# Homology groups and fundamental group need not determine homotopy groups

## Statement

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

• $\pi_1(M_1) \cong \pi_1(M_2)$, i.e., $M_1$ and $M_2$ have isomorphic Fundamental group (?)s.
• $H_k(M_1) \cong H_k(M_2)$ for each $k \ge 0$, i.e., $M_1$ and $M_2$ have isomorphic Homology group (?)s in each homology.
• There exists some $k \ge 2$ such that the Homotopy group (?)s $\pi_k(M_1)$ and $\pi_k(M_2)$ are not isomorphic.

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.