Metamath Proof Explorer


Theorem epihom

Description: An epimorphism is a morphism. (Contributed by Mario Carneiro, 2-Jan-2017)

Ref Expression
Hypotheses isepi.b
|- B = ( Base ` C )
isepi.h
|- H = ( Hom ` C )
isepi.o
|- .x. = ( comp ` C )
isepi.e
|- E = ( Epi ` C )
isepi.c
|- ( ph -> C e. Cat )
isepi.x
|- ( ph -> X e. B )
isepi.y
|- ( ph -> Y e. B )
Assertion epihom
|- ( ph -> ( X E Y ) C_ ( X H Y ) )

Proof

Step Hyp Ref Expression
1 isepi.b
 |-  B = ( Base ` C )
2 isepi.h
 |-  H = ( Hom ` C )
3 isepi.o
 |-  .x. = ( comp ` C )
4 isepi.e
 |-  E = ( Epi ` C )
5 isepi.c
 |-  ( ph -> C e. Cat )
6 isepi.x
 |-  ( ph -> X e. B )
7 isepi.y
 |-  ( ph -> Y e. B )
8 1 2 3 4 5 6 7 isepi
 |-  ( ph -> ( f e. ( X E Y ) <-> ( f e. ( X H Y ) /\ A. z e. B Fun `' ( g e. ( Y H z ) |-> ( g ( <. X , Y >. .x. z ) f ) ) ) ) )
9 simpl
 |-  ( ( f e. ( X H Y ) /\ A. z e. B Fun `' ( g e. ( Y H z ) |-> ( g ( <. X , Y >. .x. z ) f ) ) ) -> f e. ( X H Y ) )
10 8 9 syl6bi
 |-  ( ph -> ( f e. ( X E Y ) -> f e. ( X H Y ) ) )
11 10 ssrdv
 |-  ( ph -> ( X E Y ) C_ ( X H Y ) )