Theorems · Inductive type · algebraic geometry
AlgebraicGeometry.IsPreimmersion
{X Y : AlgebraicGeometry.Scheme} → (X ⟶ Y) → PropA morphism of schemes f : X ⟶ Y is a preimmersion if the underlying map of
topological spaces is an embedding and the induced morphisms of stalks are all surjective.
- Cited by
- 14 results in Mathlib
- Foundations
- Depth 99 from the axioms · uses propext, Classical.choice, Quot.sound
Around this declaration
Dashed lines are statement dependencies; solid lines are citations in proofs.
Cites2
Mathlib declarations this one mentions in its statement or cites explicitly in its proof. Plumbing is filtered out.
- Quiver.Homstatement · cited by 32,603
- AlgebraicGeometry.Schemestatement · cited by 2,540
Cited by18
Results whose statement or proof uses this declaration.
- AlgebraicGeometry.Scheme.Hom.isEmbeddingstatement · cited by 7
- AlgebraicGeometry.IsClosedImmersion.of_isPreimmersionstatement and proof · cited by 4
- AlgebraicGeometry.isPreimmersion_iffstatement and proof · cited by 3
- AlgebraicGeometry.IsPreimmersion.SpecMap_iffstatement and proof · cited by 1
- AlgebraicGeometry.IsPreimmersion.casesOnstatement and proof · cited by 1
- AlgebraicGeometry.isImmersion_iffstatement and proof · cited by 1
- AlgebraicGeometry.IsClosedImmersion.iff_isPreimmersionstatement and proof · cited by 1
- AlgebraicGeometry.IsPreimmersion.isEmbeddingstatement and proof · cited by 1
- AlgebraicGeometry.IsPreimmersion.mk_SpecMapstatement · cited by 1
- AlgebraicGeometry.IsPreimmersion.of_compstatement and proof · cited by 1
- AlgebraicGeometry.IsImmersion.casesOnstatement and proof · cited by 1
- AlgebraicGeometry.IsImmersion.recOnstatement and proof · cited by 0