Theorem f1ocof1ob2

Description: If the range of F equals the domain of G , then the composition ( G o. F ) is bijective iff F and G are both bijective. Symmetric version of f1ocof1ob including the fact that F is a surjection onto its range. (Contributed by GL and AV, 20-Sep-2024) (Proof shortened by AV, 7-Oct-2024)

	Ref	Expression
Assertion	f1ocof1ob2	⊢ ( ( 𝐹 : 𝐴 ⟶ 𝐵 ∧ 𝐺 : 𝐶 ⟶ 𝐷 ∧ ran 𝐹 = 𝐶 ) → ( ( 𝐺 ∘ 𝐹 ) : 𝐴 –1-1-onto→ 𝐷 ↔ ( 𝐹 : 𝐴 –1-1-onto→ 𝐶 ∧ 𝐺 : 𝐶 –1-1-onto→ 𝐷 ) ) )

Proof

Step	Hyp	Ref	Expression
1		f1ocof1ob	⊢ ( ( 𝐹 : 𝐴 ⟶ 𝐵 ∧ 𝐺 : 𝐶 ⟶ 𝐷 ∧ ran 𝐹 = 𝐶 ) → ( ( 𝐺 ∘ 𝐹 ) : 𝐴 –1-1-onto→ 𝐷 ↔ ( 𝐹 : 𝐴 –1-1→ 𝐶 ∧ 𝐺 : 𝐶 –1-1-onto→ 𝐷 ) ) )
2		f1f1orn	⊢ ( 𝐹 : 𝐴 –1-1→ 𝐶 → 𝐹 : 𝐴 –1-1-onto→ ran 𝐹 )
3		f1oeq3	⊢ ( ran 𝐹 = 𝐶 → ( 𝐹 : 𝐴 –1-1-onto→ ran 𝐹 ↔ 𝐹 : 𝐴 –1-1-onto→ 𝐶 ) )
4	2 3	imbitrid	⊢ ( ran 𝐹 = 𝐶 → ( 𝐹 : 𝐴 –1-1→ 𝐶 → 𝐹 : 𝐴 –1-1-onto→ 𝐶 ) )
5	4	3ad2ant3	⊢ ( ( 𝐹 : 𝐴 ⟶ 𝐵 ∧ 𝐺 : 𝐶 ⟶ 𝐷 ∧ ran 𝐹 = 𝐶 ) → ( 𝐹 : 𝐴 –1-1→ 𝐶 → 𝐹 : 𝐴 –1-1-onto→ 𝐶 ) )
6		f1of1	⊢ ( 𝐹 : 𝐴 –1-1-onto→ 𝐶 → 𝐹 : 𝐴 –1-1→ 𝐶 )
7	5 6	impbid1	⊢ ( ( 𝐹 : 𝐴 ⟶ 𝐵 ∧ 𝐺 : 𝐶 ⟶ 𝐷 ∧ ran 𝐹 = 𝐶 ) → ( 𝐹 : 𝐴 –1-1→ 𝐶 ↔ 𝐹 : 𝐴 –1-1-onto→ 𝐶 ) )
8	7	anbi1d	⊢ ( ( 𝐹 : 𝐴 ⟶ 𝐵 ∧ 𝐺 : 𝐶 ⟶ 𝐷 ∧ ran 𝐹 = 𝐶 ) → ( ( 𝐹 : 𝐴 –1-1→ 𝐶 ∧ 𝐺 : 𝐶 –1-1-onto→ 𝐷 ) ↔ ( 𝐹 : 𝐴 –1-1-onto→ 𝐶 ∧ 𝐺 : 𝐶 –1-1-onto→ 𝐷 ) ) )
9	1 8	bitrd	⊢ ( ( 𝐹 : 𝐴 ⟶ 𝐵 ∧ 𝐺 : 𝐶 ⟶ 𝐷 ∧ ran 𝐹 = 𝐶 ) → ( ( 𝐺 ∘ 𝐹 ) : 𝐴 –1-1-onto→ 𝐷 ↔ ( 𝐹 : 𝐴 –1-1-onto→ 𝐶 ∧ 𝐺 : 𝐶 –1-1-onto→ 𝐷 ) ) )