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 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

1. 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 $M_{1}=\mathbb {P} ^{3}(\mathbb {R} )\times S^{2}$ and $M_{2}=S^{3}\times \mathbb {P} ^{2}(\mathbb {R} )$. 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 $\pi _{k},k\geq 1$. 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:

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

and:

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