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Metamath Proof Explorer

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 ) ) )