Hey everyone, I am facing the following problem:


Say that a (order-preserving) poset map $f:P\to Q$ has property $(\star)$ if for all $q\_1,q\_2\in Q$ with $q\_1\leq q\_2$ and every $p\_2\in f^{-1}(q\_2)$ there exists an element $p\_1\in f^{-1}(q\_1)$ such that $p\_1\leq p\_2$.


Given such an $f$, can we already make any statements about its geometric realization $|f|:|P|\to |Q|$, which is a continuous map between topological spaces? I am especially interested in anything that relates the homotopy types of $|P|$ and $|Q|$, as in my explicit case I know that $|Q|$ is contractible and want to show that this topological property also holds for $|P|$.


In my case, $f$ is surjective and for all $q\in Q$ the simplicial complex $|f^{-1}(q)|$ is homeomorphic to a tree, thus contractible. The Quillen fiber lemma states that the contractibility of $|f^{-1}(Q\_{\leq q})|$ for all $q\in Q$ already implies that $|f|$ is a homotopy equivalence, and now I am interested in some modified version of the Quillen fiber lemma which tells me that if $f$ has property $(\star)$, then it suffices to check contractibility on the subcomplexes $|f^{-1}(q)|\subseteq |f^{-1}(Q\_{\leq q})|$.


I have thought about several counterexamples where it does *not* suffice to check contractibility on the (geometrically realized) fibers over single points in $Q$, but all of the counterexamples' poset maps did *not* have property $(\star)$.


Thanks a ton in advance.


Sebastian



Poset fiber theorems under a special assumption on the poset map?!
Unfortunately, property $(\star)$ is *not* enough. Here is an easy [counterexample](http://tinypic.com/r/a108lz/7).


