Metamath Proof Explorer


Theorem kerf1ghm

Description: A group homomorphism F is injective if and only if its kernel is the singleton { N } . (Contributed by Thierry Arnoux, 27-Oct-2017) (Proof shortened by AV, 24-Oct-2019) (Revised by Thierry Arnoux, 13-May-2023)

Ref Expression
Hypotheses kerf1ghm.a A = Base R
kerf1ghm.b B = Base S
kerf1ghm.n N = 0 R
kerf1ghm.1 0 ˙ = 0 S
Assertion kerf1ghm F R GrpHom S F : A 1-1 B F -1 0 ˙ = N

Proof

Step Hyp Ref Expression
1 kerf1ghm.a A = Base R
2 kerf1ghm.b B = Base S
3 kerf1ghm.n N = 0 R
4 kerf1ghm.1 0 ˙ = 0 S
5 simpl F R GrpHom S F : A 1-1 B x F -1 0 ˙ F R GrpHom S F : A 1-1 B
6 f1fn F : A 1-1 B F Fn A
7 6 adantl F R GrpHom S F : A 1-1 B F Fn A
8 elpreima F Fn A x F -1 0 ˙ x A F x 0 ˙
9 7 8 syl F R GrpHom S F : A 1-1 B x F -1 0 ˙ x A F x 0 ˙
10 9 biimpa F R GrpHom S F : A 1-1 B x F -1 0 ˙ x A F x 0 ˙
11 10 simpld F R GrpHom S F : A 1-1 B x F -1 0 ˙ x A
12 10 simprd F R GrpHom S F : A 1-1 B x F -1 0 ˙ F x 0 ˙
13 fvex F x V
14 13 elsn F x 0 ˙ F x = 0 ˙
15 12 14 sylib F R GrpHom S F : A 1-1 B x F -1 0 ˙ F x = 0 ˙
16 1 2 4 3 f1ghm0to0 F R GrpHom S F : A 1-1 B x A F x = 0 ˙ x = N
17 16 biimpd F R GrpHom S F : A 1-1 B x A F x = 0 ˙ x = N
18 17 3expa F R GrpHom S F : A 1-1 B x A F x = 0 ˙ x = N
19 18 imp F R GrpHom S F : A 1-1 B x A F x = 0 ˙ x = N
20 5 11 15 19 syl21anc F R GrpHom S F : A 1-1 B x F -1 0 ˙ x = N
21 20 ex F R GrpHom S F : A 1-1 B x F -1 0 ˙ x = N
22 velsn x N x = N
23 21 22 syl6ibr F R GrpHom S F : A 1-1 B x F -1 0 ˙ x N
24 23 ssrdv F R GrpHom S F : A 1-1 B F -1 0 ˙ N
25 ghmgrp1 F R GrpHom S R Grp
26 1 3 grpidcl R Grp N A
27 25 26 syl F R GrpHom S N A
28 3 4 ghmid F R GrpHom S F N = 0 ˙
29 fvex F N V
30 29 elsn F N 0 ˙ F N = 0 ˙
31 28 30 sylibr F R GrpHom S F N 0 ˙
32 1 2 ghmf F R GrpHom S F : A B
33 ffn F : A B F Fn A
34 elpreima F Fn A N F -1 0 ˙ N A F N 0 ˙
35 32 33 34 3syl F R GrpHom S N F -1 0 ˙ N A F N 0 ˙
36 27 31 35 mpbir2and F R GrpHom S N F -1 0 ˙
37 36 snssd F R GrpHom S N F -1 0 ˙
38 37 adantr F R GrpHom S F : A 1-1 B N F -1 0 ˙
39 24 38 eqssd F R GrpHom S F : A 1-1 B F -1 0 ˙ = N
40 32 adantr F R GrpHom S F -1 0 ˙ = N F : A B
41 simpl F R GrpHom S F -1 0 ˙ = N x A y A F x = F y F R GrpHom S
42 simpr2l F R GrpHom S F -1 0 ˙ = N x A y A F x = F y x A
43 simpr2r F R GrpHom S F -1 0 ˙ = N x A y A F x = F y y A
44 simpr3 F R GrpHom S F -1 0 ˙ = N x A y A F x = F y F x = F y
45 eqid F -1 0 ˙ = F -1 0 ˙
46 eqid - R = - R
47 1 4 45 46 ghmeqker F R GrpHom S x A y A F x = F y x - R y F -1 0 ˙
48 47 biimpa F R GrpHom S x A y A F x = F y x - R y F -1 0 ˙
49 41 42 43 44 48 syl31anc F R GrpHom S F -1 0 ˙ = N x A y A F x = F y x - R y F -1 0 ˙
50 simpr1 F R GrpHom S F -1 0 ˙ = N x A y A F x = F y F -1 0 ˙ = N
51 49 50 eleqtrd F R GrpHom S F -1 0 ˙ = N x A y A F x = F y x - R y N
52 ovex x - R y V
53 52 elsn x - R y N x - R y = N
54 51 53 sylib F R GrpHom S F -1 0 ˙ = N x A y A F x = F y x - R y = N
55 41 25 syl F R GrpHom S F -1 0 ˙ = N x A y A F x = F y R Grp
56 1 3 46 grpsubeq0 R Grp x A y A x - R y = N x = y
57 55 42 43 56 syl3anc F R GrpHom S F -1 0 ˙ = N x A y A F x = F y x - R y = N x = y
58 54 57 mpbid F R GrpHom S F -1 0 ˙ = N x A y A F x = F y x = y
59 58 3anassrs F R GrpHom S F -1 0 ˙ = N x A y A F x = F y x = y
60 59 ex F R GrpHom S F -1 0 ˙ = N x A y A F x = F y x = y
61 60 ralrimivva F R GrpHom S F -1 0 ˙ = N x A y A F x = F y x = y
62 dff13 F : A 1-1 B F : A B x A y A F x = F y x = y
63 40 61 62 sylanbrc F R GrpHom S F -1 0 ˙ = N F : A 1-1 B
64 39 63 impbida F R GrpHom S F : A 1-1 B F -1 0 ˙ = N