Metamath Proof Explorer


Theorem diag1f1lem

Description: The object part of the diagonal functor is 1-1 if B is non-empty. Note that ( ph -> ( M = N <-> X = Y ) ) also holds because of diag1f1 and f1fveq . (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 ≠ ∅
diag1f1lem.x ⊢ φ → X ∈ A
diag1f1lem.y ⊢ φ → Y ∈ A
diag1f1lem.m ⊢ M = 1 st ⁡ L ⁡ X
diag1f1lem.n ⊢ N = 1 st ⁡ L ⁡ Y
Assertion diag1f1lem ⊢ φ → M = N → X = Y

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 diag1f1lem.x ⊢ φ → X ∈ A
8 diag1f1lem.y ⊢ φ → Y ∈ A
9 diag1f1lem.m ⊢ M = 1 st ⁡ L ⁡ X
10 diag1f1lem.n ⊢ N = 1 st ⁡ L ⁡ Y
11 eqid ⊢ Hom ⁡ D = Hom ⁡ D
12 eqid ⊢ Id ⁡ C = Id ⁡ C
13 1 2 3 4 7 9 5 11 12 diag1a ⊢ φ → M = B × X x ∈ B , y ∈ B ⟼ x Hom ⁡ D y × Id ⁡ C ⁡ X
14 1 2 3 4 8 10 5 11 12 diag1a ⊢ φ → N = B × Y x ∈ B , y ∈ B ⟼ x Hom ⁡ D y × Id ⁡ C ⁡ Y
15 13 14 eqeq12d ⊢ φ → M = N ↔ B × X x ∈ B , y ∈ B ⟼ x Hom ⁡ D y × Id ⁡ C ⁡ X = B × Y x ∈ B , y ∈ B ⟼ x Hom ⁡ D y × Id ⁡ C ⁡ Y
16 5 fvexi ⊢ B ∈ V
17 snex ⊢ X ∈ V
18 16 17 xpex ⊢ B × X ∈ V
19 16 16 mpoex ⊢ x ∈ B , y ∈ B ⟼ x Hom ⁡ D y × Id ⁡ C ⁡ X ∈ V
20 18 19 opth1 ⊢ B × X x ∈ B , y ∈ B ⟼ x Hom ⁡ D y × Id ⁡ C ⁡ X = B × Y x ∈ B , y ∈ B ⟼ x Hom ⁡ D y × Id ⁡ C ⁡ Y → B × X = B × Y
21 xpcan ⊢ B ≠ ∅ → B × X = B × Y ↔ X = Y
22 6 21 syl ⊢ φ → B × X = B × Y ↔ X = Y
23 sneqrg ⊢ X ∈ A → X = Y → X = Y
24 7 23 syl ⊢ φ → X = Y → X = Y
25 22 24 sylbid ⊢ φ → B × X = B × Y → X = Y
26 20 25 syl5 ⊢ φ → B × X x ∈ B , y ∈ B ⟼ x Hom ⁡ D y × Id ⁡ C ⁡ X = B × Y x ∈ B , y ∈ B ⟼ x Hom ⁡ D y × Id ⁡ C ⁡ Y → X = Y
27 15 26 sylbid ⊢ φ → M = N → X = Y