Poincare duality for reduced homology

Question

Reduced homology

In my understanding, the reduced homology is better-behaved than the usual singular homology because the $0$th reduced homology

1. counts the non-trivial "closed" $0$-chain and
2. reflects the idea of "orientation is the volume form".
   (see below for detail)

Poincare duality for singular (co)homologies

On the other hand, the Poincare duality for the usual singular (co)homologies:
$$H_i(X) \cong H^{n-i}(X)$$ holds for an oriented closed manifold $X$.

Question

So I thought there should be a correspondence of the Poincare duality for the reduced homology. That is, let $\tilde{H}'_\bullet, \tilde{H}^\bullet$ be reduced homology and the "reduced cohomology" thing (I don't know what's this). Then,
$$\tilde{H}_i(X) \cong \tilde{H}^{n-i}(X)$$ holds for an oriented closed manifold $X$.
Is there anything like this?

I Googled and found a "reduced cohomology" but found nothing about the duality between them similar to the Poincare duality.

Two reasons why I think the reduced homology is better-behaved than the usual one;

1. we can regard "closed" $0$-chain as the one whose weight is $0$. So it matches well to the idea "n-th homology count the number of nontrivial(non null-homologous) closed n-chain" at $n=0$.

2. By definition of n-simplex ($n\geq 0$), we think them as n-triangle with orientation but without orientation when $n=0$. This is odd. The orientation is (an equivalence class of) the volume form, so the orientation of $0$-triangle should be the scalar field on the $0$-triangle. This naturally leads to the definition of reduced homology.

Any reference would be appreciated.

Thanks in advance.

Answer 1

No. We have $\tilde H_k(X) = H_k(X)$ for $k > 0$ and $\tilde H_0(X) \oplus \mathbb Z \approx H_0(X)$, similarly for cohomology. Thus

$$\tilde H_i(X) \approx \tilde H^{n-i}(X)$$ for all $i$ with $0 < i < n$. For $i = 0, n$ it is wrong.