English
The morphism prodComparison is natural in both arguments: F.map (prod.map f g) then prodComparison equals prod.map (F.map f) (F.map g) then prodComparison.
Русский
У натуральности prodComparison по обоим аргументам: F.map (prod.map f g) затем prodComparison равно prod.map (F.map f) (F.map g) затем prodComparison.
LaTeX
$$$ F.map (\\operatorname{prod.map} f g) \\; \\circ \\operatorname{prodComparison} F A B = \\operatorname{prod.map} (F.map f) (F.map g) \\circ \\operatorname{prodComparison} F A' B' $$$
Lean4
@[reassoc]
theorem inv_prodComparison_map_fst [IsIso (prodComparison F A B)] :
inv (prodComparison F A B) ≫ F.map prod.fst = prod.fst := by simp [IsIso.inv_comp_eq]