Homotopy groups need not determine homology groups

Statement

It is possible to have two path-connected spaces $M_{1}$ and $M_{2}$ (in fact, we can choose both $M_{1}$ and $M_{2}$ to be compact connected manifolds) such that:

For every positive integer $k$ , there is an isomorphism of groups $π_{k} (M_{1}) ≅ π_{k} (M_{2})$ between the Homotopy group (?)s.
There exists a positive integer $k$ such that the Homology group (?)s $H_{k} (M_{1})$ and $H_{k} (M_{2})$ are not isomorphic.

In particular, because Weak homotopy-equivalent topological spaces (?) have isomorphic homology groups, this gives an example of spaces that have isomorphic homotopy groups but are not weak homotopy-equivalent.

Facts used

Covering map induces isomorphisms on higher homotopy groups

Proof

Further information: Product of real projective three-dimensional space and 2-sphere, Product of 3-sphere and real projective plane

Consider the spaces $M_{1} = P^{3} (R) \times S^{2}$ and $M_{2} = S^{3} \times P^{2} (R)$ . We note that:

Both the spaces have universal cover $S^{3} \times S^{2}$ (product of 3-sphere and 2-sphere) which is a double cover, so they both have fundamental group $π_{1} (M_{1}) ≅ π_{1} (M_{2}) ≅ Z / 2 Z$ .
By fact (1), the covering maps $S^{3} \times S^{2} \to M_{1}$ and $S^{3} \times S^{2} \to M_{2}$ both induce isomorphisms on $π_{k}, k \geq 1$ . Thus, we get $π_{k} (S^{3} \times S^{2}) ≅ π_{k} (M_{1})$ and $π_{k} (S^{3} \times S^{2}) \to π_{k} (M_{2})$ and we thus get $π_{k} (M_{1}) ≅ π_{k} (M_{2})$ .

Thus, we see that all the homotopy groups match.

However, the homology groups do not. One simple way of seeing this is to note that $M_{1}$ is orientable, and therefore has $H_{5} (M_{1}) ≅ Z$ , whereas $M_{2}$ is non-orientable, so $H_{5} (M_{2}) = 0$ .

The full homology descriptions are below:

$H_{p} (R P^{3} \times S^{2}; Z) = {\begin{matrix} Z, & p = 0, 2, 5 \\ Z / 2 Z, & p = 1 \\ Z \oplus Z / 2 Z, & p = 3 \\ 0, & p = 4, p \geq 6 \end{matrix}$

and:

$H_{p} (S^{3} \times R P^{2}; Z) = {\begin{matrix} Z, & p = 0, 3 \\ Z / 2 Z, & p = 1, 4 \\ 0, & p = 2, p \geq 5 \end{matrix}$