Are the $\Bbb S^2\times \Bbb R^2$ and $\Bbb R^2\times \Bbb S^2$ homeomorphic?

Question

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:

• $\Bbb R^2\times \Bbb S^2$: 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?

Answer 1

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

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

2. 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 $X \times Y$:

1. 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.

2. 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 $[0,1]$.

Answer 2

Your descriptions indeed do not match, but that's because neither accurately describes what the product of topological spaces really is.

Let's use $\hat\times$ to denote the operation you're describing: For a space $A$ and a pointed space $B$, we define $A \mathbin{\hat\times} B$ as the result of gluing a copy of $B$ to every point of $A$.

Let's consider $\mathbb S^1$ and $\mathbb R^1$ for simplicity.

$\mathbb R^1 \mathbin{\hat\times} \mathbb S^1$ is a line with a circle glued to every point of it. That space has a lot of loops, by contracting $\mathbb R^1$ it's homotopic to an uncountable wedge sum of circles.

$\mathbb S^1 \mathbin{\hat\times} \mathbb R^1$ is a circle with a line glued to every point of it. That space is homotopic to a single circle by contracting all the copies of $\mathbb R^1$, so it's a different space than $\mathbb R^1 \mathbin{\hat\times} \mathbb S^1$.

But neither of those two spaces is homeomorphic to $\mathbb S^1 \times \mathbb R^1$, which is just a cylinder.

A different intuition of the product of two spaces, which is more accurate, would be to imagine $A \times B$ as extruding $B$ along $A$. So $\mathbb S^1 \times \mathbb R^1$ would be taking a line and extruding it along a circle, yielding a cylinder, and $\mathbb R^1 \times \mathbb S^1$ would be taking a circle and extruding it along a line, also yielding a cylinder.

Answer 3

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 $2$.