Metamath Proof Explorer


Theorem diag1f1o

Description: The object part of the diagonal functor is a bijection if D is terminal. So any functor from a terminal category is one-to-one correspondent to an object of the target base. (Contributed by Zhi Wang, 21-Oct-2025)

Ref Expression
Hypotheses diag1f1o.a A = Base C
diag1f1o.d No typesetting found for |- ( ph -> D e. TermCat ) with typecode |-
diag1f1o.c φ C Cat
diag1f1o.l L = C Δ func D
Assertion diag1f1o φ 1 st L : A 1-1 onto D Func C

Proof

Step Hyp Ref Expression
1 diag1f1o.a A = Base C
2 diag1f1o.d Could not format ( ph -> D e. TermCat ) : No typesetting found for |- ( ph -> D e. TermCat ) with typecode |-
3 diag1f1o.c φ C Cat
4 diag1f1o.l L = C Δ func D
5 2 termccd φ D Cat
6 eqid Base D = Base D
7 6 istermc2 Could not format ( D e. TermCat <-> ( D e. ThinCat /\ E! y y e. ( Base ` D ) ) ) : No typesetting found for |- ( D e. TermCat <-> ( D e. ThinCat /\ E! y y e. ( Base ` D ) ) ) with typecode |-
8 2 7 sylib φ D ThinCat ∃! y y Base D
9 8 simprd φ ∃! y y Base D
10 euex ∃! y y Base D y y Base D
11 9 10 syl φ y y Base D
12 n0 Base D y y Base D
13 11 12 sylibr φ Base D
14 4 3 5 1 6 13 diag1f1 φ 1 st L : A 1-1 D Func C
15 f1f 1 st L : A 1-1 D Func C 1 st L : A D Func C
16 14 15 syl φ 1 st L : A D Func C
17 2 6 termcbas φ y Base D = y
18 17 adantr φ k D Func C y Base D = y
19 fveq2 x = 1 st k y 1 st L x = 1 st L 1 st k y
20 19 eqeq2d x = 1 st k y k = 1 st L x k = 1 st L 1 st k y
21 2 ad2antrr Could not format ( ( ( ph /\ k e. ( D Func C ) ) /\ ( Base ` D ) = { y } ) -> D e. TermCat ) : No typesetting found for |- ( ( ( ph /\ k e. ( D Func C ) ) /\ ( Base ` D ) = { y } ) -> D e. TermCat ) with typecode |-
22 simplr φ k D Func C Base D = y k D Func C
23 vsnid y y
24 simpr φ k D Func C Base D = y Base D = y
25 23 24 eleqtrrid φ k D Func C Base D = y y Base D
26 eqid 1 st k y = 1 st k y
27 1 21 22 6 25 26 4 diag1f1olem φ k D Func C Base D = y 1 st k y A k = 1 st L 1 st k y
28 27 simpld φ k D Func C Base D = y 1 st k y A
29 27 simprd φ k D Func C Base D = y k = 1 st L 1 st k y
30 20 28 29 rspcedvdw φ k D Func C Base D = y x A k = 1 st L x
31 18 30 exlimddv φ k D Func C x A k = 1 st L x
32 31 ralrimiva φ k D Func C x A k = 1 st L x
33 dffo3 1 st L : A onto D Func C 1 st L : A D Func C k D Func C x A k = 1 st L x
34 16 32 33 sylanbrc φ 1 st L : A onto D Func C
35 df-f1o 1 st L : A 1-1 onto D Func C 1 st L : A 1-1 D Func C 1 st L : A onto D Func C
36 14 34 35 sylanbrc φ 1 st L : A 1-1 onto D Func C