> 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?

> Thanks,
> S.