In this section we briefly indicate how to construct a large family of examples of finite ordered sets that have a fixed point free automorphism and are such that all retracts have the fixed point property. This shows that despite some nice results for special classes of ordered sets the approach 3 in the introduction might not lead to a resolution of the fixed point problem, as there are too many forbidden retracts. Theorem 4.51 was revealed to the author by an anonymous referee. It also gives a negative answer to Problem 2 in . The author would like to hereby express his gratitude to this referee.
Proof: Let be a retraction, i.e., a continuous idempotent map with . Then is isomorphic to a retract of the n-dimensional unit ball which has the topological fixed point property by Brouwer's fixed point theorem. \
Let be the
truncated incidence lattice of the triangulation
K of .
Let be a nontrivial
Then there is a minimal element m of
that is not in :
Indeed otherwise r fixes all minimal elements of , which
implies for all . Then there is a
such that r(y)>y and r(p)=p for all .
Then y has exactly one upper cover and
there is a maximal element such that
has exactly one point (otherwise y has more than
one upper cover).
Thus M is the only upper bound of y.
But then every point x that is in the interior of the
topological realization of y is such that
for all small enough we have that
is homeomorphic to
the upper half space
Since is an n-dimensional manifold
without boundary this is a contradiction
to |K| being homeomorphic to .
Let be an order-preserving map. Let . For each minimal element choose a minimal element such that . For not minimal let
Then G is order-preserving on . Moreover is a retraction and induces a simplicial map on K that is a nontrivial retraction (we have shown above that the retract does not contain all minimal elements), which can be extended to a continuous retraction . Now by Lemma 4.47 has the topological fixed point property. Thus the continuous map on R[|K|] induced by G has a fixed point p. Let S be the smallest simplex in such that . Then G maps S to a sub-simplex of S, i.e., and thus G has a fixed point. Thus has a fixed point, which must be a fixed point of f. \
Proof: For each simplex S of K let A(S):=-S. By hypothesis this is well-defined. Since no simplex is equal to its antipode is a fixed point free order-preserving automorphism of . By Lemma 4.50 all nontrivial retracts of have the (order-theoretical) fixed point property. \