Metamath Proof Explorer


Theorem concl

Description: A natural transformation from a constant functor of an object maps to morphisms whose domain is the object. Therefore, the range of the second component of a cone are morphisms with a common domain. (Contributed by Zhi Wang, 13-Nov-2025)

Ref Expression
Hypotheses islmd.l 𝐿 = ( 𝐶 Δfunc 𝐷 )
islmd.a 𝐴 = ( Base ‘ 𝐶 )
islmd.n 𝑁 = ( 𝐷 Nat 𝐶 )
islmd.b 𝐵 = ( Base ‘ 𝐷 )
concl.k 𝐾 = ( ( 1st𝐿 ) ‘ 𝑋 )
concl.x ( 𝜑𝑋𝐴 )
concl.y ( 𝜑𝑌𝐵 )
concl.h 𝐻 = ( Hom ‘ 𝐶 )
concl.r ( 𝜑𝑅 ∈ ( 𝐾 𝑁 𝐹 ) )
Assertion concl ( 𝜑 → ( 𝑅𝑌 ) ∈ ( 𝑋 𝐻 ( ( 1st𝐹 ) ‘ 𝑌 ) ) )

Proof

Step Hyp Ref Expression
1 islmd.l 𝐿 = ( 𝐶 Δfunc 𝐷 )
2 islmd.a 𝐴 = ( Base ‘ 𝐶 )
3 islmd.n 𝑁 = ( 𝐷 Nat 𝐶 )
4 islmd.b 𝐵 = ( Base ‘ 𝐷 )
5 concl.k 𝐾 = ( ( 1st𝐿 ) ‘ 𝑋 )
6 concl.x ( 𝜑𝑋𝐴 )
7 concl.y ( 𝜑𝑌𝐵 )
8 concl.h 𝐻 = ( Hom ‘ 𝐶 )
9 concl.r ( 𝜑𝑅 ∈ ( 𝐾 𝑁 𝐹 ) )
10 3 9 nat1st2nd ( 𝜑𝑅 ∈ ( ⟨ ( 1st𝐾 ) , ( 2nd𝐾 ) ⟩ 𝑁 ⟨ ( 1st𝐹 ) , ( 2nd𝐹 ) ⟩ ) )
11 3 10 4 8 7 natcl ( 𝜑 → ( 𝑅𝑌 ) ∈ ( ( ( 1st𝐾 ) ‘ 𝑌 ) 𝐻 ( ( 1st𝐹 ) ‘ 𝑌 ) ) )
12 3 10 natrcl3 ( 𝜑 → ( 1st𝐹 ) ( 𝐷 Func 𝐶 ) ( 2nd𝐹 ) )
13 12 funcrcl3 ( 𝜑𝐶 ∈ Cat )
14 12 funcrcl2 ( 𝜑𝐷 ∈ Cat )
15 1 13 14 2 6 5 4 7 diag11 ( 𝜑 → ( ( 1st𝐾 ) ‘ 𝑌 ) = 𝑋 )
16 15 oveq1d ( 𝜑 → ( ( ( 1st𝐾 ) ‘ 𝑌 ) 𝐻 ( ( 1st𝐹 ) ‘ 𝑌 ) ) = ( 𝑋 𝐻 ( ( 1st𝐹 ) ‘ 𝑌 ) ) )
17 11 16 eleqtrd ( 𝜑 → ( 𝑅𝑌 ) ∈ ( 𝑋 𝐻 ( ( 1st𝐹 ) ‘ 𝑌 ) ) )