The Stacks project

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 67.23.6. We will also use without further mention Morphisms of Spaces, Lemma 67.24.1. 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 83.5.8 this means that $\overline{u}, \overline{u}'$ are in the same $R$-orbit. By Lemma 83.5.7 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. $\square$