Metamath Proof Explorer


Theorem diag1cl

Description: The constant functor of X is a functor. (Contributed by Mario Carneiro, 6-Jan-2017) (Revised by Mario Carneiro, 15-Jan-2017)

Ref Expression
Hypotheses diagval.l 𝐿 = ( 𝐶 Δfunc 𝐷 )
diagval.c ( 𝜑𝐶 ∈ Cat )
diagval.d ( 𝜑𝐷 ∈ Cat )
diag11.a 𝐴 = ( Base ‘ 𝐶 )
diag11.c ( 𝜑𝑋𝐴 )
diag11.k 𝐾 = ( ( 1st𝐿 ) ‘ 𝑋 )
Assertion diag1cl ( 𝜑𝐾 ∈ ( 𝐷 Func 𝐶 ) )

Proof

Step Hyp Ref Expression
1 diagval.l 𝐿 = ( 𝐶 Δfunc 𝐷 )
2 diagval.c ( 𝜑𝐶 ∈ Cat )
3 diagval.d ( 𝜑𝐷 ∈ Cat )
4 diag11.a 𝐴 = ( Base ‘ 𝐶 )
5 diag11.c ( 𝜑𝑋𝐴 )
6 diag11.k 𝐾 = ( ( 1st𝐿 ) ‘ 𝑋 )
7 eqid ( 𝐷 FuncCat 𝐶 ) = ( 𝐷 FuncCat 𝐶 )
8 7 fucbas ( 𝐷 Func 𝐶 ) = ( Base ‘ ( 𝐷 FuncCat 𝐶 ) )
9 relfunc Rel ( 𝐶 Func ( 𝐷 FuncCat 𝐶 ) )
10 1 2 3 7 diagcl ( 𝜑𝐿 ∈ ( 𝐶 Func ( 𝐷 FuncCat 𝐶 ) ) )
11 1st2ndbr ( ( Rel ( 𝐶 Func ( 𝐷 FuncCat 𝐶 ) ) ∧ 𝐿 ∈ ( 𝐶 Func ( 𝐷 FuncCat 𝐶 ) ) ) → ( 1st𝐿 ) ( 𝐶 Func ( 𝐷 FuncCat 𝐶 ) ) ( 2nd𝐿 ) )
12 9 10 11 sylancr ( 𝜑 → ( 1st𝐿 ) ( 𝐶 Func ( 𝐷 FuncCat 𝐶 ) ) ( 2nd𝐿 ) )
13 4 8 12 funcf1 ( 𝜑 → ( 1st𝐿 ) : 𝐴 ⟶ ( 𝐷 Func 𝐶 ) )
14 13 5 ffvelrnd ( 𝜑 → ( ( 1st𝐿 ) ‘ 𝑋 ) ∈ ( 𝐷 Func 𝐶 ) )
15 6 14 eqeltrid ( 𝜑𝐾 ∈ ( 𝐷 Func 𝐶 ) )