Directory UMM :Journals:Journal_of_mathematics:TAC:
Theory and Appli
ations of Categories,
Vol. 6, No. 5, 1999, pp. 65{76.
CONTRAVARIANT FUNCTORS ON FINITE SETS
AND STIRLING NUMBERS
For Jim Lambek
ROBERT PARE
ABSTRACT. We
hara
terize the numeri
al fun
tions whi
h arise as the
ardinalities
of
ontravariant fun
tors on nite sets, as those whi
h have a series expansion in terms
of Stirling fun
tions. We give a pro
edure for
al
ulating the
oeÆ
ients in su
h series
and a
on
rete test for determining whether a fun
tion is of this type. A number of
examples are
onsidered.
1. Introdu
tion
Let Set be the
ategory of nite sets and F : Setop ! Set a fun
tor. Su
h a fun
tor
indu
es a fun
tion 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
ardinality if and only if they are isomorphi
, f
ould be de ned by the equation f (#X ) =
#F (X ) for all nite sets X . The question we
onsider is whi
h fun
tions f arise in this
way. As the natural numbers are the
ardinalities of nite sets, and as fun
tors are more
stru
tured than arbitrary fun
tions, one might expe
t to get a ni
e
lass of numeri
al
fun
tions this way. Let us
all them
ardinal fun
tions.
For example, the fun
tion f (n) = 3n is a
ardinal fun
tion as it is the
ardinality
of
n
n
the representable fun
tor Set ( ; 3). But what about the fun
tions n ; 2 3 2 , nn , n!,
n
n , and so on? We shall determine a
riterion whi
h will help us de
ide these questions.
We shall see that we are led to
ertain
ombinatorial fun
tions, and we
an hope for
some appli
ations in that dire
tion. We present none here, but see [3℄ for appli
ations of
ategory theory to
ombinatori
s.
The results below were presented at the AMS meeting in Montreal in September 1997.
Shortly after, Andreas Blass pointed out to me the paper [2℄ by Dougherty in whi
h
similar results are obtained. Our Theorems 4.1 and 4.3 are very similar to his Proposition
2.14 and Theorem 1.3. The proofs are not very di erent but ours have a more
ategori
al
avor. Some lemmas on absolute
olimits are of independent interest. The numeri
al
examples in our paper are new and not without interest.
0
0
2
0
(2
0
)!
2
Resear
h supported by a grant from NSERC
Re
eived by the editors 1999 Mar
h 8 and, in revised form, 1999 September 13.
Published on 1999 November 30.
1991 Mathemati
s Subje
t Classi
ation: 18A22, 05A10.
Key words and phrases: Fun
tor,
ardinality, Stirling numbers.
Robert Pare 1999. Permission to
opy for private use granted.
65
2
Theory and Appli
ations of Categories, Vol. 6, No. 5
66
2. Absolute
olimits revisited
Absolute
olimits were introdu
ed in my thesis [4℄ thirty years ago. When I told Jim
Lambek, who was then my Ph.D. supervisor, about my
hara
terization of
oequalizers
whi
h are preserved by all fun
tors, and how they
ame up in Be
k's tripleability theorem
(as it was then
alled), he said \Good! Write it up. You
an
all them
." I did.
A while later, he asked \Is there a smaller
lass of fun
tors whi
h would be suÆ
ient to
test for absoluteness?" There was, namely the representables, and so my thesis began.
In this se
tion we obtain some new results on absolute
olimits in Set0 .
absolute
2.1. Lemma. Let
f1 -
A0
f2
be pushouts in
Set0
g2
? - ?
2
A
i i
A0
f2 g2
A2
B1
- ?
B
are epimorphisms. Then
f1 g1
B0
?
?
B2
A
in whi
h the f ; g
g1 -
B0
A1
A1
B1
- ?
B2
A
B
is also a pushout.
P
roof. First, we
onsider the spe
ial
ase where our pushouts are of the form
1A
g- 0
0
0
0
A
f
B
A
? - ?f
1A
A
1B0
g
?
B0
A
Let be a splitting for , = 1B . Then in
A s
0
0
s
B
1B
?
gB :
gs
A
B
f B
A
?
B
0
A
?f B
-
As
B0
A
B0
A0 g
0
A0
?f B
-
Ag
A
B
B;
the outside re
tangle is a pushout, one in whi
h the two horizontal maps are identities,
and the middle verti
al arrow is an epimorphism, so the right square is a pushout as
required.
Theory and Appli
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67
Now, for the general
ase,
onsider
A0
B0
A2
B0
A2
B2
?
?
- A1 B0
(1)
?
- A B0
(3)
- ?
A
B2
- A 1 B1
(2)
?
- A B1
(4)
- ?
A
B
where all the arrows are the obvious Cartesian produ
ts. By
artesian
losedness, ( ) B0
and A ( ) preserve
olimits, so (1) and (4) are pushouts. (2) and (3) are diagrams of the
sort dis
ussed in the previous paragraph, so they are pushouts too. The result follows by
pasting pushouts.
The proof of the above lemma goes through, with minor modi
ations
in a monoidal
losed
ategory: if the fi and gi are regular epimorphisms, then
2.2. Remark.
A0
B0
f2
g2
A2
- A 1
B1
f1
g1
?
- A
?B
B2
is a pushout.
For the rst part of the proof, let (A0
g ) = (f
B0 ). Then and
by adjointness, to and su
h that
A0
f
?
A
-
orrespond,
[B; C ℄
?[g; C ℄
[B0 ; C ℄
ommutes. As [g; C ℄ is moni
(g is epi
) and f is a regular epi, there exists a unique
= and [g; C ℄ = . Again, by adjointness,
diagonal ll-in : A ! [B; C ℄ su
h that f
this
orresponds to a unique : A
B ! C su
h that (f
B ) = and (A
g ) = .
So the required square is a pushout.
The se
ond part of the proof uses only that ( )
B0 and A
( ) preserve pushouts.
Re
all from [5℄ that a
olimit is
alled absolute if it is preserved by all fun
tors.
2.3. Proposition. Pushouts of epimorphisms in
Set0
are absolute.
roof. 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 Set0 , the representables [A; ℄ are nite
powers ( )#A, and by Lemma 1, a nite power of a pushout of epimorphisms is again a
pushout. The result follows.
P
Theory and Appli
ations of Categories, Vol. 6, No. 5
68
!!
Ri
hard Wood points out that the same proof shows that re
exive
oequalizers in Set0 are absolute. In fa
t, if f1 ; f2 : A0
A1 is a re
exive pair, then the
oequalizer of f1 and f2 is the same as their pushout, so this is a spe
ial
ase of Proposition
2.3.
2.4. Remark.
A similar result, whi
h we shall not need in the sequel, is the following:
non-empty interse
tions are absolute in Set0 . Indeed, suppose that A and B are subsets
of C and that A B = . Choose
0 A B and de ne fun
tions f : C A by
2.5. Remark.
\ 6 ;
2 \
f (
) =
and g : B
! A \ B by
g (b) =
(
(
!
2
if
A
0 otherwise
2 \
b if b A B
0 otherwise.
Then it is easily seen that
\
-
B
A
-
C
A B
?
-
A B
f
A
g
?
\
?
ommutes. Furthermore, the outside re
tangle is an absolute pullba
k (as the two horizontal arrows are identities) and the middle verti
al arrow is an absolute monomorphism
(as it is split). Thus the left square is an absolute pullba
k.
We see that non-empty pullba
ks of monomorphisms in Set0 are absolute for a relatively simple reason. One might say that the pullba
k square itself is split. This is not
the
ase for pushouts of epimorphisms in Set0 where an unbounded number (depending
on the size of the sets involved) of fun
tions may be required to express absoluteness
equationally. Pushouts of epimorphisms between in nite sets need not be absolute either.
3. The stru
ture of
ontravariant fun
tors
!
Let F : Setop
Set0 be any fun
tor. Say that (n; a) is minimal for
0
not equal to any F ( )(b) for : [n℄ [m℄, b F [m℄ with m < n.
2
!
2
F
if a
2 F [n℄ and is
x F X be any element of F . Then:
(1) There is (n; a) minimal for F and f : X
[n℄ epi
, su
h that F (f )(a) = x.
(2) If (m; b) and g : X
[m℄ also have the same properties, then m = n and there exists
Sn su
h that g = f and F ( )(b) = a.
3.1. Proposition.
2
Let
!
!
Theory and Appli
ations of Categories, Vol. 6, No. 5
69
n; f; a), n 2 N, f : X
Proof. (1) Of all the triples (
hoose one with minimal
n.
! [n℄, a 2 F [n℄ with F (f )(a) = x,
(There is at least one su
h triple, for if we let
n
be the
X , then there will be an isomorphism f : X ! [n℄ and we
an take
a = F (f 1 )(x).) Then (n; a) is minimal for F , be
ause 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 epi
it would fa
tor as g
where g : X ! [m℄ and : [m℄ ! [n℄ with m < n. Then F (g )(F ( )(a)) = F ( g )(a) =
F (f )(a) = x and again, n would not be minimal for x.
(2) Let (m; b) be minimal for F and g : X ! [m℄ epi
su
h that F (g )(b) = x. Take
ardinality of
the pushout
- [n℄
X
f
g
?
- [p℄
[
?
m℄
whi
h is absolute by Proposition 2.3. Thus
F [p℄
?
F [m℄
- F [n℄
- ?
FX
F (f )(a) = x = F (g )(b), there exists
2 F [p℄ with F ( )(
) = a and
F ( )(
) = b. As (n; a) is minimal for F , p
annot be less than n, so is an isomorphism.
Similarly, is an isomorphism. It follows that n = p = m and if = 1 2 Sn , then
f = g and F ( )(b) = F ( )F ( ) 1(b) = F ( )(
) = a.
Let An be the set of minimal elements in F [n℄, i.e.
is a pullba
k. As
An = fa 2 F [n℄j(n; a) is minimal for F g:
Then the symmetri
group Sn a
ts on the right on An by
(
Also,
a; ) 7! F ( )(a):
Sn a
ts on the left on Epi(X; [n℄), the set of epimorphisms X ! [n℄, by
; f ) 7! f:
(
Now the above proposition
an be restated as follows.
3.2. Corollary. For ea
h
X , the fun
tion
X1 A
n=0
is a bije
tion.
n
Sn
Epi(X; [n℄) ! F X
a
f
7! F (f )(a)
Theory and Appli
ations of Categories, Vol. 6, No. 5
3.3. Proposition.
A1 6= ;, then
70
Given nite Sn -sets, An , for n = 0; 1; 2; : : :, su
h that A0 =
6
; and
X
An
Sn Epi(X; [n℄)
n=0
an be made into a fun
tor G : Set0 op ! Set0 .
G(X ) =
1
roof. If n > #X , then Epi(X; [n℄) = ; so that for any xed X the in nite
oprodu
t is
essentially nite and G(X ) is a nite set. Note that S0 and S1 are both the trivial group,
so that A0 and A1 are just sets. Pi
k a0 2 A0 and a1 2 A1 . Let Y : Y ! [1℄ denote the
unique fun
tion into the terminal obje
t. It is epi if Y 6= ;.
Let g : Y ! X and (n; a
f ) 2 G(X ). De ne
P
8
>< (n; a
(fg))
G(g )(n; a
f ) = > (1; a
Y )
: (0; a
1)
1
0
if f g is epi
if f g is not epi and Y 6= ;
if f g is not epi but Y = ;:
First, G(g ) is well-de ned. Indeed, if a
f = a
f , then there is 2 Sn su
h that
f = f and a = a. Then f g is epi if and only if f g is epi and in that
ase a
(fg ) =
(a )
(f g ) = a
(f g ) = a
f g .
It is
lear from the de nition that G(1X ) = 1G(X ) . Now let h : Z ! Y . If fgh is epi,
then so is f g and
0
0
0
0
0
0
0
0
0
G(h)G(g )(n; a
f ) = G(h)(n; a
(fg )) = (n; a
(fgh)) = G(gh)(n; a
f ):
If fg is epi but f gh is not and Z =
6 ;, then
G(h)G(g )(n; a
f ) = G(h)(n; a
(fg )) = (1; a1
Z ) = G(gh)(n; a
f ):
If fg is not epi, then f gh is not either. If Z =
6 ;, then
G(h)G(g )(n; a
f ) = G(h)(1; a1
Y ) = (1; a1
Y h) = (1; a1
Z ) = G(gh)(n; a
f ):
Finally, if Z = ;, then
G(h)G(g )(n; a
f ) = (0; a0
1) = G(gh)(n; a
f ):
The above
onstru
tion is not
anoni
al so we
an hardly expe
t the
bije
tion of the previous
orollary to be natural, although we do have naturality if we
restri
t to morphisms that are epi
.
3.4. Remark.
Theory and Appli
ations of Categories, Vol. 6, No. 5
71
4. Stirling series
The Stirling numbers of the se
ond kind are the numbers ( ) of partitions of into
pie
es. They satisfy the re
urren
e relations
(
=0
(0 ) = 10 ifotherwise
( +1 )= (
1) + ( )
A table of values for ( )
an be
onstru
ted from these relations, just like Pas
al's
triangle.
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 might guess from this table that ( 2) = 2m 1 1, 1, and this is easily seen.
It
an also be seen that ( 3) = (3m 1 2m + 1) 2. For more on Stirling numbers any
basi
text on
ombinatori
s
an be
onsulted, e.g. [1℄.
: N ! N is a
ardinal fun
tion if and only if it
an be written as a
Stirling series
1
X
( )= n ( )
S m; n
m
n
S
S m
n
;n
;n
:
S m; n
nS m; n :
S m; n
n
m
S m;
m
S m;
=
4.1. Theorem. f
f m
where the
an
n=0
a S m; n
are natural numbers with the properties
= 0 ) n = 0 for all
(2) 1 = 0 ) n = 0 for all
1.
roof. Assume that is a
ardinal fun
tion
orresponding to the fun
tor and let n
be as in Corollary 3.2. As n a
ts freely on ( [ ℄),
( ( [ ℄))
( [ ℄) = n
n
Sn
but an orbit is pre
isely a quotient of with elements. Thus
#( n
Sn ( [ ℄)) = # n (# )
It then follows by Corollary 3.2 that any
ardinal fun
tion
an be written as a Stirling
series with n = # n.
If 0 = 0, then (;), whi
h is 0, is empty. Sin
e there is always a fun
tion ; ! ,
we have ( ) ! (;) = ; so that ( ) = ;. Thus all n = 0.
(1)
a0
a
n
a
a
n
P
f
F
S
A
E pi X; n
A
X
A
A
E pi X; n
E pi X; n
Orbits E pi X; n
;
n
A
S
X; n :
f
a
a
A
F
F X
F
A
F X
X
a
Theory and Appli
ations of Categories, Vol. 6, No. 5
72
Similarly, if a1 = 0 then F (1) = A1 = ;, and as there is a fun
tion 1 ! X for every
non-empty X , we have F (X ) ! F (1) = ; whi
h implies F (X ) = ;. We
on
lude that
an = 0 for all n 1.
Conversely, given any Stirling series P1n=0 anS (m; n) with a0 and a1 non-zero, we
an
hoose Sn-sets An with
ardinalities an (say with trivial
P1 a
tion). Then Proposition 3.3 will
give a fun
tor G with the right
ardinality, thus n=0 anS (m; n) is a
ardinal fun
tion.
The fun
tor F = A0 [ ; ;℄ takes the value A0 at ; and ; elsewhere, whi
h
overs the
ase where a0 or a1 are 0.
The hom fun
tor [ ; [k℄℄ : Setop0 ! Set0 gives rise to the exponential
fun
tion f (m) = km so we should be able to write km as a Stirling series. As km is the
ardinality of the set of fun
tions : [m℄ ! [k℄ and ea
h su
h fa
tors
as a
followed by a one-to-one map, we get
km = S (m; 0) + kS (m; 1) + k(k 1)S (m; 2) + k(k 1)(k 2)S (m; 3) +
This is be
ause the number of one-to-one maps from a set with n elements to one with k
is given by the
k!
k#n = k(k 1)(k 2) (k n + 1) = :
n!
P
1
Thus km = n=0 k#n S (m; n).
The additive Abelian group Z[x℄ is free with basis h1; x; x2; x3 ; : : :i. But as x#n is a
moni
polynomial of degree n, h1; x; x#2 ; x#3 ; : : :i also forms a basis. The above equation
shows that the
hange of bases matrix,
hanging from the rst to the se
ond, is given by
the Stirling numbers of the se
ond kind [S (m; n)℄. Its inverse, whi
h
hanges from the
se
ondPto the rst basis, de nes the Stirling numbers of the rst kind [s(n; m)℄. Thus
x#n = m s(n; m)xm . In parti
ular we have
(
X
n=k
s(n; m)S (m; k) = 10 ifotherwise.
(*)
m
Of
ourse, all this is well-known (see [1℄).
Let E : NN ! NN be the shift operator, (Ef )(n) = f (n + 1), and I the identity
operator. These are used in the
al
ulus of nite di eren
es, where the di eren
e operator
= E I is the main topi
of study.
With these preliminaries we
an now prove the following theorem giving the Stirling
oeÆ
ients of a
ardinal fun
tion.
P a S (m; n)
f (m) = 1
an = E #n f (0)
n=0 n
roof. E #n = E (E I )(E 2I ) (E (n 1)I ) = Pm s(n; m)E m so
P
E #n f (0) = Pm s(n; m)E m f (0)
= Pm s(n; m)fP(m)
= PmPs(n; m) k ak S (m; k)
= k m s(n; m)S (m; k)ak
= an (by (*)).
4.2. Example.
uniquely
quotient
falling power
4.3. Theorem. Let
P
. Then
.
Theory and Appli
ations of Categories, Vol. 6, No. 5
f : N ! N is a
ardinal fun
tion if and only if one of the following
4.4. Corollary.
holds:
(a)
(b)
73
f (n) = 0 for all n 1
E #n f (0) 0 for all n and f (0); f (1) 6= 0.
5. Examples
5.1. Example.
m
2
m
Asymptoti
ally, S (m; n) nn (for xed n), i.e.
!
mlim
!1
n!S (m; n)
= 1:
nm
Intuitively, if m n, a random fun
tion : [m℄ ! [n℄ is almost
ertainly onto, so the
number of quotients will be approximately the number of fun
tions [m℄ ! [n℄ divided by
the number of permutations on [n℄. The reader who is not
onvin
ed by this probabilisti
argument
an
onsult [1℄ p. 140 #10 where some hints are given.
Thus a non-
onstant polynomial never de nes a
ardinal fun
tion, for if it is nonm
onstant some an 6= 0 (n > 1) and the fun
tion nn grows faster than any polynomial.
!
5.2. Example.
Note that
m
m
2
E # n f (m) = (E
(
+1)
nI )E #n f (m) = E #n f (m + 1) nE #n f (m)
so we
an
al
ulate the values E #n f (m) re
ursively. We arrange the values in a table with
the values of f (m) in the rst row, with ea
h new entry being
al
ulated using thetwo
values above it in the previous row, like for nite di eren
es. Thus for f (m) = mm we
get:
m
n
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
2
As E # f (0) = 2, we see that f (m) = mm is not a
ardinal fun
tion. We might have
believed that it was one, as 2m mm 4m .
2
6
2
Theory and Appli
ations of Categories, Vol. 6, No. 5
5.3. Example. 2
3
m
74
2m
This is a
ardinal fun
tion as is easily seen by
onstru
ting a table as above. The values
E # f (0) turn out to be 1; 4; 10; 12; 0; 0; 0; : : :. But it is easy to
onstru
t a fun
tor with
ardinality 2 3
2 . Let : [2℄ ! [3℄ be the in
lusion. Then the pushout P in
n
of
m
m
[
; [2℄℄
[
; [3℄℄
-
?
[
; [3℄℄
- P?
has the right
ardinality.
5.4. Example.
(2m)!
m
2
5.5. Lemma. For
n m natural numbers we have
X iE # f (m
n
E # f ( m n ) = f ( m)
1
i
n
1
i):
i=0
Proof.
E # f (m n) = (E (n 1)I )E #
f (m n)
#
#
=E
f (m n + 1) (n 1)E
f (m n)
f (m n + 2) (n 2)E #
f (m n + 1) (n 1)E #
n
(n
(n
=
E#
(n
2)
(n
1)
(n
1)
1)
2)
(n
1)
f (m n)
..
.
E # f (m) 0E # f (m 1) 1E # f (m 2) (n 1)E #
f (m n)
#
#
#
= f ( m)
(1E f (m
2) + 2E f (m
3) + + (n
1)E
f (m n)):
0
=
0
1
2
f (m), then f
(n
2
5.6. Proposition. Suppose that neither
m(m+1)
(n
1
1)
1)
f (0) nor f (1) is 0 and for every m, f (m + 1)
is a
ardinal fun
tion.
Proof. We shall prove by indu
tion on
m that 0 E # f (m n) f (m) for all 0 n m.
n
m = 0, we have only n = 0 and the statement is obvious.
Assume the statement holds for m. Then
P
E # f (m + 1 n) = f (m + 1) P iE # f (m i)
f (m + 1)
if (m) (by indu
tion hypothesis)
= f (m + 1)
f (m)
f (m) (as n m + 1)
f (m + 1)
0:
It is also
lear that, as E # f (m + 1
n) = f (m + 1) (non-negative terms),
E # f (m + 1 n) f (m + 1):
For
n 1
i=0
n 1
i=0
(n 1)n
n
i
2
m(m+1)
2
n
n
This proves the indu
tive step.
Then putting
n = m we get E # f (0) 0 for all n, so f
n
is a
ardinal fun
tion.
Theory and Appli
ations of Categories, Vol. 6, No. 5
75
Consider f (m) = mm . Its values are natural numbers and
(2m + 2)! (2m + 2)(2m + 1)
f (m + 1) =
=
f (m)
2m
2
so by our proposition it is a
ardinal fun
tion.
The smallest fun
tion satisfying the
onditions of Proposition 5.6 is
(2
)!
2
+1
f (m) = (m
1)!m!=2m
1
m 1,
with f (0) = 1. This is a
ardinal fun
tion.
As
ardinal fun
tions are
losed under produ
ts, f (m) = (2m)! = 2m mm is also
ardinal.
Consider f (m) = m m .
(2
)!
2
2
f (m + 1) = (m + 1) m
> (m + 1) m
(2
+2)
2
2m
= (m + 1) (m + 1)
> m m f (m):
2
(
2m
+1)
2
So m m is a
ardinal fun
tion.
We don't know of any naturally arising fun
tor with these
ardinalities.
m
5.7. Example. m!, m
m m
in Proposition 5.6 is the best we
an do with that kind of
ondition, as
The
the following shows.
5.8. Proposition. Let : N ! N be a fun
tion with the property that any f for
whi
h f (m + 1) (m)f (m) for all m and f (0); f (1) 6= 0, is a
ardinal fun
tion. Then
(m) m m .
Proof. For natural numbers, p and q , de ne a fun
tion f by
2
(
+1)
2
(
+1)
2
f (m) =
(
0Q
if m < p
q m
(
k
)
if m p
k p
1
=
with the
onvention that an empty produ
t is 1 (so that f (p) = q ). Then f (m + 1) =
(m)f (m) if m 6= p 1 and f (m + 1) = q 0 = (m)f (m) if m = p 1. Thus f satis es
our
onditions, ex
ept
for f (0); f (1) 6= 0.
Q
m
Let g (m) = k ( (k) + 1). Then g (m + 1) = ( (m) + 1)g (m) (m)g (m) and
g (0); g (1) 6= 0. So g satis es all the
onditions. It follows that f + g does too, so it is a
ardinal fun
tion, by hypothesis, and by Corollary 4.4, E #n (f + g )(0) 0.
Now, as f (m) = 0 for all m 0 for all n 100, whi
h strongly suggests that they are
ardinal fun
tions.
Referen
es
[1℄ Comtet, Louis, Analyse
Combinatoire
, Tomes I & II, Presses Universitaires de Fran
e, Paris, 1970.
[2℄ Dougherty, Randall, \Fun
tors on the Category of Finite Sets,"
1992.
[3℄ Joyal, Andre, \Une theorie
ombinatoire des series formelles,"
1980, pp. 1-82.
TAMS
, vol. 330, Number 2, April
Advan
es in Math.
, Vol. 42, No. 1,
[4℄ Pare, Robert, \Absoluteness Properties in Category Theory," Ph.D. Thesis, M
Gill, 1969.
[5℄ Pare, Robert, \On Absolute Colimits,"
J. Alg.
19(1971), pp. 80-95.
Department of Mathemati
s and Statisti
s
Dalhousie University
Halifax, Nova S
otia
Canada B3H 3J5
Email:
paremathstat.dal.
a
http://www.ta
.mta.
a/ta
/ or by anonyftp://ftp.ta
.mta.
a/pub/ta
/html/volumes/6/n5/n5.fdvi,psg
This arti
le may be a
essed via WWW at
mous ftp at
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a
ations of Categories,
Vol. 6, No. 5, 1999, pp. 65{76.
CONTRAVARIANT FUNCTORS ON FINITE SETS
AND STIRLING NUMBERS
For Jim Lambek
ROBERT PARE
ABSTRACT. We
hara
terize the numeri
al fun
tions whi
h arise as the
ardinalities
of
ontravariant fun
tors on nite sets, as those whi
h have a series expansion in terms
of Stirling fun
tions. We give a pro
edure for
al
ulating the
oeÆ
ients in su
h series
and a
on
rete test for determining whether a fun
tion is of this type. A number of
examples are
onsidered.
1. Introdu
tion
Let Set be the
ategory of nite sets and F : Setop ! Set a fun
tor. Su
h a fun
tor
indu
es a fun
tion 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
ardinality if and only if they are isomorphi
, f
ould be de ned by the equation f (#X ) =
#F (X ) for all nite sets X . The question we
onsider is whi
h fun
tions f arise in this
way. As the natural numbers are the
ardinalities of nite sets, and as fun
tors are more
stru
tured than arbitrary fun
tions, one might expe
t to get a ni
e
lass of numeri
al
fun
tions this way. Let us
all them
ardinal fun
tions.
For example, the fun
tion f (n) = 3n is a
ardinal fun
tion as it is the
ardinality
of
n
n
the representable fun
tor Set ( ; 3). But what about the fun
tions n ; 2 3 2 , nn , n!,
n
n , and so on? We shall determine a
riterion whi
h will help us de
ide these questions.
We shall see that we are led to
ertain
ombinatorial fun
tions, and we
an hope for
some appli
ations in that dire
tion. We present none here, but see [3℄ for appli
ations of
ategory theory to
ombinatori
s.
The results below were presented at the AMS meeting in Montreal in September 1997.
Shortly after, Andreas Blass pointed out to me the paper [2℄ by Dougherty in whi
h
similar results are obtained. Our Theorems 4.1 and 4.3 are very similar to his Proposition
2.14 and Theorem 1.3. The proofs are not very di erent but ours have a more
ategori
al
avor. Some lemmas on absolute
olimits are of independent interest. The numeri
al
examples in our paper are new and not without interest.
0
0
2
0
(2
0
)!
2
Resear
h supported by a grant from NSERC
Re
eived by the editors 1999 Mar
h 8 and, in revised form, 1999 September 13.
Published on 1999 November 30.
1991 Mathemati
s Subje
t Classi
ation: 18A22, 05A10.
Key words and phrases: Fun
tor,
ardinality, Stirling numbers.
Robert Pare 1999. Permission to
opy for private use granted.
65
2
Theory and Appli
ations of Categories, Vol. 6, No. 5
66
2. Absolute
olimits revisited
Absolute
olimits were introdu
ed in my thesis [4℄ thirty years ago. When I told Jim
Lambek, who was then my Ph.D. supervisor, about my
hara
terization of
oequalizers
whi
h are preserved by all fun
tors, and how they
ame up in Be
k's tripleability theorem
(as it was then
alled), he said \Good! Write it up. You
an
all them
." I did.
A while later, he asked \Is there a smaller
lass of fun
tors whi
h would be suÆ
ient to
test for absoluteness?" There was, namely the representables, and so my thesis began.
In this se
tion we obtain some new results on absolute
olimits in Set0 .
absolute
2.1. Lemma. Let
f1 -
A0
f2
be pushouts in
Set0
g2
? - ?
2
A
i i
A0
f2 g2
A2
B1
- ?
B
are epimorphisms. Then
f1 g1
B0
?
?
B2
A
in whi
h the f ; g
g1 -
B0
A1
A1
B1
- ?
B2
A
B
is also a pushout.
P
roof. First, we
onsider the spe
ial
ase where our pushouts are of the form
1A
g- 0
0
0
0
A
f
B
A
? - ?f
1A
A
1B0
g
?
B0
A
Let be a splitting for , = 1B . Then in
A s
0
0
s
B
1B
?
gB :
gs
A
B
f B
A
?
B
0
A
?f B
-
As
B0
A
B0
A0 g
0
A0
?f B
-
Ag
A
B
B;
the outside re
tangle is a pushout, one in whi
h the two horizontal maps are identities,
and the middle verti
al arrow is an epimorphism, so the right square is a pushout as
required.
Theory and Appli
ations of Categories, Vol. 6, No. 5
67
Now, for the general
ase,
onsider
A0
B0
A2
B0
A2
B2
?
?
- A1 B0
(1)
?
- A B0
(3)
- ?
A
B2
- A 1 B1
(2)
?
- A B1
(4)
- ?
A
B
where all the arrows are the obvious Cartesian produ
ts. By
artesian
losedness, ( ) B0
and A ( ) preserve
olimits, so (1) and (4) are pushouts. (2) and (3) are diagrams of the
sort dis
ussed in the previous paragraph, so they are pushouts too. The result follows by
pasting pushouts.
The proof of the above lemma goes through, with minor modi
ations
in a monoidal
losed
ategory: if the fi and gi are regular epimorphisms, then
2.2. Remark.
A0
B0
f2
g2
A2
- A 1
B1
f1
g1
?
- A
?B
B2
is a pushout.
For the rst part of the proof, let (A0
g ) = (f
B0 ). Then and
by adjointness, to and su
h that
A0
f
?
A
-
orrespond,
[B; C ℄
?[g; C ℄
[B0 ; C ℄
ommutes. As [g; C ℄ is moni
(g is epi
) and f is a regular epi, there exists a unique
= and [g; C ℄ = . Again, by adjointness,
diagonal ll-in : A ! [B; C ℄ su
h that f
this
orresponds to a unique : A
B ! C su
h that (f
B ) = and (A
g ) = .
So the required square is a pushout.
The se
ond part of the proof uses only that ( )
B0 and A
( ) preserve pushouts.
Re
all from [5℄ that a
olimit is
alled absolute if it is preserved by all fun
tors.
2.3. Proposition. Pushouts of epimorphisms in
Set0
are absolute.
roof. 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 Set0 , the representables [A; ℄ are nite
powers ( )#A, and by Lemma 1, a nite power of a pushout of epimorphisms is again a
pushout. The result follows.
P
Theory and Appli
ations of Categories, Vol. 6, No. 5
68
!!
Ri
hard Wood points out that the same proof shows that re
exive
oequalizers in Set0 are absolute. In fa
t, if f1 ; f2 : A0
A1 is a re
exive pair, then the
oequalizer of f1 and f2 is the same as their pushout, so this is a spe
ial
ase of Proposition
2.3.
2.4. Remark.
A similar result, whi
h we shall not need in the sequel, is the following:
non-empty interse
tions are absolute in Set0 . Indeed, suppose that A and B are subsets
of C and that A B = . Choose
0 A B and de ne fun
tions f : C A by
2.5. Remark.
\ 6 ;
2 \
f (
) =
and g : B
! A \ B by
g (b) =
(
(
!
2
if
A
0 otherwise
2 \
b if b A B
0 otherwise.
Then it is easily seen that
\
-
B
A
-
C
A B
?
-
A B
f
A
g
?
\
?
ommutes. Furthermore, the outside re
tangle is an absolute pullba
k (as the two horizontal arrows are identities) and the middle verti
al arrow is an absolute monomorphism
(as it is split). Thus the left square is an absolute pullba
k.
We see that non-empty pullba
ks of monomorphisms in Set0 are absolute for a relatively simple reason. One might say that the pullba
k square itself is split. This is not
the
ase for pushouts of epimorphisms in Set0 where an unbounded number (depending
on the size of the sets involved) of fun
tions may be required to express absoluteness
equationally. Pushouts of epimorphisms between in nite sets need not be absolute either.
3. The stru
ture of
ontravariant fun
tors
!
Let F : Setop
Set0 be any fun
tor. Say that (n; a) is minimal for
0
not equal to any F ( )(b) for : [n℄ [m℄, b F [m℄ with m < n.
2
!
2
F
if a
2 F [n℄ and is
x F X be any element of F . Then:
(1) There is (n; a) minimal for F and f : X
[n℄ epi
, su
h that F (f )(a) = x.
(2) If (m; b) and g : X
[m℄ also have the same properties, then m = n and there exists
Sn su
h that g = f and F ( )(b) = a.
3.1. Proposition.
2
Let
!
!
Theory and Appli
ations of Categories, Vol. 6, No. 5
69
n; f; a), n 2 N, f : X
Proof. (1) Of all the triples (
hoose one with minimal
n.
! [n℄, a 2 F [n℄ with F (f )(a) = x,
(There is at least one su
h triple, for if we let
n
be the
X , then there will be an isomorphism f : X ! [n℄ and we
an take
a = F (f 1 )(x).) Then (n; a) is minimal for F , be
ause 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 epi
it would fa
tor as g
where g : X ! [m℄ and : [m℄ ! [n℄ with m < n. Then F (g )(F ( )(a)) = F ( g )(a) =
F (f )(a) = x and again, n would not be minimal for x.
(2) Let (m; b) be minimal for F and g : X ! [m℄ epi
su
h that F (g )(b) = x. Take
ardinality of
the pushout
- [n℄
X
f
g
?
- [p℄
[
?
m℄
whi
h is absolute by Proposition 2.3. Thus
F [p℄
?
F [m℄
- F [n℄
- ?
FX
F (f )(a) = x = F (g )(b), there exists
2 F [p℄ with F ( )(
) = a and
F ( )(
) = b. As (n; a) is minimal for F , p
annot be less than n, so is an isomorphism.
Similarly, is an isomorphism. It follows that n = p = m and if = 1 2 Sn , then
f = g and F ( )(b) = F ( )F ( ) 1(b) = F ( )(
) = a.
Let An be the set of minimal elements in F [n℄, i.e.
is a pullba
k. As
An = fa 2 F [n℄j(n; a) is minimal for F g:
Then the symmetri
group Sn a
ts on the right on An by
(
Also,
a; ) 7! F ( )(a):
Sn a
ts on the left on Epi(X; [n℄), the set of epimorphisms X ! [n℄, by
; f ) 7! f:
(
Now the above proposition
an be restated as follows.
3.2. Corollary. For ea
h
X , the fun
tion
X1 A
n=0
is a bije
tion.
n
Sn
Epi(X; [n℄) ! F X
a
f
7! F (f )(a)
Theory and Appli
ations of Categories, Vol. 6, No. 5
3.3. Proposition.
A1 6= ;, then
70
Given nite Sn -sets, An , for n = 0; 1; 2; : : :, su
h that A0 =
6
; and
X
An
Sn Epi(X; [n℄)
n=0
an be made into a fun
tor G : Set0 op ! Set0 .
G(X ) =
1
roof. If n > #X , then Epi(X; [n℄) = ; so that for any xed X the in nite
oprodu
t is
essentially nite and G(X ) is a nite set. Note that S0 and S1 are both the trivial group,
so that A0 and A1 are just sets. Pi
k a0 2 A0 and a1 2 A1 . Let Y : Y ! [1℄ denote the
unique fun
tion into the terminal obje
t. It is epi if Y 6= ;.
Let g : Y ! X and (n; a
f ) 2 G(X ). De ne
P
8
>< (n; a
(fg))
G(g )(n; a
f ) = > (1; a
Y )
: (0; a
1)
1
0
if f g is epi
if f g is not epi and Y 6= ;
if f g is not epi but Y = ;:
First, G(g ) is well-de ned. Indeed, if a
f = a
f , then there is 2 Sn su
h that
f = f and a = a. Then f g is epi if and only if f g is epi and in that
ase a
(fg ) =
(a )
(f g ) = a
(f g ) = a
f g .
It is
lear from the de nition that G(1X ) = 1G(X ) . Now let h : Z ! Y . If fgh is epi,
then so is f g and
0
0
0
0
0
0
0
0
0
G(h)G(g )(n; a
f ) = G(h)(n; a
(fg )) = (n; a
(fgh)) = G(gh)(n; a
f ):
If fg is epi but f gh is not and Z =
6 ;, then
G(h)G(g )(n; a
f ) = G(h)(n; a
(fg )) = (1; a1
Z ) = G(gh)(n; a
f ):
If fg is not epi, then f gh is not either. If Z =
6 ;, then
G(h)G(g )(n; a
f ) = G(h)(1; a1
Y ) = (1; a1
Y h) = (1; a1
Z ) = G(gh)(n; a
f ):
Finally, if Z = ;, then
G(h)G(g )(n; a
f ) = (0; a0
1) = G(gh)(n; a
f ):
The above
onstru
tion is not
anoni
al so we
an hardly expe
t the
bije
tion of the previous
orollary to be natural, although we do have naturality if we
restri
t to morphisms that are epi
.
3.4. Remark.
Theory and Appli
ations of Categories, Vol. 6, No. 5
71
4. Stirling series
The Stirling numbers of the se
ond kind are the numbers ( ) of partitions of into
pie
es. They satisfy the re
urren
e relations
(
=0
(0 ) = 10 ifotherwise
( +1 )= (
1) + ( )
A table of values for ( )
an be
onstru
ted from these relations, just like Pas
al's
triangle.
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 might guess from this table that ( 2) = 2m 1 1, 1, and this is easily seen.
It
an also be seen that ( 3) = (3m 1 2m + 1) 2. For more on Stirling numbers any
basi
text on
ombinatori
s
an be
onsulted, e.g. [1℄.
: N ! N is a
ardinal fun
tion if and only if it
an be written as a
Stirling series
1
X
( )= n ( )
S m; n
m
n
S
S m
n
;n
;n
:
S m; n
nS m; n :
S m; n
n
m
S m;
m
S m;
=
4.1. Theorem. f
f m
where the
an
n=0
a S m; n
are natural numbers with the properties
= 0 ) n = 0 for all
(2) 1 = 0 ) n = 0 for all
1.
roof. Assume that is a
ardinal fun
tion
orresponding to the fun
tor and let n
be as in Corollary 3.2. As n a
ts freely on ( [ ℄),
( ( [ ℄))
( [ ℄) = n
n
Sn
but an orbit is pre
isely a quotient of with elements. Thus
#( n
Sn ( [ ℄)) = # n (# )
It then follows by Corollary 3.2 that any
ardinal fun
tion
an be written as a Stirling
series with n = # n.
If 0 = 0, then (;), whi
h is 0, is empty. Sin
e there is always a fun
tion ; ! ,
we have ( ) ! (;) = ; so that ( ) = ;. Thus all n = 0.
(1)
a0
a
n
a
a
n
P
f
F
S
A
E pi X; n
A
X
A
A
E pi X; n
E pi X; n
Orbits E pi X; n
;
n
A
S
X; n :
f
a
a
A
F
F X
F
A
F X
X
a
Theory and Appli
ations of Categories, Vol. 6, No. 5
72
Similarly, if a1 = 0 then F (1) = A1 = ;, and as there is a fun
tion 1 ! X for every
non-empty X , we have F (X ) ! F (1) = ; whi
h implies F (X ) = ;. We
on
lude that
an = 0 for all n 1.
Conversely, given any Stirling series P1n=0 anS (m; n) with a0 and a1 non-zero, we
an
hoose Sn-sets An with
ardinalities an (say with trivial
P1 a
tion). Then Proposition 3.3 will
give a fun
tor G with the right
ardinality, thus n=0 anS (m; n) is a
ardinal fun
tion.
The fun
tor F = A0 [ ; ;℄ takes the value A0 at ; and ; elsewhere, whi
h
overs the
ase where a0 or a1 are 0.
The hom fun
tor [ ; [k℄℄ : Setop0 ! Set0 gives rise to the exponential
fun
tion f (m) = km so we should be able to write km as a Stirling series. As km is the
ardinality of the set of fun
tions : [m℄ ! [k℄ and ea
h su
h fa
tors
as a
followed by a one-to-one map, we get
km = S (m; 0) + kS (m; 1) + k(k 1)S (m; 2) + k(k 1)(k 2)S (m; 3) +
This is be
ause the number of one-to-one maps from a set with n elements to one with k
is given by the
k!
k#n = k(k 1)(k 2) (k n + 1) = :
n!
P
1
Thus km = n=0 k#n S (m; n).
The additive Abelian group Z[x℄ is free with basis h1; x; x2; x3 ; : : :i. But as x#n is a
moni
polynomial of degree n, h1; x; x#2 ; x#3 ; : : :i also forms a basis. The above equation
shows that the
hange of bases matrix,
hanging from the rst to the se
ond, is given by
the Stirling numbers of the se
ond kind [S (m; n)℄. Its inverse, whi
h
hanges from the
se
ondPto the rst basis, de nes the Stirling numbers of the rst kind [s(n; m)℄. Thus
x#n = m s(n; m)xm . In parti
ular we have
(
X
n=k
s(n; m)S (m; k) = 10 ifotherwise.
(*)
m
Of
ourse, all this is well-known (see [1℄).
Let E : NN ! NN be the shift operator, (Ef )(n) = f (n + 1), and I the identity
operator. These are used in the
al
ulus of nite di eren
es, where the di eren
e operator
= E I is the main topi
of study.
With these preliminaries we
an now prove the following theorem giving the Stirling
oeÆ
ients of a
ardinal fun
tion.
P a S (m; n)
f (m) = 1
an = E #n f (0)
n=0 n
roof. E #n = E (E I )(E 2I ) (E (n 1)I ) = Pm s(n; m)E m so
P
E #n f (0) = Pm s(n; m)E m f (0)
= Pm s(n; m)fP(m)
= PmPs(n; m) k ak S (m; k)
= k m s(n; m)S (m; k)ak
= an (by (*)).
4.2. Example.
uniquely
quotient
falling power
4.3. Theorem. Let
P
. Then
.
Theory and Appli
ations of Categories, Vol. 6, No. 5
f : N ! N is a
ardinal fun
tion if and only if one of the following
4.4. Corollary.
holds:
(a)
(b)
73
f (n) = 0 for all n 1
E #n f (0) 0 for all n and f (0); f (1) 6= 0.
5. Examples
5.1. Example.
m
2
m
Asymptoti
ally, S (m; n) nn (for xed n), i.e.
!
mlim
!1
n!S (m; n)
= 1:
nm
Intuitively, if m n, a random fun
tion : [m℄ ! [n℄ is almost
ertainly onto, so the
number of quotients will be approximately the number of fun
tions [m℄ ! [n℄ divided by
the number of permutations on [n℄. The reader who is not
onvin
ed by this probabilisti
argument
an
onsult [1℄ p. 140 #10 where some hints are given.
Thus a non-
onstant polynomial never de nes a
ardinal fun
tion, for if it is nonm
onstant some an 6= 0 (n > 1) and the fun
tion nn grows faster than any polynomial.
!
5.2. Example.
Note that
m
m
2
E # n f (m) = (E
(
+1)
nI )E #n f (m) = E #n f (m + 1) nE #n f (m)
so we
an
al
ulate the values E #n f (m) re
ursively. We arrange the values in a table with
the values of f (m) in the rst row, with ea
h new entry being
al
ulated using thetwo
values above it in the previous row, like for nite di eren
es. Thus for f (m) = mm we
get:
m
n
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
2
As E # f (0) = 2, we see that f (m) = mm is not a
ardinal fun
tion. We might have
believed that it was one, as 2m mm 4m .
2
6
2
Theory and Appli
ations of Categories, Vol. 6, No. 5
5.3. Example. 2
3
m
74
2m
This is a
ardinal fun
tion as is easily seen by
onstru
ting a table as above. The values
E # f (0) turn out to be 1; 4; 10; 12; 0; 0; 0; : : :. But it is easy to
onstru
t a fun
tor with
ardinality 2 3
2 . Let : [2℄ ! [3℄ be the in
lusion. Then the pushout P in
n
of
m
m
[
; [2℄℄
[
; [3℄℄
-
?
[
; [3℄℄
- P?
has the right
ardinality.
5.4. Example.
(2m)!
m
2
5.5. Lemma. For
n m natural numbers we have
X iE # f (m
n
E # f ( m n ) = f ( m)
1
i
n
1
i):
i=0
Proof.
E # f (m n) = (E (n 1)I )E #
f (m n)
#
#
=E
f (m n + 1) (n 1)E
f (m n)
f (m n + 2) (n 2)E #
f (m n + 1) (n 1)E #
n
(n
(n
=
E#
(n
2)
(n
1)
(n
1)
1)
2)
(n
1)
f (m n)
..
.
E # f (m) 0E # f (m 1) 1E # f (m 2) (n 1)E #
f (m n)
#
#
#
= f ( m)
(1E f (m
2) + 2E f (m
3) + + (n
1)E
f (m n)):
0
=
0
1
2
f (m), then f
(n
2
5.6. Proposition. Suppose that neither
m(m+1)
(n
1
1)
1)
f (0) nor f (1) is 0 and for every m, f (m + 1)
is a
ardinal fun
tion.
Proof. We shall prove by indu
tion on
m that 0 E # f (m n) f (m) for all 0 n m.
n
m = 0, we have only n = 0 and the statement is obvious.
Assume the statement holds for m. Then
P
E # f (m + 1 n) = f (m + 1) P iE # f (m i)
f (m + 1)
if (m) (by indu
tion hypothesis)
= f (m + 1)
f (m)
f (m) (as n m + 1)
f (m + 1)
0:
It is also
lear that, as E # f (m + 1
n) = f (m + 1) (non-negative terms),
E # f (m + 1 n) f (m + 1):
For
n 1
i=0
n 1
i=0
(n 1)n
n
i
2
m(m+1)
2
n
n
This proves the indu
tive step.
Then putting
n = m we get E # f (0) 0 for all n, so f
n
is a
ardinal fun
tion.
Theory and Appli
ations of Categories, Vol. 6, No. 5
75
Consider f (m) = mm . Its values are natural numbers and
(2m + 2)! (2m + 2)(2m + 1)
f (m + 1) =
=
f (m)
2m
2
so by our proposition it is a
ardinal fun
tion.
The smallest fun
tion satisfying the
onditions of Proposition 5.6 is
(2
)!
2
+1
f (m) = (m
1)!m!=2m
1
m 1,
with f (0) = 1. This is a
ardinal fun
tion.
As
ardinal fun
tions are
losed under produ
ts, f (m) = (2m)! = 2m mm is also
ardinal.
Consider f (m) = m m .
(2
)!
2
2
f (m + 1) = (m + 1) m
> (m + 1) m
(2
+2)
2
2m
= (m + 1) (m + 1)
> m m f (m):
2
(
2m
+1)
2
So m m is a
ardinal fun
tion.
We don't know of any naturally arising fun
tor with these
ardinalities.
m
5.7. Example. m!, m
m m
in Proposition 5.6 is the best we
an do with that kind of
ondition, as
The
the following shows.
5.8. Proposition. Let : N ! N be a fun
tion with the property that any f for
whi
h f (m + 1) (m)f (m) for all m and f (0); f (1) 6= 0, is a
ardinal fun
tion. Then
(m) m m .
Proof. For natural numbers, p and q , de ne a fun
tion f by
2
(
+1)
2
(
+1)
2
f (m) =
(
0Q
if m < p
q m
(
k
)
if m p
k p
1
=
with the
onvention that an empty produ
t is 1 (so that f (p) = q ). Then f (m + 1) =
(m)f (m) if m 6= p 1 and f (m + 1) = q 0 = (m)f (m) if m = p 1. Thus f satis es
our
onditions, ex
ept
for f (0); f (1) 6= 0.
Q
m
Let g (m) = k ( (k) + 1). Then g (m + 1) = ( (m) + 1)g (m) (m)g (m) and
g (0); g (1) 6= 0. So g satis es all the
onditions. It follows that f + g does too, so it is a
ardinal fun
tion, by hypothesis, and by Corollary 4.4, E #n (f + g )(0) 0.
Now, as f (m) = 0 for all m 0 for all n 100, whi
h strongly suggests that they are
ardinal fun
tions.
Referen
es
[1℄ Comtet, Louis, Analyse
Combinatoire
, Tomes I & II, Presses Universitaires de Fran
e, Paris, 1970.
[2℄ Dougherty, Randall, \Fun
tors on the Category of Finite Sets,"
1992.
[3℄ Joyal, Andre, \Une theorie
ombinatoire des series formelles,"
1980, pp. 1-82.
TAMS
, vol. 330, Number 2, April
Advan
es in Math.
, Vol. 42, No. 1,
[4℄ Pare, Robert, \Absoluteness Properties in Category Theory," Ph.D. Thesis, M
Gill, 1969.
[5℄ Pare, Robert, \On Absolute Colimits,"
J. Alg.
19(1971), pp. 80-95.
Department of Mathemati
s and Statisti
s
Dalhousie University
Halifax, Nova S
otia
Canada B3H 3J5
Email:
paremathstat.dal.
a
http://www.ta
.mta.
a/ta
/ or by anonyftp://ftp.ta
.mta.
a/pub/ta
/html/volumes/6/n5/n5.fdvi,psg
This arti
le may be a
essed via WWW at
mous ftp at
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