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