The Stacks Project

Tag 04XJ

Lemma 90.4.5. Let $\mathcal{X}$ be an algebraic stack. Let $\mathcal{X} = [U/R]$ be a presentation of $\mathcal{X}$, see Algebraic Stacks, Definition 84.16.5. Then the image of $|R| \to |U| \times |U|$ is an equivalence relation and $|\mathcal{X}|$ is the quotient of $|U|$ by this equivalence relation.

Proof. The assumption means that we have a smooth groupoid $(U, R, s, t, c)$ in algebraic spaces, and an equivalence $f : [U/R] \to \mathcal{X}$. We may assume $\mathcal{X} = [U/R]$. The induced morphism $p : U \to \mathcal{X}$ is smooth and surjective, see Algebraic Stacks, Lemma 84.17.2. Hence $|U| \to |\mathcal{X}|$ is surjective by Lemma 90.4.4. Note that $R = U \times_\mathcal{X} U$, see Groupoids in Spaces, Lemma 69.21.2. Hence Lemma 90.4.3 implies the map $$|R| \longrightarrow |U| \times_{|\mathcal{X}|} |U|$$ is surjective. Hence the image of $|R| \to |U| \times |U|$ is exactly the set of pairs $(u_1, u_2) \in |U| \times |U|$ such that $u_1$ and $u_2$ have the same image in $|\mathcal{X}|$. Combining these two statements we get the result of the lemma. $\square$

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\begin{lemma}
\label{lemma-points-presentation}
Let $\mathcal{X}$ be an algebraic stack.
Let $\mathcal{X} = [U/R]$ be a presentation of $\mathcal{X}$, see
Algebraic Stacks, Definition \ref{algebraic-definition-presentation}.
Then the image of $|R| \to |U| \times |U|$ is an equivalence relation
and $|\mathcal{X}|$ is the quotient of $|U|$ by this equivalence relation.
\end{lemma}

\begin{proof}
The assumption means that we have a smooth groupoid $(U, R, s, t, c)$
in algebraic spaces, and an equivalence $f : [U/R] \to \mathcal{X}$.
We may assume $\mathcal{X} = [U/R]$.
The induced morphism $p : U \to \mathcal{X}$ is smooth and surjective, see
Algebraic Stacks,
Lemma \ref{algebraic-lemma-smooth-quotient-smooth-presentation}.
Hence $|U| \to |\mathcal{X}|$ is surjective by
Lemma \ref{lemma-characterize-surjective}.
Note that $R = U \times_\mathcal{X} U$, see
Groupoids in Spaces,
Lemma \ref{spaces-groupoids-lemma-quotient-stack-2-cartesian}.
Hence
Lemma \ref{lemma-points-cartesian}
implies the map
$$|R| \longrightarrow |U| \times_{|\mathcal{X}|} |U|$$
is surjective. Hence the image of $|R| \to |U| \times |U|$ is
exactly the set of pairs $(u_1, u_2) \in |U| \times |U|$
such that $u_1$ and $u_2$ have the same image in $|\mathcal{X}|$.
Combining these two statements we get the result of the lemma.
\end{proof}

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