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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

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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

2 Answers 2

up vote 3 down vote accepted

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

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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.

Addendum: here's a reference:

Chapman, T. A. Cell-like mappings. Algebraic and geometrical methods in topology (Conf. Topological Methods in Algebraic Topology, State Univ. New York, Binghamton, N.Y., 1973), pp. 230–240. Lecture Notes in Math., Vol. 428, Springer, Berlin, 1974.

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.

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John, I was thinking about this before I posted my comment above. How do you know that all of the point inverses of |f| are contractible, given that this is true at the vertices of the simplicial complex (= elements of the poset)? In general I think that's false: |f| might not even be surjective. Take the poset (a, b, c, d) with a>c,d and b> c,d and map it to a 4-chain w<x<y<z by sending a to z, b to y, c to x, and d to w. This is the inclusion of a 4-cycle into a tetrahedron. The point inverses at the vertices are each a point. (Note: this map doesn't have Sebastian's property (*).) –  Dan Ramras Feb 19 '11 at 0:26
    
Dan is right, there are very easy examples of order-preserving maps between posets f:P->Q such that for all q in Q the geometric realization of the fiber of f over q is contractible, but the geometric realization of f has non-contractible fibers. In particular, the "Quillen-fibers" might not be contractible. Unfortunately, my posets are also not finite. –  Sebastian Feb 19 '11 at 15:19
    
I messed this up! See my second addendum. –  John Klein Feb 20 '11 at 2:00
    
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
    
@Sergey: you are right. Upon further reflection, I now think that (*) is reminiscent of the path lifting condition. –  John Klein Feb 21 '11 at 1:15

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