Metamath Proof Explorer

Theorem invfun

Description: The inverse relation is a function, which is to say that every morphism has at most one inverse. (Contributed by Mario Carneiro, 2-Jan-2017)

Ref Expression
Hypotheses invfval.b ⊢ 𝐵 = ( Base ‘ 𝐶 )
invfval.n ⊢ 𝑁 = ( Inv ‘ 𝐶 )
invfval.c ⊢ ( 𝜑 → 𝐶 ∈ Cat )
invfval.x ⊢ ( 𝜑 → 𝑋 ∈ 𝐵 )
invfval.y ⊢ ( 𝜑 → 𝑌 ∈ 𝐵 )
Assertion invfun ( 𝜑 → Fun ( 𝑋 𝑁 𝑌 ) )

Proof

Step Hyp Ref Expression
1 invfval.b ⊢ 𝐵 = ( Base ‘ 𝐶 )
2 invfval.n ⊢ 𝑁 = ( Inv ‘ 𝐶 )
3 invfval.c ⊢ ( 𝜑 → 𝐶 ∈ Cat )
4 invfval.x ⊢ ( 𝜑 → 𝑋 ∈ 𝐵 )
5 invfval.y ⊢ ( 𝜑 → 𝑌 ∈ 𝐵 )
6 eqid ⊢ ( Hom ‘ 𝐶 ) = ( Hom ‘ 𝐶 )
7 1 2 3 4 5 6 invss ⊢ ( 𝜑 → ( 𝑋 𝑁 𝑌 ) ⊆ ( ( 𝑋 ( Hom ‘ 𝐶 ) 𝑌 ) × ( 𝑌 ( Hom ‘ 𝐶 ) 𝑋 ) ) )
8 relxp ⊢ Rel ( ( 𝑋 ( Hom ‘ 𝐶 ) 𝑌 ) × ( 𝑌 ( Hom ‘ 𝐶 ) 𝑋 ) )
9 relss ⊢ ( ( 𝑋 𝑁 𝑌 ) ⊆ ( ( 𝑋 ( Hom ‘ 𝐶 ) 𝑌 ) × ( 𝑌 ( Hom ‘ 𝐶 ) 𝑋 ) ) → ( Rel ( ( 𝑋 ( Hom ‘ 𝐶 ) 𝑌 ) × ( 𝑌 ( Hom ‘ 𝐶 ) 𝑋 ) ) → Rel ( 𝑋 𝑁 𝑌 ) ) )
10 7 8 9 mpisyl ⊢ ( 𝜑 → Rel ( 𝑋 𝑁 𝑌 ) )
11 eqid ⊢ ( Sect ‘ 𝐶 ) = ( Sect ‘ 𝐶 )
12 3 adantr ⊢ ( ( 𝜑 ∧ ( 𝑓 ( 𝑋 𝑁 𝑌 ) 𝑔 ∧ 𝑓 ( 𝑋 𝑁 𝑌 ) ℎ ) ) → 𝐶 ∈ Cat )
13 5 adantr ⊢ ( ( 𝜑 ∧ ( 𝑓 ( 𝑋 𝑁 𝑌 ) 𝑔 ∧ 𝑓 ( 𝑋 𝑁 𝑌 ) ℎ ) ) → 𝑌 ∈ 𝐵 )
14 4 adantr ⊢ ( ( 𝜑 ∧ ( 𝑓 ( 𝑋 𝑁 𝑌 ) 𝑔 ∧ 𝑓 ( 𝑋 𝑁 𝑌 ) ℎ ) ) → 𝑋 ∈ 𝐵 )
15 1 2 3 4 5 11 isinv ⊢ ( 𝜑 → ( 𝑓 ( 𝑋 𝑁 𝑌 ) 𝑔 ↔ ( 𝑓 ( 𝑋 ( Sect ‘ 𝐶 ) 𝑌 ) 𝑔 ∧ 𝑔 ( 𝑌 ( Sect ‘ 𝐶 ) 𝑋 ) 𝑓 ) ) )
16 15 simplbda ⊢ ( ( 𝜑 ∧ 𝑓 ( 𝑋 𝑁 𝑌 ) 𝑔 ) → 𝑔 ( 𝑌 ( Sect ‘ 𝐶 ) 𝑋 ) 𝑓 )
17 16 adantrr ⊢ ( ( 𝜑 ∧ ( 𝑓 ( 𝑋 𝑁 𝑌 ) 𝑔 ∧ 𝑓 ( 𝑋 𝑁 𝑌 ) ℎ ) ) → 𝑔 ( 𝑌 ( Sect ‘ 𝐶 ) 𝑋 ) 𝑓 )
18 1 2 3 4 5 11 isinv ⊢ ( 𝜑 → ( 𝑓 ( 𝑋 𝑁 𝑌 ) ℎ ↔ ( 𝑓 ( 𝑋 ( Sect ‘ 𝐶 ) 𝑌 ) ℎ ∧ ℎ ( 𝑌 ( Sect ‘ 𝐶 ) 𝑋 ) 𝑓 ) ) )
19 18 simprbda ⊢ ( ( 𝜑 ∧ 𝑓 ( 𝑋 𝑁 𝑌 ) ℎ ) → 𝑓 ( 𝑋 ( Sect ‘ 𝐶 ) 𝑌 ) ℎ )
20 19 adantrl ⊢ ( ( 𝜑 ∧ ( 𝑓 ( 𝑋 𝑁 𝑌 ) 𝑔 ∧ 𝑓 ( 𝑋 𝑁 𝑌 ) ℎ ) ) → 𝑓 ( 𝑋 ( Sect ‘ 𝐶 ) 𝑌 ) ℎ )
21 1 11 12 13 14 17 20 sectcan ⊢ ( ( 𝜑 ∧ ( 𝑓 ( 𝑋 𝑁 𝑌 ) 𝑔 ∧ 𝑓 ( 𝑋 𝑁 𝑌 ) ℎ ) ) → 𝑔 = ℎ )
22 21 ex ⊢ ( 𝜑 → ( ( 𝑓 ( 𝑋 𝑁 𝑌 ) 𝑔 ∧ 𝑓 ( 𝑋 𝑁 𝑌 ) ℎ ) → 𝑔 = ℎ ) )
23 22 alrimiv ⊢ ( 𝜑 → ∀ ℎ ( ( 𝑓 ( 𝑋 𝑁 𝑌 ) 𝑔 ∧ 𝑓 ( 𝑋 𝑁 𝑌 ) ℎ ) → 𝑔 = ℎ ) )
24 23 alrimivv ⊢ ( 𝜑 → ∀ 𝑓 ∀ 𝑔 ∀ ℎ ( ( 𝑓 ( 𝑋 𝑁 𝑌 ) 𝑔 ∧ 𝑓 ( 𝑋 𝑁 𝑌 ) ℎ ) → 𝑔 = ℎ ) )
25 dffun2 ⊢ ( Fun ( 𝑋 𝑁 𝑌 ) ↔ ( Rel ( 𝑋 𝑁 𝑌 ) ∧ ∀ 𝑓 ∀ 𝑔 ∀ ℎ ( ( 𝑓 ( 𝑋 𝑁 𝑌 ) 𝑔 ∧ 𝑓 ( 𝑋 𝑁 𝑌 ) ℎ ) → 𝑔 = ℎ ) ) )
26 10 24 25 sylanbrc ⊢ ( 𝜑 → Fun ( 𝑋 𝑁 𝑌 ) )