Theorems · Inductive type · algebraic geometry
AlgebraicGeometry.Scheme.RationalMap.IsDominant
{X Y : AlgebraicGeometry.Scheme} → X.RationalMap Y → PropA rational map is dominant if some (equivalently, any) representative partial map has dominant underlying morphism.
- Cited by
- 8 results in Mathlib
- Foundations
- Depth 148 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.
- AlgebraicGeometry.Schemestatement · cited by 2,540
- AlgebraicGeometry.Scheme.RationalMapstatement · cited by 25
Cited by11
Results whose statement or proof uses this declaration.
- AlgebraicGeometry.Scheme.RationalMap.compstatement and proof · cited by 7
- AlgebraicGeometry.Scheme.RationalMap.comp_defstatement and proof · cited by 2
- AlgebraicGeometry.Scheme.RationalMap.isDominant_iffstatement and proof · cited by 1
- AlgebraicGeometry.Scheme.RationalMap.IsDominant.casesOnstatement and proof · cited by 1
- AlgebraicGeometry.Scheme.PartialMap.isDominant_toRationalMap_iffstatement · cited by 0
- AlgebraicGeometry.Scheme.RationalMap.comp_assocstatement and proof · cited by 0
- AlgebraicGeometry.Scheme.RationalMap.comp_idstatement and proof · cited by 0
- AlgebraicGeometry.Scheme.RationalMap.comp_toRationalMapstatement and proof · cited by 0
- AlgebraicGeometry.Scheme.RationalMap.IsDominant.outstatement and proof · cited by 0
- AlgebraicGeometry.Scheme.RationalMap.IsDominant.recOnstatement and proof · cited by 0
- AlgebraicGeometry.Scheme.RationalMap.comp.congr_simpstatement and proof · cited by 0