Homotopy groups need not determine homology groups

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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(\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}