open import 1Lab.HLevel.Retracts
open import 1Lab.Path.Groupoid
open import 1Lab.Univalence
open import 1Lab.HIT.S1
open import 1Lab.HLevel
open import 1Lab.Equiv
open import 1Lab.Path
open import 1Lab.Type

module 1Lab.Counterexamples.IsIso where

# is-iso is not a proposition🔗

We show that if is-iso were a proposition, then (x : A) → x ≡ x would be contractible for any choice of A. Taking A to be , we show that this can not be the case. Suppose that is-iso is a proposition.

module
_ (iso-is-prop :  {} {A B : Type } {f : A  B}  is-prop (is-iso f))
where

First we characterise the type is-iso f by showing that, if it is inhabited, then it is equivalent to the centre of A, i.e. the loop-assigning maps of A:

lemma :  {} {A B : Type } {f : A  B}  is-iso f  is-iso f  ((x : A)  x  x)
lemma {A = A} {B} {f} iiso =
EquivJ  _ f  is-iso (f .fst)  ((x : A)  x  x))
(Iso→Equiv helper)
(f , is-iso→is-equiv iiso)
where

By equivalence induction, it suffices to cover the case where f is the identity function. In that case, we can construct an isomorphism quite readily, where the proof uses our assumption iso-is-prop for convenience.

helper : Iso _ _
helper .fst iiso x =
sym (iiso .is-iso.linv x)  iiso .is-iso.rinv x
helper .snd .is-iso.inv x = iso  x  x) x  _  refl)
helper .snd .is-iso.rinv p = funext λ x  ∙-id-l _
helper .snd .is-iso.linv x = iso-is-prop _ _

We thus have that is-iso id ≃ ((x : A) → x ≡ x) - since the former is a prop (by assumption), then so is the latter:

is-prop-loops :  {} {A : Type }  is-prop ((x : A)  x  x)
is-prop-loops {A = A} = equiv→is-hlevel 1 (helper .fst) (helper .snd) iso-is-prop
where helper = lemma {f = λ (x : A)  x}
(iso  x  x)  x  refl)  x  refl))

Thus, it suffices to choose a type for which (x : A) → x ≡ x has two distinct elements. We go with the circle, :

¬is-prop-loops : is-prop ((x : )  x  x)
¬is-prop-loops prop = refl≠loop (happly (prop  x  refl) always-loop) base)