Metamath Proof Explorer


Theorem hof1fval

Description: The object part of the Hom functor is the Homf operation, which is just a functionalized version of Hom . That is, it is a two argument function, which maps X , Y to the set of morphisms from X to Y . (Contributed by Mario Carneiro, 15-Jan-2017)

Ref Expression
Hypotheses hofval.m ⊢ M = Hom F ⁡ C
hofval.c ⊢ φ → C ∈ Cat
Assertion hof1fval ⊢ φ → 1 st ⁡ M = Hom 𝑓 ⁡ C

Proof

Step Hyp Ref Expression
1 hofval.m ⊢ M = Hom F ⁡ C
2 hofval.c ⊢ φ → C ∈ Cat
3 eqid ⊢ Base C = Base C
4 eqid ⊢ Hom ⁡ C = Hom ⁡ C
5 eqid ⊢ comp ⁡ C = comp ⁡ C
6 1 2 3 4 5 hofval ⊢ φ → M = Hom 𝑓 ⁡ C x ∈ Base C × Base C , y ∈ Base C × Base C ⟼ f ∈ 1 st ⁡ y Hom ⁡ C 1 st ⁡ x , g ∈ 2 nd ⁡ x Hom ⁡ C 2 nd ⁡ y ⟼ h ∈ Hom ⁡ C ⁡ x ⟼ g x comp ⁡ C 2 nd ⁡ y h 1 st ⁡ y 1 st ⁡ x comp ⁡ C 2 nd ⁡ y f
7 fvex ⊢ Hom 𝑓 ⁡ C ∈ V
8 fvex ⊢ Base C ∈ V
9 8 8 xpex ⊢ Base C × Base C ∈ V
10 9 9 mpoex ⊢ x ∈ Base C × Base C , y ∈ Base C × Base C ⟼ f ∈ 1 st ⁡ y Hom ⁡ C 1 st ⁡ x , g ∈ 2 nd ⁡ x Hom ⁡ C 2 nd ⁡ y ⟼ h ∈ Hom ⁡ C ⁡ x ⟼ g x comp ⁡ C 2 nd ⁡ y h 1 st ⁡ y 1 st ⁡ x comp ⁡ C 2 nd ⁡ y f ∈ V
11 7 10 op1std ⊢ M = Hom 𝑓 ⁡ C x ∈ Base C × Base C , y ∈ Base C × Base C ⟼ f ∈ 1 st ⁡ y Hom ⁡ C 1 st ⁡ x , g ∈ 2 nd ⁡ x Hom ⁡ C 2 nd ⁡ y ⟼ h ∈ Hom ⁡ C ⁡ x ⟼ g x comp ⁡ C 2 nd ⁡ y h 1 st ⁡ y 1 st ⁡ x comp ⁡ C 2 nd ⁡ y f → 1 st ⁡ M = Hom 𝑓 ⁡ C
12 6 11 syl ⊢ φ → 1 st ⁡ M = Hom 𝑓 ⁡ C