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