Lemma 82.5.20. Let $B \to S$ as in Section 82.2. 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 66.23.6. We will also use without further mention Morphisms of Spaces, Lemma 66.24.1. 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 82.5.8 this means that $\overline{u}, \overline{u}'$ are in the same $R$-orbit. By Lemma 82.5.7 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.
$\square$

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