Metamath Proof Explorer


Theorem nelsubc3

Description: Remark 4.2(2) of Adamek p. 48. There exists a set satisfying all conditions for a subcategory but the existence of identity morphisms. Therefore such condition in df-subc is necessary.

Note that this theorem cheated a little bit because ` ( C |``cat J ) ` is not a category. In fact ` ( C |``cat J ) e. Cat ` is a stronger statement than the condition (d) of Definition 4.1(1) of Adamek p. 48, as stated here (see the proof of issubc3 ). To construct such a category, see setc1onsubc and cnelsubc . (Contributed by Zhi Wang, 5-Nov-2025)

Ref Expression
Assertion nelsubc3 c Cat j s j Fn s × s j cat Hom 𝑓 c ¬ x s Id c x x j x x s y s z s f x j y g y j z g x y comp c z f x j z

Proof

Step Hyp Ref Expression
1 2oex 2 𝑜 V
2 eqid SetCat 2 𝑜 = SetCat 2 𝑜
3 2 setccat 2 𝑜 V SetCat 2 𝑜 Cat
4 1 3 ax-mp SetCat 2 𝑜 Cat
5 1oex 1 𝑜 V
6 5 5 xpex 1 𝑜 × 1 𝑜 V
7 p0ex V
8 6 7 xpex 1 𝑜 × 1 𝑜 × V
9 1 a1i 2 𝑜 V
10 2 9 setcbas 2 𝑜 = Base SetCat 2 𝑜
11 10 mptru 2 𝑜 = Base SetCat 2 𝑜
12 2on0 2 𝑜
13 2on 2 𝑜 On
14 13 onordi Ord 2 𝑜
15 ordge1n0 Ord 2 𝑜 1 𝑜 2 𝑜 2 𝑜
16 14 15 ax-mp 1 𝑜 2 𝑜 2 𝑜
17 12 16 mpbir 1 𝑜 2 𝑜
18 17 a1i 1 𝑜 2 𝑜
19 1n0 1 𝑜
20 19 a1i 1 𝑜
21 eqidd 1 𝑜 × 1 𝑜 × = 1 𝑜 × 1 𝑜 ×
22 eqid Hom 𝑓 SetCat 2 𝑜 = Hom 𝑓 SetCat 2 𝑜
23 11 18 20 21 22 nelsubclem 1 𝑜 × 1 𝑜 × Fn 1 𝑜 × 1 𝑜 1 𝑜 × 1 𝑜 × cat Hom 𝑓 SetCat 2 𝑜 ¬ x 1 𝑜 Id SetCat 2 𝑜 x x 1 𝑜 × 1 𝑜 × x x 1 𝑜 y 1 𝑜 z 1 𝑜 f x 1 𝑜 × 1 𝑜 × y g y 1 𝑜 × 1 𝑜 × z g x y comp SetCat 2 𝑜 z f x 1 𝑜 × 1 𝑜 × z
24 23 mptru 1 𝑜 × 1 𝑜 × Fn 1 𝑜 × 1 𝑜 1 𝑜 × 1 𝑜 × cat Hom 𝑓 SetCat 2 𝑜 ¬ x 1 𝑜 Id SetCat 2 𝑜 x x 1 𝑜 × 1 𝑜 × x x 1 𝑜 y 1 𝑜 z 1 𝑜 f x 1 𝑜 × 1 𝑜 × y g y 1 𝑜 × 1 𝑜 × z g x y comp SetCat 2 𝑜 z f x 1 𝑜 × 1 𝑜 × z
25 4 8 5 24 nelsubc3lem c Cat j s j Fn s × s j cat Hom 𝑓 c ¬ x s Id c x x j x x s y s z s f x j y g y j z g x y comp c z f x j z