-representable-diagonal}. Hence the target is an algebraic stack by Algebraic Stacks, Lemma \ref{algebraic-lemma-smooth-surjective-morphism-implies-algebraic}.
Assume $F$ is algebraic and $\Delta : \mathcal{Y} \to \mathcal{Y} \times \mathcal{Y}$ is representable by algebraic spaces. Take a scheme $U$ over $S$ and two objects $x_1, x_2$ of $\mathcal{X}$ over $U$. We have to show that $\mathit{Isom}(x_1, x_2)$ is an algebraic space over $U$, see Algebraic Stacks, Lemma \ref{algebraic-lemma-representable-diagonal}. Set $y_i = F(x_i)$. We have a morphism of sheaves of sets $$ f : \mathit{Isom}(x_1, x_2) \to \mathit{Isom}(y_1, y_2) $$ and the target is an algebraic space by assumption. Thus it suffices to show that $f$ is representable by algebraic spaces, see Bootstrap, Lemma \ref{bootstrap-lemma-representable-by-spaces-over-space}. Thus we can choose a scheme $V$ over $U$ and an isomorphism $\beta : y_{1, V} \to y_{2, V}$ and we have to show the functor $$ (\Sch/V)_{fppf} \to \textit{Sets},\quad T/V \mapsto \{\alpha : x_{1, T} \to x_{2, T} \text{ in }\mathcal{X}_T \mid F(\alpha) = \beta|_T\} $$ is an algebraic space. Consider the objects $z_1 = (V, x_{1, V}, \text{id})$ and  $z_2 = (V, x_{2, V}, \beta)$ of $$ \mathcal{Z} = (\Sch/V)_{fppf} \times_{y_{1, V}, \mathcal{Y}} \mathcal{X} $$ Then it is straightforward to verify that the functor above is equal to $\mathit{Isom}(z_1, z_2)$ on $(\Sch/V)_{fppf}$. Hence this is an algebraic space by our assumption that $F$ is algebraic (and the definition of algebraic stacks).
We may choose a separated scheme $W'$ and a surjective \'etale morphism $W' \to W$. Hence after replacing $W$ by $W'$ we may assume that $W$ is a separated scheme. Write $f : W \to Z$ and $\pi : Z \to U$. Note that $f \circ \pi : W \to U$ is separated as $W$ is separated (see Schemes, Lemma \ref{schemes-lemma-compose-after-separated}). Let $u \in U$ be a point. Clearly it suffices to find an \'etale neighbourhood $(U', u')$ of $(U, u)$ such that a section $\sigma$ exists over $U'$. Let $z_1, \ldots, z_r$ be the points of $Z$ lying above $u$. For each $i$ choose a point $w_i \in W$ which maps to $z_i$. We may pick an \'etale neighbourhood $(U', u') \to (U, u)$ such that the conclusions of More on Morphisms, Lemma \ref{more-morphisms-lemma-etale-splits-off-quasi-finite-part-technical-variant} hold for both $Z \to U$ and the points $z_1, \ldots, z_r$ and $W \to U$ and the points $w_1, \ldots, w_r$. Hence, after replacing $(U, u)$ by $(U', u')$ and relabeling, we may assume that all the field extensions $\kappa(u) \subset \kappa(z_i)$ and $\kappa(u) \subset \kappa(w_i)$ are purely inseparable, and moreover that there exist disjoint union decompositions $$ Z = V_1 \amalg \ldots \amalg V_r \amalg A, \quad W = W_1 \amalg \ldots \amalg W_r \amalg B $$ by open and closed subschemes with $z_i \in V_i$, $w_i \in W_i$ and $V_i \to U$, $W_i \to U$ finite. After replacing $U$ by $U \setminus \pi(A)$ we may assume that $A = \emptyset$, i.e., $Z = V_1 \amalg \ldots \amalg V_r$. After replacing $W_i$ by $W_i \cap f^{-1}(V_i)$ and $B$ by $B \cup \bigcup W_i \cap f^{-1}(Z \setminus V_i)$ we may assume that $f$ maps $W_i$ into $V_i$. Then $f_i = f|_{W_i} : W_i \to V_i$ is a morphism of schemes finite over $U$, hence finite (see Morphisms, Lemma \ref{morphisms-lemma-finite-permanence}). It is also \'etale (by assumption), $f_i^{-1}(\{z_i\}) = w_i$, and induces an isomorphism of residue fields $\kappa(z_i) = \kappa(w_i)$ (because both are purely inseparable extensions of $\kappa(u)$ and $\kappa(z_i) \subset \kappa(w_i)$ is separable as $f$ is \'etale). Hence by \'Etale Morphisms, Lemma \ref{etale-lemma-finite-etale-one-point} we see that $f_i$ is an isomorphism in a neighbourhood $V_i'$ of $z_i$. Since $\pi : Z \to U$ is closed, after shrinking $U$, we may assume that $W_i \to V_i$ is an isomorphism. This proves the lemma.
Let $T$ be an algebraic space over $S$. Let $\{T_i \to T\}$ be an fppf covering of $T$ (as in Topologies on Spaces, Section \ref{spaces-topologies-section-fppf}). Suppose that $(a_i, b_i) \in \mathit{Mor}_B(Z, X)(T_i)$ such that $(a_i, b_i)|_{T_i \times_T T_j} = (a_j, b_j)|_{T_i \times_T T_j}$ for all $i, j$. Then by Descent on Spaces, Lemma \ref{spaces-descent-lemma-fpqc-universal-effective-epimorphisms} there exists a unique morphism $a : T \to B$ such that $a_i$ is the composition of $T_i \to T$ and $a$. Then $\{T_i \times_{a_i, B} Z \to T \times_{a, B} Z\}$ is an fppf covering too and the same lemma implies there exists a unique morphism $b : T \times_{a, B} Z \to T \times_{a, B} X$ such that $b_i$ is the composition of $T_i \times_{a_i, B} Z \to T \times_{a, B} Z$ and $b$. Hence $(a, b) \in \mathit{Mor}_B(Z, X)(T)$ restricts to $(a_i, b_i)$ over $T_i$ for all $i$. \medskip\noindent Note that the result of the preceding paragraph in particular implies (1). \medskip\noindent Let $T$ be an algebraic space over $S$. In order to prove (2) we will construct mutually inverse maps between the displayed sets. In the following when we say ``pair'' we mean a pair $(a, b)$ fitting into (\ref{equation-hom}). \medskip\noindent Let $v : T \to \mathit{Mor}_B(Z, X)$ be a natural transformation. Choose a scheme $U$ and a surjective \'etale morphism $p : U \to T$. Then $v(p) \in \mathit{Mor}_B(Z, X)(U)$ corresponds to a pair $(a_U, b_U)$ over $U$. Let $R = U \times_T U$ with projections $t, s : R \to U$. As $v$ is a transformation of functors we see that the pullbacks of $(a_U, b_U)$ by $s$ and $t$ agree. Hence, since $\{U \to T\}$ is an fppf covering, we may apply the result of the first paragraph that deduce that there exists a unique pair $(a, b)$ over $T$. \medskip\noindent Conversely, let $(a, b)$ be a pair over $T$. Let $U \to T$, $R = U \times_T U$, and $t, s : R \to U$ be as above. Then the restriction $(a, b)|_U$ gives rise to a transformation of functors $v : h_U \to \mathit{Mor}_B(Z, X)$ by the Yoneda lemma (Categories, Lemma \ref{categories-lemma-yoneda}). As the two pullbacks $s^*(a, b)|_U$ and $t^*(a, b)|_U$ are equal, we see that $v$ coequalizes the two maps $h_t, h_s : h_R \to h_U$. Since $T = U/R$ is the fppf quotient sheaf by Spaces, Lemma \ref{spaces-lemma-space-presentation} and since $\mathit{Mor}_B(Z, X)$ is an fppf sheaf by (1) we conclude that $v$ factors through a map $T \to \mathit{Mor}_B(Z, X)$. \medskip\noindent We omit the verification that the two constructions above are mutually inverse.
Let $U$ be a scheme and let $\xi = (a, b)$ be an element of $\mathit{Mor}_B(Z, X)(U)$. We have to prove that the functor $$ h_U \times_{\xi, \mathit{Mor}_B(Z, X)} \mathit{Mor}_B(Z, X') $$ is representable by an algebraic space \'etale over $U$. Set $Z_U = U \times_{a, B} Z$ and $W = Z_U \times_{b, X} X'$. Then $W \to Z_U \to U$ is as in Lemma \ref{lemma-space-of-sections} and the sheaf $F$ defined there is identified with the fibre product displayed above. Hence the first assertion of the lemma. The second assertion follows from this and Lemma \ref{lemma-surjection-space-of-sections} which guarantees that $F \to U$ is surjective in the situation above.
Let $T$ be an algebraic space over $S$. Let $\{T_i \to T\}$ be an fppf covering of $T$ (as in Topologies on Spaces, Section \ref{spaces-topologies-section-fppf}). Suppose that $(a_i, b_i) \in \text{Res}_{Z/B}(X)(T_i)$ such that $(a_i, b_i)|_{T_i \times_T T_j} = (a_j, b_j)|_{T_i \times_T T_j}$ for all $i, j$. Then by Descent on Spaces, Lemma \ref{spaces-descent-lemma-fpqc-universal-effective-epimorphisms} there exists a unique morphism $a : T \to B$ such that $a_i$ is the composition of $T_i \to T$ and $a$. Then $\{T_i \times_{a_i, B} Z \to T \times_{a, B} Z\}$ is an fppf covering too and the same lemma implies there exists a unique morphism $b : T \times_{a, B} Z \to X$ such that $b_i$ is the composition of $T_i \times_{a_i, B} Z \to T \times_{a, B} Z$ and $b$. Hence $(a, b) \in \text{Res}_{Z/B}(X)(T)$ restricts to $(a_i, b_i)$ over $T_i$ for all $i$. \medskip\noindent Note that the result of the preceding paragraph in particular implies (1). \medskip\noindent Let $T$ be an algebraic space over $S$. In order to prove (2) we will construct mutually inverse maps between the displayed sets. In the following when we say ``pair'' we mean a pair $(a, b)$ fitting into (\ref{equation-pairs}). \medskip\noindent Let $v : T \to \text{Res}_{Z/B}(X)$ be a natural transformation. Choose a scheme $U$ and a surjective \'etale morphism $p : U \to T$. Then $v(p) \in \text{Res}_{Z/B}(X)(U)$ corresponds to a pair $(a_U, b_U)$ over $U$. Let $R = U \times_T U$ with projections $t, s : R \to U$. As $v$ is a transformation of functors we see that the pullbacks of $(a_U, b_U)$ by $s$ and $t$ agree. Hence, since $\{U \to T\}$ is an fppf covering, we may apply the result of the first paragraph that deduce that there exists a unique pair $(a, b)$ over $T$. \medskip\noindent Conversely, let $(a, b)$ be a pair over $T$. Let $U \to T$, $R = U \times_T U$, and $t, s : R \to U$ be as above. Then the restriction $(a, b)|_U$ gives rise to a transformation of functors $v : h_U \to \text{Res}_{Z/B}(X)$ by the Yoneda lemma (Categories, Lemma \ref{categories-lemma-yoneda}). As the two pullbacks $s^*(a, b)|_U$ and $t^*(a, b)|_U$ are equal, we see that $v$ coequalizes the two maps $h_t, h_s : h_R \to h_U$. Since $T = U/R$ is the fppf quotient sheaf by Spaces, Lemma \ref{spaces-lemma-space-presentation} and since $\text{Res}_{Z/B}(X)$ is an fppf sheaf by (1) we conclude that $v$ factors through a map $T \to \text{Res}_{Z/B}(X)$. \medskip\noindent We omit the verification that the two constructions above are mutually inverse.
Let $U$ be a scheme and let $\xi = (a, b)$ be an element of $\text{Res}_{Z/B}(X)(U)$. We have to prove that the functor $$ h_U \times_{\xi, \text{Res}_{Z/B}(X)} \text{Res}_{Z/B}(X') $$ is representable by an algebraic space \'etale over $U$. Set $Z_U = U \times_{a, B} Z$ and $W = Z_U \times_{b, X} X'$. Then $W \to Z_U \to U$ is as in Lemma \ref{lemma-space-of-sections} and the sheaf $F$ defined there is identified with the fibre product displayed above. Hence the first assertion of the lemma. The second assertion follows from this and Lemma \ref{lemma-surjection-space-of-sections} which guarantees that $F \to U$ is surjective in the situation above.
We map the object $(U, Z, y', x', \alpha')$ to the object $(U, Z, H(y'), G(x'), \gamma \star \text{id}_H \star \alpha')$ where $\star$ denotes horizontal composition of $2$-morphisms, see Categories, Definition \ref{categories-definition-horizontal-composition}. To a morphism $(f, g, b, a) : (U_1, Z_1, y_1', x_1', \alpha_1') \to (U_2, Z_2, y_2', x_2', \alpha_2')$ we assign $(f, g, H(b), G(a))$. We omit the verification that this defines a functor between categories over $(\Sch/S)_{fppf}$.
We get a $2$-commutative diagram by Lemma \ref{lemma-map-hilbert} and hence we get a $1$-morphism (i.e., a functor) $$ \mathcal{H}_d(\mathcal{X}'/\mathcal{Y}') \longrightarrow \mathcal{Y}' \times_\mathcal{Y} \mathcal{H}_d(\mathcal{X}/\mathcal{Y}) $$ We indicate why this functor is essentially surjective. Namely, an object of the category on the right hand side is given by a scheme $U$ over $S$, an object $y'$ of $\mathcal{Y}'_U$, an object $(U, Z, y, x, \alpha)$ of $\mathcal{H}_d(\mathcal{X}/\mathcal{Y})$ over $U$ and an isomorphism $H(y') \to y$ in $\mathcal{Y}_U$. The assumption means exactly that there exists an object $x'$ of $\mathcal{X}'_Z$ such that there exist isomorphisms $G(x') \cong x$ and $\alpha' : y'|_Z \to F'(x')$ compatible with $\alpha$. Then we see that $(U, Z, y', x', \alpha')$ is an object of $\mathcal{H}_d(\mathcal{X}'/\mathcal{Y}')$ over $U$. Details omitted.
Let $U$ be a scheme and let $\xi = (U, Z, y, x, \alpha)$ be an object of $\mathcal{H}_d(\mathcal{X}/\mathcal{Y})$ over $U$. We have to prove that the $2$-fibre product \begin{equation} \label{equation-to-show} (\Sch/U)_{fppf} \times_{\xi, \mathcal{H}_d(\mathcal{X}/\mathcal{Y})} \mathcal{H}_d(\mathcal{X}'/\mathcal{Y}) \end{equation} is representable by an algebraic space \'etale over $U$. An object of this over $U'$ corresponds to an object $x'$ in the fibre category of $\mathcal{X}'$ over $Z_{U'}$ such that $G(x') \cong x|_{Z_{U'}}$. By assumption the $2$-fibre product $$ (\Sch/Z)_{fppf} \times_{x, \mathcal{X}} \mathcal{X}' $$ is representable by an algebraic space $W$ such that the projection $W \to Z$ is \'etale. Then (\ref{equation-to-show}) is representable by the algebraic space $F$ parametrizing sections of $W \to Z$ over $U$ introduced in Lemma \ref{lemma-space-of-sections}. Since $F \to U$ is \'etale we conclude that $\mathcal{H}_d(\mathcal{X}'/\mathcal{Y}) \to \mathcal{H}_d(\mathcal{X}/\mathcal{Y})$ is representable by algebraic spaces and \'etale. Finally, if $\mathcal{X}' \to \mathcal{X}$ is surjective also, then $W \to Z$ is surjective, and hence $F \to U$ is surjective by Lemma \ref{lemma-surjection-space-of-sections}. Thus in this case $\mathcal{H}_d(\mathcal{X}'/\mathcal{Y}) \to \mathcal{H}_d(\mathcal{X}/\mathcal{Y})$ is also surjective.
Set $\mathcal{X}'' = \mathcal{Y}' \times_\mathcal{Y} \mathcal{X}$. By Lemma \ref{lemma-etale-permanence} the $1$-morphism $\mathcal{X}' \to \mathcal{X}''$ is representable by algebraic spaces and \'etale (in particular the condition in the second statement of the lemma that $\mathcal{X}' \to \mathcal{X}''$ be surjective makes sense). We obtain a $2$-commutative diagram $$ \xymatrix{ \mathcal{X}' \ar[r] \ar[d] & \mathcal{X}'' \ar[r] \ar[d] & \mathcal{X} \ar[d] \\ \mathcal{Y}' \ar[r] & \mathcal{Y}' \ar[r] & \mathcal{Y} } $$ It follows from Lemma \ref{lemma-cartesian-map-hilbert} that $\mathcal{H}_d(\mathcal{X}''/\mathcal{Y}')$ is the base change of $\mathcal{H}_d(\mathcal{X}/\mathcal{Y})$ by $\mathcal{Y}' \to \mathcal{Y}$. In particular we see that $\mathcal{H}_d(\mathcal{X}''/\mathcal{Y}') \to \mathcal{H}_d(\mathcal{X}/\mathcal{Y})$ is representable by algebraic spaces and \'etale, see Algebraic Stacks, Lemma \ref{algebraic-lemma-base-change-representable-transformations-property}. Moreover, it is also surjective if $H$ is.  Hence if we can show that the result holds for the left square in the diagram, then we're done. In this way we reduce to the case where $\mathcal{Y}' = \mathcal{Y}$ which is the content of Lemma \ref{lemma-etale-covering-hilbert}.
Let $U$ be a scheme and let $\xi = (U, Z, p, x, 1)$ be an object of $\mathcal{H}_d(\mathcal{X}) = \mathcal{H}_d(\mathcal{X}/S)$ over $U$. Here $p$ is just the structure morphism of $U$. The fifth component $1$ exists and is unique since everything is over $S$. Also, let $y$ be an object of $\mathcal{Y}$ over $U$. We have to show the $2$-fibre product \begin{equation} \label{equation-res-isom} (\Sch/U)_{fppf} \times_{\xi \times y, \mathcal{H}_d(\mathcal{X}) \times \mathcal{Y}} \mathcal{H}_d(\mathcal{X}/\mathcal{Y}) \end{equation} is representable by an algebraic space. To explain why this is so we introduce $$ I = \mathit{Isom}_\mathcal{Y}(y|_Z, F(x)) $$ which is an algebraic space over $Z$ by assumption. Let $a : U' \to U$ be a scheme over $U$. What does it mean to give an object of the fibre category of (\ref{equation-res-isom}) over $U'$? Well, it means that we have an object $\xi' = (U', Z', y', x', \alpha')$ of $\mathcal{H}_d(\mathcal{X}/\mathcal{Y})$ over $U'$ and isomorphisms $(U', Z', p', x', 1) \cong (U, Z, p, x, 1)|_{U'}$ and $y' \cong y|_{U'}$. Thus $\xi'$ is isomorphic to $(U', U' \times_{a, U} Z, a^*y, x|_{U' \times_{a, U} Z}, \alpha)$ for some morphism $$ \alpha : a^*y|_{U' \times_{a, U} Z} \longrightarrow F(x|_{U' \times_{a, U} Z}) $$ in the fibre category of $\mathcal{Y}$ over $U' \times_{a, U} Z$. Hence we can view $\alpha$ as a morphism $b : U' \times_{a, U} Z \to I$. In this way we see that (\ref{equation-res-isom}) is representable by $\text{Res}_{Z/U}(I)$ which is an algebraic space by Proposition \ref{proposition-restriction-of-scalars-algebraic-space}.
Let $U$ be a scheme and let $\xi = (U, Z, y, x, \alpha)$ be an object of $\mathcal{H}_d(\mathcal{X}/\mathcal{Y})$ over $U$. We have to prove that the $2$-fibre product \begin{equation} \label{equation-to-show-again} (\Sch/U)_{fppf} \times_{\xi, \mathcal{H}_d(\mathcal{X}/\mathcal{Y})} \mathcal{H}_d(\mathcal{X}'/\mathcal{Y}) \end{equation} is representable by an algebraic space \'etale over $U$. An object of this over $a : U' \to U$ corresponds to an object $x'$ of $\mathcal{X}'$ over $U' \times_{a, U} Z$ such that $G(x') \cong x|_{U' \times_{a, U} Z}$. By assumption the $2$-fibre product $$ (\Sch/Z)_{fppf} \times_{x, \mathcal{X}} \mathcal{X}' $$ is representable by an algebraic space $X$ over $Z$. It follows that (\ref{equation-to-show-again}) is representable by $\text{Res}_{Z/U}(X)$, which is an algebraic space by Proposition \ref{proposition-restriction-of-scalars-algebraic-space}.
The $1$-morphism $$ \mathcal{H}_d(X) \longrightarrow \mathcal{H}_d $$ is representable by algebraic spaces according to Lemma \ref{lemma-representable-on-top}. The stack $\mathcal{H}_d$ is an algebraic stack according to Proposition \ref{proposition-finite-hilbert-point}. Hence $\mathcal{H}_d(X)$ is an algebraic stack by Algebraic Stacks, Lemma \ref{algebraic-lemma-representable-morphism-to-algebraic}.
Let $\Xi = (U, Z, y, x, \alpha)$ be an object of $\mathcal{H}_d$. It follows from the remark following (\ref{equation-relative-map}) that the pullback of $\Xi$ by $U' \to U$ belongs to $\mathcal{H}_{d, lci}(\mathcal{X}/\mathcal{Y})$ if and only if the base change of $x_\alpha$ is unramified and a local complete intersection morphism. Note that $Z \to U$ is finite locally free (hence flat, locally of finite presentation and universally closed) and that $X_y \to U$ is flat and locally of finite presentation by our assumption on $F$. Then More on Morphisms of Spaces, Lemmas \ref{spaces-more-morphisms-lemma-where-unramified} and \ref{spaces-more-morphisms-lemma-where-lci} imply exists an open subscheme $W \subset U$ such that a morphism $U' \to U$ factors through $W$ if and only if the base change of $x_\alpha$ via $U' \to U$ is unramified and a local complete intersection morphism. This implies that $$ (\Sch/U)_{fppf} \times_{\Xi, \mathcal{H}_d(\mathcal{X}/\mathcal{Y})} \mathcal{H}_{d, lci}(\mathcal{X}/\mathcal{Y}) $$ is representable by $W$. Hence the final statement of the lemma holds. The first statement (that $\mathcal{H}_{d, lci}(\mathcal{X}/\mathcal{Y})$ is a stack in groupoids) follows from this and Algebraic Stacks, Lemma \ref{algebraic-lemma-open-fibred-category-is-algebraic}.
By More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-unramified-lci} the conormal sheaf $\mathcal{C}_{Z/X}$ of the unramified morphism $Z \to X$ is a finite locally free $\mathcal{O}_Z$-module and by More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-transitivity-conormal-lci} we have an exact sequence $$ 0 \to i^*\mathcal{C}_{X/X'} \to \mathcal{C}_{Z/X'} \to \mathcal{C}_{Z/X} \to 0 $$ of conormal sheaves. Since $Z$ is affine this sequence is split. Choose a splitting $$ \mathcal{C}_{Z/X'} = i^*\mathcal{C}_{X/X'} \oplus \mathcal{C}_{Z/X} $$ Let $Z \subset Z''$ be the universal first order thickening of $Z$ over $X'$ (see More on Morphisms of Spaces, Section \ref{spaces-more-morphisms-section-universal-thickening}). Denote $\mathcal{I} \subset \mathcal{O}_{Z''}$ the quasi-coherent sheaf of ideals corresponding to $Z \subset Z''$. By definition we have $\mathcal{C}_{Z/X'}$ is $\mathcal{I}$ viewed as a sheaf on $Z$. Hence the splitting above determines a splitting $$ \mathcal{I} = i^*\mathcal{C}_{X/X'} \oplus \mathcal{C}_{Z/X} $$ Let $Z' \subset Z''$ be the closed subscheme cut out by $\mathcal{C}_{Z/X} \subset \mathcal{I}$ viewed as a quasi-coherent sheaf of ideals on $Z''$. It is clear that $Z'$ is a first order thickening of $Z$ and that we obtain a commutative diagram of first order thickenings as in the statement of the lemma. \medskip\noindent Since $X' \to U'$ is flat and since $X = U \times_{U'} X'$ we see that $\mathcal{C}_{X/X'}$ is the pullback of $\mathcal{C}_{U/U'}$ to $X$, see More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-deform}. Note that by construction $\mathcal{C}_{Z/Z'} = i^*\mathcal{C}_{X/X'}$ hence we conclude that $\mathcal{C}_{Z/Z'}$ is isomorphic to the pullback of $\mathcal{C}_{U/U'}$ to $Z$. Applying More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-deform} once again (or its analogue for schemes, see More on Morphisms, Lemma \ref{more-morphisms-lemma-deform}) we conclude that $Z' \to U'$ is flat and that $Z = U \times_{U'} Z'$. Finally, More on Morphisms, Lemma \ref{more-morphisms-lemma-deform-property} shows that $Z' \to U'$ is finite locally free of degree $d$.
We have to show the following: Given \begin{enumerate} \item an object $(U, Z, y, x, \alpha)$ of $\mathcal{H}_{d, lci}(\mathcal{X}/\mathcal{Y})$ over an affine scheme $U$, \item a first order thickening $U \subset U'$, and \item an object $y'$ of $\mathcal{Y}$ over $U'$ such that $y'|_U = y$, \end{enumerate} then there exists an object $(U', Z', y', x', \alpha')$ of $\mathcal{H}_{d, lci}(\mathcal{X}/\mathcal{Y})$ over $U'$ with $Z = U \times_{U'} Z'$, with $x = x'|_Z$, and with $\alpha = \alpha'|_U$. Namely, the last two equalities will take care of the commutativity of (\ref{equation-formally-smooth}). \medskip\noindent Consider the morphism $x_\alpha : Z \to X_y$ constructed in Equation (\ref{equation-relative-map}). Denote similarly $X'_{y'}$ the algebraic space over $U'$ representing the $2$-fibre product $(\Sch/U')_{fppf} \times_{y', \mathcal{Y}, F} \mathcal{X}$. By assumption the morphism $X'_{y'} \to U'$ is flat (and locally of finite presentation). As $y'|_U = y$ we see that $X_y = U \times_{U'} X'_{y'}$. Hence we may apply Lemma \ref{lemma-lci-unobstructed} to find $Z' \to U'$ finite locally free of degree $d$ with $Z = U \times_{U'} Z'$ and with $Z' \to X'_{y'}$ extending $x_\alpha$. By construction the morphism $Z' \to X'_{y'}$ corresponds to a pair $(x', \alpha')$. It is clear that $(U', Z', y', x', \alpha')$ is an object of $\mathcal{H}_d(\mathcal{X}/\mathcal{Y})$ over $U'$ with $Z = U \times_{U'} Z'$, with $x = x'|_Z$, and with $\alpha = \alpha'|_U$. As we've seen in Lemma \ref{lemma-lci-locus-stack-in-groupoids} that $\mathcal{H}_{d, lci}(\mathcal{X}/\mathcal{Y}) \subset \mathcal{H}_d(\mathcal{X}/\mathcal{Y})$ is an ``open substack'' it follows that $(U', Z', y', x', \alpha')$ is an object of $\mathcal{H}_{d, lci}(\mathcal{X}/\mathcal{Y})$ as desired.