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