Metamath Proof Explorer


Theorem focofob

Description: If the domain of a function G equals the range of a function F , then the composition ( G o. F ) is surjective iff G and F as function to the domain of G are both surjective. Symmetric version of fnfocofob including the fact that F is a surjection onto its range. (Contributed by GL and AV, 20-Sep-2024) (Proof shortened by AV, 29-Sep-2024)

Ref Expression
Assertion focofob ⊢ F : A ⟶ B ∧ G : C ⟶ D ∧ ran ⁡ F = C → G ∘ F : A ⟶ onto D ↔ F : A ⟶ onto C ∧ G : C ⟶ onto D

Proof

Step Hyp Ref Expression
1 ffn ⊢ F : A ⟶ B → F Fn A
2 fnfocofob ⊢ F Fn A ∧ G : C ⟶ D ∧ ran ⁡ F = C → G ∘ F : A ⟶ onto D ↔ G : C ⟶ onto D
3 1 2 syl3an1 ⊢ F : A ⟶ B ∧ G : C ⟶ D ∧ ran ⁡ F = C → G ∘ F : A ⟶ onto D ↔ G : C ⟶ onto D
4 dffn4 ⊢ F Fn A ↔ F : A ⟶ onto ran ⁡ F
5 1 4 sylib ⊢ F : A ⟶ B → F : A ⟶ onto ran ⁡ F
6 5 3ad2ant1 ⊢ F : A ⟶ B ∧ G : C ⟶ D ∧ ran ⁡ F = C → F : A ⟶ onto ran ⁡ F
7 foeq3 ⊢ ran ⁡ F = C → F : A ⟶ onto ran ⁡ F ↔ F : A ⟶ onto C
8 7 3ad2ant3 ⊢ F : A ⟶ B ∧ G : C ⟶ D ∧ ran ⁡ F = C → F : A ⟶ onto ran ⁡ F ↔ F : A ⟶ onto C
9 6 8 mpbid ⊢ F : A ⟶ B ∧ G : C ⟶ D ∧ ran ⁡ F = C → F : A ⟶ onto C
10 9 biantrurd ⊢ F : A ⟶ B ∧ G : C ⟶ D ∧ ran ⁡ F = C → G : C ⟶ onto D ↔ F : A ⟶ onto C ∧ G : C ⟶ onto D
11 3 10 bitrd ⊢ F : A ⟶ B ∧ G : C ⟶ D ∧ ran ⁡ F = C → G ∘ F : A ⟶ onto D ↔ F : A ⟶ onto C ∧ G : C ⟶ onto D