open import Cat.Prelude
open import Cat.Gaunt

import Cat.Reasoning

module Cat.Instances.Free where


# Graphs and free categoriesπ

A graph (really, an 1) is given by a set of vertices and, for each pair of elements a set of edges from to Thatβs it: a set and a family of sets over Really, for our purposes, graphs by themselves are not very interesting: their utility comes in constructing new categories.

record Graph o β : Type (lsuc o β lsuc β) where
field
vert : Set o
edge : β£ vert β£ β β£ vert β£ β Set β


Given a graph we construct a strict category in the following manner:

• The objects of are the vertices of
• The morphisms in are given by finite paths in Finite paths are defined by the following indexed-inductive type
module _ {o β} (G : Graph o β) where
private module G = Graph G

  data Path-in : β£ G.vert β£ β β£ G.vert β£ β Type (o β β) where
nil  : β {a} β Path-in a a
cons : β {a b c} β β£ G.edge a b β£ β Path-in b c β Path-in a c


That is: a path is either empty, in which case it starts and ends at itself (these are the identity morphisms), or we can form a path from by starting with a path and precomposing with an edge Much of the code below is dedicated to characterising the identity types between paths. Indeed, to construct a category we must show that paths in form a set.

We have a couple of options here:

• We can construct paths by recursion, on their length: Defining a βpath from to of length β by recursion on and then defining a path from to as being a pair where and is a path of length

• We can define paths by induction, as is done above.

The former approach makes it easy to show that paths form a set: we can directly construct the set of paths, by recursion, then project the underlying type if necessary. But working with these paths is very inconvenient, since we have to deal with explicit identities between the endpoints. The latter approach makes defining functions on paths easy, but showing that they are a set is fairly involved. Letβs see how to do it.

The first thing weβll need is a predicate expressing that a path really encodes the empty path, and we have an identity of vertices We can do this by recursion: If is nil, then we can take this to be the unit type, otherwise itβs the bottom type.

  is-nil : β {a b} β Path-in a b β Type (o β β)
is-nil nil        = Lift _ β€
is-nil (cons _ _) = Lift _ β₯


Weβd like to define a relation standing for an identification of paths But observe what happens in the case where weβve built up a path by adding an edge: We know that the edges start at and the inner paths end at but the inner vertex may vary!

Weβll need to package an identification in the relation for and so, weβll have to encode for a path over some identification of their start points. Thatβs why we have path-codep and not βpath-codeβ. A value in codes for a path over

  path-codep
: β (a : I β β£ G.vert β£) {c}
β Path-in (a i0) c
β Path-in (a i1) c
β Type (o β β)


Note that in the case where Agda refines to be definitionally and we can no longer match on the right-hand-side path Thatβs where the is-nil predicate comes in: We say that is equal to if is-nil holds. Of course, a cons and a nil can never be equal.

  path-codep a nil ys = is-nil ys
path-codep a (cons x xs) nil = Lift _ β₯


The recursive case constructs an identification of cons cells as a triple consisting of an identification between their intermediate vertices, and over that data, an identification between the added edges, and a code for an identification between the tails.

  path-codep a {c} (cons {b = b} x xs) (cons {b = b'} y ys) =
Ξ£[ bs β (b β‘ b') ]
(PathP (Ξ» i β β£ G.edge (a i) (bs i) β£) x y Γ path-codep (Ξ» i β bs i) xs ys)


By recursion on the paths and the code for an equality, we can show that if we have a code for an identification, we can indeed compute an identification. The most involved case is actually when the lists are empty, in which case we must show that is-nil(xs)2 implies that but it must be over an arbitrary identification 3. Fortunately, vertices in a graph live in a set, so is reflexivity.

  path-encode
: β (a : I β β£ G.vert β£) {c} xs ys
β path-codep a xs ys
β PathP (Ξ» i β Path-in (a i) c) xs ys
path-encode a (cons x xs) (cons y ys) (p , q , r) i =
cons {a = a i} {b = p i} (q i) $path-encode (Ξ» i β p i) xs ys r i path-encode a nil ys p = lemma (Ξ» i β a (~ i)) ys p where lemma : β {a b} (p : a β‘ b) (q : Path-in a b) β is-nil q β PathP (Ξ» i β Path-in (p (~ i)) b) nil q lemma {a = a} p nil (lift lower) = to-pathp$
subst (Ξ» e β Path-in e a) (sym p) nil β‘β¨ (Ξ» i β subst (Ξ» e β Path-in e a) (G.vert .is-tr a a (sym p) refl i) nil) β©β‘
subst (Ξ» e β Path-in e a) refl nil    β‘β¨ transport-refl _ β©β‘
nil                                   β
lemma _ (cons x p) ()


The next step is to show that codes for identifications between paths live in a proposition; But this is immediate by their construction: in every case, we can show that they are either literally a proposition (the base case) or built out of propositions: this last case is inductive.

  path-codep-is-prop
: β (a : I β β£ G.vert β£) {b}
β (p : Path-in (a i0) b) (q : Path-in (a i1) b) β is-prop (path-codep a p q)
path-codep-is-prop a nil xs x y = is-nil-is-prop xs x y where
is-nil-is-prop : β {a b} (xs : Path-in a b) β is-prop (is-nil xs)
is-nil-is-prop nil x y = refl
path-codep-is-prop a (cons h t) (cons h' t') (p , q , r) (p' , q' , r') =
Ξ£-pathp (G.vert .is-tr _ _ _ _) $Ξ£-pathp (is-propβpathp (Ξ» i β PathP-is-hlevel' 1 (G.edge _ _ .is-tr) _ _) q q') (is-propβpathp (Ξ» i β path-codep-is-prop (Ξ» j β G.vert .is-tr _ _ p p' i j) t t') r r')  And finally, by proving that there is a code for the reflexivity path, we can show that we have an identity system in the type of paths from to given by their codes. Since these codes are propositions, and identity systems give a characterisation of a typeβs identity types, we conclude that paths between a pair of vertices live in a set!  path-codep-refl : β {a b} (p : Path-in a b) β path-codep (Ξ» i β a) p p path-codep-refl nil = lift tt path-codep-refl (cons x p) = refl , refl , path-codep-refl p path-identity-system : β {a b} β is-identity-system {A = Path-in a b} (path-codep (Ξ» i β a)) path-codep-refl path-identity-system = set-identity-system (path-codep-is-prop Ξ» i β _) (path-encode _ _ _) path-is-set : β {a b} β is-set (Path-in a b) path-is-set {a = a} = identity-systemβhlevel 1 path-identity-system$
path-codep-is-prop Ξ» i β a

  path-decode
: β {a b} {xs ys : Path-in a b}
β xs β‘ ys
β path-codep (Ξ» _ β a) xs ys
path-decode = Equiv.from (identity-system-gives-path path-identity-system)


## The path categoryπ

By comparison, constructing the actual precategory of paths is almost trivial. The composition operation, concatenation, is defined by recursion over the left-hand-side path. This is definitionally unital on the left.

  _++_ : β {a b c} β Path-in a b β Path-in b c β Path-in a c
nil ++ ys = ys
cons x xs ++ ys = cons x (xs ++ ys)


Right unit and associativity are proven by induction.

  ++-idr : β {a b} (xs : Path-in a b) β xs ++ nil β‘ xs
++-idr nil         = refl
++-idr (cons x xs) = ap (cons x) (++-idr xs)

++-assoc
: β {a b c d} (p : Path-in a b) (q : Path-in b c) (r : Path-in c d)
β (p ++ q) ++ r β‘ p ++ (q ++ r)
++-assoc nil q r        = refl
++-assoc (cons x p) q r = ap (cons x) (++-assoc p q r)


And thatβs it! Note that we must compose paths backwards, since the type of the concatenation operation and the type of morphism composition are mismatched (theyβre reversed).

  open Precategory
Path-category : Precategory o (o β β)
Path-category .Ob = β£ G.vert β£
Path-category .Hom = Path-in
Path-category .Hom-set _ _ = path-is-set
Path-category .id = nil
Path-category ._β_ xs ys = ys ++ xs
Path-category .idr f = refl
Path-category .idl f = ++-idr f
Path-category .assoc f g h = ++-assoc h g f


Moreover, free categories are always gaunt: they are automatically strict and, as can be seen with a bit of work, univalent. Univalence follows because any non-trivial isomorphism would have to arise as a cons, but cons can never be nil β which would be required for a composition to equal the identity.

While types prevent us from directly stating βif a map is invertible, it is nilβ, we can nevertheless pass around some equalities to make this induction acceptable.

  Path-category-is-category : is-category Path-category
Path-category-is-category = r where
module Pc = Cat.Reasoning Path-category

remβ : β {x y} (j : Pc.Isomorphism x y) β Ξ£ (x β‘ y) Ξ» p β PathP (Ξ» i β Pc.Isomorphism x (p i)) Pc.id-iso j
remβ {x = x} im = go im (im .Pc.to) refl (path-decode (im .Pc.invr)) where
go : β {y} (im : Pc.Isomorphism x y) (j' : Path-in x y) β j' β‘ im .Pc.to
β path-codep (Ξ» _ β x) (j' ++ im .Pc.from) nil
β Ξ£ (x β‘ y) Ξ» p β PathP (Ξ» i β Pc.Isomorphism x (p i)) Pc.id-iso im
go im nil p q = refl , ext p

r : is-category Path-category
r .to-path i      = remβ i .fst
r .to-path-over i = remβ i .snd

Path-category-is-gaunt : is-gaunt Path-category
Path-category-is-gaunt = record
{ has-category = Path-category-is-category
; has-strict   = hlevel 2
}


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