I am trying to understand Poincare duality at a concrete example.
Consider X as the boundary of the embedded standard 2-simplex, i.e. X
looks like this
           a
         /   \
        /     \
       v       v
       b --- > c
where the opposite edge of a,b,c gets the name A,B,C respectively.
This is an orientated 1-manifold without boundary which is
homeomorphic to S^1. To calculate the homology with integer
coefficients one has to consider the chain complex
0 ----> Z[A,B,C] --D--> Z[a,b,c] ---->0
where D=[[0,-1,1],[-1,0,1],[-1,1,0]] so a basis of the kernel of D is
for example (A,-B,C) and H_1(X)=Z.
Now one obtains the dual cell structure (which is a simplicial complex
too in this case) by taking the center of the edge A as a new 0-
simplex a', b' and c' the same way and an edge connecting for example
a' and b' is named C' (because it runs over the vertex c). The new
simplicial complex (let's call it X' even X and X' are the same
topological spaces) yields a canonical orientation a' > b' > c'.
Now I like to show that there is a map f_* of cell complexes
0 ----> Z[A,B,C] --D--> Z[a,b,c] ------>0
|           | f_1           | f_0       |
v           v               v           v
0 ----> Z[a',b',c'] --D'-> Z[A',B',C'] -->0
which means that f_0D=D'f_1. From this I can go further to show that
the two complexes are chain-homotopic.

But here is my problem: What is the map D'? Every time I have seen
cohomology the co-chain complex which have occurred has been of the
form hom(C_*,Z) where C_* has been a normal chain complex. What can I
doo here? I can't see how to interpretate the elements a' or A' (for
example) as a map ?-->Z...? Can anybody help me to see how the
relation to the cohomology occurs in my above example?