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Metamath Proof Explorer

Theorem invss

Description: The inverse relation is a relation between morphisms F : X --> Y and their inverses G : Y --> X . (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 )
invss.x ⊢ ( 𝜑 → 𝑋 ∈ 𝐵 )
invss.y ⊢ ( 𝜑 → 𝑌 ∈ 𝐵 )
invss.h ⊢ 𝐻 = ( Hom ‘ 𝐶 )
Assertion invss ( 𝜑 → ( 𝑋 𝑁 𝑌 ) ⊆ ( ( 𝑋 𝐻 𝑌 ) × ( 𝑌 𝐻 𝑋 ) ) )

Proof

Step Hyp Ref Expression
1 invfval.b ⊢ 𝐵 = ( Base ‘ 𝐶 )
2 invfval.n ⊢ 𝑁 = ( Inv ‘ 𝐶 )
3 invfval.c ⊢ ( 𝜑 → 𝐶 ∈ Cat )
4 invss.x ⊢ ( 𝜑 → 𝑋 ∈ 𝐵 )
5 invss.y ⊢ ( 𝜑 → 𝑌 ∈ 𝐵 )
6 invss.h ⊢ 𝐻 = ( Hom ‘ 𝐶 )
7 eqid ⊢ ( Sect ‘ 𝐶 ) = ( Sect ‘ 𝐶 )
8 1 2 3 4 5 7 invfval ⊢ ( 𝜑 → ( 𝑋 𝑁 𝑌 ) = ( ( 𝑋 ( Sect ‘ 𝐶 ) 𝑌 ) ∩ ◡ ( 𝑌 ( Sect ‘ 𝐶 ) 𝑋 ) ) )
9 inss1 ⊢ ( ( 𝑋 ( Sect ‘ 𝐶 ) 𝑌 ) ∩ ◡ ( 𝑌 ( Sect ‘ 𝐶 ) 𝑋 ) ) ⊆ ( 𝑋 ( Sect ‘ 𝐶 ) 𝑌 )
10 8 9 eqsstrdi ⊢ ( 𝜑 → ( 𝑋 𝑁 𝑌 ) ⊆ ( 𝑋 ( Sect ‘ 𝐶 ) 𝑌 ) )
11 eqid ⊢ ( comp ‘ 𝐶 ) = ( comp ‘ 𝐶 )
12 eqid ⊢ ( Id ‘ 𝐶 ) = ( Id ‘ 𝐶 )
13 1 6 11 12 7 3 4 5 sectss ⊢ ( 𝜑 → ( 𝑋 ( Sect ‘ 𝐶 ) 𝑌 ) ⊆ ( ( 𝑋 𝐻 𝑌 ) × ( 𝑌 𝐻 𝑋 ) ) )
14 10 13 sstrd ⊢ ( 𝜑 → ( 𝑋 𝑁 𝑌 ) ⊆ ( ( 𝑋 𝐻 𝑌 ) × ( 𝑌 𝐻 𝑋 ) ) )