Lemma 97.11.5. Let $\mathcal{Z}$ be an algebraic stack. Assume $\mathcal{Z}$ satisfies the equivalent conditions of Lemma 97.11.2. 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 97.11.3. 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 97.8.4. 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 64.28.7). We see that $\mathcal{Z}' = [U/R]$ is an algebraic stack by Criteria for Representability, Theorem 94.17.2. By Algebraic Stacks, Lemma 91.16.1 we obtain a canonical morphism

which is fully faithful. Hence this morphism is representable by algebraic spaces, see Algebraic Stacks, Lemma 91.15.2 and a monomorphism, see Lemma 97.8.4. By Criteria for Representability, Lemma 94.17.1 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 97.11.3. By Algebraic Stacks, Lemma 91.12.4 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 97.4.3. Hence the two projections are isomorphisms by Lemma 97.11.4 and we win. $\square$

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