Metamath Proof Explorer


Theorem elpi1i

Description: The elements of the fundamental group. (Contributed by Jeff Madsen, 19-Jun-2010) (Revised by Mario Carneiro, 10-Jul-2015)

Ref Expression
Hypotheses elpi1.g ⊢ G = J π 1 Y
elpi1.b ⊢ B = Base G
elpi1.1 ⊢ φ → J ∈ TopOn ⁡ X
elpi1.2 ⊢ φ → Y ∈ X
elpi1i.3 ⊢ φ → F ∈ II Cn J
elpi1i.4 ⊢ φ → F ⁡ 0 = Y
elpi1i.5 ⊢ φ → F ⁡ 1 = Y
Assertion elpi1i ⊢ φ → F ≃ ph ⁡ J ∈ B

Proof

Step Hyp Ref Expression
1 elpi1.g ⊢ G = J π 1 Y
2 elpi1.b ⊢ B = Base G
3 elpi1.1 ⊢ φ → J ∈ TopOn ⁡ X
4 elpi1.2 ⊢ φ → Y ∈ X
5 elpi1i.3 ⊢ φ → F ∈ II Cn J
6 elpi1i.4 ⊢ φ → F ⁡ 0 = Y
7 elpi1i.5 ⊢ φ → F ⁡ 1 = Y
8 eceq1 ⊢ f = F → f ≃ ph ⁡ J = F ≃ ph ⁡ J
9 8 eqcomd ⊢ f = F → F ≃ ph ⁡ J = f ≃ ph ⁡ J
10 9 biantrud ⊢ f = F → f ⁡ 0 = Y ∧ f ⁡ 1 = Y ↔ f ⁡ 0 = Y ∧ f ⁡ 1 = Y ∧ F ≃ ph ⁡ J = f ≃ ph ⁡ J
11 fveq1 ⊢ f = F → f ⁡ 0 = F ⁡ 0
12 11 eqeq1d ⊢ f = F → f ⁡ 0 = Y ↔ F ⁡ 0 = Y
13 fveq1 ⊢ f = F → f ⁡ 1 = F ⁡ 1
14 13 eqeq1d ⊢ f = F → f ⁡ 1 = Y ↔ F ⁡ 1 = Y
15 12 14 anbi12d ⊢ f = F → f ⁡ 0 = Y ∧ f ⁡ 1 = Y ↔ F ⁡ 0 = Y ∧ F ⁡ 1 = Y
16 10 15 bitr3d ⊢ f = F → f ⁡ 0 = Y ∧ f ⁡ 1 = Y ∧ F ≃ ph ⁡ J = f ≃ ph ⁡ J ↔ F ⁡ 0 = Y ∧ F ⁡ 1 = Y
17 16 rspcev ⊢ F ∈ II Cn J ∧ F ⁡ 0 = Y ∧ F ⁡ 1 = Y → ∃ f ∈ II Cn J f ⁡ 0 = Y ∧ f ⁡ 1 = Y ∧ F ≃ ph ⁡ J = f ≃ ph ⁡ J
18 5 6 7 17 syl12anc ⊢ φ → ∃ f ∈ II Cn J f ⁡ 0 = Y ∧ f ⁡ 1 = Y ∧ F ≃ ph ⁡ J = f ≃ ph ⁡ J
19 1 2 3 4 elpi1 ⊢ φ → F ≃ ph ⁡ J ∈ B ↔ ∃ f ∈ II Cn J f ⁡ 0 = Y ∧ f ⁡ 1 = Y ∧ F ≃ ph ⁡ J = f ≃ ph ⁡ J
20 18 19 mpbird ⊢ φ → F ≃ ph ⁡ J ∈ B