Identification of product of two spheres from wedge of two spheres $\mathbf {The \ Problem \ is}:$ Show that $S^2×S^2$ can be obtained by attaching a $4-$cell to $W=S^2\vee S^2.$
$\mathbf {My \ approach}:$ Actually, I am a beginner in learning CW complexes .
I was thinking to draw a pushout such that $S^3$ is attached to $W$ at the wedge point $*.$
But, I can't imagine properly , somehow; I am trying to glue $S^3$ to previous skeleton .
A small hint with a motivation in solving these type of problems is warmly appreciated, thanks in advance .
 A: In general, there is a way to give a product of two $CW$-complexes a $CW$-structure as seen in Hacther's appendix. The cells of the product $CW$-structure are the products of the cells and the characteristic maps are the product of the characteristic maps. This gives $S^2\times S^2$ a CW-structure with one $0$-cell, two $2$-cells, and one $4$-cell. The $2$-skeleton then becomes $S^2\vee S^2$ and the characteristic map for the $4$-cell is given by $$D^4\cong D^2\times D^2 \longrightarrow S^2\times S^2$$
sending the first disk to the first $2$-sphere with the boundary to the basepoint and the second disk to the second $2$-sphere with the boundary to the basepoint. The attaching map will then be the restriction of this map to $\partial D^2\times D^2 \cup D^2\times \partial D^2=\partial D^4$.
Note that one has to be a bit careful doing this for a general product of $CW$-complexes. If $A$ and $B$ are $CW$-complexes, then the product $CW$-structure on the set $A\times B$ will topologize it with the Kelly product topology and not the standard product topology. That is, the Kellyfication of the standard product topology. You can read about compactly generated spaces if you are interrested. However, those two topologies agree if the $CW$-structures on $A$ and $B$ consists of countably many cells.
