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 Δ func D
islmd.a ⊢ A = Base C
islmd.n ⊢ N = D Nat C
islmd.b ⊢ B = Base D
concl.k ⊢ K = 1 st ⁡ L ⁡ X
concl.x ⊢ φ → X ∈ A
concl.y ⊢ φ → Y ∈ B
concl.h ⊢ H = Hom ⁡ C
concl.r ⊢ φ → R ∈ K N F
Assertion concl ⊢ φ → R ⁡ Y ∈ X H 1 st ⁡ F ⁡ Y

Proof

Step Hyp Ref Expression
1 islmd.l ⊢ L = C Δ func D
2 islmd.a ⊢ A = Base C
3 islmd.n ⊢ N = D Nat C
4 islmd.b ⊢ B = Base D
5 concl.k ⊢ K = 1 st ⁡ L ⁡ X
6 concl.x ⊢ φ → X ∈ A
7 concl.y ⊢ φ → Y ∈ B
8 concl.h ⊢ H = Hom ⁡ C
9 concl.r ⊢ φ → R ∈ K N F
10 3 9 nat1st2nd ⊢ φ → R ∈ 1 st ⁡ K 2 nd ⁡ K N 1 st ⁡ F 2 nd ⁡ F
11 3 10 4 8 7 natcl ⊢ φ → R ⁡ Y ∈ 1 st ⁡ K ⁡ Y H 1 st ⁡ F ⁡ Y
12 3 10 natrcl3 ⊢ φ → 1 st ⁡ F D Func C 2 nd ⁡ F
13 12 funcrcl3 ⊢ φ → C ∈ Cat
14 12 funcrcl2 ⊢ φ → D ∈ Cat
15 1 13 14 2 6 5 4 7 diag11 ⊢ φ → 1 st ⁡ K ⁡ Y = X
16 15 oveq1d ⊢ φ → 1 st ⁡ K ⁡ Y H 1 st ⁡ F ⁡ Y = X H 1 st ⁡ F ⁡ Y
17 11 16 eleqtrd ⊢ φ → R ⁡ Y ∈ X H 1 st ⁡ F ⁡ Y