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?

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    $\begingroup$ I could be wrong, but I think your descriptions of those two spaces are incorrect. $\endgroup$ – mathworker21 Aug 2 '19 at 9:31
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    $\begingroup$ 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. $\endgroup$ – nicomezi Aug 2 '19 at 9:33
  • $\begingroup$ @nicomezi, I know that. I want to imagine that why this two description are as same?Not Algebraically. $\endgroup$ – C.F.G Aug 2 '19 at 9:36
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    $\begingroup$ 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 ? $\endgroup$ – nicomezi Aug 2 '19 at 9:40
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    $\begingroup$ 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 $\endgroup$ – 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.


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 $[0,1]$.

  • $\begingroup$ 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? $\endgroup$ – 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.

  • $\begingroup$ Very nicely explained. Wish I could give more than one up-vote. $\endgroup$ – 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 $2$.


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