English
The counitInv on e.functor X ⊗ e.functor Y composed with e.functor.map(δ e.inverse (e.functor.obj X) (e.functor.obj Y)) equals μ e.functor X Y composed with e.functor.map(unit morphisms at X and Y).
Русский
counitInv на e.functor.obj X ⊗ e.functor.obj Y, далее δ через e.inverse, равно μ e.functor X Y, затем отображение через unit по X и Y.
LaTeX
$$$e.counitInv.app (e.functor.obj X \\otimes e.functor.obj Y) \\\\circ e.functor.map (\\\\delta e.inverse (e.functor.obj X) (e.functor.obj Y)) = μ e.functor X Y \\\\circ e.functor.map (e.unitIso.hom.app X \\otimes_? e.unitIso.hom.app Y)$$$
Lean4
@[reassoc (attr := simp)]
theorem map_η_comp_η : e.functor.map (η e.inverse) ≫ η e.functor = e.counit.app (𝟙_ D) :=
e.toAdjunction.map_η_comp_η