Metamath Proof Explorer


Theorem 2fcoidinvd

Description: Show that a function is the inverse of a function if their compositions are the identity functions. (Contributed by Mario Carneiro, 21-Mar-2015) (Revised by AV, 15-Dec-2019)

Ref Expression
Hypotheses fcof1od.f
|- ( ph -> F : A --> B )
fcof1od.g
|- ( ph -> G : B --> A )
fcof1od.a
|- ( ph -> ( G o. F ) = ( _I |` A ) )
fcof1od.b
|- ( ph -> ( F o. G ) = ( _I |` B ) )
Assertion 2fcoidinvd
|- ( ph -> `' F = G )

Proof

Step Hyp Ref Expression
1 fcof1od.f
 |-  ( ph -> F : A --> B )
2 fcof1od.g
 |-  ( ph -> G : B --> A )
3 fcof1od.a
 |-  ( ph -> ( G o. F ) = ( _I |` A ) )
4 fcof1od.b
 |-  ( ph -> ( F o. G ) = ( _I |` B ) )
5 1 2 3 4 fcof1od
 |-  ( ph -> F : A -1-1-onto-> B )
6 5 2 4 fcof1oinvd
 |-  ( ph -> `' F = G )