Definition 101.44.1. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. We say $f$ is a local complete intersection morphism or Koszul if the equivalent conditions of Lemma 101.16.1 hold with $\mathcal{P} = \text{local complete intersection}$.
101.44 Local complete intersection morphisms
The property “being a local complete intersection morphism” of morphisms of algebraic spaces is smooth local on the source-and-target, see Descent on Spaces, Lemma 74.20.4 and More on Morphisms of Spaces, Lemmas 76.48.9 and 76.48.10. By Lemma 101.16.1 above, we may define what it means for a morphism of algebraic spaces to be a local complete intersection morphism as follows and it agrees with the already existing notion defined in More on Morphisms of Spaces, Section 76.48 when both source and target are algebraic spaces.
Lemma 101.44.2. The composition of local complete intersection morphisms is a local complete intersection.
Proof. Combine Remark 101.16.3 with More on Morphisms of Spaces, Lemma 76.48.5. $\square$
Lemma 101.44.3. A flat base change of a local complete intersection morphism is a local complete intersection morphism.
Proof. Omitted. Hint: Argue exactly as in Remark 101.16.4 (but only for flat $\mathcal{Y}' \to \mathcal{Y}$) using More on Morphisms of Spaces, Lemma 76.48.4. $\square$
Lemma 101.44.4. Let be a commutative diagram of morphisms of algebraic stacks. Assume $\mathcal{Y} \to \mathcal{Z}$ is smooth and $\mathcal{X} \to \mathcal{Z}$ is a local complete intersection morphism. Then $f : \mathcal{X} \to \mathcal{Y}$ is a local complete intersection morphism.
Proof. Choose a scheme $W$ and a surjective smooth morphism $W \to \mathcal{Z}$. Choose a scheme $V$ and a surjective smooth morphism $V \to W \times _\mathcal {Z} \mathcal{Y}$. Choose a scheme $U$ and a surjective smooth morphism $U \to V \times _\mathcal {Y} \mathcal{X}$. Then $U \to W$ is a local complete intersection morphism of schemes and $V \to W$ is a smooth morphism of schemes. By the result for schemes (More on Morphisms, Lemma 37.62.10) we conclude that $U \to V$ is a local complete intersection morphism. By definition this means that $f$ is a local complete intersection morphism. $\square$
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