Dgdrxrt

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 $mathbbZ_2$.

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)

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 $C_0(K) = mathbbF_2^2$ and $C_1(K) = mathbbF_2$. There is no triangulation $K^vee$ such that $C^0(K^vee) = mathbbF_2^2$ and $C^-1(K^vee) = mathbbF_2$.