Metamath Proof Explorer


Theorem diag1f1

Description: The object part of the diagonal functor is 1-1 if B is non-empty. (Contributed by Zhi Wang, 19-Oct-2025)

Ref Expression
Hypotheses diag1f1.l ⊢ L = C Δ func D
diag1f1.c ⊢ φ → C ∈ Cat
diag1f1.d ⊢ φ → D ∈ Cat
diag1f1.a ⊢ A = Base C
diag1f1.b ⊢ B = Base D
diag1f1.0 ⊢ φ → B ≠ ∅
Assertion diag1f1 ⊢ φ → 1 st ⁡ L : A ⟶ 1-1 D Func C

Proof

Step Hyp Ref Expression
1 diag1f1.l ⊢ L = C Δ func D
2 diag1f1.c ⊢ φ → C ∈ Cat
3 diag1f1.d ⊢ φ → D ∈ Cat
4 diag1f1.a ⊢ A = Base C
5 diag1f1.b ⊢ B = Base D
6 diag1f1.0 ⊢ φ → B ≠ ∅
7 eqid ⊢ D FuncCat C = D FuncCat C
8 7 fucbas ⊢ D Func C = Base D FuncCat C
9 1 2 3 7 diagcl ⊢ φ → L ∈ C Func D FuncCat C
10 9 func1st2nd ⊢ φ → 1 st ⁡ L C Func D FuncCat C 2 nd ⁡ L
11 4 8 10 funcf1 ⊢ φ → 1 st ⁡ L : A ⟶ D Func C
12 2 adantr ⊢ φ ∧ x ∈ A ∧ y ∈ A → C ∈ Cat
13 3 adantr ⊢ φ ∧ x ∈ A ∧ y ∈ A → D ∈ Cat
14 6 adantr ⊢ φ ∧ x ∈ A ∧ y ∈ A → B ≠ ∅
15 simprl ⊢ φ ∧ x ∈ A ∧ y ∈ A → x ∈ A
16 simprr ⊢ φ ∧ x ∈ A ∧ y ∈ A → y ∈ A
17 eqid ⊢ 1 st ⁡ L ⁡ x = 1 st ⁡ L ⁡ x
18 eqid ⊢ 1 st ⁡ L ⁡ y = 1 st ⁡ L ⁡ y
19 1 12 13 4 5 14 15 16 17 18 diag1f1lem ⊢ φ ∧ x ∈ A ∧ y ∈ A → 1 st ⁡ L ⁡ x = 1 st ⁡ L ⁡ y → x = y
20 19 ralrimivva ⊢ φ → ∀ x ∈ A ∀ y ∈ A 1 st ⁡ L ⁡ x = 1 st ⁡ L ⁡ y → x = y
21 dff13 ⊢ 1 st ⁡ L : A ⟶ 1-1 D Func C ↔ 1 st ⁡ L : A ⟶ D Func C ∧ ∀ x ∈ A ∀ y ∈ A 1 st ⁡ L ⁡ x = 1 st ⁡ L ⁡ y → x = y
22 11 20 21 sylanbrc ⊢ φ → 1 st ⁡ L : A ⟶ 1-1 D Func C