module Data.Sum.Properties where
As warmup, we have that both constructors are embeddings:
inl-inj : {x y : A} β inl {B = B} x β‘ inl y β x β‘ y inl-inj {A = A} {x = x} path = ap f path where f : A β B β A f (inl x) = x f (inr _) = x inr-inj : {A : Type b} {x y : B} β inr {A = A} x β‘ inr y β x β‘ y inr-inj {B = B} {x = x} path = ap f path where f : A β B β B f (inl _) = x f (inr x) = x inlβ inr : {A : Type a} {B : Type b} {x : A} {y : B} β Β¬ inl x β‘ inr y inlβ inr path = subst (Ξ» { (inl x) β β€ ; (inr x) β β₯ }) path tt inrβ inl : {A : Type a} {B : Type b} {x : A} {y : B} β Β¬ inr x β‘ inl y inrβ inl path = inlβ inr (sym path)
Closure under h-levelsπ
If
and
are
for
then so is their coproduct. The way we prove this is by characterising
the entire path space of A β B
in terms of the path spaces
for A
and B
, using a recursive definition:
module βPath where Code : A β B β A β B β Type (level-of A β level-of B) Code {B = B} (inl x) (inl y) = Lift (level-of B) (x β‘ y) Code (inl x) (inr y) = Lift _ β₯ Code (inr x) (inl y) = Lift _ β₯ Code {A = A} (inr x) (inr y) = Lift (level-of A) (x β‘ y)
Given a Code
for a path in
A β B
, we can turn it into a legitimate path. Agda
automatically lets us ignore the cases where the Code
computes to the empty type
.
decode : {x y : A β B} β Code x y β x β‘ y decode {x = inl x} {y = inl xβ} code = ap inl (lower code) decode {x = inr x} {y = inr xβ} code = ap inr (lower code)
In the inverse direction, we have a procedure for turning paths into codes:
encode : {x y : A β B} β x β‘ y β Code x y encode {x = inl x} {y = inl y} path = lift (inl-inj path) encode {x = inl x} {y = inr y} path = absurd (inlβ inr path) encode {x = inr x} {y = inl y} path = absurd (inrβ inl path) encode {x = inr x} {y = inr y} path = lift (inr-inj path)
Now we must establish that encode
and decode
are inverses. In the one
direction, we can use path induction:
decode-encode : {x y : A β B} (p : x β‘ y) β decode (encode p) β‘ p decode-encode = J (Ξ» _ p β decode (encode p) β‘ p) d-e-refl where d-e-refl : {x : A β B} β decode (encode (Ξ» i β x)) β‘ (Ξ» i β x) d-e-refl {x = inl x} = refl d-e-refl {x = inr x} = refl
In the other direction, the proof is by case analysis, and everything
computes wonderfully to make the right-hand sides fillable by refl
:
encode-decode : {x y : A β B} (p : Code x y) β encode (decode p) β‘ p encode-decode {x = inl x} {y = inl y} p = refl encode-decode {x = inr x} {y = inr y} p = refl
Thus, we have an equivalence between codes for paths in
A β B
and actual paths A β B
. Since
Code
has a nice computational
structure, we can establish its h-level by induction:
CodeβPath : {x y : A β B} β (x β‘ y) β Code x y CodeβPath = IsoβEquiv (encode , iso decode encode-decode decode-encode)
open βPath Code-is-hlevel : {x y : A β B} {n : Nat} β is-hlevel A (2 + n) β is-hlevel B (2 + n) β is-hlevel (Code x y) (suc n) Code-is-hlevel {x = inl x} {inl y} {n} ahl bhl = Lift-is-hlevel (suc n) (ahl x y) Code-is-hlevel {x = inr x} {inr y} {n} ahl bhl = Lift-is-hlevel (suc n) (bhl x y)
In the two cases where x
and y
match, we
can use the fact that Lift preserves h-levels
and the
assumption that A
(or B
) have the given
h-level.
Code-is-hlevel {x = inl x} {inr y} {n} ahl bhl = Lift-is-hlevel (suc n) (is-propβis-hlevel-suc Ξ» x β absurd x) Code-is-hlevel {x = inr x} {inl y} {n} ahl bhl = Lift-is-hlevel (suc n) (is-propβis-hlevel-suc Ξ» x β absurd x)
In the mismatched cases, we use the fact that propositions have any successor h-level
to prove that β₯
is also at the same h-level as
A
and B
. Thus, we have:
β-is-hlevel : (n : Nat) β is-hlevel A (2 + n) β is-hlevel B (2 + n) β is-hlevel (A β B) (2 + n) β-is-hlevel n ahl bhl x y = Equivβis-hlevel (1 + n) CodeβPath (Code-is-hlevel ahl bhl) instance H-Level-β : β {n} β¦ _ : 2 β€ n β¦ β¦ _ : H-Level A n β¦ β¦ _ : H-Level B n β¦ β H-Level (A β B) n H-Level-β {n = suc (suc n)} β¦ sβ€s (sβ€s p) β¦ = hlevel-instance $ β-is-hlevel _ (hlevel (2 + n)) (hlevel (2 + n))
Note that, in general, being a proposition and contractible are not preserved
under coproducts. Consider the case where (A, a)
and
(B, b)
are both contractible (this generalises to
propositions): Then their coproduct has two distinct points,
inl a
and inr b
. However, the coproduct of
disjoint propositions is a proposition:
disjoint-β-is-prop : is-prop A β is-prop B β Β¬ (A Γ B) β is-prop (A β B) disjoint-β-is-prop Ap Bp notab (inl x) (inl y) = ap inl (Ap x y) disjoint-β-is-prop Ap Bp notab (inl x) (inr y) = absurd (notab (x , y)) disjoint-β-is-prop Ap Bp notab (inr x) (inl y) = absurd (notab (y , x)) disjoint-β-is-prop Ap Bp notab (inr x) (inr y) = ap inr (Bp x y)
Discretenessπ
If A
and B
are discrete, then so is
A β B
.
instance Discrete-β : β¦ _ : Discrete A β¦ β¦ _ : Discrete B β¦ β Discrete (A β B) Discrete-β {x = inl x} {inl y} = invmap (ap inl) inl-inj (x β‘? y) Discrete-β {x = inl x} {inr y} = no inlβ inr Discrete-β {x = inr x} {inl y} = no inrβ inl Discrete-β {x = inr x} {inr y} = invmap (ap inr) inr-inj (x β‘? y)