Metamath Proof Explorer


Theorem diag12

Description: Value of the constant functor at a morphism. (Contributed by Mario Carneiro, 6-Jan-2017) (Revised by Mario Carneiro, 15-Jan-2017)

Ref Expression
Hypotheses diagval.l L = C Δ func D
diagval.c φ C Cat
diagval.d φ D Cat
diag11.a A = Base C
diag11.c φ X A
diag11.k K = 1 st L X
diag11.b B = Base D
diag11.y φ Y B
diag12.j J = Hom D
diag12.i 1 ˙ = Id C
diag12.z φ Z B
diag12.f φ F Y J Z
Assertion diag12 φ Y 2 nd K Z F = 1 ˙ X

Proof

Step Hyp Ref Expression
1 diagval.l L = C Δ func D
2 diagval.c φ C Cat
3 diagval.d φ D Cat
4 diag11.a A = Base C
5 diag11.c φ X A
6 diag11.k K = 1 st L X
7 diag11.b B = Base D
8 diag11.y φ Y B
9 diag12.j J = Hom D
10 diag12.i 1 ˙ = Id C
11 diag12.z φ Z B
12 diag12.f φ F Y J Z
13 1 2 3 diagval φ L = C D curry F C 1 st F D
14 13 fveq2d φ 1 st L = 1 st C D curry F C 1 st F D
15 14 fveq1d φ 1 st L X = 1 st C D curry F C 1 st F D X
16 6 15 eqtrid φ K = 1 st C D curry F C 1 st F D X
17 16 fveq2d φ 2 nd K = 2 nd 1 st C D curry F C 1 st F D X
18 17 oveqd φ Y 2 nd K Z = Y 2 nd 1 st C D curry F C 1 st F D X Z
19 18 fveq1d φ Y 2 nd K Z F = Y 2 nd 1 st C D curry F C 1 st F D X Z F
20 eqid C D curry F C 1 st F D = C D curry F C 1 st F D
21 eqid C × c D = C × c D
22 eqid C 1 st F D = C 1 st F D
23 21 2 3 22 1stfcl φ C 1 st F D C × c D Func C
24 eqid 1 st C D curry F C 1 st F D X = 1 st C D curry F C 1 st F D X
25 20 4 2 3 23 7 5 24 8 9 10 11 12 curf12 φ Y 2 nd 1 st C D curry F C 1 st F D X Z F = 1 ˙ X X Y 2 nd C 1 st F D X Z F
26 df-ov 1 ˙ X X Y 2 nd C 1 st F D X Z F = X Y 2 nd C 1 st F D X Z 1 ˙ X F
27 21 4 7 xpcbas A × B = Base C × c D
28 eqid Hom C × c D = Hom C × c D
29 5 8 opelxpd φ X Y A × B
30 5 11 opelxpd φ X Z A × B
31 21 27 28 2 3 22 29 30 1stf2 φ X Y 2 nd C 1 st F D X Z = 1 st X Y Hom C × c D X Z
32 31 fveq1d φ X Y 2 nd C 1 st F D X Z 1 ˙ X F = 1 st X Y Hom C × c D X Z 1 ˙ X F
33 26 32 eqtrid φ 1 ˙ X X Y 2 nd C 1 st F D X Z F = 1 st X Y Hom C × c D X Z 1 ˙ X F
34 eqid Hom C = Hom C
35 4 34 10 2 5 catidcl φ 1 ˙ X X Hom C X
36 35 12 opelxpd φ 1 ˙ X F X Hom C X × Y J Z
37 21 4 7 34 9 5 8 5 11 28 xpchom2 φ X Y Hom C × c D X Z = X Hom C X × Y J Z
38 36 37 eleqtrrd φ 1 ˙ X F X Y Hom C × c D X Z
39 38 fvresd φ 1 st X Y Hom C × c D X Z 1 ˙ X F = 1 st 1 ˙ X F
40 op1stg 1 ˙ X X Hom C X F Y J Z 1 st 1 ˙ X F = 1 ˙ X
41 35 12 40 syl2anc φ 1 st 1 ˙ X F = 1 ˙ X
42 33 39 41 3eqtrd φ 1 ˙ X X Y 2 nd C 1 st F D X Z F = 1 ˙ X
43 19 25 42 3eqtrd φ Y 2 nd K Z F = 1 ˙ X