## 81.5 Quotients as orbit spaces

Let $j = (t, s) : R \to U \times _ B U$ be a pre-relation. If $j$ is a pre-equivalence relation, then loosely speaking the “orbits” of $R$ on $U$ are the subsets $t(s^{-1}(\{ u\} ))$ of $U$. However, if $j$ is just a pre-relation, then we need to take the equivalence relation generated by $R$.

Definition 81.5.1. Let $S$ be a scheme, and let $B$ be an algebraic space over $S$. Let $j : R \to U \times _ B U$ be a pre-relation over $B$. If $u \in |U|$, then the orbit, or more precisely the $R$-orbit of $u$ is

$O_ u = \left\{ u' \in |U|\ : \begin{matrix} \exists n \geq 1, \ \exists u_0, \ldots , u_ n \in |U|\text{ such that } u_0 = u \text{ and } u_ n = u' \\ \text{and for all }i \in \{ 0, \ldots , n - 1\} \text{ either } u_ i = u_{i + 1}\text{ or } \\ \exists r \in |R|, \ s(r) = u_ i, t(r) = u_{i + 1} \text{ or } \\ \exists r \in |R|, \ t(r) = u_ i, s(r) = u_{i + 1} \end{matrix} \right\}$

It is clear that these are the equivalence classes of an equivalence relation, i.e., we have $u' \in O_ u$ if and only if $u \in O_{u'}$. The following lemma is a reformulation of Groupoids in Spaces, Lemma 76.4.4.

Lemma 81.5.2. Let $B \to S$ as in Section 81.2. Let $j : R \to U \times _ B U$ be a pre-equivalence relation of algebraic spaces over $B$. Then

$O_ u = \{ u' \in |U| \text{ such that } \exists r \in |R|, \ s(r) = u, \ t(r) = u'\} .$

Proof. By the aforementioned Groupoids in Spaces, Lemma 76.4.4 we see that the orbits $O_ u$ as defined in the lemma give a disjoint union decomposition of $|U|$. Thus we see they are equal to the orbits as defined in Definition 81.5.1. $\square$

Lemma 81.5.3. In the situation of Definition 81.5.1. Let $\phi : U \to X$ be an $R$-invariant morphism of algebraic spaces over $B$. Then $|\phi | : |U| \to |X|$ is constant on the orbits.

Proof. To see this we just have to show that $\phi (u) = \phi (u')$ for all $u, u' \in |U|$ such that there exists an $r \in |R|$ such that $s(r) = u$ and $t(r) = u'$. And this is clear since $\phi$ equalizes $s$ and $t$. $\square$

There are several problems with considering the orbits $O_ u \subset |U|$ as a tool for singling out properties of quotient maps. One issue is the following. Suppose that $\mathop{\mathrm{Spec}}(k) \to B$ is a geometric point of $B$. Consider the canonical map

$U(k) \longrightarrow |U|.$

Then it is usually not the case that the equivalence classes of the equivalence relation generated by $j(R(k)) \subset U(k) \times U(k)$ are the inverse images of the orbits $O_ u \subset |U|$. A silly example is to take $S = B = \mathop{\mathrm{Spec}}(\mathbf{Z})$, $U = R = \mathop{\mathrm{Spec}}(k)$ with $s = t = \text{id}_ k$. Then $|U| = |R|$ is a single point but $U(k)/R(k)$ is enormous. A more interesting example is to take $S = B = \mathop{\mathrm{Spec}}(\mathbf{Q})$, choose some of number fields $K \subset L$, and set $U = \mathop{\mathrm{Spec}}(L)$ and $R = \mathop{\mathrm{Spec}}(L \otimes _ K L)$ with obvious maps $s, t : R \to U$. In this case $|U|$ still has just one point, but the quotient

$U(k)/R(k) = \mathop{\mathrm{Hom}}\nolimits (K, k)$

consists of more than one element. We conclude from both examples that if $U \to X$ is an $R$-invariant map and if we want it to “separate orbits” we get a much stronger and interesting notion by considering the induced maps $U(k) \to X(k)$ and ask that those maps separate orbits.

There is an issue with this too. Namely, suppose that $S = B = \mathop{\mathrm{Spec}}(\mathbf{R})$, $U = \mathop{\mathrm{Spec}}(\mathbf{C})$, and $R = \mathop{\mathrm{Spec}}(\mathbf{C}) \amalg \mathop{\mathrm{Spec}}(K)$ for some field extension $\sigma : \mathbf{C} \to K$. Let the maps $s, t$ be given by the identity on the component $\mathop{\mathrm{Spec}}(\mathbf{C})$, but by $\sigma , \sigma \circ \tau$ on the second component where $\tau$ is complex conjugation. If $K$ is a nontrivial extension of $\mathbf{C}$, then the two points $1, \tau \in U(\mathbf{C})$ are not equivalent under $j(R(\mathbf{C}))$. But after choosing an extension $\mathbf{C} \subset \Omega$ of sufficiently large cardinality (for example larger than the cardinality of $K$) then the images of $1, \tau \in U(\mathbf{C})$ in $U(\Omega )$ do become equivalent! It seems intuitively clear that this happens either because $s, t : R \to U$ are not locally of finite type or because the cardinality of the field $k$ is not large enough.

Keeping this in mind we make the following definition.

Definition 81.5.4. Let $S$ be a scheme, and let $B$ be an algebraic space over $S$. Let $j : R \to U \times _ B U$ be a pre-relation over $B$. Let $\mathop{\mathrm{Spec}}(k) \to B$ be a geometric point of $B$.

1. We say $\overline{u}, \overline{u}' \in U(k)$ are weakly $R$-equivalent if they are in the same equivalence class for the equivalence relation generated by the relation $j(R(k)) \subset U(k) \times U(k)$.

2. We say $\overline{u}, \overline{u}' \in U(k)$ are $R$-equivalent if for some overfield $k \subset \Omega$ the images in $U(\Omega )$ are weakly $R$-equivalent.

3. The weak orbit, or more precisely the weak $R$-orbit of $\overline{u} \in U(k)$ is set of all elements of $U(k)$ which are weakly $R$-equivalent to $\overline{u}$.

4. The orbit, or more precisely the $R$-orbit of $\overline{u} \in U(k)$ is set of all elements of $U(k)$ which are $R$-equivalent to $\overline{u}$.

It turns out that in good cases orbits and weak orbits agree, see Lemma 81.5.7. The following lemma illustrates the difference in the special case of a pre-equivalence relation.

Lemma 81.5.5. Let $S$ be a scheme, and let $B$ be an algebraic space over $S$. Let $\mathop{\mathrm{Spec}}(k) \to B$ be a geometric point of $B$. Let $j : R \to U \times _ B U$ be a pre-equivalence relation over $B$. In this case the weak orbit of $\overline{u} \in U(k)$ is simply

$\{ \overline{u}' \in U(k) \text{ such that } \exists \overline{r} \in R(k), \ s(\overline{r}) = \overline{u}, \ t(\overline{r}) = \overline{u}' \}$

and the orbit of $\overline{u} \in U(k)$ is

$\{ \overline{u}' \in U(k) : \exists \text{ field extension }k \subset K, \ \exists \ r \in R(K), \ s(r) = \overline{u}, \ t(r) = \overline{u}'\}$

Proof. This is true because by definition of a pre-equivalence relation the image $j(R(k)) \subset U(k) \times U(k)$ is an equivalence relation. $\square$

Let us describe the recipe for turning any pre-relation into a pre-equivalence relation. We will use the morphisms

81.5.5.1
$$\label{groupoids-quotients-equation-list} \begin{matrix} j_{diag} & : & U & \longrightarrow & U \times _ B U, & u & \longmapsto & (u, u) \\ j_{flip} & : & R & \longrightarrow & U \times _ B U, & r & \longmapsto & (s(r), t(r)) \\ j_{comp} & : & R \times _{s, U, t} R & \longrightarrow & U \times _ B U, & (r, r') & \longmapsto & (t(r), s(r')) \end{matrix}$$

We define $j_1 = (t_1, s_1) : R_1 \to U \times _ B U$ to be the morphism

$j \amalg j_{diag} \amalg j_{flip} : R \amalg U \amalg R \longrightarrow U \times _ B U$

with notation as in Equation (81.5.5.1). For $n > 1$ we set

$j_ n = (t_ n, s_ n) : R_ n = R_1 \times _{s_1, U, t_{n - 1}} R_{n - 1} \longrightarrow U \times _ B U$

where $t_ n$ comes from $t_1$ precomposed with projection onto $R_1$ and $s_ n$ comes from $s_{n - 1}$ precomposed with projection onto $R_{n - 1}$. Finally, we denote

$j_\infty = (t_\infty , s_\infty ) : R_\infty = \coprod \nolimits _{n \geq 1} R_ n \longrightarrow U \times _ B U.$

Lemma 81.5.6. Let $S$ be a scheme, and let $B$ be an algebraic space over $S$. Let $j : R \to U \times _ B U$ be a pre-relation over $B$. Then $j_\infty : R_\infty \to U \times _ B U$ is a pre-equivalence relation over $B$. Moreover

1. $\phi : U \to X$ is $R$-invariant if and only if it is $R_\infty$-invariant,

2. the canonical map of quotient sheaves $U/R \to U/R_\infty$ (see Groupoids in Spaces, Section 76.18) is an isomorphism,

3. weak $R$-orbits agree with weak $R_\infty$-orbits,

4. $R$-orbits agree with $R_\infty$-orbits,

5. if $s, t$ are locally of finite type, then $s_\infty$, $t_\infty$ are locally of finite type,

6. add more here as needed.

Proof. Omitted. Hint for (5): Any property of $s, t$ which is stable under composition and stable under base change, and Zariski local on the source will be inherited by $s_\infty , t_\infty$. $\square$

Lemma 81.5.7. Let $S$ be a scheme, and let $B$ be an algebraic space over $S$. Let $j : R \to U \times _ B U$ be a pre-relation over $B$. Let $\mathop{\mathrm{Spec}}(k) \to B$ be a geometric point of $B$.

1. If $s, t : R \to U$ are locally of finite type then weak $R$-equivalence on $U(k)$ agrees with $R$-equivalence, and weak $R$-orbits agree with $R$-orbits on $U(k)$.

2. If $k$ has sufficiently large cardinality then weak $R$-equivalence on $U(k)$ agrees with $R$-equivalence, and weak $R$-orbits agree with $R$-orbits on $U(k)$.

Proof. We first prove (1). Assume $s, t$ locally of finite type. By Lemma 81.5.6 we may assume that $R$ is a pre-equivalence relation. Let $k$ be an algebraically closed field over $B$. Suppose $\overline{u}, \overline{u}' \in U(k)$ are $R$-equivalent. Then for some extension field $k \subset \Omega$ there exists a point $\overline{r} \in R(\Omega )$ mapping to $(\overline{u}, \overline{u}') \in (U \times _ B U)(\Omega )$, see Lemma 81.5.5. Hence

$Z = R \times _{j, U \times _ B U, (\overline{u}, \overline{u}')} \mathop{\mathrm{Spec}}(k)$

is nonempty. As $s$ is locally of finite type we see that also $j$ is locally of finite type, see Morphisms of Spaces, Lemma 65.23.6. This implies $Z$ is a nonempty algebraic space locally of finite type over the algebraically closed field $k$ (use Morphisms of Spaces, Lemma 65.23.3). Thus $Z$ has a $k$-valued point, see Morphisms of Spaces, Lemma 65.24.1. Hence we conclude there exists a $\overline{r} \in R(k)$ with $j(\overline{r}) = (\overline{u}, \overline{u}')$, and we conclude that $\overline{u}, \overline{u}'$ are $R$-equivalent as desired.

The proof of part (2) is the same, except that it uses Morphisms of Spaces, Lemma 65.24.2 instead of Morphisms of Spaces, Lemma 65.24.1. This shows that the assertion holds as soon as $|k| > \lambda (R)$ with $\lambda (R)$ as introduced just above Morphisms of Spaces, Lemma 65.24.1. $\square$

In the following definition we use the terminology “$k$ is a field over $B$” to mean that $\mathop{\mathrm{Spec}}(k)$ comes equipped with a morphism $\mathop{\mathrm{Spec}}(k) \to B$.

Definition 81.5.8. Let $S$ be a scheme, and let $B$ be an algebraic space over $S$. Let $j : R \to U \times _ B U$ be a pre-relation over $B$.

1. We say $\phi : U \to X$ is set-theoretically $R$-invariant if and only if the map $U(k) \to X(k)$ equalizes the two maps $s, t : R(k) \to U(k)$ for every algebraically closed field $k$ over $B$.

2. We say $\phi : U \to X$ separates orbits, or separates $R$-orbits if it is set-theoretically $R$-invariant and $\phi (\overline{u}) = \phi (\overline{u}')$ in $X(k)$ implies that $\overline{u}, \overline{u}' \in U(k)$ are in the same orbit for every algebraically closed field $k$ over $B$.

In Example 81.5.12 we show that being set-theoretically invariant is “too weak” a notion in the category of algebraic spaces. A more geometric reformulation of what it means to be set-theoretically invariant or to separate orbits is in Lemma 81.5.17.

Lemma 81.5.9. In the situation of Definition 81.5.8. A morphism $\phi : U \to X$ is set-theoretically $R$-invariant if and only if for any algebraically closed field $k$ over $B$ the map $U(k) \to X(k)$ is constant on orbits.

Proof. This is true because the condition is supposed to hold for all algebraically closed fields over $B$. $\square$

Lemma 81.5.10. In the situation of Definition 81.5.8. An invariant morphism is set-theoretically invariant.

Proof. This is immediate from the definitions. $\square$

Lemma 81.5.11. In the situation of Definition 81.5.8. Let $\phi : U \to X$ be a morphism of algebraic spaces over $B$. Assume

1. $\phi$ is set-theoretically $R$-invariant,

2. $R$ is reduced, and

3. $X$ is locally separated over $B$.

Then $\phi$ is $R$-invariant.

Proof. Consider the equalizer

$Z = R \times _{(\phi , \phi ) \circ j, X \times _ B X, \Delta _{X/B}} X$

algebraic space. Then $Z \to R$ is an immersion by assumption (3). By assumption (1) $|Z| \to |R|$ is surjective. This implies that $Z \to R$ is a bijective closed immersion (use Schemes, Lemma 26.10.4) and by assumption (2) we conclude that $Z = R$. $\square$

Example 81.5.12. There exist reduced quasi-separated algebraic spaces $X$, $Y$ and a pair of morphisms $a, b : Y \to X$ which agree on all $k$-valued points but are not equal. To get an example take $Y = \mathop{\mathrm{Spec}}(k[[x]])$ and

$X = \mathbf{A}^1_ k \Big/ \big (\Delta \amalg \{ (x, -x) \mid x \not= 0\} \big )$

the algebraic space of Spaces, Example 63.14.1. The two morphisms $a, b : Y \to X$ come from the two maps $x \mapsto x$ and $x \mapsto -x$ from $Y$ to $\mathbf{A}^1_ k = \mathop{\mathrm{Spec}}(k[x])$. On the generic point the two maps are the same because on the open part $x \not= 0$ of the space $X$ the functions $x$ and $-x$ are equal. On the closed point the maps are obviously the same. It is also true that $a \not= b$. This implies that Lemma 81.5.11 does not hold with assumption (3) replaced by the assumption that $X$ be quasi-separated. Namely, consider the diagram

$\xymatrix{ Y \ar[d]_{-1} \ar[r]_1 & Y \ar[d]^ a \\ Y \ar[r]^ a & X }$

then the composition $a \circ (-1) = b$. Hence we can set $R = Y$, $U = Y$, $s = 1$, $t = -1$, $\phi = a$ to get an example of a set-theoretically invariant morphism which is not invariant.

The example above is instructive because the map $Y \to X$ even separates orbits. It shows that in the category of algebraic spaces there are simply too many set-theoretically invariant morphisms lying around. Next, let us define what it means for $R$ to be a set-theoretic equivalence relation, while remembering that we need to allow for field extensions to make this work correctly.

Definition 81.5.13. Let $S$ be a scheme, and let $B$ be an algebraic space over $S$. Let $j : R \to U \times _ B U$ be a pre-relation over $B$.

1. We say $j$ is a set-theoretic pre-equivalence relation if

$\overline{u} \sim _ R \overline{u}' \Leftrightarrow \begin{matrix} \exists \text{ field extension }k \subset K, \ \exists \ r \in R(K), \\ s(r) = \overline{u}, \ t(r) = \overline{u}' \end{matrix}$

defines an equivalence relation on $U(k)$ for all algebraically closed fields $k$ over $B$.

2. We say $j$ is a set-theoretic equivalence relation if $j$ is universally injective and a set-theoretic pre-equivalence relation.

Let us reformulate this in more geometric terms.

Lemma 81.5.14. In the situation of Definition 81.5.13. The following are equivalent:

1. The morphism $j$ is a set-theoretic pre-equivalence relation.

2. The subset $j(|R|) \subset |U \times _ B U|$ contains the image of $|j'|$ for any of the morphisms $j'$ as in Equation (81.5.5.1).

3. For every algebraically closed field $k$ over $B$ of sufficiently large cardinality the subset $j(R(k)) \subset U(k) \times U(k)$ is an equivalence relation.

If $s, t$ are locally of finite type these are also equivalent to

1. For every algebraically closed field $k$ over $B$ the subset $j(R(k)) \subset U(k) \times U(k)$ is an equivalence relation.

Proof. Assume (2). Let $k$ be an algebraically closed field over $B$. We are going to show that $\sim _ R$ is an equivalence relation. Suppose that $\overline{u}_ i : \mathop{\mathrm{Spec}}(k) \to U$, $i = 1, 2$ are $k$-valued points of $U$. Suppose that $(\overline{u}_1, \overline{u}_2)$ is the image of a $K$-valued point $r \in R(K)$. Consider the solid commutative diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K') \ar@{..>}[r] \ar@{..>}[d] & \mathop{\mathrm{Spec}}(k) \ar[d]_{(\overline{u}_2, \overline{u}_1)} & \mathop{\mathrm{Spec}}(K) \ar[d] \ar[l] \\ R \ar[r]^-j & U \times _ B U & R \ar[l]_-{j_{flip}} }$

We also denote $r \in |R|$ the image of $r$. By assumption the image of $|j_{flip}|$ is contained in the image of $|j|$, in other words there exists a $r' \in |R|$ such that $|j|(r') = |j_{flip}|(r)$. But note that $(\overline{u}_2, \overline{u}_1)$ is in the equivalence class that defines $|j|(r')$ (by the commutativity of the solid part of the diagram). This means there exists a field extension $k \subset K'$ and a morphism $r' : \mathop{\mathrm{Spec}}(K) \to R$ (abusively denoted $r'$ as well) with $j \circ r' = (\overline{u}_2, \overline{u}_1) \circ i$ where $i : \mathop{\mathrm{Spec}}(K') \to \mathop{\mathrm{Spec}}(K)$ is the obvious map. In other words the dotted part of the diagram commutes. This proves that $\sim _ R$ is a symmetric relation on $U(k)$. In the similar way, using that the image of $|j_{diag}|$ is contained in the image of $|j|$ we see that $\sim _ R$ is reflexive (details omitted).

To show that $\sim _ R$ is transitive assume given $\overline{u}_ i : \mathop{\mathrm{Spec}}(k) \to U$, $i = 1, 2, 3$ and field extensions $k \subset K_ i$ and points $r_ i : \mathop{\mathrm{Spec}}(K_ i) \to R$, $i = 1, 2$ such that $j(r_1) = (\overline{u}_1, \overline{u}_2)$ and $j(r_1) = (\overline{u}_2, \overline{u}_3)$. Then we may choose a commutative diagram of fields

$\xymatrix{ K & K_2 \ar[l] \\ K_1 \ar[u] & k \ar[l] \ar[u] }$

and we may think of $r_1, r_2 \in R(K)$. We consider the commutative solid diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K') \ar@{..>}[r] \ar@{..>}[d] & \mathop{\mathrm{Spec}}(k) \ar[d]_{(\overline{u}_1, \overline{u}_3)} & \mathop{\mathrm{Spec}}(K) \ar[d]^{(r_1, r_2)} \ar[l] \\ R \ar[r]^-j & U \times _ B U & R \times _{s, U, t} R \ar[l]_-{j_{comp}} }$

By exactly the same reasoning as in the first part of the proof, but this time using that $|j_{comp}|((r_1, r_2))$ is in the image of $|j|$, we conclude that a field $K'$ and dotted arrows exist making the diagram commute. This proves that $\sim _ R$ is transitive and concludes the proof that (2) implies (1).

Assume (1) and let $k$ be an algebraically closed field over $B$ whose cardinality is larger than $\lambda (R)$, see Morphisms of Spaces, Lemma 65.24.2. Suppose that $\overline{u} \sim _ R \overline{u}'$ with $\overline{u}, \overline{u}' \in U(k)$. By assumption there exists a point in $|R|$ mapping to $(\overline{u}, \overline{u}') \in |U \times _ B U|$. Hence by Morphisms of Spaces, Lemma 65.24.2 we conclude there exists an $\overline{r} \in R(k)$ with $j(\overline{r}) = (\overline{u}, \overline{u}')$. In this way we see that (1) implies (3).

Assume (3). Let us show that $\mathop{\mathrm{Im}}(|j_{comp}|) \subset \mathop{\mathrm{Im}}(|j|)$. Pick any point $c \in |R \times _{s, U, t} R|$. We may represent this by a morphism $\overline{c} : \mathop{\mathrm{Spec}}(k) \to R \times _{s, U, t} R$, with $k$ over $B$ having sufficiently large cardinality. By assumption we see that $j_{comp}(\overline{c}) \in U(k) \times U(k) = (U \times _ B U)(k)$ is also the image $j(\overline{r})$ for some $\overline{r} \in R(k)$. Hence $j_{comp}(c) = j(r)$ in $|U \times _ B U|$ as desired (with $r \in |R|$ the equivalence class of $\overline{r}$). The same argument shows also that $\mathop{\mathrm{Im}}(|j_{diag}|) \subset \mathop{\mathrm{Im}}(|j|)$ and $\mathop{\mathrm{Im}}(|j_{flip}|) \subset \mathop{\mathrm{Im}}(|j|)$ (details omitted). In this way we see that (3) implies (2). At this point we have shown that (1), (2) and (3) are all equivalent.

It is clear that (4) implies (3) (without any assumptions on $s$, $t$). To finish the proof of the lemma we show that (1) implies (4) if $s, t$ are locally of finite type. Namely, let $k$ be an algebraically closed field over $B$. Suppose that $\overline{u} \sim _ R \overline{u}'$ with $\overline{u}, \overline{u}' \in U(k)$. By assumption the algebraic space $Z = R \times _{j, U \times _ B U, (\overline{u}, \overline{u}')} \mathop{\mathrm{Spec}}(k)$ is nonempty. On the other hand, since $j = (t, s)$ is locally of finite type the morphism $Z \to \mathop{\mathrm{Spec}}(k)$ is locally of finite type as well (use Morphisms of Spaces, Lemmas 65.23.6 and 65.23.3). Hence $Z$ has a $k$ point by Morphisms of Spaces, Lemma 65.24.1 and we conclude that $(\overline{u}, \overline{u}') \in j(R(k))$ as desired. This finishes the proof of the lemma. $\square$

Lemma 81.5.15. In the situation of Definition 81.5.13. The following are equivalent:

1. The morphism $j$ is a set-theoretic equivalence relation.

2. The morphism $j$ is universally injective and $j(|R|) \subset |U \times _ B U|$ contains the image of $|j'|$ for any of the morphisms $j'$ as in Equation (81.5.5.1).

3. For every algebraically closed field $k$ over $B$ of sufficiently large cardinality the map $j : R(k) \to U(k) \times U(k)$ is injective and its image is an equivalence relation.

If $j$ is decent, or locally separated, or quasi-separated these are also equivalent to

1. For every algebraically closed field $k$ over $B$ the map $j : R(k) \to U(k) \times U(k)$ is injective and its image is an equivalence relation.

Proof. The implications (1) $\Rightarrow$ (2) and (2) $\Rightarrow$ (3) follow from Lemma 81.5.14 and the definitions. The same lemma shows that (3) implies $j$ is a set-theoretic pre-equivalence relation. But of course condition (3) also implies that $j$ is universally injective, see Morphisms of Spaces, Lemma 65.19.2, so that $j$ is indeed a set-theoretic equivalence relation. At this point we know that (1), (2), (3) are all equivalent.

Condition (4) implies (3) without any further hypotheses on $j$. Assume $j$ is decent, or locally separated, or quasi-separated and the equivalent conditions (1), (2), (3) hold. By More on Morphisms of Spaces, Lemma 74.3.4 we see that $j$ is radicial. Let $k$ be any algebraically closed field over $B$. Let $\overline{u}, \overline{u}' \in U(k)$ with $\overline{u} \sim _ R \overline{u}'$. We see that $R \times _{U \times _ B U, (\overline{u}, \overline{u}')} \mathop{\mathrm{Spec}}(k)$ is nonempty. Hence, as $j$ is radicial, its reduction is the spectrum of a field purely inseparable over $k$. As $k = \overline{k}$ we see that it is the spectrum of $k$. Whence a point $\overline{r} \in R(k)$ with $t(\overline{r}) = \overline{u}$ and $s(\overline{r}) = \overline{u}'$ as desired. $\square$

Lemma 81.5.16. Let $S$ be a scheme, and let $B$ be an algebraic space over $S$. Let $j : R \to U \times _ B U$ be a pre-relation over $B$.

1. If $j$ is a pre-equivalence relation, then $j$ is a set-theoretic pre-equivalence relation. This holds in particular when $j$ comes from a groupoid in algebraic spaces, or from an action of a group algebraic space on $U$.

2. If $j$ is an equivalence relation, then $j$ is a set-theoretic equivalence relation.

Proof. Omitted. $\square$

Lemma 81.5.17. Let $B \to S$ be as in Section 81.2. Let $j : R \to U \times _ B U$ be a pre-relation. Let $\phi : U \to X$ be a morphism of algebraic spaces over $B$. Consider the diagram

$\xymatrix{ (U \times _ X U) \times _{(U \times _ B U)} R \ar[d]^ q \ar[r]_-p & R \ar[d]^ j \\ U \times _ X U \ar[r]^ c & U \times _ B U }$

Then we have:

1. The morphism $\phi$ is set-theoretically invariant if and only if $p$ is surjective.

2. If $j$ is a set-theoretic pre-equivalence relation then $\phi$ separates orbits if and only if $p$ and $q$ are surjective.

3. If $p$ and $q$ are surjective, then $j$ is a set-theoretic pre-equivalence relation (and $\phi$ separates orbits).

4. If $\phi$ is $R$-invariant and $j$ is a set-theoretic pre-equivalence relation, then $\phi$ separates orbits if and only if the induced morphism $R \to U \times _ X U$ is surjective.

Proof. Assume $\phi$ is set-theoretically invariant. This means that for any algebraically closed field $k$ over $B$ and any $\overline{r} \in R(k)$ we have $\phi (s(\overline{r})) = \phi (t(\overline{r}))$. Hence $((\phi (t(\overline{r})), \phi (s(\overline{r}))), \overline{r})$ defines a point in the fibre product mapping to $\overline{r}$ via $p$. This shows that $p$ is surjective. Conversely, assume $p$ is surjective. Pick $\overline{r} \in R(k)$. As $p$ is surjective, we can find a field extension $k \subset K$ and a $K$-valued point $\tilde r$ of the fibre product with $p(\tilde r) = \overline{r}$. Then $q(\tilde r) \in U \times _ X U$ maps to $(t(\overline{r}), s(\overline{r}))$ in $U \times _ B U$ and we conclude that $\phi (s(\overline{r})) = \phi (t(\overline{r}))$. This proves that $\phi$ is set-theoretically invariant.

The proofs of (2), (3), and (4) are omitted. Hint: Assume $k$ is an algebraically closed field over $B$ of large cardinality. Consider the associated diagram of sets

$\xymatrix{ (U(k) \times _{X(k)} U(k)) \times _{U(k) \times U(k)} R(k) \ar[d]^ q \ar[r]_-p & R(k) \ar[d]^ j \\ U(k) \times _{X(k)} U(k) \ar[r]^ c & U(k) \times U(k) }$

By the lemmas above the equivalences posed in (2), (3), and (4) become set-theoretic questions related to the diagram we just displayed, using that surjectivity translates into surjectivity on $k$-valued points by Morphisms of Spaces, Lemma 65.24.2. $\square$

Because we have seen above that the notion of a set-theoretically invariant morphism is a rather weak one in the category of algebraic spaces, we define an orbit space for a pre-relation as follows.

Definition 81.5.18. Let $B \to S$ as in Section 81.2. Let $j : R \to U \times _ B U$ be a pre-relation. We say $\phi : U \to X$ is an orbit space for $R$ if

1. $\phi$ is $R$-invariant,

2. $\phi$ separates $R$-orbits, and

3. $\phi$ is surjective.

The definition of separating $R$-orbits involves a discussion of points with values in algebraically closed fields. But as we've seen in many cases this just corresponds to the surjectivity of certain canonically associated morphisms of algebraic spaces. We summarize some of the discussion above in the following characterization of orbit spaces.

Lemma 81.5.19. Let $B \to S$ as in Section 81.2. Let $j : R \to U \times _ B U$ be a set-theoretic pre-equivalence relation. A morphism $\phi : U \to X$ is an orbit space for $R$ if and only if

1. $\phi \circ s = \phi \circ t$, i.e., $\phi$ is invariant,

2. the induced morphism $(t, s) : R \to U \times _ X U$ is surjective, and

3. the morphism $\phi : U \to X$ is surjective.

This characterization applies for example if $j$ is a pre-equivalence relation, or comes from a groupoid in algebraic spaces over $B$, or comes from the action of a group algebraic space over $B$ on $U$.

Proof. Follows immediately from Lemma 81.5.17 part (4). $\square$

In the following lemma it is (probably) not good enough to assume just that the morphisms $s, t$ are locally of finite type. The reason is that it may happen that some map $\phi : U \to X$ is an orbit space, yet is not locally of finite type. In that case $U(k) \to X(k)$ may not be surjective for all algebraically closed fields $k$ over $B$.

Lemma 81.5.20. Let $B \to S$ as in Section 81.2. Let $j = (t, s) : R \to U \times _ B U$ be a pre-relation. Assume $R, U$ are locally of finite type over $B$. Let $\phi : U \to X$ be an $R$-invariant morphism of algebraic spaces over $B$. Then $\phi$ is an orbit space for $R$ if and only if the natural map

$U(k)/\big (\text{equivalence relation generated by }j(R(k))\big ) \longrightarrow X(k)$

is bijective for all algebraically closed fields $k$ over $B$.

Proof. Note that since $U$, $R$ are locally of finite type over $B$ all of the morphisms $s, t, j, \phi$ are locally of finite type, see Morphisms of Spaces, Lemma 65.23.6. We will also use without further mention Morphisms of Spaces, Lemma 65.24.1. Assume $\phi$ is an orbit space. Let $k$ be any algebraically closed field over $B$. Let $\overline{x} \in X(k)$. Consider $U \times _{\phi , X, \overline{x}} \mathop{\mathrm{Spec}}(k)$. This is a nonempty algebraic space which is locally of finite type over $k$. Hence it has a $k$-valued point. This shows the displayed map of the lemma is surjective. Suppose that $\overline{u}, \overline{u}' \in U(k)$ map to the same element of $X(k)$. By Definition 81.5.8 this means that $\overline{u}, \overline{u}'$ are in the same $R$-orbit. By Lemma 81.5.7 this means that they are equivalent under the equivalence relation generated by $j(R(k))$. Thus the displayed morphism is injective.

Conversely, assume the displayed map is bijective for all algebraically closed fields $k$ over $B$. This condition clearly implies that $\phi$ is surjective. We have already assumed that $\phi$ is $R$-invariant. Finally, the injectivity of all the displayed maps implies that $\phi$ separates orbits. Hence $\phi$ is an orbit space. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).