Theorems · Inductive type · category theory
CategoryTheory.MorphismProperty.HasPushoutsAlong
{C : Type u} →
[inst : CategoryTheory.Category.{v, u} C] → CategoryTheory.MorphismProperty C → {X Y : C} → (X ⟶ Y) → PropP.HasPushoutsAlong f states that for any morphism satisfying P with the same domain
as f, the pushout of that morphism along f exists.
- Cited by
- 12 results in Mathlib
- Foundations
- Depth 3 from the axioms · uses no axioms
- Assumes
- CategoryTheory.Category
Around this declaration
Dashed lines are statement dependencies; solid lines are citations in proofs.
Cites3
Mathlib declarations this one mentions in its statement or cites explicitly in its proof. Plumbing is filtered out.
- CategoryTheory.Categorystatement · cited by 32,673
- Quiver.Homstatement · cited by 32,603
- CategoryTheory.MorphismPropertystatement · cited by 2,179
Cited by20
Results whose statement or proof uses this declaration.
- CategoryTheory.MorphismProperty.Under.pushoutstatement and proof · cited by 15
- CategoryTheory.MorphismProperty.Under.mapPushoutAdjstatement and proof · cited by 3
- CategoryTheory.MorphismProperty.HasPushoutsAlong.hasPushoutstatement and proof · cited by 2
- CategoryTheory.MorphismProperty.Under.pushoutCompstatement and proof · cited by 2
- CategoryTheory.MorphismProperty.Under.pushoutCongrstatement and proof · cited by 2
- CategoryTheory.MorphismProperty.Under.pushoutCongr_hom_app_left_fststatement and proof · cited by 1
- CategoryTheory.MorphismProperty.isLeftAdjoint_pushoutstatement and proof · cited by 0
- CategoryTheory.MorphismProperty.HasPushoutsAgainst.casesOnstatement and proof · cited by 0
- CategoryTheory.MorphismProperty.HasPushoutsAgainst.hasPushoutsAlongstatement · cited by 0
- CategoryTheory.MorphismProperty.HasPushoutsAgainst.recOnstatement and proof · cited by 0
- CategoryTheory.MorphismProperty.HasPushoutsAlong.casesOnstatement and proof · cited by 0
- CategoryTheory.MorphismProperty.HasPushoutsAlong.recOnstatement and proof · cited by 0