Let $\pi\colon X\rightarrow Y$ be a proper map of topological spaces. Let's assume that both $X$ and $Y$ are paracompact, Hausdorff and locally weakly contractible. Then is it enough to conclude that for any point $y\in Y$ there is a base of open neighborhoods $U\subset Y$ such that the preimage $\pi^{1}(U)$ is weakly homotopy equivalent to $\pi^{1}(y)$? And if not, what is the natural generality in which this would be true?
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3$\begingroup$ I suspect any space filling curve $[0,1] \rightarrow [0,1] \times [0,1]$ is a counterexample as stated. $\endgroup$– Connor MalinMay 25 at 22:01

2$\begingroup$ If $\pi$ is definable in the sense of ominimality, then we have a stratification theorem that breaks Y into locally closed subsets over which $\pi$ is isomorphic to a projection. For example, any analytic map between real vector spaces is definable in the analytic structure of the reals. Maybe in this case we also get the homotopy property you mention? $\endgroup$– John WiltshireGordonMay 25 at 22:49

$\begingroup$ @ConnorMalin Ah, good point! and I guess the standard Peano curve is given by a proper map in particular.. $\endgroup$– user42024May 25 at 22:58

$\begingroup$ @JohnWiltshireGordon actually yes, I was really hoping that at least for analytic maps between analytic subsets of R^n this should be true, though I still didn't have any concrete argument in mind. this stratification theorem seems nice; is it also possible to see how the preimages of these strata glue inside X? like e.g. would it be true that I have a continuous "specialization" map from the fiber over bigger stratum to a smaller one..? $\endgroup$– user42024May 25 at 23:16

$\begingroup$ @user42024 I'm not an expert in ominimality, just a casual admirer. But I'm guessing that something very strong will be truethere will be Matherstyle control data (as in a Whitney stratification) all of which is definable, and gives fully compatible definable homotopies on any sufficiently small neighborhood of a point. I also expect that this has already been worked out! Not quite sure what to google, though. $\endgroup$– John WiltshireGordonMay 25 at 23:36
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