Metamath Proof Explorer


Theorem funcfn2

Description: The morphism part of a functor is a function. (Contributed by Mario Carneiro, 3-Jan-2017)

Ref Expression
Hypotheses funcfn2.b ⊢ 𝐵 = ( Base ‘ 𝐷 )
funcfn2.f ⊢ ( 𝜑 → 𝐹 ( 𝐷 Func 𝐸 ) 𝐺 )
Assertion funcfn2 ( 𝜑 → 𝐺 Fn ( 𝐵 × 𝐵 ) )

Proof

Step Hyp Ref Expression
1 funcfn2.b ⊢ 𝐵 = ( Base ‘ 𝐷 )
2 funcfn2.f ⊢ ( 𝜑 → 𝐹 ( 𝐷 Func 𝐸 ) 𝐺 )
3 eqid ⊢ ( Hom ‘ 𝐷 ) = ( Hom ‘ 𝐷 )
4 eqid ⊢ ( Hom ‘ 𝐸 ) = ( Hom ‘ 𝐸 )
5 1 3 4 2 funcixp ⊢ ( 𝜑 → 𝐺 ∈ X 𝑥 ∈ ( 𝐵 × 𝐵 ) ( ( ( 𝐹 ‘ ( 1st ‘ 𝑥 ) ) ( Hom ‘ 𝐸 ) ( 𝐹 ‘ ( 2nd ‘ 𝑥 ) ) ) ↑m ( ( Hom ‘ 𝐷 ) ‘ 𝑥 ) ) )
6 ixpfn ⊢ ( 𝐺 ∈ X 𝑥 ∈ ( 𝐵 × 𝐵 ) ( ( ( 𝐹 ‘ ( 1st ‘ 𝑥 ) ) ( Hom ‘ 𝐸 ) ( 𝐹 ‘ ( 2nd ‘ 𝑥 ) ) ) ↑m ( ( Hom ‘ 𝐷 ) ‘ 𝑥 ) ) → 𝐺 Fn ( 𝐵 × 𝐵 ) )
7 5 6 syl ⊢ ( 𝜑 → 𝐺 Fn ( 𝐵 × 𝐵 ) )