There is the section of an argument in a topology textbook:
- $ S^2 \vee S^2$ is the 3-skeleton of $S^2 \times S^2$ (as in CW-complex)
- $ \Sigma (S^2 \times S^2) \simeq \Sigma (S^2 \vee S^2) \vee S^5$
- homotopy retraction $\Sigma (S^2 \times S^2) \to \Sigma (S^2 \vee S^2) $
Here $\Sigma$ is the suspension and we have that $\Sigma X = S^1 \wedge X$ : $$ S^1 \wedge (S^2 \times S^2) \simeq \big( S^1 \wedge ( S^2 \vee S^2 ) \big) \vee S^5$$ I guess the the question is about the relation between the cartesian product $\times$ and the wedge sum $\vee$.
These spaces are difficult to visualize since the dimension count is $2 + 2 + 1 = 5$ dimensions.
