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Since it's fairly easy to come up with a two spaces that have different homotopy groups but the same homology groups ($S^2\times S^4$ and $\mathbb{C}\textrm{P}^3$). Are there any nice examples of spaces going the other way around? Are there any obvious ways to approach a problem like this?

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Related: – Grigory M Jan 15 '12 at 16:39
It's worth noting that $S^2\times S^4$ and $\mathbb{CP}^3$ have the same cohomology groups as well, but they have different cohomology rings. – Michael Albanese Jun 21 '15 at 23:16
up vote 7 down vote accepted

Standard example is $\mathbb RP^2\times S^3$ and $\mathbb RP^3\times S^2$ (they have same homotopy groups since they both have $\pi_1=\mathbb Z/2$ and the universal cover is in both cases $S^2\times S^3$).

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Consider $X=S^1\vee S^3$ and its double cover $X_2$ i.e, attach two copy of $S^3$ one in north pole and one in south pole of $S^1$. Then $\pi_1(X) =\mathbb{Z} = \pi_1(X_2)$. And covering map induced isomorphism in $\pi_n$ for all $n\geq 2$. But they are not homotopically equivalent/ they have different homology groups since their Eular Characteristics are different.

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