hcju/stacksqa huggingface dataset:
To see this we extend the bottom of the diagram (\ref{equation-pull}) as follows $$ \xymatrix{ Y \ar[r] \ar[d] & R \times_{s, U, t} R \ar@<1ex>[r]^-c \ar@<-1ex>[r]_-{\text{pr}_0} \ar[d]_{\text{pr}_1} & R \ar[r]^t \ar[d]^s & U \\ U \ar[r]_\delta & R \ar@<1ex>[r]^s \ar@<-1ex>[r]_t & U } $$ where the left square is cartesian and this is our definition of $Y$; we will not need to know more about $Y$. There is a similar diagram with similar properties obtained by base change to $U_0$ everywhere. We are trying to show that $\text{id}_U = s \circ \delta$ and $f = t \circ \delta$ induce the same maps on conormal sheaves. Since $s$ is flat and surjective, it suffices to prove the same thing for the two compositions $a, b : Y \to R$ along the top row. Observe that $a_0 = b_0$ and that one of $a$ and $b$ is an isomorphism as we know that $s \circ \delta$ is an isomorphism. Therefore the two morphisms $a, b : Y \to R$ are morphisms between algebraic spaces flat over $U$ (via the morphism $t : R \to U$ and the morphism $t \circ a = t \circ b : Y \to U$). This implies what we want. Namely, by the compatibility with compositions in More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-conormal-functorial-more} we conclude that both maps $a_0^*\mathcal{C}_{R_0/R} \to \mathcal{C}_{Y_0/Y}$ fit into a commutative diagram $$ \xymatrix{ a_0^*\mathcal{C}_{R_0/R} \ar[rr] & & \mathcal{C}_{Y_0/Y} \\ a_0^*t_0^*\mathcal{C}_{U_0/U} \ar[u] \ar@{=}[rr] & & (t_0 \circ a_0)^*\mathcal{C}_{U_0/U} \ar[u] } $$ whose vertical arrows are isomorphisms by More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-deform}. Thus the lemma holds.
Let $\delta_1, \delta_2 \in \Gamma_0$ correspond to $\theta_1, \theta_2$ as above and the composition $\delta = \delta_1 \circ \delta_2$ in $\Gamma_0$ correspond to $\theta$. We have to show that $\theta = \theta_1 + \theta_2$. Recall (More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-action-by-derivations}) that $\theta_1, \theta_2, \theta$ correspond to derivations $D_1, D_2, D : e_0^{-1}\mathcal{O}_{R_0} \to \mathcal{C}_{U_0/U}$ given by $D_1 = \theta_1 \circ \text{d}_{R_0/U_0}$ and so on. It suffices to check that $D = D_1 + D_2$. \medskip\noindent We may check equality on stalks. Let $\overline{u}$ be a geometric point of $U$ and let us use the local rings $A, B, C$ introduced in Section \ref{section-local}. The morphisms $\delta_i$ correspond to ring maps $\delta_i : B \to A$. Let $K \subset A$ be the ideal of square zero such that $A/K = \mathcal{O}_{U_0, \overline{u}}$. In other words, $K$ is the stalk of $\mathcal{C}_{U_0/U}$ at $\overline{u}$. The fact that $\delta_i \in \Gamma_0$ means exactly that $\delta_i(I) \subset K$. The derivation $D_i$ is just the map $\delta_i - e : B \to A$. Since $B = s(A) \oplus I$ we see that $D_i$ is determined by its restriction to $I$ and that this is just given by $\delta_i|_I$. Moreover $D_i$ and hence $\delta_i$ annihilates $I^2$ because $I = \Ker(I)$. \medskip\noindent To finish the proof we observe that $\delta$ corresponds to the composition $$ B \to C = (B \otimes_{s, A, t} B)^h_{\mathfrak m_B \otimes B + B \otimes \mathfrak m_B} \to A $$ where the first arrow is $c$ and the second arrow is determined by the rule $b_1 \otimes b_2 \mapsto \delta_2(t(\delta_1(b_1))) \delta_2(b_2)$ as follows from (\ref{equation-composition}). By Lemma \ref{lemma-first-order-structure-c} we see that an element $\zeta$ of $I$ maps to $\zeta \otimes 1 + 1 \otimes \zeta$ plus higher order terms. Hence we conclude that $$ D(\zeta) = (\delta_2 \circ t)\left(D_1(\zeta)\right) + D_2(\zeta) $$ However, by Lemma \ref{lemma-idenity-on-conormal} the action of $\delta_2 \circ t$ on $K = \mathcal{C}_{U_0/U, \overline{u}}$ is the identity and we win.
The existence and properties of the open $W \subset U$ are described in Descent on Spaces, Lemma \ref{spaces-descent-lemma-largest-open-of-the-base}. In Diagram (\ref{equation-diagram}) let $W_1 \subset R$ be the maximal open subscheme over which the morphism $\text{pr}_1 : R \times_{s, U, t} R \to R$ has property $\mathcal{P}$. It follows from the aforementioned Descent on Spaces, Lemma \ref{spaces-descent-lemma-largest-open-of-the-base} and the assumption that $\{s : R \to U\}$ and $\{t : R \to U\}$ are coverings for the $\tau$-topology that $t^{-1}(W) = W_1 = s^{-1}(W)$ as desired.
The existence and properties of the open $W \subset U$ are described in Descent on Spaces, Lemma \ref{spaces-descent-lemma-largest-open-of-the-base}. The morphism $$ G \times_{U, t} R \longrightarrow R \times_{s, U} G, \quad (g, r) \longmapsto (r, r^{-1} \circ g \circ r) $$ is an isomorphism of algebraic spaces over $R$ (where $\circ$ denotes composition in the groupoid). Hence $s^{-1}(W) = t^{-1}(W)$ by the properties of $W$ proved in the aforementioned Descent on Spaces, Lemma \ref{spaces-descent-lemma-largest-open-of-the-base}.
The property of being locally of finite type is stable under composition and arbitrary base change, see Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-composition-finite-type} and \ref{spaces-morphisms-lemma-base-change-finite-type}. Hence (1) is clear from Diagram (\ref{equation-restriction}). For the other cases, see Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-composition-finite-presentation}, \ref{spaces-morphisms-lemma-base-change-finite-presentation}, \ref{spaces-morphisms-lemma-composition-flat}, and \ref{spaces-morphisms-lemma-base-change-flat}.
The composition is isomorphic to the projection map $\text{pr}_1 : R \times_{t, U, t} R \to R$ by Diagram (\ref{equation-pull}). The projection is flat as a base change of the flat morphism $t$ and open by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-space-over-field-universally-open}. The second assertion follows immediately from the first because $m$ matches $c$ in (\ref{equation-groupoid-from-group}).
We first prove the second assertion. By Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-group-scheme-separated} we have to show that $e : S \to G$ is a closed immersion. This follows from Decent Spaces, Lemma \ref{decent-spaces-lemma-finite-residue-field-extension-finite}. \medskip\noindent Next, we prove the first assertion. To do this we may replace $B$ by $S$. By the paragraph above the stabilizer group scheme $G \to U$ is separated. By Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-diagonal} the morphism $j = (t, s) : R \to U \times_S U$ is separated. As $U$ is the spectrum of a field the scheme $U \times_S U$ is affine (by the construction of fibre products in Schemes, Section \ref{schemes-section-fibre-products}). Hence $R$ is separated, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-separated-over-separated}.
The morphism $U' \to U$ equals $\Spec(k') \to \Spec(k)$, hence is affine, surjective and flat. The morphisms $s, t : R \to U$ and the morphism $U' \to U$ are universally open by Morphisms, Lemma \ref{morphisms-lemma-scheme-over-field-universally-open}. Since $R$ is not empty and $U$ is the spectrum of a field the morphisms $s, t : R \to U$ are surjective and flat. Then you conclude by using Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-base-change-surjective}, \ref{spaces-morphisms-lemma-composition-surjective}, \ref{spaces-morphisms-lemma-composition-open}, \ref{spaces-morphisms-lemma-base-change-affine}, \ref{spaces-morphisms-lemma-composition-affine}, \ref{spaces-morphisms-lemma-base-change-flat}, and \ref{spaces-morphisms-lemma-composition-flat}.
Let's represent $r$ by a morphism $r : \Spec(K) \to R$ for some field $K$. To prove the lemma we have to find an algebraically closed field $k'$ and a commutative diagram $$ \xymatrix{ k' & k' \ar[l]^1 & \\ k' \ar[u]^\tau & K \ar[lu]^\sigma & k \ar[l]^-s \ar[lu]_i \\ & k \ar[lu]^i \ar[u]_t } $$ where $s, t : k \to K$ are the field maps coming from $s \circ r$ and $t \circ r$. In the proof of More on Groupoids, Lemma \ref{more-groupoids-lemma-groupoid-on-field-explain-points} it is shown how to construct such a diagram.
Let $r \in |R|$. We have to show that $R$ is geometrically unibranch at $r$. Combining Lemma \ref{lemma-restrict-groupoid-on-field} with Descent on Spaces, Lemma \ref{spaces-descent-lemma-descend-unibranch} we see that it suffices to prove this in case $k$ is algebraically closed and $r$ comes from a morphism $r : \Spec(k) \to R$ such that $s \circ r$ and $t \circ r$ are automorphisms of $\Spec(k)$. By Lemma \ref{lemma-groupoid-on-field-move-point} we reduce to the case that $r = e$ is the identity of $R$ and $k$ is algebraically closed. \medskip\noindent Assume $r = e$ and $k$ is algebraically closed. Let $A = \mathcal{O}_{R, e}$ be the \'etale local ring of $R$ at $e$ and let $C = \mathcal{O}_{R \times_{s, U, t} R, (e, e)}$ be the \'etale local ring of $R \times_{s, U, t} R$ at $(e, e)$. By More on Algebra, Lemma \ref{more-algebra-lemma-minimal-primes-tensor-strictly-henselian} the minimal prime ideals $\mathfrak q$ of $C$ correspond $1$-to-$1$ to pairs of minimal primes $\mathfrak p, \mathfrak p' \subset A$. On the other hand, the composition law induces a flat ring map $$ \xymatrix{ A \ar[r]_{c^\sharp} & C & \mathfrak q \\ & A \otimes_{s^\sharp, k, t^\sharp} A \ar[u] & \mathfrak p \otimes A + A \otimes \mathfrak p' \ar@{|}[u] } $$ Note that $(c^\sharp)^{-1}(\mathfrak q)$ contains both $\mathfrak p$ and $\mathfrak p'$ as the diagrams $$ \xymatrix{ A \ar[r]_{c^\sharp} & C \\ A \otimes_{s^\sharp, k} k \ar[u] & A \otimes_{s^\sharp, k, t^\sharp} A \ar[l]_{1 \otimes e^\sharp} \ar[u] } \quad\quad \xymatrix{ A \ar[r]_{c^\sharp} & C \\ k \otimes_{k, t^\sharp} A \ar[u] & A \otimes_{s^\sharp, k, t^\sharp} A \ar[l]_{e^\sharp \otimes 1} \ar[u] } $$ commute by (\ref{equation-diagram}). Since $c^\sharp$ is flat (as $c$ is a flat morphism by Lemma \ref{lemma-groupoid-on-field-open-multiplication}), we see that $(c^\sharp)^{-1}(\mathfrak q)$ is a minimal prime of $A$. Hence $\mathfrak p = (c^\sharp)^{-1}(\mathfrak q) = \mathfrak p'$.
Let $r \in |R|$. Denote $\text{trdeg}(r/_{\!\! s}k)$ the transcendence degree of $r$ over $\Spec(k)$ via $s$. Choose an \'etale morphism $\varphi : V \to R$ where $V$ is a scheme and $v \in V$ mapping to $r$. Using the definitions mentioned above the lemma we see that $$ \dim_r(R) = \dim_v(V) = \dim(\mathcal{O}_{V, v}) + \text{trdeg}_{s(k)}(\kappa(v)) = \dim(\mathcal{O}_{R, \overline{r}}) + \text{trdeg}(r/_{\!\! s}k) $$ and similarly for $t$ (the second equality by Morphisms, Lemma \ref{morphisms-lemma-dimension-fibre-at-a-point}). Hence we see that $\text{trdeg}(r/_{\!\! s}k) = \text{trdeg}(r/_{\!\! t}k)$, i.e., (2) holds. \medskip\noindent Let $k \subset k'$ be a field extension. Note that the restriction $R'$ of $R$ to $\Spec(k')$ (see Lemma \ref{lemma-restrict-groupoid-on-field}) is obtained from $R$ by two base changes by morphisms of fields. Thus Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-dimension-fibre-after-base-change} shows the dimension of $R$ at a point is unchanged by this operation. Hence in order to prove (1) we may assume, by Lemma \ref{lemma-groupoid-on-field-explain-points}, that $r$ is represented by a morphism $r : \Spec(k) \to R$ such that both $s \circ r$ and $t \circ r$ are automorphisms of $\Spec(k)$. In this case there exists an automorphism $R \to R$ which maps $r$ to $e$ (Lemma \ref{lemma-groupoid-on-field-move-point}). Hence we see that $\dim_r(R) = \dim_e(R)$ for any $r$. By definition this means that $\dim_r(R) = \dim(R)$. \medskip\noindent Part (3) is a formal consequence of the results obtained in the discussion above.
Since $G$ and $R$ are equidimensional (see Lemmas \ref{lemma-groupoid-on-field-locally-finite-type-dimension} and \ref{lemma-group-over-field-locally-finite-type-dimension}) it suffices to prove that $\dim_e(R) = \dim_e(G)$. Let $V$ be an affine scheme, $v \in V$, and let $\varphi : V \to R$ be an \'etale morphism of schemes such that $\varphi(v) = e$. Note that $V$ is a Noetherian scheme as $s \circ \varphi$ is locally of finite type as a composition of morphisms locally of finite type and as $V$ is quasi-compact (use Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-composition-finite-type}, \ref{spaces-morphisms-lemma-etale-locally-finite-presentation}, and \ref{spaces-morphisms-lemma-finite-presentation-finite-type} and Morphisms, Lemma \ref{morphisms-lemma-finite-type-noetherian}). Hence $V$ is locally connected (see Properties, Lemma \ref{properties-lemma-Noetherian-topology} and Topology, Lemma \ref{topology-lemma-locally-Noetherian-locally-connected}). Thus we may replace $V$ by the connected component containing $v$ (it is still affine as it is an open and closed subscheme of $V$). Set $T = V_{red}$ equal to the reduction of $V$. Consider the two morphisms $a, b : T \to \Spec(k)$ given by $a = s \circ \varphi|_T$ and $b = t \circ \varphi|_T$. Note that $a, b$ induce the same field map $k \to \kappa(v)$ because $\varphi(v) = e$! Let $k_a \subset \Gamma(T, \mathcal{O}_T)$ be the integral closure of $a^\sharp(k) \subset \Gamma(T, \mathcal{O}_T)$. Similarly, let $k_b \subset \Gamma(T, \mathcal{O}_T)$ be the integral closure of $b^\sharp(k) \subset \Gamma(T, \mathcal{O}_T)$. By Varieties, Proposition \ref{varieties-proposition-unique-base-field} we see that $k_a = k_b$. Thus we obtain the following commutative diagram $$ \xymatrix{ k \ar[rd]^a \ar[rrrd] \\ & k_a = k_b \ar[r] & \Gamma(T, \mathcal{O}_T) \ar[r] & \kappa(v) \\ k \ar[ru]_b \ar[rrru] } $$ As discussed above the long arrows are equal. Since $k_a = k_b \to \kappa(v)$ is injective we conclude that the two morphisms $a$ and $b$ agree. Hence $T \to R$ factors through $G$. It follows that $R_{red} = G_{red}$ in an open neighbourhood of $e$ which certainly implies that $\dim_e(R) = \dim_e(G)$.
By Spaces over Fields, Lemma \ref{spaces-over-fields-lemma-when-scheme-after-base-change} it suffices to show that $G_K$ is a scheme for some field extension $K/k$. Denote $G_K' \subset G_K$ the schematic locus of $G_K$ as in Properties of Spaces, Lemma \ref{spaces-properties-lemma-subscheme}. By Properties of Spaces, Proposition \ref{spaces-properties-proposition-locally-quasi-separated-open-dense-scheme} we see that $G_K' \subset G_K$ is dense open, in particular not empty. Choose a scheme $U$ and a surjective \'etale morphism $U \to G$. By Varieties, Lemma \ref{varieties-lemma-make-Jacobson} if $K$ is an algebraically closed field of large enough transcendence degree, then $U_K$ is a Jacobson scheme and every closed point of $U_K$ is $K$-rational. Hence $G_K'$ has a $K$-rational point and it suffices to show that every $K$-rational point of $G_K$ is in $G_K'$. If $g \in G_K(K)$ is a $K$-rational point and $g' \in G_K'(K)$ a $K$-rational point in the schematic locus, then we see that $g$ is in the image of $G_K'$ under the automorphism $$ G_K \longrightarrow G_K,\quad h \longmapsto g(g')^{-1}h $$ of $G_K$. Since automorphisms of $G_K$ as an algebraic space preserve $G_K'$, we conclude that $g \in G_K'$ as desired.
Set $R = X \times_Z X$. By (2) we see that $Z = X/R$ as sheaves. Also (2) implies that the inverse image of $V$ in $R$ is scheme theoretically dense in $R$ (Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-flat-morphism-scheme-theoretically-dense-open}). The we see that the two compositions $R \to X \to Y$ are equal by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-equality-of-morphisms}. The lemma follows.
We may assume $k$ is algebraically closed (details omitted); we only do this so we may argue using rational points, but the reader can work around this if she/he so desires. In the proof products are over $k$. The automorphism group algebraic space of $(\mathbf{P}^1_k)^n$ contains $G = (\text{GL}_{2, k})^n$. If $C \subset (\mathbf{P}^1_k)^n$ is a closed subvariety (in particular irreducible over $k$) which is mapped to a point, then we can apply More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-flat-proper-family-cannot-collapse-fibre} to the morphism $$ G \times C \to G \times Z,\quad (g, c) \mapsto (g, f(g \cdot c)) $$ over $G$. Hence $g(C)$ is mapped to a point for $g \in G(k)$ lying in a Zariski open $U \subset G$. Suppose $x = (x_1, \ldots, x_n)$, $y = (y_1, \ldots, y_n)$ are $k$-valued points of $(\mathbf{P}^1_k)^n$. Let $I \subset \{1, \ldots, n\}$ be the set of indices $i$ such that $x_i = y_i$. Then $$ \{g(x) \mid g(y) = y,\ g \in U(k)\} $$ is Zariski dense in the fibre of the projection $\pi_I : (\mathbf{P}^1_k)^n \to \prod_{i \in I} \mathbf{P}^1_k$ (exercise). Hence if $x, y \in C(k)$ are distinct, we conclude that $f$ maps the whole fibre of $\pi_I$ containing $x, y$ to a single point. Moreover, the $U(k)$-orbit of $C$ meets a Zariski open set of fibres of $\pi_I$. By Lemma \ref{lemma-factor-through-over-open} the morphism $f$ factors through $\pi_I$. After repeating this process finitely many times we reach the stage where all fibres of $f$ over $k$ points are finite. In this case $f$ is finite by More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-proper-finite-fibre-finite-in-neighbourhood} and the fact that $k$ points are dense in $Z$ (Spaces over Fields, Lemma \ref{spaces-over-fields-lemma-smooth-separable-closed-points-dense}).
Assume $f$ is nonconstant. Consider the morphisms $$ \mathbf{P}^1_k \times_{\Spec(k)} \ldots \times_{\Spec(k)} \mathbf{P}^1_k \longrightarrow G, \quad (t_1, \ldots, t_n) \longmapsto f(g_1) \ldots f(g_n) $$ where on the right hand side we use multiplication in the group. By Lemma \ref{lemma-quotient-power-P1} and the assumption that $f$ is nonconstant this morphism is finite onto its image. Hence $\dim(G) \geq n$ for all $n$, which is impossible by Lemma \ref{lemma-group-over-field-locally-finite-type-dimension} and the fact that $G$ is locally of finite type over $k$.
Let $T$ be an algebraic space over $S$. Let $\{T_i \to T\}$ be an fppf covering (by algebraic spaces). Let $s_i = (a_i, Z_i)$ be pairs over $T_i$ satisfying \ref{equation-finite-conditions} such that we have $s_i|_{T_i \times_T T_j} = s_j|_{T_i \times_T T_j}$. First, this implies in particular that $a_i$ and $a_j$ define the same morphism $T_i \times_T T_j \to Y$. By Descent on Spaces, Lemma \ref{spaces-descent-lemma-fpqc-universal-effective-epimorphisms} we deduce that there exists a unique morphism $a : T \to Y$ such that $a_i$ equals the composition $T_i \to T \to Y$. Second, this implies that $Z_i \subset T_i \times_Y X$ are open subspaces whose inverse images in $(T_i \times_T T_j) \times_Y X$ are equal. Since $\{T_i \times_Y X \to T \times_Y X\}$ is an fppf covering we deduce that there exists a unique open subspace $Z \subset T \times_Y X$ which restricts back to $Z_i$ over $T_i$, see Descent on Spaces, Lemma \ref{spaces-descent-lemma-open-fpqc-covering}. We claim that the projection $Z \to T$ is finite. This follows as being finite is local for the fpqc topology, see Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-finite}. \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 satisfying conditions \ref{equation-finite-conditions}. \medskip\noindent Let $v : T \to (X/Y)_{fin}$ be a natural transformation. Choose a scheme $U$ and a surjective \'etale morphism $p : U \to T$. Then $v(p) \in (X/Y)_{fin}(U)$ corresponds to a pair $(a_U, Z_U)$ 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 $(a_U, Z_U)$ 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 $(a, Z)$ over $T$. \medskip\noindent Conversely, let $(a, Z)$ 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 $(a, Z)|_U$ gives rise to a transformation of functors $v : h_U \to (X/Y)_{fin}$ by the Yoneda lemma (Categories, Lemma \ref{categories-lemma-yoneda}). As the two pullbacks $s^*(a, Z)|_U$ and $t^*(a, Z)|_U$ 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 $(X/Y)_{fin}$ is an fppf sheaf by (1) we conclude that $v$ factors through a map $T \to (X/Y)_{fin}$. \medskip\noindent We omit the verification that the two constructions above are mutually inverse.
Lemma \ref{lemma-finite-open} gives us an injective map $(X'/Y)_{fin} \to (X/Y)_{fin}$. Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-locally-finite-type-quasi-finite-part} assures us that formation of $X'$ commutes with base change. Hence everything comes down to proving that if $Z \subset X$ is an open subspace such that $f|_Z : Z \to Y$ is finite, then $Z \subset X'$. This is true because a finite morphism is locally quasi-finite, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-finite-quasi-finite}.
A finite morphism of algebraic spaces is universally closed by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-finite-proper}. Since $f$ is separated so is the morphism $T \times_Y X \to T$, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-separated}. Thus the closedness of $Z$ follows from Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-universally-closed-permanence}.
The second statement follows from the first as the absolute diagonal is the composition of the relative diagonal and a base change of the diagonal of $Y$ (which is representable by schemes), see Spaces, Section \ref{spaces-section-representable}. To prove the first assertion we have to show the following: Given a scheme $T$ and two pairs $(a, Z_1)$ and $(a, Z_2)$ over $T$ with identical first component satisfying \ref{equation-finite-conditions} there is an open subscheme $V \subset T$ with the following property: For any morphism of schemes $h : T' \to T$ we have $$ h(T') \subset V \Leftrightarrow \Big(T' \times_T Z_1 = T' \times_T Z_2 \text{ as subspaces of }T' \times_Y X\Big) $$ Let us construct $V$. Note that $Z_1 \cap Z_2$ is open in $Z_1$ and in $Z_2$. Since $\text{pr}_0|_{Z_i} : Z_i \to T$ is finite, hence proper (see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-finite-proper}) we see that $$ E = \text{pr}_0|_{Z_1}\left(Z_1 \setminus Z_1 \cap Z_2)\right) \cup \text{pr}_0|_{Z_2}\left(Z_2 \setminus Z_1 \cap Z_2)\right) $$ is closed in $T$. Now it is clear that $V = T \setminus E$ works.
This is formal from the description of the diagonal in Lemma \ref{lemma-finite-diagonal} but we write it out since it is an important step in the development of the theory. We have to check that for any scheme $T$ over $S$ and a morphism $T \to (X/Y)_{fin}$ the projection map $$ T \times_{(X/Y)_{fin}} U \longrightarrow T $$ is \'etale. Note that $$ T \times_{(X/Y)_{fin}} U = (X/Y)_{fin} \times_{((X/Y)_{fin} \times_Y (X/Y)_{fin})} (T \times_Y U) $$ Applying the result of Lemma \ref{lemma-finite-diagonal} we see that $T \times_{(X/Y)_{fin}} U$ is represented by an open subscheme of $T \times_Y U$. As the projection $T \times_Y U \to T$ is \'etale by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-etale} we conclude.