Metamath Proof Explorer


Theorem cofidf2a

Description: If " F is a section of G " in a category of small categories (in a universe), then the morphism part of F is injective, and the morphism part of G is surjective in the image of F . (Contributed by Zhi Wang, 15-Nov-2025)

Ref Expression
Hypotheses cofidvala.i I = id func D
cofidvala.b B = Base D
cofidvala.f φ F D Func E
cofidvala.g φ G E Func D
cofidvala.o φ G func F = I
cofidvala.h H = Hom D
cofidf2a.j J = Hom E
cofidf2a.x φ X B
cofidf2a.y φ Y B
Assertion cofidf2a φ X 2 nd F Y : X H Y 1-1 1 st F X J 1 st F Y 1 st F X 2 nd G 1 st F Y : 1 st F X J 1 st F Y onto X H Y

Proof

Step Hyp Ref Expression
1 cofidvala.i I = id func D
2 cofidvala.b B = Base D
3 cofidvala.f φ F D Func E
4 cofidvala.g φ G E Func D
5 cofidvala.o φ G func F = I
6 cofidvala.h H = Hom D
7 cofidf2a.j J = Hom E
8 cofidf2a.x φ X B
9 cofidf2a.y φ Y B
10 3 func1st2nd φ 1 st F D Func E 2 nd F
11 2 6 7 10 8 9 funcf2 φ X 2 nd F Y : X H Y 1 st F X J 1 st F Y
12 5 fveq2d φ 2 nd G func F = 2 nd I
13 12 oveqd φ X 2 nd G func F Y = X 2 nd I Y
14 2 3 4 8 9 cofu2nd φ X 2 nd G func F Y = 1 st F X 2 nd G 1 st F Y X 2 nd F Y
15 10 funcrcl2 φ D Cat
16 1 2 15 6 8 9 idfu2nd φ X 2 nd I Y = I X H Y
17 13 14 16 3eqtr3d φ 1 st F X 2 nd G 1 st F Y X 2 nd F Y = I X H Y
18 fcof1 X 2 nd F Y : X H Y 1 st F X J 1 st F Y 1 st F X 2 nd G 1 st F Y X 2 nd F Y = I X H Y X 2 nd F Y : X H Y 1-1 1 st F X J 1 st F Y
19 11 17 18 syl2anc φ X 2 nd F Y : X H Y 1-1 1 st F X J 1 st F Y
20 1 2 8 3 4 5 cofid1a φ 1 st G 1 st F X = X
21 1 2 9 3 4 5 cofid1a φ 1 st G 1 st F Y = Y
22 20 21 oveq12d φ 1 st G 1 st F X H 1 st G 1 st F Y = X H Y
23 eqid Base E = Base E
24 4 func1st2nd φ 1 st G E Func D 2 nd G
25 2 23 10 funcf1 φ 1 st F : B Base E
26 25 8 ffvelcdmd φ 1 st F X Base E
27 25 9 ffvelcdmd φ 1 st F Y Base E
28 23 7 6 24 26 27 funcf2 φ 1 st F X 2 nd G 1 st F Y : 1 st F X J 1 st F Y 1 st G 1 st F X H 1 st G 1 st F Y
29 22 28 feq3dd φ 1 st F X 2 nd G 1 st F Y : 1 st F X J 1 st F Y X H Y
30 fcofo 1 st F X 2 nd G 1 st F Y : 1 st F X J 1 st F Y X H Y X 2 nd F Y : X H Y 1 st F X J 1 st F Y 1 st F X 2 nd G 1 st F Y X 2 nd F Y = I X H Y 1 st F X 2 nd G 1 st F Y : 1 st F X J 1 st F Y onto X H Y
31 29 11 17 30 syl3anc φ 1 st F X 2 nd G 1 st F Y : 1 st F X J 1 st F Y onto X H Y
32 19 31 jca φ X 2 nd F Y : X H Y 1-1 1 st F X J 1 st F Y 1 st F X 2 nd G 1 st F Y : 1 st F X J 1 st F Y onto X H Y