Metamath Proof Explorer


Theorem funcsetc1o

Description: Value of the functor to the trivial category. The converse is also true because F would be the empty set if C were not a category; and the empty set cannot equal an ordered pair of two sets. (Contributed by Zhi Wang, 22-Oct-2025)

Ref Expression
Hypotheses funcsetc1o.1 1 ˙ = SetCat 1 𝑜
funcsetc1o.f F = 1 st 1 ˙ Δ func C
funcsetc1o.c φ C Cat
funcsetc1o.b B = Base C
funcsetc1o.h H = Hom C
Assertion funcsetc1o φ F = B × 1 𝑜 x B , y B x H y × 1 𝑜

Proof

Step Hyp Ref Expression
1 funcsetc1o.1 1 ˙ = SetCat 1 𝑜
2 funcsetc1o.f F = 1 st 1 ˙ Δ func C
3 funcsetc1o.c φ C Cat
4 funcsetc1o.b B = Base C
5 funcsetc1o.h H = Hom C
6 eqid 1 ˙ Δ func C = 1 ˙ Δ func C
7 setc1oterm Could not format ( SetCat ` 1o ) e. TermCat : No typesetting found for |- ( SetCat ` 1o ) e. TermCat with typecode |-
8 1 7 eqeltri Could not format .1. e. TermCat : No typesetting found for |- .1. e. TermCat with typecode |-
9 8 a1i Could not format ( ph -> .1. e. TermCat ) : No typesetting found for |- ( ph -> .1. e. TermCat ) with typecode |-
10 9 termccd φ 1 ˙ Cat
11 1 setc1obas 1 𝑜 = Base 1 ˙
12 0lt1o 1 𝑜
13 12 a1i φ 1 𝑜
14 eqid Id 1 ˙ = Id 1 ˙
15 6 10 3 11 13 2 4 5 14 diag1a φ F = B × x B , y B x H y × Id 1 ˙
16 df1o2 1 𝑜 =
17 16 xpeq2i B × 1 𝑜 = B ×
18 1 14 setc1oid Id 1 ˙ =
19 18 sneqi Id 1 ˙ =
20 16 19 eqtr4i 1 𝑜 = Id 1 ˙
21 20 xpeq2i x H y × 1 𝑜 = x H y × Id 1 ˙
22 21 a1i x B y B x H y × 1 𝑜 = x H y × Id 1 ˙
23 22 mpoeq3ia x B , y B x H y × 1 𝑜 = x B , y B x H y × Id 1 ˙
24 17 23 opeq12i B × 1 𝑜 x B , y B x H y × 1 𝑜 = B × x B , y B x H y × Id 1 ˙
25 15 24 eqtr4di φ F = B × 1 𝑜 x B , y B x H y × 1 𝑜