Metamath Proof Explorer


Theorem precofcl

Description: The pre-composition functor as a transposed curry of the functor composition bifunctor is a functor. (Contributed by Zhi Wang, 11-Oct-2025)

Ref Expression
Hypotheses precofval.q 𝑄 = ( 𝐶 FuncCat 𝐷 )
precofval.r 𝑅 = ( 𝐷 FuncCat 𝐸 )
precofval.o ( 𝜑 = ( ⟨ 𝑄 , 𝑅 ⟩ curryF ( ( ⟨ 𝐶 , 𝐷 ⟩ ∘F 𝐸 ) ∘func ( 𝑄 swapF 𝑅 ) ) ) )
precofval.f ( 𝜑𝐹 ∈ ( 𝐶 Func 𝐷 ) )
precofval.e ( 𝜑𝐸 ∈ Cat )
precofval.k ( 𝜑𝐾 = ( ( 1st ) ‘ 𝐹 ) )
precofcl.s 𝑆 = ( 𝐶 FuncCat 𝐸 )
Assertion precofcl ( 𝜑𝐾 ∈ ( 𝑅 Func 𝑆 ) )

Proof

Step Hyp Ref Expression
1 precofval.q 𝑄 = ( 𝐶 FuncCat 𝐷 )
2 precofval.r 𝑅 = ( 𝐷 FuncCat 𝐸 )
3 precofval.o ( 𝜑 = ( ⟨ 𝑄 , 𝑅 ⟩ curryF ( ( ⟨ 𝐶 , 𝐷 ⟩ ∘F 𝐸 ) ∘func ( 𝑄 swapF 𝑅 ) ) ) )
4 precofval.f ( 𝜑𝐹 ∈ ( 𝐶 Func 𝐷 ) )
5 precofval.e ( 𝜑𝐸 ∈ Cat )
6 precofval.k ( 𝜑𝐾 = ( ( 1st ) ‘ 𝐹 ) )
7 precofcl.s 𝑆 = ( 𝐶 FuncCat 𝐸 )
8 1 fucbas ( 𝐶 Func 𝐷 ) = ( Base ‘ 𝑄 )
9 4 func1st2nd ( 𝜑 → ( 1st𝐹 ) ( 𝐶 Func 𝐷 ) ( 2nd𝐹 ) )
10 9 funcrcl2 ( 𝜑𝐶 ∈ Cat )
11 9 funcrcl3 ( 𝜑𝐷 ∈ Cat )
12 1 10 11 fuccat ( 𝜑𝑄 ∈ Cat )
13 2 11 5 fuccat ( 𝜑𝑅 ∈ Cat )
14 2 1 oveq12i ( 𝑅 ×c 𝑄 ) = ( ( 𝐷 FuncCat 𝐸 ) ×c ( 𝐶 FuncCat 𝐷 ) )
15 14 7 10 11 5 fucofunca ( 𝜑 → ( ⟨ 𝐶 , 𝐷 ⟩ ∘F 𝐸 ) ∈ ( ( 𝑅 ×c 𝑄 ) Func 𝑆 ) )
16 3 8 12 13 15 4 6 tposcurf1cl ( 𝜑𝐾 ∈ ( 𝑅 Func 𝑆 ) )