Metamath Proof Explorer


Theorem idfth

Description: The inclusion functor is a faithful functor. (Contributed by Zhi Wang, 10-Nov-2025)

Ref Expression
Hypothesis idfth.i I = id func C
Assertion idfth I D Func E I D Faith E

Proof

Step Hyp Ref Expression
1 idfth.i I = id func C
2 relfunc Rel D Func E
3 1st2nd Rel D Func E I D Func E I = 1 st I 2 nd I
4 2 3 mpan I D Func E I = 1 st I 2 nd I
5 id I D Func E I D Func E
6 5 func1st2nd I D Func E 1 st I D Func E 2 nd I
7 f1oi I x Hom D y : x Hom D y 1-1 onto x Hom D y
8 dff1o3 I x Hom D y : x Hom D y 1-1 onto x Hom D y I x Hom D y : x Hom D y onto x Hom D y Fun I x Hom D y -1
9 7 8 mpbi I x Hom D y : x Hom D y onto x Hom D y Fun I x Hom D y -1
10 9 simpri Fun I x Hom D y -1
11 simpl I D Func E x Base D y Base D I D Func E
12 eqidd I D Func E x Base D y Base D Base D = Base D
13 simprl I D Func E x Base D y Base D x Base D
14 simprr I D Func E x Base D y Base D y Base D
15 eqidd I D Func E x Base D y Base D x Hom D y = x Hom D y
16 1 11 12 13 14 15 idfu2nda I D Func E x Base D y Base D x 2 nd I y = I x Hom D y
17 16 cnveqd I D Func E x Base D y Base D x 2 nd I y -1 = I x Hom D y -1
18 17 funeqd I D Func E x Base D y Base D Fun x 2 nd I y -1 Fun I x Hom D y -1
19 10 18 mpbiri I D Func E x Base D y Base D Fun x 2 nd I y -1
20 19 ralrimivva I D Func E x Base D y Base D Fun x 2 nd I y -1
21 eqid Base D = Base D
22 21 isfth 1 st I D Faith E 2 nd I 1 st I D Func E 2 nd I x Base D y Base D Fun x 2 nd I y -1
23 6 20 22 sylanbrc I D Func E 1 st I D Faith E 2 nd I
24 df-br 1 st I D Faith E 2 nd I 1 st I 2 nd I D Faith E
25 23 24 sylib I D Func E 1 st I 2 nd I D Faith E
26 4 25 eqeltrd I D Func E I D Faith E