The Stacks project

40.11 Morphisms of groupoids on fields

This section studies morphisms between groupoids on fields. This is slightly more general, but very akin to, studying morphisms of groupschemes over a field.

Situation 40.11.1. Let $S$ be a scheme. Let $U = \mathop{\mathrm{Spec}}(k)$ be a scheme over $S$ with $k$ a field. Let $(U, R_1, s_1, t_1, c_1)$, $(U, R_2, s_2, t_2, c_2)$ be groupoid schemes over $S$ with identical first component. Let $a : R_1 \to R_2$ be a morphism such that $(\text{id}_ U, a)$ defines a morphism of groupoid schemes over $S$, see Groupoids, Definition 39.13.1. In particular, the following diagrams commute

\[ \vcenter { \xymatrix{ R_1 \ar[rrd]^{t_1} \ar[rdd]_{s_1} \ar[rd]_ a \\ & R_2 \ar[d]^{t_2} \ar[r]_{s_2} & U \\ & U } } \quad \quad \vcenter { \xymatrix{ R_1 \times _{s_1, U, t_1} R_1 \ar[r]_-{c_1} \ar[d]_{a \times a} & R_1 \ar[d]^ a \\ R_2 \times _{s_2, U, t_2} R_2 \ar[r]^-{c_2} & R_2 } } \]

The following lemma is a generalization of Groupoids, Lemma 39.7.7.

Lemma 40.11.2. Notation and assumptions as in Situation 40.11.1. If $a(R_1)$ is open in $R_2$, then $a(R_1)$ is closed in $R_2$.

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

\[ \xymatrix{ R'_1 \ar[r]_{a'} \ar[d]_{p_1} & R'_2 \ar[d]^{p_2} \\ R_1 \ar[r]^ a & R_2 } \]

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. $\square$

Lemma 40.11.3. Notation and assumptions as in Situation 40.11.1. Let $Z \subset R_2$ be the reduced closed subscheme (see Schemes, Definition 26.12.5) whose underlying topological space is the closure of the image of $a : R_1 \to R_2$. Then $c_2(Z \times _{s_2, U, t_2} Z) \subset Z$ set theoretically.

Proof. Consider the commutative diagram

\[ \xymatrix{ R_1 \times _{s_1, U, t_1} R_1 \ar[r] \ar[d] & R_1 \ar[d] \\ R_2 \times _{s_2, U, t_2} R_2 \ar[r] & R_2 } \]

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. $\square$

Lemma 40.11.4. Notation and assumptions as in Situation 40.11.1. Assume that $k$ is perfect. Let $Z \subset R_2$ be the reduced closed subscheme (see Schemes, Definition 26.12.5) whose underlying topological space is the closure of the image of $a : R_1 \to R_2$. Then

\[ (U, Z, s_2|_ Z, t_2|_ Z, c_2|_ Z) \]

is a groupoid scheme over $S$.

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. $\square$

Lemma 40.11.5. Notation and assumptions as in Situation 40.11.1. If the image $a(R_1)$ is a locally closed subset of $R_2$ then it is a closed subset.

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 $a$. $\square$

Lemma 40.11.6. Notation and assumptions as in Situation 40.11.1. Assume that $a : R_1 \to R_2$ is a quasi-compact morphism. Let $Z \subset R_2$ be the scheme theoretic image (see Morphisms, Definition 29.6.2) of $a : R_1 \to R_2$. Then

\[ (U, Z, s_2|_ Z, t_2|_ Z, c_2|_ Z) \]

is a groupoid scheme over $S$.

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

\[ \xymatrix{ R_1 \times _{s_1, U, t_1} R_1 \ar[r] \ar[d]^{a \times a} & R_1 \ar[d] \\ R_2 \times _{s_2, U, t_2} R_2 \ar[r] & R_2 } \]

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. $\square$

Lemma 40.11.7. Let $S$ be a scheme. Let $(U, R, s, t, c)$ be a groupoid scheme over $S$. Assume $U$ is the spectrum of a field. Let $Z \subset U \times _ S U$ be the reduced closed subscheme (see Schemes, Definition 26.12.5) whose underlying topological space is the closure of the image of $j = (t, s) : R \to U \times _ S U$. Then $\text{pr}_{02}(Z \times _{\text{pr}_1, U, \text{pr}_0} Z) \subset Z$ set theoretically.

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

\[ \xymatrix{ R_ s \times _ k {}_ tR \ar[r]^ c \ar[d]_{j_ s \times {}_ tj} & R \ar[d]^ j \\ U^2_ s \times _ k {}_ tU^2 \ar[r] & U \times _ S U } \]

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. $\square$

Lemma 40.11.8. Let $S$ be a scheme. Let $(U, R, s, t, c)$ be a groupoid scheme over $S$. Assume $U$ is the spectrum of a perfect field. Let $Z \subset U \times _ S U$ be the reduced closed subscheme (see Schemes, Definition 26.12.5) whose underlying topological space is the closure of the image of $j = (t, s) : R \to U \times _ S U$. Then

\[ (U, Z, \text{pr}_0|_ Z, \text{pr}_1|_ Z, \text{pr}_{02}|_{Z \times _{\text{pr}_1, U, \text{pr}_0} Z}) \]

is a groupoid scheme over $S$.

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. $\square$

Lemma 40.11.9. Let $S$ be a scheme. Let $(U, R, s, t, c)$ be a groupoid scheme over $S$. Assume $U$ is the spectrum of a field and assume $R$ is quasi-compact (equivalently $s, t$ are quasi-compact). Let $Z \subset U \times _ S U$ be the scheme theoretic image (see Morphisms, Definition 29.6.2) of $j = (t, s) : R \to U \times _ S U$. Then

\[ (U, Z, \text{pr}_0|_ Z, \text{pr}_1|_ Z, \text{pr}_{02}|_{Z \times _{\text{pr}_1, U, \text{pr}_0} Z}) \]

is a groupoid scheme over $S$.

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

\[ \xymatrix{ R_ s \times _ k {}_ tR \ar[r]^ c \ar[d]_{j_ s \times {}_ tj} & R \ar[d]^ j \\ U^2_ s \times _ k {}_ tU^2 \ar[r] & U \times _ S U } \]

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. $\square$


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