Metamath Proof Explorer


Theorem diag2

Description: Value of the diagonal functor at a morphism. (Contributed by Mario Carneiro, 7-Jan-2017)

Ref Expression
Hypotheses diag2.l L = C Δ func D
diag2.a A = Base C
diag2.b B = Base D
diag2.h H = Hom C
diag2.c φ C Cat
diag2.d φ D Cat
diag2.x φ X A
diag2.y φ Y A
diag2.f φ F X H Y
Assertion diag2 φ X 2 nd L Y F = B × F

Proof

Step Hyp Ref Expression
1 diag2.l L = C Δ func D
2 diag2.a A = Base C
3 diag2.b B = Base D
4 diag2.h H = Hom C
5 diag2.c φ C Cat
6 diag2.d φ D Cat
7 diag2.x φ X A
8 diag2.y φ Y A
9 diag2.f φ F X H Y
10 1 5 6 diagval φ L = C D curry F C 1 st F D
11 10 fveq2d φ 2 nd L = 2 nd C D curry F C 1 st F D
12 11 oveqd φ X 2 nd L Y = X 2 nd C D curry F C 1 st F D Y
13 12 fveq1d φ X 2 nd L Y F = X 2 nd C D curry F C 1 st F D Y F
14 eqid C D curry F C 1 st F D = C D curry F C 1 st F D
15 eqid C × c D = C × c D
16 eqid C 1 st F D = C 1 st F D
17 15 5 6 16 1stfcl φ C 1 st F D C × c D Func C
18 eqid Id D = Id D
19 eqid X 2 nd C D curry F C 1 st F D Y F = X 2 nd C D curry F C 1 st F D Y F
20 14 2 5 6 17 3 4 18 7 8 9 19 curf2 φ X 2 nd C D curry F C 1 st F D Y F = x B F X x 2 nd C 1 st F D Y x Id D x
21 15 2 3 xpcbas A × B = Base C × c D
22 eqid Hom C × c D = Hom C × c D
23 5 adantr φ x B C Cat
24 6 adantr φ x B D Cat
25 opelxpi X A x B X x A × B
26 7 25 sylan φ x B X x A × B
27 opelxpi Y A x B Y x A × B
28 8 27 sylan φ x B Y x A × B
29 15 21 22 23 24 16 26 28 1stf2 φ x B X x 2 nd C 1 st F D Y x = 1 st X x Hom C × c D Y x
30 29 oveqd φ x B F X x 2 nd C 1 st F D Y x Id D x = F 1 st X x Hom C × c D Y x Id D x
31 df-ov F 1 st X x Hom C × c D Y x Id D x = 1 st X x Hom C × c D Y x F Id D x
32 9 adantr φ x B F X H Y
33 eqid Hom D = Hom D
34 simpr φ x B x B
35 3 33 18 24 34 catidcl φ x B Id D x x Hom D x
36 32 35 opelxpd φ x B F Id D x X H Y × x Hom D x
37 7 adantr φ x B X A
38 8 adantr φ x B Y A
39 15 2 3 4 33 37 34 38 34 22 xpchom2 φ x B X x Hom C × c D Y x = X H Y × x Hom D x
40 36 39 eleqtrrd φ x B F Id D x X x Hom C × c D Y x
41 40 fvresd φ x B 1 st X x Hom C × c D Y x F Id D x = 1 st F Id D x
42 31 41 syl5eq φ x B F 1 st X x Hom C × c D Y x Id D x = 1 st F Id D x
43 op1stg F X H Y Id D x x Hom D x 1 st F Id D x = F
44 9 35 43 syl2an2r φ x B 1 st F Id D x = F
45 30 42 44 3eqtrd φ x B F X x 2 nd C 1 st F D Y x Id D x = F
46 45 mpteq2dva φ x B F X x 2 nd C 1 st F D Y x Id D x = x B F
47 fconstmpt B × F = x B F
48 46 47 eqtr4di φ x B F X x 2 nd C 1 st F D Y x Id D x = B × F
49 13 20 48 3eqtrd φ X 2 nd L Y F = B × F