## 77.4 Equivalence relations

Please refer to Groupoids, Section 39.3 for notation.

Definition 77.4.1. Let $B \to S$ as in Section 77.3. Let $U$ be an algebraic space over $B$.

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

A *relation* on $U$ over $B$ is a monomorphism $j : R \to U \times _ B U$ of algebraic spaces over $B$.

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

We say a morphism $R \to U \times _ B U$ of algebraic spaces over $B$ is an *equivalence relation on $U$ over $B$* if and only if for every $T$ over $B$ the $T$-valued points of $R$ define an equivalence relation on the set of $T$-valued points of $U$.

In other words, an equivalence relation is a pre-equivalence relation such that $j$ is a relation.

Lemma 77.4.2. Let $B \to S$ as in Section 77.3. Let $U$ be an algebraic space over $B$. Let $j : R \to U \times _ B U$ be a pre-relation. Let $g : U' \to U$ be a morphism of algebraic spaces over $B$. Finally, set

\[ R' = (U' \times _ B U')\times _{U \times _ B U} R \xrightarrow {j'} U' \times _ B U' \]

Then $j'$ is a pre-relation on $U'$ over $B$. If $j$ is a relation, then $j'$ is a relation. If $j$ is a pre-equivalence relation, then $j'$ is a pre-equivalence relation. If $j$ is an equivalence relation, then $j'$ is an equivalence relation.

**Proof.**
Omitted.
$\square$

Definition 77.4.3. Let $B \to S$ as in Section 77.3. Let $U$ be an algebraic space over $B$. Let $j : R \to U \times _ B U$ be a pre-relation. Let $g : U' \to U$ be a morphism of algebraic spaces over $B$. The pre-relation $j' : R' \to U' \times _ B U'$ of Lemma 77.4.2 is called the *restriction*, or *pullback* of the pre-relation $j$ to $U'$. In this situation we sometimes write $R' = R|_{U'}$.

Lemma 77.4.4. Let $B \to S$ as in Section 77.3. Let $j : R \to U \times _ B U$ be a pre-relation of algebraic spaces over $B$. Consider the relation on $|U|$ defined by the rule

\[ x \sim y \Leftrightarrow \exists \ r \in |R| : t(r) = x, s(r) = y. \]

If $j$ is a pre-equivalence relation then this is an equivalence relation.

**Proof.**
Suppose that $x \sim y$ and $y \sim z$. Pick $r \in |R|$ with $t(r) = x$, $s(r) = y$ and pick $r' \in |R|$ with $t(r') = y$, $s(r') = z$. We may pick a field $K$ such that $r$ and $r'$ can be represented by morphisms $r, r' : \mathop{\mathrm{Spec}}(K) \to R$ with $s \circ r = t \circ r'$. Denote $x = t \circ r$, $y = s \circ r = t \circ r'$, and $z = s \circ r'$, so $x, y, z : \mathop{\mathrm{Spec}}(K) \to U$. By construction $(x, y) \in j(R(K))$ and $(y, z) \in j(R(K))$. Since $j$ is a pre-equivalence relation we see that also $(x, z) \in j(R(K))$. This clearly implies that $x \sim z$.

The proof that $\sim $ is reflexive and symmetric is omitted.
$\square$

## Comments (0)