show that if a simply connected space has $H_2(X) = \mathbb{Z} \oplus \mathbb{Z}$ then it is homotopic to $S^1 \bigwedge S^2$

Question

The problem is as follows

show that if a simply connected CW-complex has $H_2(X) = \mathbb{Z} \oplus \mathbb{Z}$ and $H_i(X) = 0$ for all $i \neq 2$ then it is homotopy equivalent to $S^1 \bigwedge S^2$

I have no idea where to begin. I tried to start out with $S^2 \bigwedge S^1$ but it didn't work.I tried to explicitly xonstruxt such a CW complex but failed.

Any help or hint is appreciated

Answer 1

This is not correct. First of all the smash product $S^1\wedge S^2$ is homeomorphic to $S^3$. And therefore $H_2(S^3)=0$.

Secondly, it is not a simple matter of replacing the smash product $\wedge$ with the wedge sum $\vee$. Because $H_2(S^1\vee S^2)=\mathbb{Z}$. And besides $S^1\vee S^2$ is not simply connected.

So the correct statement is: it is homotopy equivalent to $S^2\vee S^2$.

And the reason for that is that we are dealing with a Moore space of type $M(\mathbb{Z}\oplus \mathbb{Z},2)$. And Moore spaces are unique (up to homotopy equivalence) among CW complexes. This result can be found for example in Allen Hatcher's "Algebraic Topology", exercise 4.34.