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Two pointed, connected CW complexes with the same homotopy groups need not be homotopy equivalent (Are there two non-homotopy equivalent spaces with equal homotopy groups?). Moreover, having the same homotopy and homology groups is also not enough (Spaces with same homotopy and homology groups that are not homotopy equivalent?).

Question: Suppose that two pointed, connected CW complexes have homotopy equivalent $n$-Postnikov sections for every $n \geq 1$. Are the spaces homotopy equivalent?

I expect the answer to the question to be negative, but I'm having a hard time finding a counterexample. So far, I know that such a counterexample would have both spaces have the same homology and homotopy groups, and that both spaces would need to have nontrivial homology and homotopy groups in arbitrarily high dimension.

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This is a pretty well-known phenomenon, linked with phantom maps.

One of the first existence results was Brayton Gray's paper

Spaces of the same $n$-type, for all $n$, Topology 5 (1966) 241--243

Clarence Wilkerson classified the spaces of the same $n$-type for all $n$ in

Classification of spaces of the same $n$-type for all $n$, Proc. Amer. Math. Soc., 60 (1976) 279--285 (1977)

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    $\begingroup$ Wow, thanks! I guess this also proves that I failed to do the obvious googling, namely "spaces of the same n-type for all n." Let me add that in the paper by Adams, cited in both of the references, a very simple counter-example is given, and it is also proven that the answer is positive if the spaces have finite homotopy groups. $\endgroup$ – luisl Jul 30 at 17:46

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