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.

**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
, with .
Then
occurs once
(positive or negative)
in
, but by assumption
it does not occur in
Therefore
occurs in an even
number, say 2*k*, of the
( )
and it occurs such that
in *k*
of the
the summand
is positive
and in the other *k* boundaries
it is negative.
If
is a positive summand of
, then
is a negative summand of
and
vice versa.
Thus no multiple of
occurs in
by the observed cancellation.
The conclusion follows.
\

**Proof:**
Assume ,
.
If *q*=1 assume
,
if assume
.

First, we show that every homology class of *q*-dimensional cycles
has a representative that does not contain *v*.
Assume .
Let
be a representative of the homology class
,
with .
If *t*=0, we are done, so we can assume .
By Lemma 4.53,
is a (*q*-1)-cycle of *G* and hence also of
.
Since
there are *q*-simplexes
, ,
of
such that
But then

Thus

Thus
forms a *q*-dimensional cycle with
vertices in .
Finally *C* and *C*' are homologous, since

In case *q*=1,
let
be a representative of the homology class
,
with .
*t* must be even, say *t*=2*k* and
*k* of the must be 0, the rest 1
(otherwise the boundary of *C* contains a multiple of [*v*]).
Since
, we have by Lemma 4.19
that
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
.

If we can show now that for any *q*-cycles
*C*,*C*' that do not contain *v* we have
iff
then the map that maps
every homology class in
to the corresponding homology
class in 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 `` " is trivial.
For the other direction let
with
with .
Then

with .
By Lemma 4.53
is a *q*-cycle and since
there must be (*q*+1)-dimensional simplexes in
such that

But then

with the cancellation in the last step happening since is a cycle in . 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.