English
There is a canonical linear isomorphism between M and the free module on its basis index.
Русский
Существует каноническое линейное изоморфизм между M и свободной модулем на его индексном базисе.
LaTeX
$$$$ M \cong_R (\mathrm{ChooseBasisIndex}(R,M) \to_0 R) $$$$
Lean4
/-- The isomorphism `M ≃ₗ[R] (ChooseBasisIndex R M →₀ R)`. -/
@[deprecated Module.Free.chooseBasis (since := "2025-08-01")]
noncomputable def repr : M ≃ₗ[R] ChooseBasisIndex R M →₀ R :=
(chooseBasis R M).repr