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Let $K$ be a simplicial complex of dimension $n$, $M$ be a topological manifold, and $f \colon |K| \to M$ be a continuous map. Let $X$ be an embedded manifold in $M$ of codimension $n$, such that $f(|K|) \cap X$ is a finite set of good points (i.e., a neighborhood of each point homeomorphic to a disc where $f(|K|)$ and $X$ intersect as planes in general position), and their preimages lie in $|K \setminus sk_{n-1} K|$.

Denote by $PD([X])$ the Poincaré dual of $X$ cohomology class lying in $H^n(M, \mathbb{Z}_2)$. It seems correct that the pullback $f^*PD([X])$ in $H^n(K, \mathbb{Z}_2)$ is represented, for example, by a cochain that associates to each $n$-simplex $A$ the count of points in $f(A) \cap X$ modulo two.

Is this statement true? Perhaps there is a formal proof of it somewhere, or did I miss something, and it can be proved in just a few lines?

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Yes, this is a special case of the first theorem of Section 3.6 of WHITNEY STRATIFIED CHAINS AND COCHAINS.

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  • $\begingroup$ I think a version of this is also in Bredon's Geometry and Topology text. $\endgroup$ Commented Nov 24, 2023 at 2:23
  • $\begingroup$ Thank you! It seems that the theorem you refer to is stated only in the smooth category, and its connection with a smooth structure on the manifold is quite strong, e.g., I assume the stratification uses the smooth structure. I tried to look at 'A geometric approach to homology theory' cited in the paper but couldn't find something relevant to my question, at least at first glance. $\endgroup$
    – Alex
    Commented Nov 27, 2023 at 15:15
  • $\begingroup$ @RyanBudney Thank you! I believe you're referring to Chapter 6, correct? I've been reviewing it but, unfortunately, haven't found the version you mentioned. Would you kindly provide additional details? I'm grateful for your help! $\endgroup$
    – Alex
    Commented Nov 27, 2023 at 15:22
  • $\begingroup$ @Alex Apologies, I missed the topological manifold assumption. I think if you replace $M$ by a regular neighborhood of Euclidean space and then apply the theorem you might be able to extract your statement, but I have not checked. $\endgroup$ Commented Nov 27, 2023 at 18:46

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