Suppose we have a simplicial combinatorial manifold (just a triangulated manifold) and its Poincaré dual cell complex.
Corresponding homology simplicial and homology cell complexes are quasi-isomorphic of cause. But this quasi-isomormohism as it usually quoted from Solomon Lefschetz "Alg Topology" book is transcendental. Can we have a reasonable formula for such a quasi-isomorphism (for homology over a good field at least)?
Update
1) sorry for Russian math-slang use of "transcendental" here "transcendental"="non-constructive"
2) among the others one motivation is to see really the Poincare duality for simplicial chains and cochains of the given triangulation of a manifold.
UUpdate
The problem has a nice very canonical solution, with Laplas operator Green function heat kerenel etc. It allows to solve some problems. The preprint(s) are in preparation.
Best Answer
Both the cell complex, $C$, and the dual cell complex $C'$ are refined by the first barycentric subdivision $BC$. There are maps $C \to BC$ and $C' \to BC$, sending a cell $\sigma$ to the sum of all cells of the same dimension contained in $\sigma$; these maps are both quasi-isomorphisms.
So, if you allow me to formally invert quasi-isomorphisms, I'm done.
Is the question whether there is an honest map of chain complexes between $C$ and $C'$, without subdividing?
UPDATE Here is something you can do, and something you can't do.
With $C$ and $BC$ as above, and $r : C \to BC$ the refinement map, there is a homotopy inverse $s: BC \to C$. (More precisely, $C \to BC \to C$ is the identity, and $BC \to C \to BC$ is homotopic to the identity.) Working the same trick with $r' : C' \to BC$, we get quasi-isomorphisms between $C$ and $C'$ which are homotopy inverse to each other. As you will see, however, this construction is very nongeometric and inelegant.
Construction: Let $q:BC \to Q$ be the cokernel of $C \to BC$. An easy computation checks that each $Q_i$ is free. Since $C \to BC$ is a quasi-isomorphism, $Q$ is exact. An exact complex of free $\mathbb{Z}$ modules must be isomorphic to a direct sum of complexes of the form $\cdots \to 0 \to \mathbb{Z} \to \mathbb{Z} \to 0 \to \cdots$. Choose such a decomposition of $Q$, so $Q_i = A_{i+1} \oplus A_{i}$ and the map $Q_i \to Q_{i-1}$ is the projection onto $A_{i}$.
Now, consider the map $q_i^{-1}(A_i) \to A_i$ in degree $i$. This is surjective, and $A_i$ is free, so choose a section $p^1_i$. We also define a map $p^2_i$ from the $A_{i+1}$ summand of $Q_{i}$ to $BC_i$ by $p^2_i = d p^1_{i+1} d^{-1}$. In this way, we get maps $p_i = p^1_i \oplus p^2_i: Q_{i} \to BC_i$ which give a map of chain complexes.
We note that $qp: Q \to Q$ is the identity. Therefore, $1-pq$, a map from $BC \to BC$, lands in the subcomplex $C$ and gives a section $s:BC \to C$. Proof of the claim about homotopies will be provided on request.
On the other hand, here is something you can't do: Get the quasi-isomorphism to respect the symmetries of your original space. For example, let $C$ be the chain complex of the cube, and $C'$ the chain complex of the octahedron. I claim that there is no quasi-isomorphism $C \to C'$ which commutes with the group $S_4$ of orientation preserving symmetries.
Consider what would happen in degree $0$. A vertex of the cube must be sent to some linear combination of the vertices of the octahedron. By symmetry, it must be set to $$a (\mbox{sum of the "near" vertices}) + b (\mbox{sum of the "far" vertices})$$ for some integers $a$ and $b$. But then the map on $H_0$ is multiplication by $3(a+b)$, and cannot be $1$.
I imagine you want something stronger then my first answer, but weaker than my second. I am not sure what it it, though.