Metamath Proof Explorer


Theorem natideq

Description: A natural transformation is an identity exactly when all its components are identities. This also covers an empty object set. (Contributed by Zhi Wang, 14-Sep-2026)

Ref Expression
Hypotheses natideq.q ⊢ Q = C FuncCat D
natideq.n ⊢ N = C Nat D
natideq.b ⊢ B = Base C
natideq.i ⊢ I = Id ⁡ Q
natideq.j ⊢ J = Id ⁡ D
natideq.a ⊢ φ → A ∈ F N F
Assertion natideq ⊢ φ → A = I ⁡ F ↔ ∀ x ∈ B A ⁡ x = J ⁡ 1 st ⁡ F ⁡ x

Proof

Step Hyp Ref Expression
1 natideq.q ⊢ Q = C FuncCat D
2 natideq.n ⊢ N = C Nat D
3 natideq.b ⊢ B = Base C
4 natideq.i ⊢ I = Id ⁡ Q
5 natideq.j ⊢ J = Id ⁡ D
6 natideq.a ⊢ φ → A ∈ F N F
7 2 natrcl ⊢ A ∈ F N F → F ∈ C Func D ∧ F ∈ C Func D
8 6 7 syl ⊢ φ → F ∈ C Func D ∧ F ∈ C Func D
9 8 simpld ⊢ φ → F ∈ C Func D
10 1 4 5 9 fucid ⊢ φ → I ⁡ F = J ∘ 1 st ⁡ F
11 1 2 5 9 fucidcl ⊢ φ → J ∘ 1 st ⁡ F ∈ F N F
12 10 11 eqeltrd ⊢ φ → I ⁡ F ∈ F N F
13 2 3 6 12 nateq ⊢ φ → A = I ⁡ F ↔ ∀ x ∈ B A ⁡ x = I ⁡ F ⁡ x
14 10 adantr ⊢ φ ∧ x ∈ B → I ⁡ F = J ∘ 1 st ⁡ F
15 14 fveq1d ⊢ φ ∧ x ∈ B → I ⁡ F ⁡ x = J ∘ 1 st ⁡ F ⁡ x
16 eqid ⊢ Base D = Base D
17 9 func1st2nd ⊢ φ → 1 st ⁡ F C Func D 2 nd ⁡ F
18 3 16 17 funcf1 ⊢ φ → 1 st ⁡ F : B ⟶ Base D
19 18 adantr ⊢ φ ∧ x ∈ B → 1 st ⁡ F : B ⟶ Base D
20 simpr ⊢ φ ∧ x ∈ B → x ∈ B
21 19 20 fvco3d ⊢ φ ∧ x ∈ B → J ∘ 1 st ⁡ F ⁡ x = J ⁡ 1 st ⁡ F ⁡ x
22 15 21 eqtrd ⊢ φ ∧ x ∈ B → I ⁡ F ⁡ x = J ⁡ 1 st ⁡ F ⁡ x
23 22 eqeq2d ⊢ φ ∧ x ∈ B → A ⁡ x = I ⁡ F ⁡ x ↔ A ⁡ x = J ⁡ 1 st ⁡ F ⁡ x
24 23 ralbidva ⊢ φ → ∀ x ∈ B A ⁡ x = I ⁡ F ⁡ x ↔ ∀ x ∈ B A ⁡ x = J ⁡ 1 st ⁡ F ⁡ x
25 13 24 bitrd ⊢ φ → A = I ⁡ F ↔ ∀ x ∈ B A ⁡ x = J ⁡ 1 st ⁡ F ⁡ x