Lemma 83.5.17 (049Y)—The Stacks project

Lemma 83.5.17. Let $B \to S$ be as in Section 83.2. 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:

The morphism $\phi$ is set-theoretically invariant if and only if $p$ is surjective.
If $j$ is a set-theoretic pre-equivalence relation then $\phi$ separates orbits if and only if $p$ and $q$ are surjective.
If $p$ and $q$ are surjective, then $j$ is a set-theoretic pre-equivalence relation (and $\phi$ separates orbits).
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 $K/k$ 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 67.24.2. $\square$

The Stacks project

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