The Stacks project

Proof. Let $r_2 \in R_2$ be a point in the closure of $a(R_1)$. We want to show $r_2 \in a(R_1)$. Pick $k \subset k'$ and $r_2' \in R'_2$ adapted to $(U, R_2, s_2, t_2, c_2)$ and $r_2$ as in Lemma 40.10.5. Let $R_ i'$ be the restriction of $R_ i$ via the morphism $U' = \mathop{\mathrm{Spec}}(k') \to U = \mathop{\mathrm{Spec}}(k)$. Let $a' : R'_1 \to R_2'$ be the base change of $a$. The diagram

is a fibre square. Hence the image of $a'$ is the inverse image of the image of $a$ via the morphism $p_2 : R'_2 \to R_2$. By Lemma 40.10.4 the map $p_2$ is surjective and open. Hence by Topology, Lemma 5.6.4 we see that $r_2'$ is in the closure of $a'(R'_1)$. This means that we may assume that $r_2 \in R_2$ has the property that the maps $k \to \kappa (r_2)$ induced by $s_2$ and $t_2$ are isomorphisms.

In this case we can use Lemma 40.10.6. This lemma implies $c(r_2, a(R_1))$ is an open neighbourhood of $r_2$. Hence $a(R_1) \cap c(r_2, a(R_1)) \not= \emptyset$ as we assumed that $r_2$ was a point of the closure of $a(R_1)$. Using the inverse of $R_2$ and $R_1$ we see this means $c_2(a(R_1), a(R_1))$ contains $r_2$. As $c_2(a(R_1), a(R_1)) \subset a(c_1(R_1, R_1)) = a(R_1)$ we conclude $r_2 \in a(R_1)$ as desired. $\square$

Proof. Consider the commutative diagram

By Varieties, Lemma 33.24.2 the closure of the image of the left vertical arrow is (set theoretically) $Z \times _{s_2, U, t_2} Z$. Hence the result follows. $\square$

Proof. We first explain why the statement makes sense. Since $U$ is the spectrum of a perfect field $k$, the scheme $Z$ is geometrically reduced over $k$ (via either projection), see Varieties, Lemma 33.6.3. Hence the scheme $Z \times _{s_2, U, t_2} Z \subset Z$ is reduced, see Varieties, Lemma 33.6.7. Hence by Lemma 40.11.3 we see that $c$ induces a morphism $Z \times _{s_2, U, t_2} Z \to Z$. Finally, it is clear that $e_2$ factors through $Z$ and that the map $i_2 : R_2 \to R_2$ preserves $Z$. Since the morphisms of the septuple $(U, R_2, s_2, t_2, c_2, e_2, i_2)$ satisfies the axioms of a groupoid, it follows that after restricting to $Z$ they satisfy the axioms. $\square$

Proof. Let $k \subset k'$ be a perfect closure of the field $k$. Let $R_ i'$ be the restriction of $R_ i$ via the morphism $U' = \mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k)$. Note that the morphisms $R_ i' \to R_ i$ are universal homeomorphisms as compositions of base changes of the universal homeomorphism $U' \to U$ (see diagram in statement of Lemma 40.10.4). Hence it suffices to prove that $a'(R_1')$ is closed in $R_2'$. In other words, we may assume that $k$ is perfect.

If $k$ is perfect, then the closure of the image is a groupoid scheme $Z \subset R_2$, by Lemma 40.11.4. By the same lemma applied to $\text{id}_{R_1} : R_1 \to R_1$ we see that $(R_2)_{red}$ is a groupoid scheme. Thus we may apply Lemma 40.11.2 to the morphism $a|_{(R_2)_{red}} : (R_2)_{red} \to Z$ to conclude that $Z$ equals the image of $a$. $\square$

Proof. The main difficulty is to show that $c_2|_{Z \times _{s_2, U, t_2} Z}$ maps into $Z$. Consider the commutative diagram

By Varieties, Lemma 33.24.3 we see that the scheme theoretic image of $a \times a$ is $Z \times _{s_2, U, t_2} Z$. By the commutativity of the diagram we conclude that $Z \times _{s_2, U, t_2} Z$ maps into $Z$ by the bottom horizontal arrow. As in the proof of Lemma 40.11.4 it is also true that $i_2(Z) \subset Z$ and that $e_2$ factors through $Z$. Hence we conclude as in the proof of that lemma. $\square$

Proof. As $(U, U \times _ S U, \text{pr}_1, \text{pr}_0, \text{pr}_{02})$ is a groupoid scheme over $S$ this is a special case of Lemma 40.11.3. But we can also prove it directly as follows.

Write $U = \mathop{\mathrm{Spec}}(k)$. Denote $R_ s$ (resp. $Z_ s$, resp. $U^2_ s$) the scheme $R$ (resp. $Z$, resp. $U \times _ S U$) viewed as a scheme over $k$ via $s$ (resp. $\text{pr}_1|_ Z$, resp. $\text{pr}_1$). Similarly, denote ${}_ tR$ (resp. ${}_ tZ$, resp. ${}_ tU^2$) the scheme $R$ (resp. $Z$, resp. $U \times _ S U$) viewed as a scheme over $k$ via $t$ (resp. $\text{pr}_0|_ Z$, resp. $\text{pr}_0$). The morphism $j$ induces morphisms of schemes $j_ s : R_ s \to U^2_ s$ and ${}_ tj : {}_ tR \to {}_ tU^2$ over $k$. Consider the commutative diagram

By Varieties, Lemma 33.24.2 we see that the closure of the image of $j_ s \times {}_ tj$ is $Z_ s \times _ k {}_ tZ$. By the commutativity of the diagram we conclude that $Z_ s \times _ k {}_ tZ$ maps into $Z$ by the bottom horizontal arrow. $\square$

Proof. As $(U, U \times _ S U, \text{pr}_1, \text{pr}_0, \text{pr}_{02})$ is a groupoid scheme over $S$ this is a special case of Lemma 40.11.4. But we can also prove it directly as follows.

We first explain why the statement makes sense. Since $U$ is the spectrum of a perfect field $k$, the scheme $Z$ is geometrically reduced over $k$ (via either projection), see Varieties, Lemma 33.6.3. Hence the scheme $Z \times _{\text{pr}_1, U, \text{pr}_0} Z \subset Z$ is reduced, see Varieties, Lemma 33.6.7. Hence by Lemma 40.11.7 we see that $\text{pr}_{02}$ induces a morphism $Z \times _{\text{pr}_1, U, \text{pr}_0} Z \to Z$. Finally, it is clear that $\Delta _{U/S}$ factors through $Z$ and that the map $\sigma : U \times _ S U \to U \times _ S U$, $(x, y) \mapsto (y, x)$ preserves $Z$. Since $(U, U \times _ S U, \text{pr}_0, \text{pr}_1, \text{pr}_{02}, \Delta _{U/S}, \sigma )$ satisfies the axioms of a groupoid, it follows that after restricting to $Z$ they satisfy the axioms. $\square$

Proof. As $(U, U \times _ S U, \text{pr}_1, \text{pr}_0, \text{pr}_{02})$ is a groupoid scheme over $S$ this is a special case of Lemma 40.11.6. But we can also prove it directly as follows.

The main difficulty is to show that $\text{pr}_{02}|_{Z \times _{\text{pr}_1, U, \text{pr}_0} Z}$ maps into $Z$. Write $U = \mathop{\mathrm{Spec}}(k)$. Denote $R_ s$ (resp. $Z_ s$, resp. $U^2_ s$) the scheme $R$ (resp. $Z$, resp. $U \times _ S U$) viewed as a scheme over $k$ via $s$ (resp. $\text{pr}_1|_ Z$, resp. $\text{pr}_1$). Similarly, denote ${}_ tR$ (resp. ${}_ tZ$, resp. ${}_ tU^2$) the scheme $R$ (resp. $Z$, resp. $U \times _ S U$) viewed as a scheme over $k$ via $t$ (resp. $\text{pr}_0|_ Z$, resp. $\text{pr}_0$). The morphism $j$ induces morphisms of schemes $j_ s : R_ s \to U^2_ s$ and ${}_ tj : {}_ tR \to {}_ tU^2$ over $k$. Consider the commutative diagram

By Varieties, Lemma 33.24.3 we see that the scheme theoretic image of $j_ s \times {}_ tj$ is $Z_ s \times _ k {}_ tZ$. By the commutativity of the diagram we conclude that $Z_ s \times _ k {}_ tZ$ maps into $Z$ by the bottom horizontal arrow. As in the proof of Lemma 40.11.8 it is also true that $\sigma (Z) \subset Z$ and that $\Delta _{U/S}$ factors through $Z$. Hence we conclude as in the proof of that lemma. $\square$