Lemma 100.11.5. Let $\mathcal{Z}$ be an algebraic stack. Assume $\mathcal{Z}$ satisfies the equivalent conditions of Lemma 100.11.2. Then there exists a unique strictly full subcategory $\mathcal{Z}' \subset \mathcal{Z}$ such that $\mathcal{Z}'$ is an algebraic stack which satisfies the equivalent conditions of Lemma 100.11.3. The inclusion morphism $\mathcal{Z}' \to \mathcal{Z}$ is a monomorphism of algebraic stacks.

**Proof.**
The last part is immediate from the first part and Lemma 100.8.4. Pick a field $k$ and a morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{Z}$ which is surjective, flat, and locally of finite type. Set $U = \mathop{\mathrm{Spec}}(k)$ and $R = U \times _\mathcal {Z} U$. The projections $s, t : R \to U$ are locally of finite type. Since $U$ is the spectrum of a field, it follows that $s, t$ are flat and locally of finite presentation (by Morphisms of Spaces, Lemma 67.28.7). We see that $\mathcal{Z}' = [U/R]$ is an algebraic stack by Criteria for Representability, Theorem 97.17.2. By Algebraic Stacks, Lemma 94.16.1 we obtain a canonical morphism

which is fully faithful. Hence this morphism is representable by algebraic spaces, see Algebraic Stacks, Lemma 94.15.2 and a monomorphism, see Lemma 100.8.4. By Criteria for Representability, Lemma 97.17.1 the morphism $U \to \mathcal{Z}'$ is surjective, flat, and locally of finite presentation. Hence $\mathcal{Z}'$ is an algebraic stack which satisfies the equivalent conditions of Lemma 100.11.3. By Algebraic Stacks, Lemma 94.12.4 we may replace $\mathcal{Z}'$ by its essential image in $\mathcal{Z}$. Hence we have proved all the assertions of the lemma except for the uniqueness of $\mathcal{Z}' \subset \mathcal{Z}$. Suppose that $\mathcal{Z}'' \subset \mathcal{Z}$ is a second such algebraic stack. Then the projections

are monomorphisms. The algebraic stack in the middle is nonempty by Lemma 100.4.3. Hence the two projections are isomorphisms by Lemma 100.11.4 and we win. $\square$

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