Theorems · Theorem · category theory
CategoryTheory.IsCofiltered.eq_condition_assoc
∀ {C : Type u} [inst : CategoryTheory.Category.{v, u} C] [inst_1 : CategoryTheory.IsCofilteredOrEmpty C] {j j' : C}
(f f' : j ⟶ j') {Z : C} (h : j' ⟶ Z),
CategoryTheory.CategoryStruct.comp (CategoryTheory.IsCofiltered.eqHom f f') (CategoryTheory.CategoryStruct.comp f h) =
CategoryTheory.CategoryStruct.comp (CategoryTheory.IsCofiltered.eqHom f f')
(CategoryTheory.CategoryStruct.comp f' h)eq_condition f f', for morphisms f f' : j ⟶ j', is the proof that
eqHom f f' ≫ f = eqHom f f' ≫ f'.
- Defined in
- Mathlib.CategoryTheory.Filtered.Basic
- Cited by
- 0 results in Mathlib
- Foundations
- Depth 13 from the axioms · uses Classical.choice, Quot.sound
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Mathlib declarations this one mentions in its statement or cites explicitly in its proof. Plumbing is filtered out.
- CategoryTheory.Categorystatement and proof · cited by 32,673
- Quiver.Homstatement and proof · cited by 32,603
- CategoryTheory.CategoryStruct.compstatement · cited by 17,999
- CategoryTheory.Category.assocproof · cited by 6,433
- CategoryTheory.IsCofilteredOrEmptystatement and proof · cited by 55
- CategoryTheory.IsCofiltered.eqstatement · cited by 8
- CategoryTheory.IsCofiltered.eqHomstatement and proof · cited by 6
- CategoryTheory.IsCofiltered.eq_conditionproof · cited by 4
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