# 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:

• $$\Bbb R^2\times \Bbb S^2$$: 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?

• I could be wrong, but I think your descriptions of those two spaces are incorrect. – mathworker21 Aug 2 '19 at 9:31
• Just prove that the function going from $\Bbb R^2\times \Bbb S^2$ to $\Bbb S^2\times \Bbb R^2$ and defined as $(x,y) \mapsto (y,x)$ is a homeomorphism. – nicomezi Aug 2 '19 at 9:33
• @nicomezi, I know that. I want to imagine that why this two description are as same?Not Algebraically. – C.F.G Aug 2 '19 at 9:36
• You can give a lot of sense to "to imagine", I guess I have to ask you to be more precise. Do you want a geometrical interpretation ? – nicomezi Aug 2 '19 at 9:40
• Well it's a matter of referential : take your second picture and move around the sphere; you get a bunch of planes. Now in every movement around the sphere you mane, pretend the planes you get are alo the same : the sphere has to be moving - you get the first description of one plane woth a bunch of spheres – Maxime Ramzi Aug 2 '19 at 9:46

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 $$[0,1]$$.

• first, thanks for your answer. but, I don't see theme as isolated bags. According to your explanation, How you justify this paradox that in first case we have 1 plan with many spheres and in second case 1 sphere with many plans? – C.F.G Aug 3 '19 at 4:09

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.

• Very nicely explained. Wish I could give more than one up-vote. – JonathanZ supports MonicaC Aug 3 '19 at 4:28

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 $$2$$.