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Let's say we are given two finite simplicial complexes, which I will suggestively call $E$ and $B$. We'd like an algorithm for the following decision problem:

Does there exist a simplicial map $p:E \to B$ and a simplicial complex $F$ so that (a) the inverse image $p^{-1}(\sigma)$ of every simplex $\sigma \in B$ is homotopy-equivalent to $F$, and moreover, (b) every inclusion $p^{-1}(\sigma) \hookrightarrow p^{-1}(\tau)$ where $\sigma$ is a face of $\tau$ is a homotopy-equivalence?

For those of you licking their lips at the prospect of constructing some awful word-problem type obstruction, please note that I'd also be delighted if one could answer the question above replacing all instances of homotopy with homology.

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    $\begingroup$ I suppose in general (the hard version of) your question is difficult. For example, if $E$ and $B$ are $K(\pi,1)$'s your question is asking whether $E$ covers $B$ or not, which for the groups amounts to asking whether or not one group can be found as a subgroup of another -- the embedding problem for groups. Although this is not what you're asking, if you were to go further and ask for a genuine fibre bundle, that is a very hard problem. For high-dimensional manifolds there is a (difficult to compute) complete obstruction called the Farrell Fibering Theorem. $\endgroup$
    – Ryan Budney
    Commented Sep 5, 2014 at 16:30
  • $\begingroup$ @RyanBudney, since he is asking for a simplicial map, I think it is not quite the same thing. You could have that $E$ covers $B$ but that you have to do some subdivision to make the covering simplicial. $\endgroup$
    – Benjamin Steinberg
    Commented Sep 5, 2014 at 18:09
  • $\begingroup$ Sorry for being silly, but it seems to me that the homology part of the question can be solved by a stupid algorithm. Since everything is finite, you can enumerate all the simplicial maps, enumerate all the inclusions of simplices - and homology isomorphisms can be decided. My algebraic topology is not strong enough for anything slicker... $\endgroup$
    – Matthias Wendt
    Commented Sep 5, 2014 at 18:15
  • $\begingroup$ Another small point: I would probably prefer calling a map as in the question a quasifibration (as in the works of Dold and Thom). Fibrations are defined by a lifting property and it is quite rare that a map between finite simplicial complexes satisfies the full Kan lifting property. $\endgroup$
    – Matthias Wendt
    Commented Sep 5, 2014 at 18:50
  • $\begingroup$ Thanks for the various comments. I realize that the question might get hard: for instance, even the special case where $F$ is fixed to be a point requires testing whether a simplicial map realizes homotopy equivalence between $B$ and $E$. While various weaker versions could be "solved" by just enumerating all the simplicial maps, maybe there is something less expensive that one might be able to say in special cases. $\endgroup$
    – Vidit Nanda
    Commented Sep 5, 2014 at 19:21

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