CONTRAVARIANT FUNCTORS ON FINITE SETS
AND STIRLING NUMBERS
For Jim Lambek
ROBERTPAR
E
ABSTRACT. Weharaterizethenumerialfuntionswhihariseastheardinalities
ofontravariantfuntorsonnite sets,asthose whih haveaseriesexpansioninterms
of Stirlingfuntions. Wegiveaproedure foralulatingtheoeÆientsin suhseries
and a onrete test for determining whether afuntion is of this type. A number of
examplesareonsidered.
1. Introdution
Let Set
0
be the ategory of nite sets and F : Set op
0
! Set
0
a funtor. Suh a funtor
indues a funtion on the natural numbers f : N ! N by f(n) = #F[n℄ where #
represents ardinality and [n℄ is the set f0;1;2;:::;n 1g. As two sets have the same
ardinalityifandonlyiftheyareisomorphi,f ouldbedenedbytheequationf(#X)=
#F(X) for allnite sets X. The question we onsider is whih funtions f arise in this
way. As the naturalnumbers are the ardinalitiesof nite sets, and as funtorsare more
strutured than arbitrary funtions, one might expet to get a nie lass of numerial
funtions this way. Letus allthem ardinalfuntions.
For example, the funtion f(n) = 3 n
is a ardinal funtion as it is the ardinality of
therepresentablefuntorSet
0
( ;3). Butwhataboutthefuntionsn 2
;23 n
2 n
,
2n
n
,n!,
(2n)!
2 n
,and soon? Weshalldetermine ariterionwhihwillhelpusdeidethese questions.
We shall see that we are ledto ertain ombinatorialfuntions, and we an hopefor
some appliations inthat diretion. We present none here, but see [3℄ for appliationsof
ategory theory toombinatoris.
The resultsbelowwere presentedattheAMSmeetinginMontrealinSeptember1997.
Shortly after, Andreas Blass pointed out to me the paper [2℄ by Dougherty in whih
similarresultsareobtained. OurTheorems4.1and 4.3areverysimilartohisProposition
2.14 andTheorem 1.3. Theproofsare notverydierentbut ourshaveamore ategorial
avor. Some lemmas on absolute olimits are of independent interest. The numerial
examples inour paper are new and not without interest.
ResearhsupportedbyagrantfromNSERC
Reeivedbytheeditors1999Marh8and,inrevisedform,1999September13.
Publishedon1999November30.
1991MathematisSubjetClassiation: 18A22,05A10.
Keywordsandphrases: Funtor,ardinality,Stirlingnumbers.
2. Absolute olimits revisited
Absolute olimits were introdued in my thesis [4℄ thirty years ago. When I told Jim
Lambek, who was then my Ph.D. supervisor, about my haraterization of oequalizers
whihare preserved by allfuntors,andhowthey ameupinBek'stripleabilitytheorem
(as it was then alled), he said \Good! Write it up. You an allthem absolute." I did.
A while later, he asked \Is there a smaller lass of funtors whih would be suÆient to
test for absoluteness?" There was,namely the representables, and so my thesis began.
In this setion we obtain some new results onabsoluteolimits inSet
0 .
2.1. Lemma. Let
A
0
A
1
A
2
A
-? ?
f1
f
2
B
0
B
1
B
2
B
-? ?
g1
g
2
be pushouts in Set
0
in whih the f
i ;g
i
are epimorphisms. Then
A
0 B
0
A
1 B
1
A
2 B
2
AB
-? ?
f
1 g
1
f
2 g
2
is also a pushout.
Proof. First, weonsider the speial ase where our pushouts are of the form
A
0
A
0
A A
-? ?
1
A
0
1A
f f
B
0
B
B
0
B :
-? ?
g
g 1
B
0
1
B
Let s be asplitting forg,gs=1
B
. Then in
A
0
B A
0 B
0
A
0 B
AB
AB
0
AB;
-? ? ?
A
0 s
As
A0g
Ag
fB fB0 fB
the outside retangle is a pushout, one in whih the two horizontal maps are identities,
Now, for the general ase, onsider
A
0 B
0
A
1 B
0
A
1 B
1
A
2 B
0
AB
0
AB
1
A
2 B
2
AB
2
AB
-
--
--
-? ? ?
? ? ?
(1) (2)
(3) (4)
whereallthe arrows arethe obviousCartesianproduts. Byartesian losedness,()B
0
and A()preserveolimits,so(1)and(4) are pushouts. (2)and(3)arediagramsofthe
sort disussed inthe previous paragraph,so they are pushouts too. The resultfollows by
pasting pushouts.
2.2. Remark. The proof of the above lemma goes through, with minor modiations
in amonoidallosed ategory: if the f
i and g
i
are regular epimorphisms,then
A
0 B
0
A
1 B
1
A
2 B
2
AB
-? ?
f1g1
f2g2
is apushout.
For the rst part of the proof, let(A
0
g)= (f B
0
). Then and orrespond,
by adjointness, to
and
suh that
A
0
[B;C℄
A
[B
0 ;C℄
-? ?
f [g;C℄
ommutes. As [g;C℄ is moni (g is epi) and f is a regular epi, there exists a unique
diagonal ll-in
: A! [B;C℄ suh that
f =
and [g;C℄
=
. Again, by adjointness,
this orresponds to a unique :AB ! C suh that (f B) = and (Ag)= .
Sothe required square isa pushout.
The seond part of the proof uses onlythat ( )B
0
and A( )preserve pushouts.
Reall from [5℄that a olimitis alled absolute if itis preserved by all funtors.
2.3. Proposition. Pushouts of epimorphisms in Set
0
are absolute.
Proof. One of the basi results of [5℄ is that a olimit is absolute if and only if it is
preserved by all representables. In the ase of Set
0
, the representables [A; ℄ are nite
powers ( ) #A
2.4. Remark. Rihard Wood points out that the same proof shows that reexive
o-equalizers in Set
0
are absolute. In fat, if f
1 ;f
2 : A
0 !
! A
1
is a reexive pair, then the
oequalizeroff
1 andf
2
isthesameastheirpushout,sothisisaspeialaseofProposition
2.3.
2.5. Remark. A similar result, whih we shall not need in the sequel, is the following:
non-empty intersetions are absolute inSet
0
. Indeed, suppose that A and B are subsets
of C and that A\B 6=;. Choose
0
2A\B and denefuntions f :C !A by
f()= (
if 2A
0
otherwise
and g :B !A\B by
g(b)= (
b if b2A\B
0
otherwise.
Then itis easilyseen that
A\B B A\B
A C A
-? ? ?
g
f
ommutes. Furthermore, the outside retangle is an absolute pullbak (as the two
hori-zontalarrows are identities)and the middlevertial arrow isanabsolutemonomorphism
(as it is split). Thus the left square is anabsolutepullbak.
We see that non-empty pullbaks of monomorphisms in Set
0
are absolute for a
rela-tively simple reason. One might say that the pullbak square itself is split. This is not
the ase for pushouts of epimorphisms in Set
0
where an unbounded number (depending
on the size of the sets involved) of funtions may be required to express absoluteness
equationally. Pushoutsofepimorphismsbetweeninnitesetsneed notbeabsoluteeither.
3. The struture of ontravariant funtors
Let F :Set op
0
!Set
0
be any funtor. Say that (n;a) is minimal for Fif a2 F[n℄ and is
not equal toany F( )(b) for :[n℄