> <mariano.suarezalva...@gmail.com> wrote:
> > On 5 jul, 14:24, sanchopanch...@web.de wrote:

> > > Hello,

> > > 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.

> > Poincaré duality is an isomorphism from homology
> > to cohomology. What you are doing is constructing
> > a map between the complexes which compute *homology*
> > for two related triangulations.

> > I doubt I'd be able to explain how Poincaré duality
> > works in ASCII. The very besy you can do is get
> > a textbook and follow the construction.
> > Greenberg's book has a very nice discussion, for
> > example.

> > -- m

> Thank you for your answer. Anyway, I don't think that my construction
> from above is very far from what is done in the Poincare duality
> construction. Some textbook of mine gives a similar example with S^2
> and wikipedia (http://en.wikipedia.org/wiki/
> Poincaré_duality#Dual_cell_structures) starts the same way as I have
> done as far as I have understood that. My problem from above rises in
> the wikipedia article here:

> "Then one can show that the boundary operator
> C_k(X) \to C_{k-1}(X)
> is the transpose of the boundary operator
> C_{n-k+1}(Y)=Hom_Z(C_{k-1}(X),Z) \to C_{n-k}(Y)=Hom_Z(C_k(X),Z)."

> If I can answer that problem and get to know what the boundary map D'
> is, I can show that the two complexes from above are chain homotopic
> because they are both chain homotopic to the complex of the
> barycentric subdivision. I am pretty confident, that the second
> complex my be easily seen as a simplicial cochain complex.

> I don't want to proof anything of the Poincare duality theorem, I'd
> just like to understand this 'nearly obvious' (like the statement fom
> wikipedia) things written everywhere which are sadly not obvious to
> me.