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Given a closed $n$-manifold $M$, Poincare duality equips us with an isomorphism: $$H_k(M)\cong H^{n-k}(M)$$ Here I'm speaking of singular homology with coeffecient in $\mathbb{Z}_2$.

Suppose now $M$ has a triangulation $K$. Then clearly, we have a similar Poincare isomorphism as above for $K$ as well. In fact, even more can be said: there exists a polyhedral dual structure $K^{\vee}$ of $K$ which gives an isomorphism already at the level of (co)chain complexes: $$C_k(K^{\vee})\cong C^{n-k}(K)$$ (in order to be safe let's further assume we are considering smooth manifolds so stuff as https://mathoverflow.net/questions/194297/dual-cell-structures-on-manifolds do not occur)

My question is:

Is there an isomorphism of the form $C_k(K^{\vee})\cong C^{n-k}(K)$ even when $K$ is not necessarily a triangulation of $M$, but instead only homotopy equivalent $M$

I'll remark, that I'm in fact interested in Poincare-Lefschetz duality (i.e for manifolds with boundary) but thought it is better to start from here. Furthermore, I'm in particular interested in the case where $K$ is the nerve of some "good cover" of $M$ (and thus homotopy equivalent from the nerve lemma)

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Your question is not very precise, but I would say the answer is no. Consider $M$ to be the $0$-manifold given by the singleton. Take the simplicial complex given by the simplest triangulation of $[0,1]$ (two vertices, one edge), which satisfies $M \simeq [0,1]$. Then the cellular complex of $K$ is $C_0(K) = \mathbb{F}_2^2$ and $C_1(K) = \mathbb{F}_2$. There is no triangulation $K^\vee$ such that $C^0(K^\vee) = \mathbb{F}_2^2$ and $C^{-1}(K^\vee) = \mathbb{F}_2$.

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  • $\begingroup$ I agree that my question is not very precise.. I wasn't aware of how problematic it is to construct a dual polyhedral structure to some abstract simplicial complex. Even for a triangulation of a manifold with boundary there are subtleties to deal with. I assume that's your point there right? But then I must ask, in your answer what do you mean by $K^\vee$? do you mean some standard construction of a dual complex for a triangulation of a manifold with boundary, one which gives a Lefschetz type duality? $\endgroup$ – Itamar Vigi Apr 2 at 13:49
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    $\begingroup$ @ItamarVigi For a manifold with boundary, in general you would want an isomorphism $C_*(K) \cong C^{n-*}(K^\vee, \partial K^\vee)$ (this is Poincaré–Lefschetz duality). Here in my answer, $K^\vee$ is just a notation: what I am saying is that there is no cell complex $X$ such that $C^{-1}(X) = \mathbb{F}_2$. There is never anything in negative degrees. $\endgroup$ – Najib Idrissi Apr 2 at 13:51
  • $\begingroup$ Thank you for pointing out the absurdity of what I'm looking for. $\endgroup$ – Itamar Vigi Apr 4 at 8:12

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