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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 ⊢ ( 𝜑 → 𝐴 ∈ ( ⟨ 𝐹 , 𝐺 ⟩ 𝑁 ⟨ 𝐾 , 𝐿 ⟩ ) )
natixp.b ⊢ 𝐵 = ( Base ‘ 𝐶 )
natixp.j ⊢ 𝐽 = ( Hom ‘ 𝐷 )
natcl.1 ⊢ ( 𝜑 → 𝑋 ∈ 𝐵 )
Assertion natcl ( 𝜑 → ( 𝐴 ‘ 𝑋 ) ∈ ( ( 𝐹 ‘ 𝑋 ) 𝐽 ( 𝐾 ‘ 𝑋 ) ) )

Proof

Step Hyp Ref Expression
1 natrcl.1 ⊢ 𝑁 = ( 𝐶 Nat 𝐷 )
2 natixp.2 ⊢ ( 𝜑 → 𝐴 ∈ ( ⟨ 𝐹 , 𝐺 ⟩ 𝑁 ⟨ 𝐾 , 𝐿 ⟩ ) )
3 natixp.b ⊢ 𝐵 = ( Base ‘ 𝐶 )
4 natixp.j ⊢ 𝐽 = ( Hom ‘ 𝐷 )
5 natcl.1 ⊢ ( 𝜑 → 𝑋 ∈ 𝐵 )
6 1 2 3 4 natixp ⊢ ( 𝜑 → 𝐴 ∈ X 𝑥 ∈ 𝐵 ( ( 𝐹 ‘ 𝑥 ) 𝐽 ( 𝐾 ‘ 𝑥 ) ) )
7 fveq2 ⊢ ( 𝑥 = 𝑋 → ( 𝐹 ‘ 𝑥 ) = ( 𝐹 ‘ 𝑋 ) )
8 fveq2 ⊢ ( 𝑥 = 𝑋 → ( 𝐾 ‘ 𝑥 ) = ( 𝐾 ‘ 𝑋 ) )
9 7 8 oveq12d ⊢ ( 𝑥 = 𝑋 → ( ( 𝐹 ‘ 𝑥 ) 𝐽 ( 𝐾 ‘ 𝑥 ) ) = ( ( 𝐹 ‘ 𝑋 ) 𝐽 ( 𝐾 ‘ 𝑋 ) ) )
10 9 fvixp ⊢ ( ( 𝐴 ∈ X 𝑥 ∈ 𝐵 ( ( 𝐹 ‘ 𝑥 ) 𝐽 ( 𝐾 ‘ 𝑥 ) ) ∧ 𝑋 ∈ 𝐵 ) → ( 𝐴 ‘ 𝑋 ) ∈ ( ( 𝐹 ‘ 𝑋 ) 𝐽 ( 𝐾 ‘ 𝑋 ) ) )
11 6 5 10 syl2anc ⊢ ( 𝜑 → ( 𝐴 ‘ 𝑋 ) ∈ ( ( 𝐹 ‘ 𝑋 ) 𝐽 ( 𝐾 ‘ 𝑋 ) ) )