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?

As you said, the spaces are homeomorphic. You imagine a product $X \times Y$ in two different ways:


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*A copy of $Y$ attached at each point of $X$.

*A copy of $X$ attached at each point of $Y$.
The copies $\{x\} \times Y$ are pairwise disjoint, and I guess you imagine them as "isolated bags hanging on string". However, they are not isolated, for each $y \in Y$ the collection of points $(x,y)$ with $x \in X$ forms again a string going through the bags. Thus you see that you do not have a string with isolated bags, but a web which is on a par with respect to vertical and horizontal threads.
Edited:
For any product $X \times Y$ you have two projections $p_X : X \times Y \to X, p_X(x,y) = x$, and$p_Y : X \times Y \to Y, p_Y(x,y) = y$. This gives you two directions to look at $X \times Y$:


*

*Look from $X$ at $X \times Y$. For each $x \in X$ you see the "fiber" $p_X^{-1}(x) = \{x\} \times Y$, in the case of $\mathbb R^2 \times S^2$ a sphere "attached" at each point of the plane.

*Look from $Y$ at $X \times Y$. For each $y \in Y$ you see the "fiber" $p_Y^{-1}(y) = X \times \{y\}$, in the case of $\mathbb R^2 \times S^2$ a plane "attached" at each point of the sphere.
There is no paradox. It is just a matter of perspective. Perhaps a simpler example will illustrate this. Consider the set $P = [0,1] \times \mathbb \{0,1\}$ which is a subset of the plane $\mathbb R^2$. Looking at $P$ from the left (i.e. in the direction of the $x$-axis) you see two intervals. each attached at the points $0,1$. Looking at $P$ from below (i.e. in the direction of the $y$-axis) you see a collection of two-points sets, each attached at a point of $[0,1]$. 
