Metamath Proof Explorer


Theorem imaidfu2

Description: The image of the identity functor. (Contributed by Zhi Wang, 10-Nov-2025)

Ref Expression
Hypotheses imaidfu.i I = id func C
imaidfu.d φ I D Func E
imaidfu.h H = Hom D
imaidfu.j J = Hom 𝑓 D
imaidfu.k K = x S , y S p 1 st I -1 x × 1 st I -1 y 2 nd I p H p
imaidfu2.s φ S = Base D
Assertion imaidfu2 φ J = K

Proof

Step Hyp Ref Expression
1 imaidfu.i I = id func C
2 imaidfu.d φ I D Func E
3 imaidfu.h H = Hom D
4 imaidfu.j J = Hom 𝑓 D
5 imaidfu.k K = x S , y S p 1 st I -1 x × 1 st I -1 y 2 nd I p H p
6 imaidfu2.s φ S = Base D
7 eqid x 1 st I Base D , y 1 st I Base D p 1 st I -1 x × 1 st I -1 y 2 nd I p H p = x 1 st I Base D , y 1 st I Base D p 1 st I -1 x × 1 st I -1 y 2 nd I p H p
8 eqid 1 st I Base D = 1 st I Base D
9 1 2 3 4 7 8 imaidfu φ J 1 st I Base D × 1 st I Base D = x 1 st I Base D , y 1 st I Base D p 1 st I -1 x × 1 st I -1 y 2 nd I p H p
10 eqidd φ Base D = Base D
11 1 2 10 idfu1sta φ 1 st I = I Base D
12 11 imaeq1d φ 1 st I Base D = I Base D Base D
13 ssid Base D Base D
14 resiima Base D Base D I Base D Base D = Base D
15 13 14 ax-mp I Base D Base D = Base D
16 12 15 eqtrdi φ 1 st I Base D = Base D
17 16 sqxpeqd φ 1 st I Base D × 1 st I Base D = Base D × Base D
18 17 reseq2d φ J 1 st I Base D × 1 st I Base D = J Base D × Base D
19 eqid Base D = Base D
20 4 19 homffn J Fn Base D × Base D
21 fnresdm J Fn Base D × Base D J Base D × Base D = J
22 20 21 ax-mp J Base D × Base D = J
23 18 22 eqtrdi φ J 1 st I Base D × 1 st I Base D = J
24 15 12 6 3eqtr4a φ 1 st I Base D = S
25 eqidd φ p 1 st I -1 x × 1 st I -1 y 2 nd I p H p = p 1 st I -1 x × 1 st I -1 y 2 nd I p H p
26 24 24 25 mpoeq123dv φ x 1 st I Base D , y 1 st I Base D p 1 st I -1 x × 1 st I -1 y 2 nd I p H p = x S , y S p 1 st I -1 x × 1 st I -1 y 2 nd I p H p
27 9 23 26 3eqtr3d φ J = x S , y S p 1 st I -1 x × 1 st I -1 y 2 nd I p H p
28 27 5 eqtr4di φ J = K