 
  
  
  
  
 
In May-June 1996 results that give algebraic, resp. combinatorial proofs
 
for the result that every finite truncated noncomplemented
lattice has the fixed point property
were announced by K. Baclawski in [6] and by
T. McKee and E. Prisner in [80].
We will present McKee and Prisner's approach here as their
methods are natural applications of what is discussed in this chapter.
The following is thus nothing but a translation of
the results in [80]
from their original version for modulo 2 homology to integer homology.
It becomes necessary, since the beautiful geometric approach
to homology
in [80] seems to depend on the fact that -1=1 in   , 
while the arguments generalize beyond this visualization. One just 
needs to carefully keep track of the signs at times.
 , 
while the arguments generalize beyond this visualization. One just 
needs to carefully keep track of the signs at times.
Proof: 
In case q=1, there is nothing to prove (the boundary of a 
sum of zero-dimensional simplexes is 0). 
Let   be a (q-2)-dimensional oriented simplex that
occurs in one of the (q-1)-dimensional simplexes
  
be a (q-2)-dimensional oriented simplex that
occurs in one of the (q-1)-dimensional simplexes
  , with
 , with   .
Then
 .
Then 
  occurs once 
(positive or negative)
in
 
occurs once 
(positive or negative)
in 
  , but by assumption
it does not occur in
 , but by assumption
it does not occur in
  Therefore
 
Therefore
  occurs in an even
number, say 2k, of the
 
occurs in an even
number, say 2k, of the
  (
 
(  )
and it occurs such that 
in k 
of the
 )
and it occurs such that 
in k 
of the 
  the summand
 
the summand
  is positive
and in the other k boundaries
it is negative.
If
 
is positive
and in the other k boundaries
it is negative.
If
  is a positive summand of
 
is a positive summand of
  , then
 , then
  is a negative summand of
 
is a negative summand of 
  and
vice versa.
Thus no multiple of
  and
vice versa.
Thus no multiple of
  occurs in
 
occurs in
  by the observed cancellation.
The conclusion follows. 
\
 
by the observed cancellation.
The conclusion follows. 
\
Proof: 
Assume   ,
 , 
  .
If q=1 assume
 .
If q=1 assume
  ,
if
 ,
if   assume
  assume 
  .
 .
 
First, we show that every homology class of q-dimensional cycles
has a representative that does not contain v. 
Assume   .
Let
 .
Let 
  be a representative of the homology class
 
be a representative of the homology class
  ,
with
 ,
with   . 
If t=0, we are done, so we can assume
 . 
If t=0, we are done, so we can assume   . 
By Lemma 4.53,
 . 
By Lemma 4.53, 
  is a (q-1)-cycle of G and hence also of
 
is a (q-1)-cycle of G and hence also of
  .
Since
 .
Since 
  there are q-simplexes
 
there are q-simplexes
  ,
 ,   ,
 ,
  of
 
of
  such that
  such that
  But then
 
But then
  
 
Thus
  
 
Thus 
  forms a q-dimensional cycle with
vertices in
  forms a q-dimensional cycle with
vertices in   .
Finally C and C' are homologous, since
 .
Finally C and C' are homologous, since
  
 
In case q=1, 
let 
  be a representative of the homology class
 
be a representative of the homology class 
  ,
with
 ,
with   . 
t must be even, say t=2k and 
k of the
 . 
t must be even, say t=2k and 
k of the   must be 0, the rest 1
(otherwise the boundary of C contains a multiple of [v]). 
Since
  must be 0, the rest 1
(otherwise the boundary of C contains a multiple of [v]). 
Since 
  , we have by Lemma 4.19
that
 , we have by Lemma 4.19
that
  is connected.
However then
  is connected.
However then 
  , since
in a connected graph
any sum of differences of zero-dimensional
simplexes is the boundary of a sum of 1-dimensional 
simplexes.  
Now the argument continues as in the case
 , since
in a connected graph
any sum of differences of zero-dimensional
simplexes is the boundary of a sum of 1-dimensional 
simplexes.  
Now the argument continues as in the case
  .
 . 
 
If we can show now that for any q-cycles
C,C' that do not contain v we have 
  iff
  iff
  then the map that maps 
every homology class
  then the map that maps 
every homology class   in
  in 
  to the corresponding homology 
class in
  to the corresponding homology 
class in   defined by the 
representatives of
  defined by the 
representatives of   that do not contain v
is an isomorphism and we are done. 
To do so first note that the 
direction ``
  that do not contain v
is an isomorphism and we are done. 
To do so first note that the 
direction ``  " is trivial. 
For the other direction let
 " is trivial. 
For the other direction let 
  with
  with 
  with
 
with   . 
Then
 . 
Then 
  
 
with   . 
By Lemma 4.53
 . 
By Lemma 4.53 
  is a q-cycle and since
 
is a q-cycle and since 
  there must be (q+1)-dimensional simplexes in
 
there must be (q+1)-dimensional simplexes in 
  such that
  such that 
  
 
But then
  
 
with the cancellation in the last step happening since 
  is a cycle in
  is a cycle in   .
This finishes the proof. 
\
 .
This finishes the proof. 
\
This together with the removal of escamotable points as
discussed in section 4.5
gives a proof of Baclawski and Björner's
result that is entirely algebraic.
(Contractibility of the neighborhood was needed to
guarantee removability of the point without affecting the homology.
The contractibility of escamotable graphs was a nice consequence, which
is however stronger than what was needed.
A notion ``vertex is weakly escamotable iff its pointed neighborhood
is acyclic" is strong enough for our purposes.)
 
Baclawski's argument (cf. [6]) proves a result
similar to Corollary 4.55
about structures called ``pseudo cones" using neither algebra,
nor topology. His fixed point algorithm for pseudo cones
does not use any previously known method
and as with [29] we refer the reader to the original.
 
  
  
  
 