Metamath Proof Explorer


Theorem rngokerinj

Description: Obsolete theorem, use rhmkerinj instead. A ring homomorphism is injective if and only if its kernel is zero. (Contributed by Jeff Madsen, 16-Jun-2011) (Proof modification is discouraged.) (New usage is discouraged.)

Ref Expression
Hypotheses rngkerinj.1 ⊢ G = 1 st ⁡ R
rngkerinj.2 ⊢ X = ran ⁡ G
rngkerinj.3 ⊢ W = GId ⁡ G
rngkerinj.4 ⊢ J = 1 st ⁡ S
rngkerinj.5 ⊢ Y = ran ⁡ J
rngkerinj.6 ⊢ Z = GId ⁡ J
Assertion rngokerinj ⊢ R ∈ RingOps ∧ S ∈ RingOps ∧ F ∈ R RingOpsHom S → F : X ⟶ 1-1 Y ↔ F -1 Z = W

Proof

Step Hyp Ref Expression
1 rngkerinj.1 ⊢ G = 1 st ⁡ R
2 rngkerinj.2 ⊢ X = ran ⁡ G
3 rngkerinj.3 ⊢ W = GId ⁡ G
4 rngkerinj.4 ⊢ J = 1 st ⁡ S
5 rngkerinj.5 ⊢ Y = ran ⁡ J
6 rngkerinj.6 ⊢ Z = GId ⁡ J
7 eqid ⊢ 1 st ⁡ R = 1 st ⁡ R
8 7 rngogrpo ⊢ R ∈ RingOps → 1 st ⁡ R ∈ GrpOp
9 8 3ad2ant1 ⊢ R ∈ RingOps ∧ S ∈ RingOps ∧ F ∈ R RingOpsHom S → 1 st ⁡ R ∈ GrpOp
10 eqid ⊢ 1 st ⁡ S = 1 st ⁡ S
11 10 rngogrpo ⊢ S ∈ RingOps → 1 st ⁡ S ∈ GrpOp
12 11 3ad2ant2 ⊢ R ∈ RingOps ∧ S ∈ RingOps ∧ F ∈ R RingOpsHom S → 1 st ⁡ S ∈ GrpOp
13 7 10 rngogrphom ⊢ R ∈ RingOps ∧ S ∈ RingOps ∧ F ∈ R RingOpsHom S → F ∈ 1 st ⁡ R GrpOpHom 1 st ⁡ S
14 1 rneqi ⊢ ran ⁡ G = ran ⁡ 1 st ⁡ R
15 2 14 eqtri ⊢ X = ran ⁡ 1 st ⁡ R
16 1 fveq2i ⊢ GId ⁡ G = GId ⁡ 1 st ⁡ R
17 3 16 eqtri ⊢ W = GId ⁡ 1 st ⁡ R
18 4 rneqi ⊢ ran ⁡ J = ran ⁡ 1 st ⁡ S
19 5 18 eqtri ⊢ Y = ran ⁡ 1 st ⁡ S
20 4 fveq2i ⊢ GId ⁡ J = GId ⁡ 1 st ⁡ S
21 6 20 eqtri ⊢ Z = GId ⁡ 1 st ⁡ S
22 15 17 19 21 grpokerinj ⊢ 1 st ⁡ R ∈ GrpOp ∧ 1 st ⁡ S ∈ GrpOp ∧ F ∈ 1 st ⁡ R GrpOpHom 1 st ⁡ S → F : X ⟶ 1-1 Y ↔ F -1 Z = W
23 9 12 13 22 syl3anc ⊢ R ∈ RingOps ∧ S ∈ RingOps ∧ F ∈ R RingOpsHom S → F : X ⟶ 1-1 Y ↔ F -1 Z = W