Metamath Proof Explorer


Theorem ffthf1o

Description: The morphism map of a fully faithful functor is a bijection. (Contributed by Mario Carneiro, 29-Jan-2017)

Ref Expression
Hypotheses isfth.b ⊢ B = Base C
isfth.h ⊢ H = Hom ⁡ C
isfth.j ⊢ J = Hom ⁡ D
ffthf1o.f ⊢ φ → F C Full D ∩ C Faith D G
ffthf1o.x ⊢ φ → X ∈ B
ffthf1o.y ⊢ φ → Y ∈ B
Assertion ffthf1o ⊢ φ → X G Y : X H Y ⟶ 1-1 onto F ⁡ X J F ⁡ Y

Proof

Step Hyp Ref Expression
1 isfth.b ⊢ B = Base C
2 isfth.h ⊢ H = Hom ⁡ C
3 isfth.j ⊢ J = Hom ⁡ D
4 ffthf1o.f ⊢ φ → F C Full D ∩ C Faith D G
5 ffthf1o.x ⊢ φ → X ∈ B
6 ffthf1o.y ⊢ φ → Y ∈ B
7 brin ⊢ F C Full D ∩ C Faith D G ↔ F C Full D G ∧ F C Faith D G
8 4 7 sylib ⊢ φ → F C Full D G ∧ F C Faith D G
9 8 simprd ⊢ φ → F C Faith D G
10 1 2 3 9 5 6 fthf1 ⊢ φ → X G Y : X H Y ⟶ 1-1 F ⁡ X J F ⁡ Y
11 8 simpld ⊢ φ → F C Full D G
12 1 3 2 11 5 6 fullfo ⊢ φ → X G Y : X H Y ⟶ onto F ⁡ X J F ⁡ Y
13 df-f1o ⊢ X G Y : X H Y ⟶ 1-1 onto F ⁡ X J F ⁡ Y ↔ X G Y : X H Y ⟶ 1-1 F ⁡ X J F ⁡ Y ∧ X G Y : X H Y ⟶ onto F ⁡ X J F ⁡ Y
14 10 12 13 sylanbrc ⊢ φ → X G Y : X H Y ⟶ 1-1 onto F ⁡ X J F ⁡ Y