Metamath Proof Explorer


Theorem natcl

Description: A component of a natural transformation is a morphism. (Contributed by Mario Carneiro, 6-Jan-2017)

Ref Expression
Hypotheses natrcl.1 𝑁 = ( 𝐶 Nat 𝐷 )
natixp.2 ( 𝜑𝐴 ∈ ( ⟨ 𝐹 , 𝐺𝑁𝐾 , 𝐿 ⟩ ) )
natixp.b 𝐵 = ( Base ‘ 𝐶 )
natixp.j 𝐽 = ( Hom ‘ 𝐷 )
natcl.1 ( 𝜑𝑋𝐵 )
Assertion natcl ( 𝜑 → ( 𝐴𝑋 ) ∈ ( ( 𝐹𝑋 ) 𝐽 ( 𝐾𝑋 ) ) )

Proof

Step Hyp Ref Expression
1 natrcl.1 𝑁 = ( 𝐶 Nat 𝐷 )
2 natixp.2 ( 𝜑𝐴 ∈ ( ⟨ 𝐹 , 𝐺𝑁𝐾 , 𝐿 ⟩ ) )
3 natixp.b 𝐵 = ( Base ‘ 𝐶 )
4 natixp.j 𝐽 = ( Hom ‘ 𝐷 )
5 natcl.1 ( 𝜑𝑋𝐵 )
6 1 2 3 4 natixp ( 𝜑𝐴X 𝑥𝐵 ( ( 𝐹𝑥 ) 𝐽 ( 𝐾𝑥 ) ) )
7 fveq2 ( 𝑥 = 𝑋 → ( 𝐹𝑥 ) = ( 𝐹𝑋 ) )
8 fveq2 ( 𝑥 = 𝑋 → ( 𝐾𝑥 ) = ( 𝐾𝑋 ) )
9 7 8 oveq12d ( 𝑥 = 𝑋 → ( ( 𝐹𝑥 ) 𝐽 ( 𝐾𝑥 ) ) = ( ( 𝐹𝑋 ) 𝐽 ( 𝐾𝑋 ) ) )
10 9 fvixp ( ( 𝐴X 𝑥𝐵 ( ( 𝐹𝑥 ) 𝐽 ( 𝐾𝑥 ) ) ∧ 𝑋𝐵 ) → ( 𝐴𝑋 ) ∈ ( ( 𝐹𝑋 ) 𝐽 ( 𝐾𝑋 ) ) )
11 6 5 10 syl2anc ( 𝜑 → ( 𝐴𝑋 ) ∈ ( ( 𝐹𝑋 ) 𝐽 ( 𝐾𝑋 ) ) )