Are the $\Bbb S^2\times \Bbb R^2$ and $\Bbb R^2\times \Bbb S^2$ homeomorphic? Are the $\Bbb S^2\times \Bbb R^2$ and $\Bbb R^2\times \Bbb S^2$ homeomorphic? I know that the answer is certainly yes but what is confused me is the following:


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*$\Bbb R^2\times \Bbb S^2$: Consider a plane and attach a $2$-sphere to each point of it,

*$\Bbb S^2\times \Bbb R^2$: Consider a $2$-sphere and attach  a plan $\Bbb R^2$ to each point of it.


How to justify geometrically that this two are exactly a copy of each other?
Update: How to justify this paradox: In first case we have 1 plan with many spheres and in second case 1 sphere with many plans?

Here's another way to think about the Cartesian product that I like. It's not as geometric as you're maybe looking for, but I find it's more useful for questions like these.
The product $$\prod_{j\in{J}}X_j$$
can be defined as the set of all functions $$f:J \rightarrow \bigcup_{j\in{J}}X_j $$
such that $f(j)\in x_j$ for all $j\in J$. This works even for uncountable $J$.
Thus the Cartesian product $\mathbb{R}^2\times S^2$ can be interpreted as the set of all functions from $\{1,2\}$ where ${1}$ is mapped onto the Euclidean plane and $2$ is mapped onto a 2-sphere. 'Geometrically', I like to draw the numbers $1$ and $2$, then draw a few arrows from $1$ to the plane, and $2$ to the circle. The cartesian product is this collection of arrows.
The way I 'see' this is a homeomorphism is that there's no distinguishing features between $1$ and $2$ in this context; I can simply change all the arrows leading from $1$ to lead from $2$.
