7
1
0
6
0
6
5
1+
0
7
8
9
1
3
5
2
1+
6
6 5
1 0
7 2
8 7
9
2
1+
3
5
8
9
3
1+
6
0 1
1+
6
1
3
7
8
9
5
0
2
9
5
0
2
1+
7
8
6
1
3
8
9
6
1
3
5
7
0
2
1+
7
8
9
0
2
1+
6
1
3
5
8 9 1
8 9
2 7 8
1 3 7
3
2
9
8
7
1
2
3
1+
5
6
0
7
8
9
2
3
1+
5
6
0
1
8
9
7
3
1+
5
6
0
1
2
9
7
8
5 2
3
3 4
9 5
8 6
7 0
6 1
0 7
1 9
2 8
4
4
5
6
0
1
2
3
8
7
9
6
0
1
2
3
1+
5
9
8
7
COM BIN A TOR 1 A L
MAT HEM A TIC S
YEA R
February 1969 - June 1970
SE~~ENT-PRESERVING MAps OF PARTIAL ORDERS
by
Martin Aigner*
Geert Pri ns
Department of Statistics
University of North Carolina
Department of Mathematlcs
Wayne State University
Institute of Statistics Mimeo Series No. 600.21
University of North Carolina at Chapel Hill
FEBRUARY 1970
*
The research in this report was partially sponsored by the U S. Air
Force Office of Scientific Research under contract No. AFOSR-68-1406.
Segment-Preserving Maps of Partial Orders
by
Martin Aigner *
Department of Statistics
University of North Carolina
I.
Geert Prins
Department of Mathematics
Wayne State University
INTRODUCTION.
Let
P be a partial order, then for a,b
the segment
[a,b]
as the set of all
X
E
be partial orders and suppose the function
one-one onto those of
a<b if and only
have
a,bEP
a<b iff
~(b) <
Q.
We call
if
~(a)
cjl(a),
verse partial order of P.
any segment
[a,b]
~
<
P with a
~
~ X ~
~
b we define
b.
~(b)
The map
in
Q,
Let
maps the elements of
¢ an isomorphism if for all
i.e, ,
Q ;;;:,P or
P,Q
P
a,bEP we
i f for all
Q =~ p* , where p* denotes the in-
i. e. ,
P the set
P with a
E
is called segment-preserving if for
~
{~(x):
x~[a,b]}
forms
a segment of Q.
The purpose of this note is an investigation of all possible segmentpreserving maps of any partial order
Clearly an isomorphism of
P
1
P which possesses a zero-element .
is an example of such a map and we shall
henceforth refer to isomorphisms as trivial maps.
After preliminary remarks we introduce semi-products and semi-decom-
positions of partial orders in Section III and proceed to relate them to
the structure of partial orders which admit non-trivial maps (Section IV)"
* The research in this report was partially sponsored by the U.S. Air
Force Office of Scientific Research under contract No. AFOSR-68-l406.
1
Alternately we may assume the existence of a unit-element, developing
the ensuing theory in an analogous fashion.
2
The final section is devoted to a discussion of strongly segment-preserving
maps and the determination of those partial orders for which every segmentpreserving map is of that type.
Throughout the paper the term map shall always mean a segment-preserving map.
A map will mostly be denoted by
the abbreviation
ment
[a,bJ
= Xl
¢(x)
is mapped by
= [UI,VIJ
onto the segment
[UI,VIJ.
XEP.
the set of elements of a partial order
E(P).
shall use the symbol
a~b
and
X
0
y
in
P
indicates
a~b
iff
and we shall frequently use
¢[a,bJ
for
¢
¢
P
Q
When we refer to
regardless of the ordering we
If for two partial orders
in
means the seg-
P ~
hold, we write
P, Q
Q.
E(P) ~ E(Q)
The notation
X, yare incomparable elements.
II, PRELIMINARY RESULTS.
Throughout this section
acting on
bEP
will denote a partial order,
¢
a map
P.
Proposition
all
P
1.
Let
P
have a zero-element
(or, similarly, if
¢[o,bJ
0.
= [b',O'J
If
¢[o,bJ = [o',b'J
for all
bEP),
then
for
¢
is
X <
Y
trivial.
Proof.
then
We first verify
XE[O,yJ
Xl, yl
(or
and hence
iff
XIE[OI,yIJ
are comparable in
yIE[XI,OIJ)
parable.
X oy
and hence
¢(P).
Xl
0
(or
yl
(or
X'<yl
P
If, say,
In either case
implies
yE[O,XJ).
Now the hypothesis plainly implies that
relations in
X,yEP.
XIE[yl,O'J).
Conversely
XE[O,yJ
for
Thus
¢
XIE[OI,yIJ
x, yare com-
either preserves all
or inverts all relations, and hence is an isomorphism,
3
Proposition 2.
P = [a,b] be, a segment.
Let
[a',b'],"f [b',a ' ],
(i)
cp[a,b] = [UI,V I] :f
Suppose
then
U
'II
P,
in
V
<;>
I
Proof.
(ii)
a
I
<>
(iii)
u,
V
Assume
= cp[a,b],
bI
cp (p) ,
in
are both different from
U, V are comparable.
[u,v]
and hence
If
u=a or u=b,
Proposition 3.
trivially.
Proof.
I
b·1
cp[u,v]
(or
a contradiction.
f[b',a
l
],
cp[a,b]=[a',b ' ]
being entirely analogous).
i.e.,
a'Ecp[u,b]
are comparable, then
bE[a,u],
or
= [UI,V I]
cp[v,u])
respectively.
But
contradicting (iii).
cp[a,b]:f[a ' ,b
Let us assume
cp[a,c]f[a',c ' ],
l
(1)
P = [a,b] be a segment and suppose ¢ maps
Let
Then
a,
= [a,b],
ai, b
bIEcp[a,u] which implies aE[u,b]
this means
Then
[v,u])
(or
Clearly (i) now implies (iii).
or
I
l
],
P non-
and hence (1) holds.
(the argument for the other case
By Proposition 1, there exists
cp[a,c]=[a',d ' ] with crd.
CEP
The segment
with
[a,c]
now satisfies the hypothesis of the preceding proposition, yet (l,iii) is
violated.
Corollary 1.
any
Let
P
have a zero-element.
For any segment
[a,b]~P
and
cp we have
either
cp[a,b]=[al,b
cp maps
or
[a,b]
l
]
or
\
in which case
trivially,
cp[a,b]=[u',v ' ] with
in which case
=[bl,a l ],
cp maps
I
I
U,V both different from a,b
[a,b] non-trivially.
I
)
(2)
4
Proposition 4.
Let
cp[a,b]=[u',v ' ].
Then for
P=[a,b]
cj>
be a segment and
a non-trivial map with
z E P,
z
;:;:
U,
Z
;:;:
V
=>
Z = b,
I
z
~
u,
z
~
V
~
Z = a,
I
Zl
;:;:
a' ,Z I
;:;:
b'
==;>
ZI = VI ,
ZI
~
a' ,z I
~
b'
=>
Zl
~
(3)
I
= UI.
I
)
Proof.
Assume
z'E[b',v'].
Z~u, z~v.
z'Ecj>[u,b]=[u',b ' ]
Then
But this implies
z'=b ' ,
and thus
by (2), and similarly
z=b.
The remaining state-
ments are proved in a similar fashion.
Proposition 5.
Let
cj>[a,b]p[u',v'].
z
Proof,
P=[a,b]
Then for
<>
Z <> V iff
zI
The assumption
ZOU, Z<;>V
implies
I II
I
a I , Z' <> b I .
cj>
a non-trivial map with
ZEP,
U an d
<>
Hence by (2) we infer
ZI
be a segment and
ZI~[UI
¢
aI
and
Z I <> b I .
(4)
z¢[a,u], [u,b], [a,v], [v,b].
,a'],[u' ,b'],[a',v'],[b' ,Vi],
or equivalently
The converse is now obvious.
SEMI -PRODUCTS OF PART! AL ORLeRS
I
We begin by recalling the following concepts (see e.g. [1]),
be a partial order.
i.e.,
and
E(I) ~ E(P),
y~x
then
A lower semi-ideal
I
of
P
Let
is a sub-order of
P
P,
endowed with the same partial order such that if
yEI.
Let
{P:
a
partial orders with zero-elements
the partial order with
E(~
aEA}
oa
xEI
be a family of pairwise disjoint
(aEA) •
aE AP)
a = UaE AE(P)
a
The direot swn L afA Pais
such that
p s q
in
5
~
aEA
P
a
produot
iff for some
rr
P
a
aEA
aEA
p,q E P
a
= (...
p
a
a
For all
Pa
With this agreement
Definition.
Let
p
such that
aEA
and all
in
The direot
P.
a
~
E (rr aE AP)
a '"
q
in
p EP
a a
( ..• p
with the A-tuple
iff
we henceforth iden-
... ),
a
s:f a.
for
LaE AP a -c rr aEA P.
a
{P:
a
aEA}
orders with zero-elements
lower semi-ideal
q
i .e. ,
P
... ): PEP , aEA},
a
a
tify the element
~
p
is the partial order defined on the set of all
A-tuples with coordinates from
{p
and
be a family of pairwise disjoint partial
oa
Pa 's is
A semi-produot of the
(aEA) .
I of the direct product such that
LaEAP
a
:= I:::.
a
ITaEAP "
a
Both the direct sum and the direct product clearly are examples of
semi-products.
To facilitate the reference, we shall denote semi-products,
{P
the direct sum and·the direct product of a family
rr
~E
respectively.
AP,
a
product.
We call the
by
a
rr aE AP,
a
the faotors of the
P 's
a
When we write down the factors of a product we shall use the no-
tation (e.g., in the case of two factors
U, V)
U ~ V,
U ~ V,
U ® V,
respectively.
In much the same way as for direct products, we now introduce semidecompositions of partial orders.
In defining semi-products the factors
were assumed to be pairwise disjoint.
We now identify all the zero-elements.
This clearly can be done without causing any ambiguity.
partial order with zero-element
orders containing
{P}
a
and
{P
a
aEA.
E
E(rr
P
aE
:::. P:
A p)
a
iff there exists
rrP P
aEA a
pEP
P
be a
a family of sub-
which are otherwise pairwise disjoint.
we may associate a unique semi-product
("0 po
.. )
a
all
0,
0
Let
With
P
and
by the rule
such that
p
a
~
p
for
6
P
{P 0.-cP:
The family
Defi niti on.
is called a semi-decomposition of
if there exists an order-preserving isomorphism
o(p a )
write
P = ITaEAP
= Pa
for all
and identify
a
P
a
from
Pa EP a .
and all
= o(p}
for all
Ponto
We then
pEP.
From the definition of a semi-product it readily follows that every
factor
P
a
is a lower semi-ideal of
of
P
it is clear that
IT
aEA a
p €P.
ments
a
p
Hence if
a
=
c..
IT
p
a
aE
AP,
a
.oo}
for a set
{XiEP :
a
there exists
plEP
iE]}
is the unique join of the ele-
(.oo PH'),
a
ment of the partial order
Again from the definition
then
Po.
is the maximal ele-
As a consequence we note that if
there exists
PEP
with
Xi~P
for all
ie:J t
then
with the same property.
a
With these preparations the following statement is immediate.
Lemma.
If
P = IT aEA Po.
and
Pa = ITSEB a Pat S
for
then
P = ITa ITS Pa,S'
Proposition 6.
P = IISEBQ S
for
aEA t
Proof.
Let
Let
P have zero-element
are two semi-decompositions of
P = II
and hence by the Lemma
peP
and suppose
a
is a lower semi-ideal
P.
Then
p = (H' Ps
.. )
qsEPan QS '
If
SEB,
since
is the maximal element of
qs'S,
thus
,
P=P a '
and
Po. = II SEB (Po. n QS )
... )
Since P
a
Po.
P £ IT
Q
a
SEB (Pan S )'
in
IISEBQS'
for all SEB t and hence
Pa
P E II SEB (Pan Q )' then
P =
S
p = (. .. Pa,
= II aEA Pa
atS (PanQS)'
with
unique join of the
P
Suppose
PSEP a
On the othe r hand, let
Po.
o.
in
II aeA Pat
P
a
then
below
p.
( .. ,
qs
qs
Po.
:s;
But
.,
0
)
for all
P is the
and the conclusion follows.
7
Every partial order
=P ®
P
{O}.
P
admits the trivial semi-decomposition
If there are no others,
Coro 11 ary 2.
P
is called semi-indecomposabZe.
If there exists a semi-decomposition
P :f
into semi-indecomposable factors
a
{O},
2
P
= IT aEA Pa
of
P
then this semi-decomposition
is unique (up to order).
Proposition 7.
A
= U;E)A;
Proof.
(
...
P
onto
P
= ITaEAP a
be a semi-decomposition of
is any partitioning of the index-set, then
pEP
Let
P;
Let
... ) ,
p
= ( ...
Pa
P;
= ( ...
Pa
and
where
P
IT.lE J (I]:ad\.Pa).
1 '
Coro 11 ary 3.
... )
... )
in
with
= IT;EJ
=0
tition
A
= Bur
= U®V
a(p)
=
maps
and satisfies all the requirements.
If there exists a semi-decomposition
P
(ITaEA;P a ),
f3 4 A.1 ,
for
P
= IT aEA Pa
to semi-indecomposable factors, then any semi-decomposition of
two factors
If
The function
IT aE AP a .
Pf3
P
P.
is given by
U
of the index-set
A.
= IT S€B Pf3' v = IT yEf Py
of
p
in-
Pinto
for some par-
For brevity, we shall shorten the terms semi-product, semi-decomposition
etc. to
IV.
S-product,
THE M8.IN THEOREM.
Theorem 1.
maps
2
S-decomposition etc. in the remainder;6fthe paper.
P
Let
P be a partial order with zero-element
non~trivially.
0
and suppose
<p
Then
The question of existence of a decomposition is unresolved
even in the case of direct products.
in general,
8
P = U I8l V,
Proof.
that
First we note
U, V
= {oJ.
UnV
By Proposition I and (2) it is clear
¢(U) ~ U*,
are mapped trivially with
[o,zJ ~ P
segment
(5)
is mapped trivially iff
¢(V) ~ V,
ZEU
or
ZEV.
and that a
Let
pEP
be
arbitrary and suppose
=
¢[o,pJ
where
P EU,
1
P2EV.
[p~'P~J,
(6)
We define the function
0
on
P
by
(7)
and proceed to show that
0
is the required isomorphism from
Ponto
U I8l P V.
If
(7)
pEU
0(p)
= p.
or
pEV,
then
Any pair
under (7) of some
pEP
[O,pJ
(P1,P2)
is mapped trivially, hence by (6) and
with
belongs to
Furthermore, it is clear that
0
U
I8l
P EU,
1
P
P2EV,
which is the image
since by (6)
V,
P1$P,
P, q
maps distinct elements
P2$P,
onto dis-
tinct pairs.
To prove the range of
that there exists
assume
r>o,
5>0.
pEP
0
is all of
p~r,
with
We h ave
p~5.
r,5E [
O,p,
J
If
w.Lo.g.
P1I <r
[r,pJ,
and
5
l
'
l
'
I8l
P
V
let
rEU,
5EV
be such
By our previous remarks, we may
an dh ence
= P2'
and thus
P1I <r
U
r 1 ,5
I
E
we are finished.
[P1,P2'
I
IJ
Suppose
Consider the segment
by (2)
¢
maps
[r,pJ
non-
trivially, say
( 8)
9
Pi
By (1)
q and thus if
0
a contradiction to (8).
r < P2'
by (8)
P2
Hence
and since also
0
q, we would conclude by (4)
P2 ' q
are comparable.
r < Pi'
If
we deduce by (3)
q
pi
0
P2 ,
~
r=o,
q',
then
a con-
p2<q, and thus r,P 2E[0,q]. Since Pi oq and
p' <q 1 q4 U• Similarly p <q and q'<p l imply q4V. Hence ep maps
2
2
1
'
the segment [o,q] non-trivially. Applying (3) to r 1<0 1, r'<ql (by
Therefore
tradiction.
(8»
o'<p'2'
and to
we infer
and thus
we are finished.
If
Otherwise
S<P 2 ,
S, q, P2 in place of r, P, Pi'
nally arrive at an element tEP with a(t) = (r,s).
peat the argument with
To prove
a(q) = (qi,q2)
a
and we re-
This way we fi-
p,qEP with a(p) = (P i ,P 2),
[o,q], and hence [p~,p;] ~ [q~,q;].
is order-preserving take
p~q.
and
Then
[o,p]
~
The whole
But this means
argument is clearly reversible, and the theorem follows.
V
I
STRCJJGLY SEGfv1ENT-PRESERVI NG
MAPs
I
According to Theorem 1 with every map
(5)
on
a uniquely determined ordered S-decomposition
prees this correspondence in symbols
arbitrary S-decomposition
with
ep
[ep]
P = U®V,
[ep]
= [U,V].
P
we may associate by
P = U®V
of
P.
Conversely given an
we ask whether there exists a map
A map
ep
on the partial order
P is called strongly segment-
preserving (a strong map, for short) ifthe.pre-image of eve.ry segment
ep(P)
ep
= [U,V].
Definition.
of
We ex-
has a zero-element
in
P.
Q
10
Theorem 2.
P
Let
decomposition
U®V
P =
[~(U,V)]
such that
P,
of
= [U,V].
For any S~(U,V)
there exists a unique strong map
~(U,V)
is given by
~(U,V)(X,y) ~ ~(U,V)(r,s)
r
iff
X, Y ~ s
~
(9)
r,XEU, S,yEV.
where
Proof.
Let
=~
~(U,V)
~(P)
o.
be a partial order with zero-element
P = U~V
be an arbitrary but fixed S-decomposition of
as given by (9).
[~]
is a partial order and
serving let
[(r,y),(x,s)]
~
~[(r,y),(x,s)]
and the fact that
This proves
If
~
The definition of
P
= [U,V].
{~(k,.e):
=
[~(x,y), ~(r,s)]
(r,y),(x,shp,
~
r~x, y~s
~
is segment-pre-
Then
=
r
and
clearly implies that
To prove
be a segment.
P
k
~
X, Y
~.e ~ S}
by (9)
ctnd hence
(x,y),(r,~hP.
is a map.
~
Q = [~(x,y),~(r,s)]
~(p),
is any segment of
(by (9)) and is minimal in the pre-image of
Q.
Hence
then
(r,y)EP
(r,y) = 0Q'
~
and
is a strong map.
Let
~
verify (9).
now be an arbitrary strong map with
If
(x,y),(r,s)EP
with
r~x, y~s,
[~]
= [U,V].
then
( r ,y )E P
We have to
and
(r,y) E [r,(r,s)],
(10)
(r ,y) E [ y, (X ,y )] .
Since by (6)
rl~(r,s)', (X,y)l~yl,
we deduce from (1) and (2) that the
segments in (10) are mapped trivially, and thus
11
(X,y) I
(r ,y) I
~
Conversely let
(x ,y) I~ (r, s)
(r,s) IJ
with
</>(P)
~
X'
(r ,s) I.
and consider the segment
I
0Q = (a,b).
(x,y) ,
~
~
I ,
By (6)
(a,b)'
~
Q :;: [(x ,y)
~
b' ,
(11)
a'
(a,b) ,
~
(r,s) ,
~
~
s' .
(a,b)~(x,y), (a,b)~(r,s),
Since by assumption
we infer
(x,b),(a,s)EP
and by (6)
</>[o,(x,b)J = [xl,b'J,
</>[o,(a,s)J = [al,sIJ.
Together with (11) this implies
(x,y)
~
(x,b),
(a,b)
By the minimality of
This means
(r,s)
</> = </>(U,V),
~
(a,s).
we now deduce
y=b, r=a
and thus
r~x, y~s.
and the proof is complete.
In the preceding proof it was shown that the righthand4to-lefthand
implication, spelled out in (9), holds for any map.
Furthermore it fol-
lows from the argument that we may also relax the reverse implication to
obtain the following characterization of strong maps:
A map
</>
with
[</>J = [U,V]
is strong iff
</>(x,y)
whenever
</>(r,s)
~
(a,b)
then there exists
(a,b)
~
(x,y),
such that
(a,b)
</>(x,y)
~
~
with
(12)
(r,s)
</>(a,b)
~
</>(r,s).
12
In the remainder of the paper we present characterizations of partial
P
orders which only admit strong maps.
will always denote a partial order,
a map.
cj>
Theorem 3.
P
Let
o.
have a zero-element
greater than
U,XEU,
with
P
of
U 0 x, yEV,
and elements
such that
(X,y)EP,
(u ,y HP.
while
Proof.
~
0
admits maps which are not
P = U~V
strong iff there exists as-decomposition
u, x, y
P
Let
= cj>(U,V)
P = U~V
U, x, y
and elements
satisfy the hypothesis.
be the strong map defined by (9).
~(U) <> ~(x,y).
In
~(p)
we add the relation
Let
U 0 X,
(9) implies
~(U) < ~(X,y)
and all re-
Since
lations arising transitively therefrom.
We proceed to verify that the
function
from
cj>(p)
for all
cj>:
~ ~(P),
P onto the partial order
thus constructed, is in fact a map.
The only way
fail to be segment-preserving is if there exists a segment
~
(m,n)]
P
such that
y~n, u~m.
case (9) implies
thus
(U,y)EP,
of the segment
~(u), ~(x,Y)E~(R)
Hence since
a contradiction.
[cj>(U),cj>(X,y)]
= [~(m,i),~(k,n)].
(m,n)EP,
we have
As by construction of
{U,(X,y)},
is just
cj>
the map
R
cj>
can
= [(k,i),
But in this
(m,y)EP
and
the pre-image
is not
cj>
strong.
Conversely suppose the hypothesis fails and let
[cj>]
= [U,V].
We wish to verify (12).
Suppose
(X,y)1
be any map with
cj>
~
(r,s)
I
in
cj>(P),
then by (6)
XI
~
(X ,y)
I
~
(
r, s)
I
~
SI
•
(13)
13
Case a.
(X,S)EP. 'By (6)
(X,S),
(r,s)::; (x,s),
setting
= (a,b)
(r,y)
Case b.
(x,s)4P.
X <> r,
then
ity is
r<x
or equivalently
y<S.
y::;s, r::;x.
X $ r,
First we note
trivially.
Similarly
and
hence by (13)
Thus
(x,y)::;
(r,y)EP,
and
condition (12) is satisfied.
X>'O, r>o,
contradiction.
= [XI,SI],
cjl[o,(x,s)]
y <> 5
5 $
Y and 5>0. Next if
But now by hypothesis
is impossible.
In this case
(r,y)EP
(x,s)EP,
a
Hence the final possibil-
and with
(r,y) =
(a,b)
condition (12) is satisfied.
Figure 1 depicts an example of a partial order with zero-element
which admits other than strong maps.
By Theorem 3 it is the smallest such
example.
yl
(X,y)
•
cjl
°
u .Figure 1
Let us interpret. Theorem 3 in terms of the following graphtheoretical
concept.
With an arbitrary partial order
ity-graph
y
and
in
I(P)
I(P)
x€M, yEN
E(P)
whose vertex-set
iff
x<>y
then
in
X and
P.
If
r~,
P we associate its incompar-abiZis such that
X is adjacent to
N are two components of
I(P)
yare comparable and it readily follows from
14
the definition that either
for all such pairs.
C.P
J
<
{oJ
C~
x<y or y<x for xEM, yEN.
I{P)
member of
This way we arrange the
= (C~:
iE])
such that
o (unit-element 1), then
P has a zero-element
If
constitutes a component of
({l})
C(p)
into a linear order
j <k.
iff
M<N or N<M ac-
Hence we may unambiguously define
cording to whether
components of
X < Y for all xEM and all yEN or y < X
I(P)
and hence is the first (last)
C( p) .
With these preparations we immediately deduce
Corollary 4.
Let
P have a zero-element. Then every map
strong iff for every S-decomposition
C.U
J
E
C{U),
C~
on
P is
P = U®V of P and every pair
C{V), we either have
E
~
for aU
XEC.U and aU
J
V
YEC k or {X,y)EP for no such pair.
Let
P have a zero-element 0 and let us denote the unit-element (if
it exists) of an, arbitrary sub-order
only admits strong maps.
position.
we write
If
otherwise
lK'
by
with
lp=l.
Assume
P
P = U®V be an arbitrary but fixed S-decom-
Let
MEC{U), NEC{V)
M~ N,
K:=.P
(X,y)EP for all XEM, yEN,
are such that
Mf N.
Clearly
(i)
M~
{oJ for all ME C{ U) ,
(ii)
{oJ
~
N for all N E C{V),
(14)
(iii)
M~ N and R < M=-> R ~ N,
(iv)
M~ N and S < N ='> M ~ S.
If for no choices
then
P = U~V,
M~
{oJ
E
the direct sum of
C{U), N ~
U and
V.
{oJ
E
C{V)
we have
M~N,
15
P f
Suppose
three factors.
U~V
and assume
P
can be S-decomposed into at least
Then in view of Proposition 6 at least one of
w.l.o.g.
U,
can be S-decomposed into two factors
E(S-{a})
is contained in a member of
parable with all of
U·
= 'U'
( r ,5 )
Hence
= U-{a, 'U}
Rand
S,
= , R'
5
r
and possibly
(14, iii) there exists
C(W)
gument that
N~ V
for all
xER,
XER, yEV
=>
P
(r,5)~X
then
= \.
Thus
if
{1 U},
Nl{ahC(V)
If some
C(U)
{a},
By assumption and
U'~N .. Writing
with
is com-
XERandfora.u XES.
consists of
exists.
'U
E(R-{a}) u
(r,5)faEU
for all
say
P
= R®W,
where
(by Proposition 7, this is possible) we conclude by the same ar-
W = S®V
Since
C(U).
U = R®S.
U, V,
= R®W
yf'V EV .
~
consists of
W1u{a}
yd~ =>
(x ,y)EP
for all
R ~ U'u{O}
and
(X ,y}EP = R®W
for all
WEW
W' = W-{a,'W}
{a},
XEU I
yEV.
XER, YEW' => (x ,y}EP
If
'U
P
= U®V -
for all
( , R,W) E
exists we deduce
and thus by Proposition 7
But this means
{'W} .
(X,y)EP
we infer in succession:
for all
,
and possibly
(possibly) {l},
(' U,y) EP
if
,
= U®V
for all
exists.
We summarize the results we obtained in the following statement.
Theorem 4.
(i)
(ii)
Let
P
Then
admits only the trivial maps iff it is S-indecomposab1e,
if for every S-decomposition into two factors either
or
P
(iii)
P have a zero-element.
if
P = UiV
or
P
= U®V
- {,};
if'
P
= U~V
exists, then
only admits strong maps,
P
can be S-decomposed into at least three factors, then the
converse of (ii) also holds.
By applying Corollary 3 to the preceding theorem and Theorem 2 we
obtain
16
Cora 11 ary 5.
II
aEA Pa
P
of
P
(i)
P
Let
IAI
... )
ex
3,
~
2 1A1
P
(i)
Let
<j> -1
P
factors
map on
iff
ment in
Pt
strong.
Let
<j>
Q :;;,
with
To prove (i) let
Let
P
is itself a
is a direct decomposition, and
o.
have a zero-element
<j>(p) = Q1 ®Q 2
<j>(P).
P
1
P
Then
<j>(P)
is segment-preserving on
Ut V.
P.
is strong.
[<j>] = [U,V]
Proof.
only admits strong maps,
contains a unit-element t i.e.,
for any S-decomposition of
(ii)
exists t then
IAI = 1 t
strong maps on
segment, then any S-decomposition of
thus every map on
(aEA)
P=
then the converse of (ii) also holds t
P
We note that if
P
exists, then
there are exactly
Theorem 5.
Pex ~{o}
admits only the trivial maps iff
if
(iv)
If as-decomposition
into S-indecomposable factors
( ... lp
(iii)
a.
have a zero-element
P
<j>(U) ,
P = U®V
iff
into two factors,
Q :;;,
2
<j>(V)
for any map
<j>
with
is a segment.
P = U0V
P
be any S-decomposition of
be a map with
[<j>] = [U,V]
and assume
Since then the pre'"'image of any segment
it in particular contains a zero-element
XEU, YEV
<j>
for any map
be arbitrary and suppose
Since by (9)
<j>[(k,l),(m,n)] = [(m,l)I,(k,n)'],
Q~
0Q t
<j>(P)
into two
<j> -1
is a
is a seg-
and thus
<j>
is
17
= (m,l),
we infer
X
(X,y)EP,
and thus
= (k,n).
y
= U~V
P
[U,VJ,
k
(m,n) =
= 0,
for any S-decomposition into two factors, then
Theorem 4 implies that every map is strong.
=
=
P = U~V.
Conversely if
[¢J
l
But this means
¢ is a map with
Hence if
then by (9)
¢-l[(X,y)',(r,s) IJ = [(r,y),(x,S)J,
[(X,y)I,(r,s)'J
for any segment
<jl
As to (ii) let
<jl(p) ,
~
<jl -1
and
is a map.
¢(p)
be the trivial map with
~
P*.
Since S-prod-
p* =~ Q1
ucts are only defined for partialordem with zero-elements,
possess a zero-element, and hence
P
Conversely let
As we remarked above,
Q1
=
¢(U), Q2
a unit-element.
be a segment and
¢
any map with
[¢J = [U,V].
¢ is strong and hence defined by (9).
¢{(lU'y): yEV}
=
P
must
and defining
(x1,(lU,y)I),
i t readily follows from (9) that
morphism from
<jl(p)
a
¢(P)
on
a
by
Setting
a(x,y) I
is the desired iso-
onto
We lastly apply Corollary 3 to the preceding theorem, and obtain
Corollary 6.
P = IT aEA Pa
Let
of
P
P
have a zero-element
o.
If as-decomposition
Pa~{o}
(aEA)
for any map
¢
into S-indecomposable factors
exis ts,
then
(i)
<jl -1
P
is segment-preserving on
iff
= IT aEAPa'
Q ::.
(ii)
a -
composition of
P
¢(p)
is a segment.
¢(p)
¢(p a )
for any map
is the S-de-
into S-indecomposable factors
~{o}
iff
=
18
We would like to express our thanks to G.-C. Rota and R. Stanley for
calling this problem to our attention and to A.M. Gleason, who independently
obtained similar results, for a stimulating discussion.
REFERENCE
[1]
G. Birkhoff, Lattice Theory.,
1948, 1967.
A.M,S. CoZZoq.
Publ.2~,
1940,