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Metamath Proof Explorer

Theorem fullres2c

Description: Condition for a full functor to also be a full functor into the restriction. (Contributed by Mario Carneiro, 30-Jan-2017)

Ref Expression
Hypotheses ffthres2c.a ⊢ 𝐴 = ( Base ‘ 𝐶 )
ffthres2c.e ⊢ 𝐸 = ( 𝐷 ↾s 𝑆 )
ffthres2c.d ⊢ ( 𝜑 → 𝐷 ∈ Cat )
ffthres2c.r ⊢ ( 𝜑 → 𝑆 ∈ 𝑉 )
ffthres2c.1 ⊢ ( 𝜑 → 𝐹 : 𝐴 ⟶ 𝑆 )
Assertion fullres2c ( 𝜑 → ( 𝐹 ( 𝐶 Full 𝐷 ) 𝐺 ↔ 𝐹 ( 𝐶 Full 𝐸 ) 𝐺 ) )

Proof

Step Hyp Ref Expression
1 ffthres2c.a ⊢ 𝐴 = ( Base ‘ 𝐶 )
2 ffthres2c.e ⊢ 𝐸 = ( 𝐷 ↾s 𝑆 )
3 ffthres2c.d ⊢ ( 𝜑 → 𝐷 ∈ Cat )
4 ffthres2c.r ⊢ ( 𝜑 → 𝑆 ∈ 𝑉 )
5 ffthres2c.1 ⊢ ( 𝜑 → 𝐹 : 𝐴 ⟶ 𝑆 )
6 1 2 3 4 5 funcres2c ⊢ ( 𝜑 → ( 𝐹 ( 𝐶 Func 𝐷 ) 𝐺 ↔ 𝐹 ( 𝐶 Func 𝐸 ) 𝐺 ) )
7 eqid ⊢ ( Hom ‘ 𝐷 ) = ( Hom ‘ 𝐷 )
8 2 7 resshom ⊢ ( 𝑆 ∈ 𝑉 → ( Hom ‘ 𝐷 ) = ( Hom ‘ 𝐸 ) )
9 4 8 syl ⊢ ( 𝜑 → ( Hom ‘ 𝐷 ) = ( Hom ‘ 𝐸 ) )
10 9 oveqd ⊢ ( 𝜑 → ( ( 𝐹 ‘ 𝑥 ) ( Hom ‘ 𝐷 ) ( 𝐹 ‘ 𝑦 ) ) = ( ( 𝐹 ‘ 𝑥 ) ( Hom ‘ 𝐸 ) ( 𝐹 ‘ 𝑦 ) ) )
11 10 eqeq2d ⊢ ( 𝜑 → ( ran ( 𝑥 𝐺 𝑦 ) = ( ( 𝐹 ‘ 𝑥 ) ( Hom ‘ 𝐷 ) ( 𝐹 ‘ 𝑦 ) ) ↔ ran ( 𝑥 𝐺 𝑦 ) = ( ( 𝐹 ‘ 𝑥 ) ( Hom ‘ 𝐸 ) ( 𝐹 ‘ 𝑦 ) ) ) )
12 11 2ralbidv ⊢ ( 𝜑 → ( ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐴 ran ( 𝑥 𝐺 𝑦 ) = ( ( 𝐹 ‘ 𝑥 ) ( Hom ‘ 𝐷 ) ( 𝐹 ‘ 𝑦 ) ) ↔ ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐴 ran ( 𝑥 𝐺 𝑦 ) = ( ( 𝐹 ‘ 𝑥 ) ( Hom ‘ 𝐸 ) ( 𝐹 ‘ 𝑦 ) ) ) )
13 6 12 anbi12d ⊢ ( 𝜑 → ( ( 𝐹 ( 𝐶 Func 𝐷 ) 𝐺 ∧ ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐴 ran ( 𝑥 𝐺 𝑦 ) = ( ( 𝐹 ‘ 𝑥 ) ( Hom ‘ 𝐷 ) ( 𝐹 ‘ 𝑦 ) ) ) ↔ ( 𝐹 ( 𝐶 Func 𝐸 ) 𝐺 ∧ ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐴 ran ( 𝑥 𝐺 𝑦 ) = ( ( 𝐹 ‘ 𝑥 ) ( Hom ‘ 𝐸 ) ( 𝐹 ‘ 𝑦 ) ) ) ) )
14 1 7 isfull ⊢ ( 𝐹 ( 𝐶 Full 𝐷 ) 𝐺 ↔ ( 𝐹 ( 𝐶 Func 𝐷 ) 𝐺 ∧ ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐴 ran ( 𝑥 𝐺 𝑦 ) = ( ( 𝐹 ‘ 𝑥 ) ( Hom ‘ 𝐷 ) ( 𝐹 ‘ 𝑦 ) ) ) )
15 eqid ⊢ ( Hom ‘ 𝐸 ) = ( Hom ‘ 𝐸 )
16 1 15 isfull ⊢ ( 𝐹 ( 𝐶 Full 𝐸 ) 𝐺 ↔ ( 𝐹 ( 𝐶 Func 𝐸 ) 𝐺 ∧ ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐴 ran ( 𝑥 𝐺 𝑦 ) = ( ( 𝐹 ‘ 𝑥 ) ( Hom ‘ 𝐸 ) ( 𝐹 ‘ 𝑦 ) ) ) )
17 13 14 16 3bitr4g ⊢ ( 𝜑 → ( 𝐹 ( 𝐶 Full 𝐷 ) 𝐺 ↔ 𝐹 ( 𝐶 Full 𝐸 ) 𝐺 ) )