English
For a in R and s in S, algebraMap R A a under the mk' construction equals mk (algebraMap R A a) s, i.e., the algebra map is transported along mk'.
Русский
Для a в R и s в S алгебраическая карта по mk' даёт mk (algebraMap R A a) s, то есть отображение алгебраической карты переносится вдоль mk'.
LaTeX
$$$\\mathrm{algebraMap}\\_{}(R,A)(a) = \\mathrm{mk}(\\mathrm{algebraMap}\_{R,A}(a), s)$$$
Lean4
theorem algebraMap_mk' {A : Type*} [Semiring A] [Algebra R A] (a : R) (s : S) :
algebraMap _ _ (IsLocalization.mk' T a s) = mk (algebraMap R A a) s :=
by
rw [Algebra.algebraMap_eq_smul_one]
change _ • mk _ _ = _
rw [mk'_smul_mk, Algebra.algebraMap_eq_smul_one, mul_one]