Metamath Proof Explorer


Theorem islmim2

Description: An isomorphism of left modules is a homomorphism whose converse is a homomorphism. (Contributed by Mario Carneiro, 6-May-2015)

Ref Expression
Assertion islmim2 ⊢ F ∈ R LMIso S ↔ F ∈ R LMHom S ∧ F -1 ∈ S LMHom R

Proof

Step Hyp Ref Expression
1 eqid ⊢ Base R = Base R
2 eqid ⊢ Base S = Base S
3 1 2 islmim ⊢ F ∈ R LMIso S ↔ F ∈ R LMHom S ∧ F : Base R ⟶ 1-1 onto Base S
4 1 2 lmhmf1o ⊢ F ∈ R LMHom S → F : Base R ⟶ 1-1 onto Base S ↔ F -1 ∈ S LMHom R
5 4 pm5.32i ⊢ F ∈ R LMHom S ∧ F : Base R ⟶ 1-1 onto Base S ↔ F ∈ R LMHom S ∧ F -1 ∈ S LMHom R
6 3 5 bitri ⊢ F ∈ R LMIso S ↔ F ∈ R LMHom S ∧ F -1 ∈ S LMHom R