# Is $\Bbb R^2\times\Bbb S^2$ simply connected?

The following answer states that the $$\Bbb R^2\times\Bbb S^2$$ space is simply connected:

Topology of Black Holes

However, the following post confirms that $$\Bbb R^2\times\Bbb S^2$$ is homeomorphic to $$\Bbb R^4$$ with a line removed:

Is $$\Bbb R^2\times\Bbb S^2$$ homeomorphic to $$\Bbb R^4$$ with a line removed?

Simply connected means that any path between any two points can be continuously deformed into any other path between the same points without leaving the space:

Simply Connected Space

In $$\Bbb R^4$$, I can connect any two points by two paths, one on one side of the removed line and the other on the other side of the removed line. I do not see how one path can be smoothly transitioned into another without crossing the removed line. This is evidently impossible in $$\Bbb R^3$$ with a line removed, does the presence of another dimension in $$\Bbb R^4$$ makes it possible?

Equivalently, a space is simply connected if any loop can be contracted to a point without leaving the space. Consider a loop around the removed line. If I contract this loop to a point, this point would be on the removed line and thus outside the space. Again, self evident in $$\Bbb R^3$$, does the presence of an extra dimension in $$\Bbb R^4$$ allows contracting such a loop to a point outside the removed line without crossing it?

So is $$\Bbb R^4$$ with a line removed simply connected? And is the homeomorphic $$\Bbb R^2\times\Bbb S^2$$ simply connected as well? What am I missing? Thank you!

• Interesting. If you remove a point out of $\mathbb R^3$ you can still move loops around freely. It might really be that the same holds true for $\mathbb R^4$ with a line removed... – PrudiiArca Jan 25 at 16:55
• You can sort of intuitively see it like this: imagine for simplicity that your removed line and your circle are both contained in a three-dimensional subspace $\mathbb{R}^3 \times \{0\} \subset \mathbb{R}^4$. To contract the circle you can just translate it up into $\mathbb{R}^3 \times \{1\}$ where there is no line removed. – William Jan 25 at 17:16
• In fact this sort of idea is what lets you embed the Klein Bottle into $\mathbb{R}^4$ even though it's impossible to embed into $\mathbb{R}^3$, the "extra space" lets you fix some intersection problems – William Jan 25 at 17:22
• I think you mislead yourself when you used the terms "one side of the removed line" and "other side of the removed line". This seems like an instance of the classic meaning of "begging the question", where you (mistakenly, in this case) introduce the thing you're trying to prove as an assumption. – JonathanZ supports MonicaC Jan 25 at 19:33
• @JonathanZsupportsMonicaC Sorry, I've no idea what you mean. I asked what I was missing and suggested that it may be an extra degree of freedom in a higher dimension, which it was indeed, because my argument does hold in $\mathbb{R}^3\setminus \ell$. – safesphere Jan 26 at 0:33

I'll try to avoid using $$\pi_1$$ as it wasn't mentioned in the question originally, although as Paul has mentioned there is a quick proof using $$\pi_1$$.

There are a few questions embedded in this post, so I'll try to address them one at a time. First and foremost, $$\mathbb{R}^4$$ with a line excised is simply connected. We can actually prove this using an easy to visualize case. Let's reduce our dimensions by $$1$$ across the board - let's consider the case of $$\mathbb{R}^3\setminus \{p\}$$. That is, we take three-space and remove a point. I trust your visual intuition informs you that any loop closes to a point continuously.

Indeed, one can simply imagine moving the loop far away from the point into a region of space that is far from the puncture, and then closing the loop. You can make this rigorous by some arguments using parametrizations (a good exercise).

Now, in the case of $$\mathbb{R}^4\setminus \ell$$, where $$\ell\subseteq \mathbb{R}^4$$ is a line it is no loss of generality to assume that $$\ell$$ is a linear subspace, i.e. passes through the origin. It is also no loss of generality to assume that $$\ell=\{(0,0,0,s\}: s\in \mathbb{R}\}$$, i.e. one of the axes. We have a loop $$\gamma:S^1\to \mathbb{R}^4\setminus \ell$$. Now, we can consider the homotopy taking $$\mathbb{R}^4\setminus \ell$$ to the space $$\mathbb{R}^3\setminus \{0\}$$ given by $$F:\mathbb{R}^4\setminus \ell \times [0,1]\to \mathbb{R}^4\setminus \ell$$ and $$F((w,x,y,z),t)=(w,x,y,tz).$$ At $$t=1$$ this is the identity map, while at $$t=0$$ this is the projection onto $$\mathbb{R}^3\setminus \{0\}\subseteq\mathbb{R}^4\setminus\ell.$$ In particular, this homotopy slides the loop $$\gamma$$ into a loop in $$\mathbb{R}^3\setminus \{0\}$$. The loop in $$\mathbb{R}^3\setminus \{0\}$$ is then clearly contractible by what we have said above.

Of course, this was not necessary. Once we know that $$\mathbb{R}^2\times \mathbb{S}^2\cong \mathbb{R}^4\setminus \ell$$, we can conclude that both are simply connected or neither because simple connectedness is preserved under homeomorphism. To see that $$\mathbb{R}^2\times \mathbb{S}^2$$ is simply connected, just use an idea similar to the above to exhibit a homotopy $$\mathbb{R}^2\times \mathbb{S}^2\simeq \mathbb{S}^2$$.

If a space $$X$$ is simply connected, then each space $$Y$$ which is homeomorphic to $$X$$ is also simply connected. A stronger result is that if $$Y$$ is homotopy equivalent to $$X$$, then $$Y$$ is simply connected. See Proving that the fundamental groups of two spaces with same homotopy type are isomorphic .

Now $$\mathbb R^2 \times S^2$$ is homotopy equivalent to $$S^2$$. But it is well-known that $$S^2$$ is simply connected.

You can also invoke the fact that $$\pi_1(A \times B) \approx \pi_1(A) \times \pi_1(B)$$. But $$\pi_1(\mathbb R^2) = 0$$ and $$\pi_1(\mathbb S^2) = 0$$.

From the point of view of Category theory, $$π_1$$ is functorial, and has a right adjoint, the classifying space functor. As a result, it respects products. Thus $$π_1(\Bbb R^2×S^2)\congπ_1(\Bbb R^2)×π_1(S^2)\cong e×e\cong e$$.

• given that OP seems rather new to algebraic topology, it might be better to give a more down-to-earth reason for why $\pi_1$ respects products – William Jan 25 at 19:11
• @William Point taken. – Chris Custer Jan 25 at 19:18
• I get the idea of the "Distributive Property" (if it is the right term) of a functorial, but I'll have to google "functorial" along with every other word in your answer to actually understand it. Still, it is a good start and direction. Thank you! +1 – safesphere Jan 25 at 23:58
• The notion of functors originated with algebraic topology, where groups are attached to topological spaces. Imo, every mathematician should know some category theory. The work of Maclane and Eilenberg on the subject can be viewed as a continuation of work of Noether. Do investigate. – Chris Custer Jan 26 at 0:24