English
In a filtered category, for morphisms f,f' : j ⟶ j', the coequalizing maps satisfy f ≫ coeqHom f f' = f' ≫ coeqHom f f'.
Русский
В фильтрованной категории равенство мコピー f и f' композитивно с коэквационным гомоморфизмом: f ≫ coeqHom f f' = f' ≫ coeqHom f f'.
LaTeX
$$$f \gg coeqHom f f' = f' \gg coeqHom f f'$$$
Lean4
/-- `coeq_condition f f'`, for morphisms `f f' : j ⟶ j'`, is the proof that
`f ≫ coeqHom f f' = f' ≫ coeqHom f f'`.
-/
@[reassoc] -- Not `@[simp]` as it does not fire.
theorem coeq_condition {j j' : C} (f f' : j ⟶ j') : f ≫ coeqHom f f' = f' ≫ coeqHom f f' :=
(IsFilteredOrEmpty.cocone_maps f f').choose_spec.choose_spec