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Hey everyone,

is there a good way to determine whether a map of (the topological realizations of) posets is a fibration without explicitely proving that it has the homotopy lifting property?


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The question doesn't make sense unless you explain what the meaning of fibration of posets is. – John Klein Feb 13 '11 at 4:08
Well, by a fibration of posets I mean a fibration between their canonical topological realizations. – Robert Feb 13 '11 at 13:23

What you're looking for might be Quillen's Theorem B. Roughly speaking, it says that if all the "combinatorial homotopy fibers" are homotopy equivalent (via base change along morphisms in the poset), then the combinatorial homotopy fibers are weakly equivalent to the honest homotopy fibers. It originally appeared in Higher Algebraic K-theory, I (Lecture Notes in Math 341).

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@Dan: Robert is asking for a fibration, not a quasi-fibration. Theorem "B" of Quillen is a statement about a functor inducing a quasi-fibration after taking classifying spaces. – John Klein Feb 13 '11 at 4:09
Yeah, I messed that up! The conclusion is just that the combinatorial homotopy fibers are equivalent to the homotopy fibers. I will edit. – Dan Ramras Feb 13 '11 at 6:54
I have been looking at Theorem B, but I need more than information about the homotopy types of the fibers. I actually need to know that my map has the homotopy lifting property. – Robert Feb 13 '11 at 13:25
I've never seen results that actually guarantee that a map of posets has the homotopy lifting property after applying geometric realization. Unfortunately, it seems like a lot to ask. Maybe you should add some discussion of your motivation to the question, so we can see why you need homotopy lifting. – Dan Ramras Feb 13 '11 at 17:10

I just though about this for another minute and it seems like I posted the first part of my question too early. If $f^{-1}(q)$ is contractible, then $f^{-1}(Q_{\leq q})$ is homotopy equivalent to a poset with maximal element, namely the contracted fiber over $q$. Thus, $f^{-1}(Q_{\leq q})$ is contractible.

However, I do not know an answer to the second problem yet.

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Please disregard this part. – Robert Feb 13 '11 at 13:25

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