Table Of ContentGordonBrown
Spring2015
RepresentableFunctorsandtheYonedaLemma
1. TheHomandYonedaFunctors
In these notes, every category C will be locally small: for any pair of objects x,y ∈ C, the hom-class
Hom (x,y) is in fact a set, and thus an object of the category Set. To remind us of this, we’ll refer to
C
thehom-classesashom-setsinstead. ThesubscriptC herewillbeomittedwhentheunderlyingcategoryC is
understood. ThedefinitionoflocallysmallimmediatelysuggestsapairoffunctorsC →Set.
Definition: The covariant hom-functor Hom(x,−) : C → Set associated to the object x ∈ C sends an object
y ∈ C to the hom-set Hom(x,y), as the notation suggests, and a morphism f : y → z to the following map
betweenhom-sets.
Hom(x,y) → Hom(x,z)
g (cid:55)→ f ◦g
We’llrefertothismorphismassignmentaspostcompositionwithf. Allofthesedatamaybeencodedasinthe
followingdiagram.
y > Hom(x,y)
f f◦−
∨
∨
z > Hom(x,z)
Similarlydefinedisthecontravarianthom-functorHom(−,x):C →Set.
y > Hom(y,x)
∧
f −◦f
∨
z > Hom(z,x)
Themorphismassignmenthereiscalledprecompositionwithf.
For categories C and D, let [C,D] be the corresponding functor category, whose objects are functors C → D
andwhosemorphismsarenaturaltransformationsbetweenthem.1 Notsurprisingly,assigningtoeveryobject
x ∈ C its covariant hom-functor Hom(x,−) defines a functor Y : C → [C,Set].2 This functor is called the
Yoneda functor after Nobuo Yoneda (1930-1966), a Japanese computer scientist and mathematician. What
issurprisingisthatY iscontravariant!3 Boththeseclaimsareeasilychecked,andinterestinglyenough,the
second boils down to the fact that the components of a natural transformation Hom(x,−) → Hom(y,−)
havetocomefromprecomposition.
1ThealternativenotationforthisfunctorcategoryisSetC,butIpersonallyprefer[C,Set].
2OfcourseonecanalsodefineafunctorC→[C,Set]assigningtoeachobjectx∈Citscontravarianthom-functorHom(−,x).
3AndtheversionofY assigningtoanobjectitscontravarianthom-functoriscovariant!
1
2. RepresentableFunctors
Recallthatanaturalisomorphismisanaturaltransformationwhoseeverycomponentisanisomorphism.
Definition: AfunctorF : C → Setisrepresentableifthereexistsanobjectx ∈ C andanaturalisomorphism
Hom(x,−) ∼= F. Thismeansforeverymorphismf : y → z inC, thereexistisomorphismsα andα inSet
y z
(setbijections)sothatthefollowingdiagramcommutes.
Hom(x,y) α>y F(y)
f◦− F(f)
∨ ∨
Hom(x,z) > F(z)
αz
InthiscasexiscalledtherepresentingobjectandHom(x,−)iscalledarepresentationofF.
Examples:
(1) TheidentityfunctorSet→Setisrepresentedbythesingletonset{x }.Indeed,forasetmapf :S →
0
T,weconsiderthefollowingdiagram.
Hom({x },S) αS> S
0
f◦− f
∨ ∨
Hom({x },T) > T
0
αT
Anyg ∈Hom({x },S)consistsofoneelementg(x )=s ∈S,andsimilarlyforanyh∈Hom({x },T)
0 0 0 0
withh(x )=t ∈T.Therefore,definingα (g)=s andα (h)=t willmakethediagramcommute:
0 0 S 0 T 0
g > g(x )=s
0 0
∨ ∨
f ◦g > (f ◦g)(x )=f(s )
0 0
Notethatthereason{x }wastherepresentingobjectherewasexactlybecauseeverymorphismorigi-
0
natingthereiscompletelydeterminedbyasinglechoiceofelementinthedestination.
(2) TheforgetfulfunctorGrp→SetisrepresentedbythegroupZofintegersunderadditionforthesame
reasonasinexample(1).
(3) The constant functor C → Set assigning every object of C to the singleton set {x } is representable if
0
and only if C has an initial object, in which case the representing object is the initial object. Indeed,
supposing this functor is represented by an object i ∈ C, we’d have for any object y ∈ C a component
α : Hom(i,y) → {x } which is an isomorphism in Set. Since {x } consists of one and only element,
y 0 0
so must Hom(i,y). This argument makes it immediately clear that representability of this constant
functorbyacontravarianthom-functorisequivalenttoC havingaterminalobject.
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3. TheYonedaLemmaandCorollaries
Let Nat(F,G) denote the collection of natural transformations F → G between two functors F,G : C → D.
In other words Nat(F,G) is just a more helpful notation for Hom (F,G), the hom-class of morphisms
[C,D]
F →Ginthefunctorcategory[C,D].
Lemma: (Yoneda)Foreveryobjectx∈C andfunctorF :C →Set,wehaveabijectionofsets
Nat(Hom(x,−),F)∼=F(x).4
Proof. One crucial diagram will illustrate. Supposing α : Hom(x,−) → F is a natural transformation and
f :x→y isamorphisminC,we’dhavethefollowingcommutativesquare.
Hom(x,x) α>x F(x)
f◦− F(f)
∨ ∨
Hom(x,y) > F(y)
αy
Denotingtheelementα (1 )∈F(x)bysforshort,wecanfollowtheidentitymorphism1 around.
x x x
1 > s
x
∨ ∨
f ◦1 =f > α (f)=F(f)(s)
x y
Thebottom-rightequalityisbytheassumptionofnaturalityforα,andwithitthefollowingcorrespondence
suggests itself. Given a natural transformation α : Hom(x,−) → F, associate to it the element α (1 ) = s
x x
of the set F(x). Conversely, given an element s ∈ F(x), associate to it the natural transformation α whose
component α for an arbitrary object y ∈ C and morphism g : y → z in C is defined by setting α (g) =
y y
F(g)(s).Thiswillthenforceα (1 )=F(1 )(s)=1 (s)=stobetrue.
x x x F(x)
It’s not hard to convince yourself that these maps are inverses of each other, but we still have to show
thatourdefinitionforthecomponentsofαgivenanelements∈F(x)makeitintoanaturaltransformation.
Tothisend,letg :y →z beamorphisminC. We’dliketodemonstratecommutativityofthefollowing.
Hom(x,y) α>y F(y)
g◦− F(g)
∨ ∨
Hom(x,z) > F(z)
αz
Lettingh∈Hom(x,y),weseethat
h > F(h)(s)
∨ ∨
g◦h > F(g◦h)(s)=F(g)◦F(h)(s)
Equalityhere,duetoF beingafunctor,establishesnaturalityforαandcompletestheproof. (cid:4)
4TheYonedalemmaactuallysaysevenmore,namelythatthisisomorphismisnaturalinbothxandF. Theseresultsaretoofar
afieldandnotimportantenoughtodiscusshere.
3
RecallthatafunctorF :C →D iscalledfaithfulifforeverypairofobjectsx,y ∈C,themap
Hom (x,y) → Hom (F(x),F(y))
C D
f (cid:55)→ F(f)
is injective, full if the map is surjective, and fully faithful is the map is bijective (F is full and faithful). It’s
a basic exercise to prove that if F(f) : F(x) → F(y) is an isomorphism in D and F if fully faithful, then
f :x→y isanisomorphisminC (inparticularxandy areisomorphic).
Corollary: The Yoneda functor Y : C → [C,Set] which sends an object x to the covariant hom-functor
Hom(x,−)isfullyfaithful.5 ItisthusrightfullycalledtheYonedaembedding.
Proof. Foreverypairofobjectsx,y ∈C,wehavebytheYonedalemmaanisomorphism
Nat(Hom(x,−),Hom(y,−))∼=Hom(y,x).6
InlightofthedefinitionsofY,Nat(Hom(x,−),Hom(y,−)),andfullyfaithful,thisobtains. (cid:4)
Corollary: Representingobjectsareuniqueuptoisomorphism.
Proof. SupposethefunctorF :C →Setisrepresentedbyboththeobjectsxandy ofC. Theninthefunctor
category, Hom(x,−) and Hom(y,−) are isomorphic because they’re both isomorphisc to F. Since these are
theimagesofxandy undertheYonedaembedding,xandy mustbeisomorphicinC. (cid:4)
TheYonedalemmaappliesquitenicelytogroups. RecallthatagroupGmaybethoughtofasacategoryG
withoneobjectGinwhicheverymorphism,labelledbyanelementg ofG,isanisomorphism.7 Likewisea
grouphomomorphismφ:G→H andafunctorΦ:G→Hamounttoexactlythesamething.
Abstracting further from group theory to category theory, a G-set is just a functor G → Set, which as-
signs to the single object G of G a set X and to every morphism g : G → G an element of the permutation
group Perm(X), a set bijection X → X. A G-map is then just a natural transformation α between two
such G-setsX andY, whichamountstoasinglecomponent(setmap) α : X → Y withthepropertythat
G
(α ◦g)x=(g◦α )(x)whereg denotestheisomorphismX →X assignedtothegroupelementg ∈G.
G G
Corollary: (Cayley)EveryfinitegroupGisisomorphictoasubgroupofasymmetricgroup.
Proof. Let G be a finite group with n elements. The Yoneda embedding G → [G,Set] assigns to the object
G its covariant hom-functor Hom(G,−), a G-set which represents the group G acting on its underlying
set by left-multiplication. Postcomposinig this with the forgetful functor [G,Set] → Set, which is faithful,
we get an embedding (the composition of faithful functors is faithful) G → Set. This sends the object G
to its underlying set and embeds Hom(G,G) into its group of permutations Perm(G). But Hom(G,G) is
by definition a group, isomorphic to G, and Perm(G) is isomorphic to the symmetric group S . That this
n
amounts to a homomorphism of groups comes from the fact that the full subcategory of Set consisting of
the single object Perm(G) is a group in the same way that G is a group, and this functor between them
establishesagrouphomomorphismG→Perm(G). (cid:4)
5AsistheYonedaembeddingsendinganobjectxtothecontravarianthom-functorHom(−,x).
6Thexandyareallowedto/supposedtoswitchplacesherebecausethisversionofY you’llrememberiscontravariant!
7Seethenotesfrommyfirstcategoriestalkformoreonthisifnecessary.
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