Let $X$ be a CW complex with given cohomologies $H^n(X; \mathbb{Z}) = G$, $H^m(X; \mathbb{Z}) = H$ and other reduced cohomologies are zero. Which additional algebraic information/structures do I need to identify the homotopy type of $X$? Similar question for three or more nonzero cohomologies. Can the answer be similar to the answer to the dual question about Postnikov invariants?
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$\begingroup$ You can also have a look into Arkowitz - Introduction to Homotopy Theory, Chapter 7. $\endgroup$– Lennart MeierCommented Jan 20, 2014 at 5:09
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2 Answers
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This isn't really an answer but it's a bit long for a comment.
Do you really mean cohomology rather than homology? The dual of a Postnikov tower is usually considered to be a homology decomposition, building a space up one homology group at a time. A space with only two nonzero cohomology groups could have up to four nonzero homology groups, by the universal coefficient theorem. Incidentally, another reference for basics on homology decompositions in addition to the Baues book in Mark Grant's answer is my algebraic topology book starting on page 464.
Two-stage Postnikov systems were studied a lot in the 60s and 70s, partly because of their connections with secondary cohomology operations. Among two-stage homology decompositions are the so-called "two-cell complexes" $X = S^n \cup e^m$ whose classification is the same as computing $\pi_{m-1}S^n$ modulo the effect of composing with reflections of $S^{m-1}$ and $S^n$. In the general case $S^{m-1}$ and $S^n$ are replaced by Moore spaces, so one is considering homotopy groups with coefficients. There should be a significant literature on this, but I'm not very familiar with it. I believe that Joseph Neisendorfer has written quite a bit about this.
It seems that things become considerably more complicated for three-stage Postnikov systems, and probably the same is true for homology decompositions.
answered Jan 20, 2014 at 19:51
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$\begingroup$ So, if I understand well, even for two-stage homology decompositions the explicit answer is not known (because homotopy groups of S^n are not known)? And therefore Postnikov systems are much better for computations than dual ones? $\endgroup$ Commented Jan 20, 2014 at 20:33
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The short answer is yes, there is a theory dual to the Postnikov decomposition, in which a simply-connected space is built out of Moore spaces by successive cofibrations. This is all worked out in detail in the book of H-J. Baues, Homotopy Type and Homology (see in particular chapters 2.6 and 2.7).
answered Jan 19, 2014 at 22:26