CONTRAVARIANT FUNCTORS ON FINITE SETS
AND STIRLING NUMBERS
For Jim Lambek
ROBERTPAR
E
ABSTRACT. Wecharacterizethenumericalfunctionswhichariseasthecardinalities
ofcontravariantfunctorsonnite sets,asthose which haveaseriesexpansioninterms
of Stirlingfunctions. Wegiveaprocedure forcalculatingthecoeÆcientsin suchseries
and a concrete test for determining whether afunction is of this type. A number of
examplesareconsidered.
1. Introduction
Let Set
0
be the category of nite sets and F : Set op
0
! Set
0
a functor. Such a functor
induces a function on the natural numbers f : N ! N by f(n) = #F[n] where #
represents cardinality and [n] is the set f0;1;2;:::;n 1g. As two sets have the same
cardinalityifandonlyiftheyareisomorphic,f couldbedenedbytheequationf(#X)=
#F(X) for allnite sets X. The question we consider is which functions f arise in this
way. As the naturalnumbers are the cardinalitiesof nite sets, and as functorsare more
structured than arbitrary functions, one might expect to get a nice class of numerical
functions this way. Letus callthem cardinal functions.
For example, the function f(n) = 3 n
is a cardinal function as it is the cardinality of
therepresentablefunctorSet
0
( ;3). Butwhataboutthefunctionsn 2
;23 n
2 n
,
2n
n
,n!,
(2n)!
2 n
,and soon? Weshalldetermine acriterionwhichwillhelpusdecidethese questions.
We shall see that we are ledto certain combinatorialfunctions, and we can hopefor
some applications inthat direction. We present none here, but see [3] for applicationsof
category theory tocombinatorics.
The resultsbelowwere presentedattheAMSmeetinginMontrealinSeptember1997.
Shortly after, Andreas Blass pointed out to me the paper [2] by Dougherty in which
similarresultsareobtained. OurTheorems4.1and 4.3areverysimilartohisProposition
2.14 andTheorem 1.3. Theproofsare notverydierentbut ourshaveamore categorical
avor. Some lemmas on absolute colimits are of independent interest. The numerical
examples inour paper are new and not without interest.
ResearchsupportedbyagrantfromNSERC
Receivedbytheeditors1999March8and,inrevisedform,1999September13.
Publishedon1999November30.
1991MathematicsSubjectClassication: 18A22,05A10.
Keywordsandphrases: Functor,cardinality,Stirlingnumbers.
c
RobertPare1999. Permissionto copyforprivateusegranted.
2. Absolute colimits revisited
Absolute colimits were introduced in my thesis [4] thirty years ago. When I told Jim
Lambek, who was then my Ph.D. supervisor, about my characterization of coequalizers
whichare preserved by allfunctors,andhowthey cameupinBeck'stripleabilitytheorem
(as it was then called), he said \Good! Write it up. You can callthem absolute." I did.
A while later, he asked \Is there a smaller class of functors which would be suÆcient to
test for absoluteness?" There was,namely the representables, and so my thesis began.
In this section we obtain some new results onabsolutecolimits 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 which 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, weconsider the special case 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 rectangle is a pushout, one in which the two horizontal maps are identities,
and the middle vertical arrow is an epimorphism, so the right square is a pushout as
Now, for the general case, consider
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 obviousCartesianproducts. Bycartesian closedness,()B
0
and A()preservecolimits,so(1)and(4) are pushouts. (2)and(3)arediagramsofthe
sort discussed 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 modications
in amonoidalclosed category: 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 correspond,
by adjointness, to
and
such that
A
0
[B;C]
A
[B
0
;C]
-
-
? ?
f [g;C]
commutes. As [g;C] is monic (g is epic) and f is a regular epi, there exists a unique
diagonal ll-in
: A! [B;C] such that
f =
and [g;C]
=
. Again, by adjointness,
this corresponds to a unique :AB ! C such that (f B) = and (Ag)= .
Sothe required square isa pushout.
The second part of the proof uses onlythat ( )B
0
and A( )preserve pushouts.
Recall from [5]that a colimitis called absolute if itis preserved by all functors.
2.3. Proposition. Pushouts of epimorphisms in Set
0
are absolute.
Proof. One of the basic results of [5] is that a colimit is absolute if and only if it is
preserved by all representables. In the case of Set
0
, the representables [A; ] are nite
powers ( )
#A
, and by Lemma 1, a nite power of a pushout of epimorphisms is again a
2.4. Remark. Richard Wood points out that the same proof shows that reexive co-
equalizers in Set
0
are absolute. In fact, if f
1
;f
2 : A
0
!
! A
1
is a reexive pair, then the
coequalizeroff
1 andf
2
isthesameastheirpushout,sothisisaspecialcaseofProposition
2.3.
2.5. Remark. A similar result, which we shall not need in the sequel, is the following:
non-empty intersections are absolute inSet
0
. Indeed, suppose that A and B are subsets
of C and that A\B 6=;. Choose c
0
2A\B and denefunctions f :C !A by
f(c)= (
c if c2A
c
0
otherwise
and g :B !A\B by
g(b)= (
b if b2A\B
c
0
otherwise.
Then itis easilyseen that
A\B B A\B
A C A
-
-
-
-
? ? ?
g
f
commutes. Furthermore, the outside rectangle is an absolute pullback (as the two hori-
zontalarrows are identities)and the middlevertical arrow isanabsolutemonomorphism
(asit is split). Thus the left square is anabsolutepullback.
We see that non-empty pullbacks of monomorphisms in Set
0
are absolute for a rela-
tively simple reason. One might say that the pullback square itself is split. This is not
the case for pushouts of epimorphisms in Set
0
where an unbounded number (depending
on the size of the sets involved) of functions may be required to express absoluteness
equationally. Pushoutsofepimorphismsbetweeninnitesetsneed notbeabsoluteeither.
3. The structure of contravariant functors
Let F :Set op
0
!Set
0
be any functor. Say that (n;a) is minimal for Fif a2 F[n] and is
not equal toany F( )(b) for :[n]![m], b2F[m] with m<n.
3.1. Proposition. Let x2FX be any element of F. Then:
(1) There is (n;a) minimal for F and f :X ![n] epic, suchthat F(f)(a)=x.
(2) If (m;b) and g :X ![m] also have the same properties, then m=n and there exists
2S such that g =f and F()(b)=a.
Proof. (1) Of all the triples (n;f;a), n 2 N, f : X ! [n], a 2 F[n] with F(f)(a) =x,
choose one with minimal n. (There is at least one such triple, for if we let n be the
cardinality of X, then there will be an isomorphism f : X ! [n] and we can take
a =F(f 1
)(x).) Then (n;a) is minimal for F, because if there were : [n] ! [m] and
b 2 F[m] with F( )(b) = a and m < n, then F( f)(b) = F(f)F( )(b) = F(f)(a) =x,
and n would not have been minimal for x. Also,if f were not epic it would factor as g
where g : X ! [m] and : [m] ! [n] with m < n. Then F(g)(F( )(a))= F( g)(a) =
F(f)(a)=xand again, n would not be minimalfor x.
(2) Let (m;b) be minimal for F and g : X ! [m] epic such that F(g)(b) = x. Take
the pushout
X
[n]
[m] [p]
-
-
? ?
f
g
whichis absoluteby Proposition 2.3. Thus
F[p] F[n]
F[m]
FX -
-
? ?
is a pullback. As F(f)(a) = x = F(g)(b), there exists c 2 F[p] with F( )(c) = a and
F()(c)=b. As (n;a) isminimalforF,pcannot beless than n, so isanisomorphism.
Similarly, is an isomorphism. It follows that n = p = m and if = 1
2 S
n , then
f =g and F()(b)=F( )F() 1
(b)=F( )(c) =a.
Let A
n
bethe set of minimalelements inF[n], i.e.
A
n
=fa2F[n]j(n;a) isminimal for Fg:
Then the symmetricgroup S
n
acts onthe right onA
n by
(a;)7!F()(a):
Also,S
n
acts onthe left onEpi(X;[n]), the set of epimorphismsX ![n], by
(;f)7!f:
Now the above propositioncan berestated asfollows.
3.2. Corollary. For each X, the function
1
X
n=0 A
n
Sn
Epi(X;[n]) !FX
af 7!F(f)(a)
3.3. Proposition. Given nite S
n
-sets, A
n
, for n = 0;1;2;:::, such that A
0
6= ; and
A
1
6=;, then
G(X)= 1
X
n=0 A
n
S
n
Epi(X;[n])
can be made into a functor G:Set
0 op
!Set
0 .
Proof. Ifn >#X,then Epi(X;[n])=; sothat for anyxed X the innitecoproduct is
essentially niteand G(X) isaniteset. Note thatS
0
and S
1
are both the trivialgroup,
so that A
0
and A
1
are just sets. Pick a
0 2A
0
and a
1 2A
1
. Let
Y
: Y ! [1]denote the
unique function intothe terminalobject. It is epiif Y 6=;.
Let g :Y !X and (n;af)2G(X). Dene
G(g)(n;af)= 8
>
<
>
:
(n;a(fg)) if fg is epi
(1;a
1
Y
) if fg is not epiand Y 6=;
(0;a
0
1) if fg is not epibut Y =;:
First, G(g) is well-dened. Indeed, if a f = a 0
f 0
, then there is 2 S
n
such that
f 0
=f and a 0
=a. Then fg is epiif and only if f 0
g is epiand inthat case a(fg)=
(a 0
)(fg)=a 0
(fg)=a 0
f 0
g.
It isclear fromthe denition that G(1
X )=1
G(X)
. Now let h:Z !Y. Iffgh is epi,
then so isfg and
G(h)G(g)(n;af)=G(h)(n;a(fg))=(n;a(fgh))=G(gh)(n;af):
If fg is epibut fgh isnot and Z 6=;,then
G(h)G(g)(n;af)=G(h)(n;a(fg))=(1;a
1
Z
)=G(gh)(n;af):
If fg is not epi, then fgh isnot either. If Z 6=;,then
G(h)G(g)(n;af)=G(h)(1;a
1
Y
)=(1;a
1
Y
h)=(1;a
1
Z
)=G(gh)(n;af):
Finally,if Z =;, then
G(h)G(g)(n;af)=(0;a
0
1)=G(gh)(n;af):
3.4. Remark. The above construction is not canonical so we can hardly expect the
bijection of the previous corollary to be natural, although we do have naturality if we
4. Stirling series
The Stirling numbers of the second kind are the numbers S(m;n) of partitionsof m into
n pieces. They satisfy the recurrencerelations
S(0;n)= (
1 if n=0
0 otherwise:
S(m+1;n)=S(m;n 1)+nS(m;n):
A table of values for S(m;n) can be constructed from these relations, just like Pascal's
triangle.
n m
@
@
0 1 2 3 4 5
0 1 0 0 0 0 0
1 0 1 1 1 1 1
2 0 0 1 3 7 15
3 0 0 0 1 6 25
4 0 0 0 0 1 10
5 0 0 0 0 0 1
One mightguess from this table that S(m;2)=2 m 1
1,m 1, and this is easilyseen.
It can alsobeseen that S(m;3)=(3 m 1
2 m
+1)=2. Formore onStirlingnumbers any
basic text oncombinatorics can be consulted, e.g. [1].
4.1. Theorem. f : N ! N is a cardinal function if and only if it can be written as a
Stirlingseries
f(m)= 1
X
n=0 a
n
S(m;n)
where the a
n
are natural numberswith the properties
(1) a
0
=0)a
n
=0 for all n
(2) a
1
=0)a
n
=0 for all n1.
Proof. Assume that f is a cardinal function corresponding to the functor F and let A
n
be as inCorollary3.2. As S
n
acts freely onEpi(X;[n]),
A
n
S
n
Epi(X;[n])
=A
n
Orbits(Epi(X;[n]));
but an orbit isprecisely a quotient of X with n elements. Thus
#(A
n
S
n
Epi(X;[n]))=#A
n
S(#X;n):
It then follows by Corollary 3.2that any cardinalfunction f can bewritten asa Stirling
series with a
n
=#A
n .
If a
0
=0, then F(;),which is A
0
, is empty. Since there is always a function ;! X,
we haveF(X)!F(;)=; sothat F(X)=;. Thusall a =0.
Similarly, if a
1
=0 then F(1)= A
1
= ;, and as there is a function 1! X for every
non-empty X, we have F(X) ! F(1) = ; which implies F(X) = ;. We conclude that
a
n
=0 for alln 1.
Conversely, given any Stirlingseries P
1
n=0 a
n
S(m;n) with a
0 and a
1
non-zero, we can
chooseS
n
-setsA
n
withcardinalitiesa
n
(saywithtrivialaction). ThenProposition3.3will
givea functor Gwith the rightcardinality,thus P
1
n=0 a
n
S(m;n) is acardinal function.
The functorF =A
0
[ ;;]takesthe value A
0
at;and ;elsewhere, whichcovers the
case where a
0 or a
1 are 0.
4.2. Example. The hom functor [ ;[k]] : Set op
0
! Set
0
gives rise to the exponential
function f(m) = k m
so we should be able to write k m
as a Stirling series. As k m
is the
cardinality of the set of functions : [m] ! [k] and each such factors uniquely as a
quotient followed by aone-to-one map, we get
k m
=S(m;0)+kS(m;1)+k(k 1)S(m;2)+k(k 1)(k 2)S(m;3)+
This isbecausethe numberof one-to-onemaps froma set with n elements toone with k
is given by the falling power
k
#n
=k(k 1)(k 2)(k n+1)= k!
n!
:
Thusk m
= P
1
n=0 k
#n
S(m;n).
The additive Abelian group Z[x] is free with basis h1;x;x 2
;x 3
;:::i. But as x
#n
is a
monic polynomial of degree n, h1;x;x
#2
;x
#3
;:::i alsoforms a basis. The above equation
shows that the changeof bases matrix, changing fromthe rst to the second, isgiven by
the Stirling numbers of the second kind [S(m;n)]. Its inverse, which changes from the
second to the rst basis, denes the Stirling numbers of the rst kind [s(n;m)]. Thus
x
#n
= P
m
s(n;m)x m
. In particular we have
X
m
s(n;m)S(m;k)= (*)
(
1 if n=k
0 otherwise.
Of course, allthis is well-known (see [1]).
Let E : N N
! N
N
be the shift operator, (Ef)(n) = f(n+1), and I the identity
operator. Theseareusedinthecalculusofnitedierences, wherethedierenceoperator
=E I is the main topic of study.
With these preliminaries we can now prove the following theorem giving the Stirling
coeÆcients of a cardinalfunction.
4.3. Theorem. Let f(m)= P
1
n=0 a
n
S(m;n). Then a
n
=E
#n
f(0).
Proof. E
#n
=E(E I)(E 2I)(E (n 1)I)= P
m
s(n;m)E m
so
E
#n
f(0) = P
m
s(n;m)E m
f(0)
= P
m
s(n;m)f(m)
= P
m
s(n;m) P
k a
k
S(m;k)
= P
k P
m
s(n;m)S(m;k)a
k
= a (by (*)).
4.4. Corollary. f : N! N is a cardinal function if and only if one of the following
holds:
(a) f(n)=0 for all n 1
(b) E
#n
f(0) 0 for all n and f(0);f(1)6=0.
5. Examples
5.1. Example. m 2
Asymptotically,S(m;n) n
m
n!
(for xed n), i.e.
lim
m!1
n!S(m;n)
n m
=1:
Intuitively, if m n, a random function : [m] ! [n] is almost certainly onto, so the
number of quotients willbeapproximatelythe number of functions[m]![n] divided by
the numberofpermutationson[n]. The readerwhoisnot convincedbythis probabilistic
argument can consult[1] p.140 #10 where some hints are given.
Thus a non-constant polynomial never denes a cardinal function, for if it is non-
constant some a
n
6=0 (n>1)and the function n
m
n!
grows faster than any polynomial.
5.2. Example.
2m
m
Note that
E
#(n+1)
f(m)=(E nI)E
#n
f(m)=E
#n
f(m+1) nE
#n
f(m)
sowecancalculatethevalues E
#n
f(m)recursively. Wearrangethevalues inatablewith
the values of f(m) in the rst row, with each new entry being calculated using the two
values above it in the previous row, like for nite dierences. Thus for f(m) =
2m
m
we
get:
n m
@
@
0 1 2 3 4 5 6
0 1 2 6 20 70 252 924
1 2 6 20 70 252 924
2 4 14 50 182 672
3 6 22 82 308
4 4 16 62
5 0 -2
6 -2
As E
#6
f(0) = 2, we see that f(m) =
2m
m
is not a cardinal function. We might have
believed that it wasone, as 2 m
2m
4 m
.
5.3. Example. 23 2
Thisisacardinalfunctionasiseasilyseenbyconstructingatableasabove. Thevalues
ofE
#n
f(0)turnout tobe1;4;10;12;0;0;0;:::. But itiseasytoconstruct afunctorwith
cardinality 23 m
2 m
. Let:[2]![3]be the inclusion. Then the pushout P in
[ ;[2]] [ ;[3]]
[ ;[3]]
P -
-
? ?
has the right cardinality.
5.4. Example.
(2m)!
2 m
5.5. Lemma. For n m natural numbers we have
E
#n
f(m n)=f(m) n 1
X
i=0 iE
#i
f(m 1 i):
Proof.
E
#n
f(m n) =(E (n 1)I)E
#(n 1)
f(m n)
=E
#(n 1)
f(m n+1) (n 1)E
#(n 1)
f(m n)
=E
#(n 2)
f(m n+2) (n 2)E
#(n 2)
f(m n+1) (n 1)E
#(n 1)
f(m n)
.
.
.
=E
#0
f(m) 0E
#0
f(m 1) 1E
#1
f(m 2) (n 1)E
#(n 1)
f(m n)
=f(m) (1E
#1
f(m 2)+2E
#2
f(m 3)++(n 1)E
#(n 1)
f(m n)):
5.6. Proposition. Suppose thatneitherf(0) norf(1)is0andforevery m,f(m+1)
m(m+1)
2
f(m), thenf is a cardinal function.
Proof. Weshallprovebyinductiononmthat0E
#n
f(m n)f(m)forall0nm.
For m=0, we have only n=0 and the statement isobvious.
Assume the statement holds for m. Then
E
#n
f(m+1 n) = f(m+1) P
n 1
i=0 iE
#i
f(m i)
f(m+1) P
n 1
i=0
if(m) (by induction hypothesis)
= f(m+1)
(n 1)n
2
f(m)
f(m+1)
m(m+1)
2
f(m) (as nm+1)
0:
It is alsoclear that, asE
#n
f(m+1 n)=f(m+1) (non-negativeterms),
E
#n
f(m+1 n)f(m+1):
This provesthe inductive step.
#n
Consider f(m)= (2m)!
2 m
. Its values are naturalnumbers and
f(m+1)=
(2m+2)!
2 m+1
=
(2m+2)(2m+1)
2
f(m)
so by our propositionit is acardinal function.
The smallest function satisfyingthe conditions of Proposition 5.6is
f(m)=(m 1)!m!=2 m 1
m1,
with f(0)=1. This is acardinal function.
As cardinal functions are closed under products, f(m) = (2m)! = 2 m
(2m)!
2 m
is also
cardinal.
Consider f(m)=m 2m
.
f(m+1) = (m+1) (2m+2)
= (m+1) 2
(m+1) 2m
> (m+1) 2
m 2m
>
(m+1)m
2
f(m):
Som 2m
is acardinal function.
We don'tknowof any naturally arising functor with these cardinalities.
5.7. Example. m!, m m
The
m(m+1)
2
in Proposition 5.6 is the best we can do with that kind of condition, as
the followingshows.
5.8. Proposition. Let : N ! N be a function with the property that any f for
which f(m+1) (m)f(m) for all m and f(0);f(1)6= 0, is a cardinal function. Then
(m)
m(m+1)
2 .
Proof. Fornatural numbers, pand q,dene a function f by
f(m)= (
0 if m<p
q Q
m 1
k=p
(k) if mp
with the convention that an empty product is 1 (so that f(p) = q). Then f(m+1) =
(m)f(m)if m6=p 1and f(m+1)=q0=(m)f(m) ifm =p 1. Thus f satises
our conditions, except for f(0);f(1)6=0.
Let g(m) = Q
m 1
k=0
((k)+1). Then g(m+1) = ((m)+1)g(m) (m)g(m) and
g(0);g(1)6=0. So g satises all the conditions. It follows that f +g does too, so it is a
cardinalfunction, by hypothesis, and by Corollary 4.4, E
#n
(f +g)(0)0.
Now,asf(m)=0forallm<p,allthedierencesE
#n
f(m n)=0for0n m<p.
Then E
#n
f(p n) = f(p) P
n 1
i=0 iE
#i
f(p 1 i) = f(p) = q for all 0 n p.
Consequently,
E
#(p+1)
f(0) = f(p+1) P
p
i=1 iE
#i
f(p i)
= q(p) P
p
i=1 iq
= q((p)
p(p+1)
2 ):
Now, E
#(p+1)
(f+g)(0) =E
#(p+1)
f(0)+E
#(p+1)
g(0)=q((p)
p(p+1)
2
)+E
#(p+1)
g(0)which
is 0for allq. Thus (p)
p(p+1)
0.
The function f(m) = m! clearly doesn't satisfy the conditions of Proposition 5.6.
Some hand calculations suggest that it might be cardinal, but using Maple, we see that
E
#12
f(0)= 519;312, soit isn't.
On the other hand, again using Maple, we see that for f(m) = m m
and f(m) =
2 m
m!;E
#n
f(0) >0 for alln 100, which strongly suggests that they are cardinal func-
tions.
References
[1] Comtet,Louis,AnalyseCombinatoire,TomesI&II,PressesUniversitairesdeFrance,Paris,1970.
[2] Dougherty,Randall,\FunctorsontheCategoryofFiniteSets,"TAMS,vol.330,Number2,April
1992.
[3] Joyal,Andre,\Unetheoriecombinatoiredesseriesformelles,"Advancesin Math.,Vol.42,No.1,
1980,pp.1-82.
[4] Pare,Robert,\AbsolutenessPropertiesinCategoryTheory,"Ph.D.Thesis,McGill,1969.
[5] Pare,Robert,\OnAbsoluteColimits,"J. Alg. 19(1971),pp.80-95.
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