Homology groups and fundamental group need not determine homotopy groups

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Statement

It is possible to have two path-connected spaces M_1 and M_2 such that:

In fact, we can choose both M_1 and M_2 to be Simply connected space (?)s, i.e., they both have trivial fundamental group.

Facts used

  1. Fundamental group of wedge sum relative to basepoints with neighborhoods that deformation retract to them is free product of fundamental groups

Proof

Further information: wedge sum of 2-sphere and 4-sphere, complex projective plane

Let M_1 = S^2 \vee S^4 be the wedge sum of the 2-sphere S^2 and the 4-sphere S^4 and M_2 = \mathbb{P}^2(\mathbb{C}). They both have the same homology groups:

H_p(M_1;\mathbb{Z}) = H_p(M_2;\mathbb{Z}) = \left\lbrace\begin{array}{rl} \mathbb{Z}, & \qquad p = 0,2,4 \\ 0, & \qquad \text{otherwise}\\\end{array}\right.

Also, they are both simply connected spaces (M_1 is simply connected by Fact (1), for M_2, see homology of complex projective space).

However, they do not have the same isomorphism class of \pi_4: \pi_4(M_1) is nontrivial whereas \pi_4(M_2) is trivial. For this, note that:

  • There is a retraction S^2 \vee S^4 \to S^4 that sends all points in the S^4 piece to themselves and all points in the S^2 piece to the point of wedging. This retraction induces a retraction on each homotopy group, so \pi_4(S^2 \vee S^4) \to \pi_4(S^4) is a retraction. In particular, since \pi_4(S^4) \cong \mathbb{Z} is nontrivial, \pi_4(S^2 \vee S^4) is also nontrivial.
  • \pi_4(\mathbb{P}^2(\mathbb{C})) is the trivial group, based on the homotopy of complex projective space.