mapId_eq — Mathlib

English

Mapping by identity yields the identity functor: map(id_Y) = id_{Under Y}.

Русский

отображение через идентичность дает тождественный функтор: map(id_Y) = id_{Under Y}.

LaTeX

$$map(\\mathrm{id} _Y) = \\mathrm{id} (\\mathrm{Under}\\, Y)$$

Lean4

/-- Mapping by the identity morphism is just the identity functor. -/
theorem mapId_eq (Y : T) : map (𝟙 Y) = 𝟭 _ := by
  fapply Functor.ext
  · intro x
    dsimp [Under, Under.map, Comma.mapLeft]
    simp only [Category.id_comp]
    exact rfl
  · intro x y u
    dsimp [Under, Under.map, Comma.mapLeft]
    simp

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