Metamath Proof Explorer


Theorem concom

Description: A cone to a diagram commutes with the diagram. (Contributed by Zhi Wang, 13-Nov-2025)

Ref Expression
Hypotheses islmd.l L = C Δ func D
islmd.a A = Base C
islmd.n N = D Nat C
islmd.b B = Base D
concl.k K = 1 st L X
concl.x φ X A
concl.y φ Y B
concom.z φ Z B
concom.m φ M Y J Z
concom.j J = Hom D
concom.o · ˙ = comp C
concom.r φ R K N F
Assertion concom φ R Z = Y 2 nd F Z M X 1 st F Y · ˙ 1 st F Z R Y

Proof

Step Hyp Ref Expression
1 islmd.l L = C Δ func D
2 islmd.a A = Base C
3 islmd.n N = D Nat C
4 islmd.b B = Base D
5 concl.k K = 1 st L X
6 concl.x φ X A
7 concl.y φ Y B
8 concom.z φ Z B
9 concom.m φ M Y J Z
10 concom.j J = Hom D
11 concom.o · ˙ = comp C
12 concom.r φ R K N F
13 3 12 nat1st2nd φ R 1 st K 2 nd K N 1 st F 2 nd F
14 3 13 4 10 11 7 8 9 nati φ R Z 1 st K Y 1 st K Z · ˙ 1 st F Z Y 2 nd K Z M = Y 2 nd F Z M 1 st K Y 1 st F Y · ˙ 1 st F Z R Y
15 3 13 natrcl3 φ 1 st F D Func C 2 nd F
16 15 funcrcl3 φ C Cat
17 15 funcrcl2 φ D Cat
18 1 16 17 2 6 5 4 7 diag11 φ 1 st K Y = X
19 1 16 17 2 6 5 4 8 diag11 φ 1 st K Z = X
20 18 19 opeq12d φ 1 st K Y 1 st K Z = X X
21 20 oveq1d φ 1 st K Y 1 st K Z · ˙ 1 st F Z = X X · ˙ 1 st F Z
22 eqidd φ R Z = R Z
23 eqid Id C = Id C
24 1 16 17 2 6 5 4 7 10 23 8 9 diag12 φ Y 2 nd K Z M = Id C X
25 21 22 24 oveq123d φ R Z 1 st K Y 1 st K Z · ˙ 1 st F Z Y 2 nd K Z M = R Z X X · ˙ 1 st F Z Id C X
26 eqid Hom C = Hom C
27 4 2 15 funcf1 φ 1 st F : B A
28 27 8 ffvelcdmd φ 1 st F Z A
29 1 2 3 4 5 6 8 26 12 concl φ R Z X Hom C 1 st F Z
30 2 26 23 16 6 11 28 29 catrid φ R Z X X · ˙ 1 st F Z Id C X = R Z
31 25 30 eqtrd φ R Z 1 st K Y 1 st K Z · ˙ 1 st F Z Y 2 nd K Z M = R Z
32 18 opeq1d φ 1 st K Y 1 st F Y = X 1 st F Y
33 32 oveq1d φ 1 st K Y 1 st F Y · ˙ 1 st F Z = X 1 st F Y · ˙ 1 st F Z
34 33 oveqd φ Y 2 nd F Z M 1 st K Y 1 st F Y · ˙ 1 st F Z R Y = Y 2 nd F Z M X 1 st F Y · ˙ 1 st F Z R Y
35 14 31 34 3eqtr3d φ R Z = Y 2 nd F Z M X 1 st F Y · ˙ 1 st F Z R Y