Metamath Proof Explorer


Theorem lmimlmhm

Description: An isomorphism of modules is a homomorphism. (Contributed by Stefan O'Rear, 21-Jan-2015)

Ref Expression
Assertion lmimlmhm ( 𝐹 ∈ ( 𝑅 LMIso 𝑆 ) → 𝐹 ∈ ( 𝑅 LMHom 𝑆 ) )

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 3 simplbi ( 𝐹 ∈ ( 𝑅 LMIso 𝑆 ) → 𝐹 ∈ ( 𝑅 LMHom 𝑆 ) )