# Intersection map giving rise to Poincaré duality

Let $$M$$ be a smoothly triangulated compact $$d$$-dimensional manifold. Consider the subcomplex $$C_*^{\pitchfork T}(M)$$ of smooth singular chains which are transverse to the triangulation. An inductive chain homotopy construction establishes that these are quasi-isomorphic to all smooth, and thus all singular, chains.

Define the intersection map $$I : C_n^{\pitchfork T}(M; R) \to C^{d-n}_\Delta(M; R)$$ (the latter being simplicial cochains arising from the triangulation) by sending $$\sigma : \Delta^d \to M$$ to the cochain whose value on the an element of the triangulation whose characteristic map is $$\iota : \Delta^{d-n} \to M$$ is the count of the zero manifold given by the pullback of $$\sigma$$ and $$\iota$$. Here either $$R$$ is $$\mathbb{Z}/2$$ or $$M$$ must be oriented and the count is with the usual signs, and one uses some version (such as this) of transversality for manifolds with corners.

Fun exercise: with appropriate signs, $$I$$ is a map of chain complexes. (Hint: as in the proof that degree as defined by counting preimages is homotopy invariant, this relies on the classification of one-manifolds.) Poincaré duality implies that the domain and range of $$I$$ are quasi-isomorphic.

Question: why is $$I$$ a quasi-isomorphism?

I think I can prove this, but only in the mod-two setting, by using Thom's seminal work on bordism and Quillen's elementary approach to cobordism (just the definitions of his "elementary" paper - not the main results, which to me are quite deep despite the title of the paper). But there must be a more direct argument, which covers the oriented case as well, and it seems like this should be in the literature somewhere - from the 1940's maybe?

(Motivation: Greg Friedman, Anibal Medina and I have what we think is a new approach to questions such as Do chains and cochains know the same thing about the manifold? through vector field flows, and would like to build on existing knowledge of the interplay between intersection and duality.)

• It must be a local argument, as $M$ is a manifold. I think this sketch works: take a cover by open balls $\{U_i\}$, and subdivide your triangulation to be subordinate to the cover. Then consider $C_{\Delta, c}^\ast(U, \Omega)$, cochains with support compactly contained in $U$, twisted by the orientation sheaf $\Omega$ (the twisting by $\Omega$ is what makes this the true target of $I$). Check quasi-isomorphism locally; standard homological algebra then shows the global result. Dec 11, 2020 at 17:05
• Thanks. Yes, I had thought about compactly supported chains, but I would prefer to not have to subdivide. But maybe we can find a "cover subordinate to the triangulation" in a sense. Dec 11, 2020 at 17:32
• If one takes the open stars of all simplices of T, this produces a good open cover U of M by open balls, and T is subordinate to U. This may be the cover you're looking for. Dec 11, 2020 at 23:00