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Let $K$ be a PL triangulation of a closed manifold and $f: K\to\mathbb R^k$ be a PL map. Equivalently, $f$ is a map that becomes linear on every simplex after subdividing. Suppose $v$ is a vertex in the interior of $f(K)$.

  1. Is $f^{-1}(v)$ is a locally flat submanifold?
  2. If so, does it have a PL trivial normal bundle?

I don't think this is answered in any standard PL topology book. All I found is a lemma on p.94 in Kirby-Siebemann's book which says that $f$ is a trivial bundle over the interior of every top-dimensional simplex in $f(K)$.

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  • $\begingroup$ The answer to (1) has to be no, doesn't it? If the only condition is that $f$ is linear on some subdivision, you could have all sorts of degenerate behaviour: the map could collapse some cells of the subdivision. If the union of the collapses was some non-manifold you could arrange $f^{-1}(v)$ to be something like a tree. Perhaps I've missed an assumption in your question. $\endgroup$
    – Ryan Budney
    Commented May 18, 2020 at 6:45
  • $\begingroup$ @RyanBudney: I am also skeptical, and want to see more examples. For example, I think $\mathbb RP^2$ has a PL map onto $[-1,1]$ such that the preimage of $\{0\}$ consists of two circles: one with trivial normal bundle, and the other one with nontrivial normal bundle. The motivation for all this is that I am reading of paper where to the best of my understanding (1)-(2) are claimed as correct. $\endgroup$
    – Igor Belegradek
    Commented May 18, 2020 at 11:30
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    $\begingroup$ You will want some transversality conditions for this to hold; see my answer here. $\endgroup$
    – Moishe Kohan
    Commented May 18, 2020 at 21:49
  • $\begingroup$ @MoisheKohan: thank you, I was aware of Armstrong-Zeeman's paper in Topology, where they filled the details of their BullAMS announcement, and the transversality for maps isn't there (they treat what they call graph transversality). However, in a subsequent paper of Armstrong "Transversality for polyhedra" he does give a general transversality statement which implies (1). I need to think more to see if it also gives (2). $\endgroup$
    – Igor Belegradek
    Commented May 18, 2020 at 23:32

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