module Cat.Displayed.Cartesian.Discrete where

# Discrete fibrations🔗

A **discrete fibration** is a displayed category whose
fibres are all *discrete
categories*: thin, univalent groupoids. Since thin, univalent
groupoids are sets, a discrete fibration over
$\mathcal{B}$
is an alternate way of encoding a presheaf over
$\mathcal{B}$,
i.e., a functor
$\mathcal{B}^{\mathrm{op}}\to\mathbf{Sets}$.
Here, we identify a purely fibrational property that picks out the
discrete fibrations among the displayed categories, without talking
about the fibres directly.

A discrete fibration is a displayed category such that each type of displayed objects is a set, and such that, for each right corner

there is a contractible space of objects $x'$ over $x$ equipped with maps $x' \to_f y'$.

module _ {o ℓ o′ ℓ′} {B : Precategory o ℓ} (E : Displayed B o′ ℓ′) where private module B = Cat.Reasoning B module E = Displayed E open Cat.Displayed.Morphism E open Cat.Displayed.Reasoning E

record Discrete-fibration : Type (o ⊔ ℓ ⊔ o′ ⊔ ℓ′) where field fibre-set : ∀ x → is-set E.Ob[ x ] lifts : ∀ {x y} (f : B.Hom x y) (y′ : E.Ob[ y ]) → is-contr (Σ[ x′ ∈ E.Ob[ x ] ] E.Hom[ f ] x′ y′)

## Discrete fibrations are Cartesian🔗

To prove that discrete fibrations deserve the name discrete
*fibrations*, we prove that any discrete fibration is a Cartesian
fibration. By assumption, every right corner has a unique lift, which is
in particular a lift: we just have to show that the lift is
Cartesian.

discrete→cartesian : Discrete-fibration → Cartesian-fibration E discrete→cartesian disc = r where open Discrete-fibration disc r : Cartesian-fibration E r .has-lift f y′ .x′ = lifts f y′ .centre .fst r .has-lift f y′ .lifting = lifts f y′ .centre .snd

So suppose we have an open diagram

where $f' : a' \to b'$ is the unique lift of $f$ along $b'$. We need to find a map $u' \to_m a'$. Observe that we have a right corner (with vertices $u$ and $a'$ over $a$), so that we an object $u_2$ over $u$ and map $l : u_2 \to_m a'$. Initially, this looks like it might not help, but observe that $u' \xrightarrow{h'}_{f \circ m} b'$ and $u_2 \xrightarrow{l}_{u} a' \xrightarrow{f'}_f b'$ are lifts of the right corner with base given by $u \to a \to b$, so that by uniqueness, $u2 = u'$: thus, we can use $l$ as our map $u' \to a'$.

r .has-lift f y′ .cartesian .universal {u} {u′} m h′ = subst (λ x → E.Hom[ m ] x (lifts f y′ .centre .fst)) (ap fst $ is-contr→is-prop (lifts (f B.∘ m) y′) (_ , lifts f y′ .centre .snd E.∘′ lifts m _ .centre .snd) (u′ , h′)) (lifts m (lifts f y′ .centre .fst) .centre .snd) r .has-lift f y′ .cartesian .commutes m h′ = Σ-inj-set (fibre-set _) $ is-contr→is-prop (lifts (f B.∘ m) y′) _ _ r .has-lift f y′ .cartesian .unique {u} {u′} {m} m′ x = Σ-inj-set (fibre-set u) $ is-contr→is-prop (lifts m _) (u′ , m′) (u′ , _)

## Fibres of discrete fibrations🔗

Let
$x$
be an object of
$\mathcal{B}$.
Let us ponder the fibre
$\mathcal{E}^*(x)$:
we know that it is strict, since by assumption there is a *set*
of objects over
$x$.
Let us show also that it is thin: imagine that we have two parallel,
vertical arrows
$f, g : a \to_{\mathrm{id}_{}} b$.
These assemble into a diagram like

whence we see that $(a', f)$ and $(a', g)$ are both lifts for the lower corner given by lifting the identity map along $b'$ — so, since lifts are unique, we have $f = g$.

discrete→thin-fibres : ∀ x → Discrete-fibration → ∀ {a b} → is-prop (Fibre E x .Precategory.Hom a b) discrete→thin-fibres x disc {a} {b} f g = Σ-inj-set (fibre-set x) $ is-contr→is-prop (lifts B.id b) (a , f) (a , g) where open Discrete-fibration disc

## Morphisms in Discrete Fibrations🔗

If $\mathcal{E}$ is a discrete fibration, then the only vertical morphisms are identities. This follows directly from lifts being contractible.

discrete→vertical-id : Discrete-fibration → ∀ {x} {x″ : E.Ob[ x ]} (f′ : Σ[ x′ ∈ E.Ob[ x ] ] (E.Hom[ B.id ] x′ x″)) → (x″ , E.id′) ≡ f′ discrete→vertical-id disc {x″ = x″} f′ = sym (lifts B.id _ .paths (x″ , E.id′)) ∙ lifts B.id x″ .paths f′ where open Discrete-fibration disc

We can use this fact in conjunction with the fact that all fibres are thin to show that every vertical morphism in a discrete fibration is invertible.

discrete→vertical-invertible : Discrete-fibration → ∀ {x} {x′ x″ : E.Ob[ x ]} → (f′ : E.Hom[ B.id ] x′ x″) → is-invertible↓ f′ discrete→vertical-invertible disc {x′ = x′} {x″ = x″} f′ = make-invertible↓ (subst (λ x′ → E.Hom[ B.id ] x″ x′) x″≡x′ E.id′) (to-pathp (discrete→thin-fibres _ disc _ _)) (to-pathp (discrete→thin-fibres _ disc _ _)) where x″≡x′ : x″ ≡ x′ x″≡x′ = ap fst (discrete→vertical-id disc (x′ , f′))

## Discrete Fibrations are Presheaves🔗

As noted earlier, a discrete fibration over $\mathcal{B}$ encodes the same data as a presheaf on $\mathcal{B}$. First, let us show that we can construct a presheaf from a discrete fibration.

discrete→presheaf : ∀ {o′ ℓ′} (E : Displayed B o′ ℓ′) → Discrete-fibration E → Functor (B ^op) (Sets o′) discrete→presheaf {o′ = o′} E disc = psh where module E = Displayed E open Discrete-fibration disc

For each object in
$X : \mathcal{B}$,
we take the set of objects
$E$
that lie over
$X$.
The functorial action of `f : Hom X Y`

is then constructed by
taking the domain of the lift of `f`

. Functoriality follows
by uniqueness of the lifts.

psh : Functor (B ^op) (Sets o′) psh .Functor.F₀ X = el E.Ob[ X ] (fibre-set X) psh .Functor.F₁ f X′ = lifts f X′ .centre .fst psh .Functor.F-id = funext λ X′ → ap fst (lifts B.id X′ .paths (X′ , E.id′)) psh .Functor.F-∘ {X} {Y} {Z} f g = funext λ X′ → let Y′ : E.Ob[ Y ] Y′ = lifts g X′ .centre .fst g′ : E.Hom[ g ] Y′ X′ g′ = lifts g X′ .centre .snd Z′ : E.Ob[ Z ] Z′ = lifts f Y′ .centre .fst f′ : E.Hom[ f ] Z′ Y′ f′ = lifts f Y′ .centre .snd in ap fst (lifts (g B.∘ f) X′ .paths (Z′ , (g′ E.∘′ f′ )))

To construct a discrete fibration from a presheaf $P$, we take the (displayed) category of elements of $P$. This is a natural choice, as it encodes the same data as $P$, just rendered down into a soup of points and bits of functions. Discreteness follows immediately from the contractibilty of singletons.

presheaf→discrete : ∀ {κ} → Functor (B ^op) (Sets κ) → Σ[ E ∈ Displayed B κ κ ] Discrete-fibration E presheaf→discrete {κ = κ} P = ∫ B P , discrete where module P = Functor P discrete : Discrete-fibration (∫ B P) discrete .Discrete-fibration.fibre-set X = P.₀ X .is-tr discrete .Discrete-fibration.lifts f P[Y] = contr (P.₁ f P[Y] , refl) Singleton-is-contr

We conclude by proving that the two maps defined above are, in fact,
inverses. Most of the complexity is in characterising paths between displayed
categories, but that doesn’t mean that the proof here is entirely
trivial, either. Well, first, we note that one direction *is*
trivial: modulo differences in the proofs of functoriality, which do not
matter for identity, turning a functor into a discrete fibration and
back is the identity.

open is-iso presheaf≃discrete : ∀ {κ} → is-iso (presheaf→discrete {κ = κ}) presheaf≃discrete .inv (d , f) = discrete→presheaf d f presheaf≃discrete .linv x = Functor-path (λ _ → n-path refl) λ _ → refl

The other direction is where the complication lies. Given a discrete fibration $P \mathrel{\htmlClass{liesover}{\hspace{1.366em}}} X$, how do we show that $\int P \equiv P$? Well, by the aforementioned characterisation of paths in displayed categories, it’ll suffice to construct a functor $(\int P) \to P$ (lying over the identity), then show that this functor has an invertible action on objects, and an invertible action on morphisms.

presheaf≃discrete .rinv (P , p-disc) = Σ-prop-path hl ∫≡dx where open Discrete-fibration p-disc open Displayed-functor open Displayed P

The functor will send an object $c \mathrel{\htmlClass{liesover}{\hspace{1.366em}}} x$ to that same object $c$; This is readily seen to be invertible. But the action on morphisms is slightly more complicated. Recall that, since $P$ is a discrete fibration, every span $b' \mathrel{\htmlClass{liesover}{\hspace{1.366em}}} b \xleftarrow{f} a$ has a contractible space of Cartesian lifts $(a', f')$. Our functor must, given objects $a'', b'$, a map $f : a \to b$, and a proof that $a'' = a'$, produce a map $a'' \to_f b$ — so we can take the canonical $f' : a' \to_f b$ and transport it over the given $a'' = a'$.

pieces : Displayed-functor (∫ B (discrete→presheaf P p-disc)) P Id pieces .F₀′ x = x pieces .F₁′ {f = f} {a′} {b′} x = subst (λ e → Hom[ f ] e b′) x $ lifts f b′ .centre .snd

This transport *threatens* to throw a spanner in the works,
since it is an equation between objects (over
$a$).
But since
$P$
is a discrete fibration, the space of objects over
$a$
is a set, so this equation *can’t* ruin our day. Directly from
the uniqueness of
$(a', f')$
we conclude that we’ve put together a functor.

pieces .F-id′ = from-pathp (ap snd (lifts _ _ .paths _)) pieces .F-∘′ {f = f} {g} {a′} {b′} {c′} {f′} {g′} = ap (λ e → subst (λ e → Hom[ f B.∘ g ] e c′) e (lifts _ _ .centre .snd)) (fibre-set _ _ _ _ _) ∙ from-pathp (ap snd (lifts _ _ .paths _))

It remains to show that, given a map $a'' \to b$ (and the rest of the data $a$, $b$, $f : a \to b$, $b' \mathrel{\htmlClass{liesover}{\hspace{1.366em}}} b$), we can recover a proof that $a''$ is the chosen lift $a'$. But again, lifts are unique, so this is immediate.

∫≡dx : ∫ B (discrete→presheaf P p-disc) ≡ P ∫≡dx = Displayed-path pieces (λ _ → id-equiv) (is-iso→is-equiv p) where p : ∀ {a b} {f : B.Hom a b} {a′} {b′} → is-iso (pieces .F₁′ {f = f} {a′} {b′}) p .inv f = ap fst $ lifts _ _ .paths (_ , f) p .rinv p = from-pathp (ap snd (lifts _ _ .paths _)) p .linv p = fibre-set _ _ _ _ _

We must additionally show that the witness that $P$ is a discrete fibration will survive a round-trip through the type of presheaves, but this witness lives in a proposition (it is a pair of propositions), so it survives automatically.

private unquoteDecl eqv = declare-record-iso eqv (quote Discrete-fibration) hl : ∀ x → is-prop _ hl x = is-hlevel≃ 1 (Iso→Equiv eqv) $ ×-is-hlevel 1 (Π-is-hlevel 1 λ _ → is-hlevel-is-prop 2) hlevel!