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 $$\pi_k(M_1) \cong \pi_k(M_2)$$ between the fact about::homotopy groups.
 * There exists a positive integer $$k$$ such that the fact about::homology groups $$H_k(M_1)$$ and $$H_k(M_2)$$ are not isomorphic.

In particular, because fact about::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

 * 1) uses::Covering map induces isomorphisms on higher homotopy groups

Proof
Consider the spaces $$M_1 = \mathbb{P}^3(\R) \times S^2$$ and $$M_2 = S^3 \times \mathbb{P}^2(\R)$$. 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 $$\pi_1(M_1) \cong \pi_1(M_2) \cong \mathbb{Z}/2\mathbb{Z}$$.
 * 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 $$\pi_k, k \ge 1$$. Thus, we get $$\pi_k(S^3 \times S^2) \cong \pi_k(M_1)$$ and $$\pi_k(S^3 \times S^2) \to \pi_k(M_2)$$ and we thus get $$\pi_k(M_1) \cong \pi_k(M_2)$$.

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 $$H_5(M_1) \cong \mathbb{Z}$$, whereas $$M_2$$ is non-orientable, so $$H_5(M_2) = 0$$.

The full homology descriptions are below:

$$H_p(\R\mathbb{P}^3 \times S^2; \mathbb{Z}) = \lbrace \begin{array}{rl} \mathbb{Z}, & p = 0,2,5\\ \mathbb{Z}/2\mathbb{Z}, & p = 1 \\ \mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}, & p = 3\\ 0, & p = 4, p \ge 6 \\\end{array}$$

and:

$$H_p(S^3 \times \R\mathbb{P}^2; \mathbb{Z}) = \lbrace\begin{array}{rl} \mathbb{Z}, & p = 0,3 \\ \mathbb{Z}/2\mathbb{Z}, & p = 1,4 \\ 0, & p = 2, p \ge 5 \\\end{array}$$