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