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I am reading the proof of Theorem Poincaré duality in "principles in AG" of Griffith.

They constructed "dual cell decomposition" of a polyhedra decomposition of manifold $M$ and the cochain complex of this dual cells.

I don't know why this cohomology group of this cochain complex is the singular cohomology group of $M$?

Someone help me give answer? Thanks in advance.

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Given a cell $\epsilon$ of $M$ there is precisely one dual cell that intersects $\epsilon$. That is the key idea for how you define the map from the chain complex to the cochain complex. If dealing with integer coefficients you have to be careful about orientations. – Ryan Budney Mar 26 '11 at 20:52
See this thread:… Also, your question is more appropriate for the math.stackexchange site since it's standard material from coursework. – Ryan Budney Mar 26 '11 at 21:14
Hatcher's "Algebraic Topology" book has a nice explanation (starting on p232). – Mark Grant Mar 27 '11 at 7:29
up vote 3 down vote accepted

For any CW complex $X$ one defines a chain complex $C_*(X)$: choose an orientation of each cell; the group $C_n(X)$ is the free abelian group with a basis whose elements correspond to the $n$-cells of $X$ and the differential $C_n(X)\to C_{n-1}(X)$ is defined by $c\mapsto \sum_{c'\subset\partial c} (c,c')c'$ where $(c,c')$ is the incidence number of $c$ and $c'\subset\partial c$ defined as follows.

By the definition of a CW complex one can extend the homeo of an open $n$-ball to $c$ to a map of the closed ball; compose the restriction of this map to the boundary $S^{n-1}$ of the closed ball with the map $X_{n-1}\to S^{n-1}$ obtained by collapsing all cells of the $n-1$ skeleton of $X$ but $c'$ to a point; the incidence number of $c$ and $c'$ is the degree of the resulting map $S^{n-1}\to S^{n-1}$ where the first sphere is oriented using the "outgoing normal first" rule and the orientation of the second one is induced from $c'$.

This generalizes the chain complex of a simplicial set. The homlogy of $C_*(X)$ is isomorphic to the singular homology of $X$, see e.g. Hatcher, Algebraic topology, p. 137 (freely available online) or Milnor, Stasheff, Characteristic classes, Appendix A.

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My question is why this cohomogy group of this cochain complex is the singular cohomogy group of M. I have tried to check four axioms for cohomogy theory of a topological space X in Spanier's book (algebraic topology). However, i haven't finished to check them. – vu viet Mar 15 '10 at 9:00

The cochain complex of the "dual cell decomposition" can actually be identified with the dual cochain complex.

To specify a linear map on the free module spanned by the cells of the decomposition is the same thing as assigning a scalar to each cell of the decomposition. If we assume the decomposition to be finite, then the module of cochains is therefore canonically isomorphic to the module spanned by the cells of the dual decomposition (as well as those of the original decomposition but it is less interesting).

It is clear that the degrees match, and it is easy to check that the codifferentials are also identified. The cohomology of both cochain complexes are therefore identified as well, as the underlying complexes are isomorphic.

If the original cell decomposition is not finite, then instead of considering the module spanned by the dual cells, one considers instead of a direct sum the cartesian product.

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