The Stacks project

Definition 39.3.1. Let $S$ be a scheme. Let $U$ be a scheme over $S$.

  1. A pre-relation on $U$ over $S$ is any morphism of schemes $j : R \to U \times _ S U$. In this case we set $t = \text{pr}_0 \circ j$ and $s = \text{pr}_1 \circ j$, so that $j = (t, s)$.

  2. A relation on $U$ over $S$ is a monomorphism of schemes $j : R \to U \times _ S U$.

  3. A pre-equivalence relation is a pre-relation $j : R \to U \times _ S U$ such that the image of $j : R(T) \to U(T) \times U(T)$ is an equivalence relation for all $T/S$.

  4. We say a morphism $R \to U \times _ S U$ of schemes is an equivalence relation on $U$ over $S$ if and only if for every scheme $T$ over $S$ the $T$-valued points of $R$ define an equivalence relation on the set of $T$-valued points of $U$.

Comments (2)

Comment #6527 by Jeroen Hekking on

Is there a typo in (4)? I guess an equivalence relation is a pre-equivalence relation which is a relation. The latter seems to be missing, or I'm missing something obvious.

Comment #6528 by on

Sorry, but what would be the typo? The condition as currently formulated in (4) is equivalent to (2) + (3) as currently formulated. Right?

There are also:

  • 6 comment(s) on Section 39.3: Equivalence relations

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Your email address will not be published.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 022P. Beware of the difference between the letter 'O' and the digit '0'.