The Stacks project

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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 04A0. Beware of the difference between the letter 'O' and the digit '0'.