# Poset fiber theorems under a special assumption on the poset map?!

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 $(\star)$.

Sebastian

-
Have you looked at the article "Poset Fiber Theorem" by Bjorner-Wachs-Welker (Trans. AMS, vol 357 no. 5, 2004)? They discuss a result of Eric Babson (see p. 1884 of the article) that might give what you need. At least it looks related. – Dan Ramras Feb 18 '11 at 17:19
Dan, thanks a lot. It really does look like what I'm searching for. Maybe both extra condition are even equivalent or something. I will think about it for a while and get back to everyone. (A better source for the lemma might be "B. Sturmfels and G. Ziegler: Extension Spaces of Oriented Matroids", Preprint, Lemma 3.2.) – Sebastian Feb 18 '11 at 19:42

Unfortunately, property $(\star)$ is not enough. Here is an easy counterexample.

-
However, something still confuses me. My counterexample seems to also be a counterexample to Lemma 3.2 in vs24.kobv.de/documents-zib/61/SC-91-11.pdf, which is the result of Eric Babson mentioned above. So I guess the empty set is not considered contractible? – Sebastian Feb 19 '11 at 16:34
The empty set is not contractible (=homotopy equivalent to a point). But in your example, everything is contractible: $|P|$ and $|Q|$ and both point-inverses. To what assertion is it a counterexample? – Sergey Melikhov Feb 19 '11 at 16:55
Wait, $|P|$ in the image linked above is not contractible. (Did you look at the link a while ago? I had to exchange the images...) – Sebastian Feb 19 '11 at 17:06
But you are obviously right, the empty set is certainly not contractible. So I guess my question is answered - thank you everyone! – Sebastian Feb 19 '11 at 17:09
Indeed, the new example works. May I wonder why did you need condition (*) and if you know any literature about it? I'm also using this condition; I'm calling maps satisfying it "closed" because that is what the condition amounts to with respect to the order topology. – Sergey Melikhov Feb 19 '11 at 20:34

In the following, I shall assume that the posets $P$ and $Q$ are finite.

Then it is at least true from the condition that $f^{-1}(q)$ is contractible for all $q \in Q$ that the map $|f|$ is a homology isomorphism, by the Vietoris-Begle theorem (http://en.wikipedia.org/wiki/Vietoris%E2%80%93Begle_mapping_theorem). This is a very old result.

(Notice: I do not even require your condition ($\ast$))

In fact it is even true that your map is a homotopy equivalence (under the assumption that point inverses are contractible) since $|f|$ is a simplicial map of simplicial complexes which has contractible point inverses. This is what people call a "simple map" in simple homotopy theory. It's a basic result to the subject that a simple map of poyhedra is a homotopy equivalence. This result dates from the late 1960s I think, perhaps the 1970s.

Second Addendum: What I wrote above is pure bunk. We also need to know that $|f|^{-1}(x)$ is constractible for all points $x$---not just the vertices. Here's what I think to be the case: Suppose one has the additional condition that $f^{-1}(\sigma)$ is contractible, where $\sigma = x_0 \le x_1 \le \cdots \le x_k$ is any finite chain. Then it looks to me as if the additional condition will guarantee that $|f|$ is a homotopy equivalence.
Now it seems to me that your condition ($\ast$) amounts to no more than the statement that $f^{-1}(x_0 \le x_1)$ is non-empty. So my condition will imply yours, but not vice-versa.
I don't think that "condition ($∗$) amounts to no more than the statement that $f^{-1}(x_0\le x_1)$ is non-empty" for each chain $(x_0\le x_1)$, because starting with any map $f$ that does not satisfy ($*$), you can enlarge its domain by taking coproduct with a copy of the target (coproduct of posets probably has some standard name; it corresponds to disjoint union on the level of order complexes), and extend $f$ by the identity. Did you mean some other statement? – Sergey Melikhov Feb 20 '11 at 13:10