$\mathbb{R}^n - B[0,r]$ is simply connected if $n>2$

Question

Question:$\mathbb{R}^n - B[0,r]$ is simply connected $\iff$ $n>2$.

I have to prove or disprove.

I know prove that for $n \in \{1,2\}$, $\mathbb{R}^n - B[0,r]$ is not simly connected. So I want prove that for $n>2$ it is simply connected.

I know that $\mathbb{R}^n - B[0,r]$ is homeomorphic to $S^n \times \mathbb{R}$. I wanna prove that $S^n$ and $\mathbb{R}$ are simply connected and conlcude, but I don't know how do it by definition. Is it true?

Definition: $X$ is simply connected if any map $f\colon S^1 \to X$ continuous has extension $\overline{f}\colon D^2 \to X$ also continuous.

Answer 1

Since $\mathbb{R}$ is convex, $\mathbb{R}$ is simply connected.

Let $n\geqslant 2$, using your definition, you can show that $\mathbb{S}^n$ is simply connected if and only if each loop on $\mathbb{S}^n$ is homotopic to a constant path. Let $\gamma :[0,1]\to\mathbb{S}^n$ be a continuous function such that $\gamma(0)=\gamma(1)$ and let distinguish the following cases:

• If $\gamma$ avoid a point on $\mathbb{S}^n$, let say $p$, using a stereographic projection of pole $p$, one has: $$\gamma([0,1])\subseteq\mathbb{S}^n\setminus\{p\}\cong\mathbb{R}^{n}.$$ $\mathbb{R}^{n}$ is convex and hence is simply connected. Therefore, $\mathbb{S}^n\setminus\{p\}$ is simply connected and $\gamma$ is homotopic to a constant path.

• If $\gamma$ does not avoid any point, let us construct a loop $\gamma'$ homotopic to $\gamma$ that avoid a point on $\mathbb{S}^n$, using the first point, $\gamma'$ will be homotopic to a constant path, which concludes. In fact, we are going to construct $\gamma'$ such that $\gamma'([0,1])$ has empty interior. By uniform continuity, there exists $N$ such that: $$\forall t,t',|t-t'|\leqslant\frac{1}{N}\Rightarrow|\gamma(t)-\gamma(t')|<\frac{1}{2}.$$ Let define $\gamma''$ by: $$\forall j,{\gamma''}_{\mid\left[\frac{j}{N},\frac{j+1}{N}\right]}=\left[\gamma\left(\frac{j}{N}\right),\gamma\left(\frac{j+1}{N}\right)\right].$$ $\gamma''$ is the loop get by joining $\displaystyle\gamma\left(\frac{j}{N}\right)$ and $\displaystyle\gamma\left(\frac{j+1}{N}\right)$ with a straight line in $\mathbb{R}^{n+1}$. Now let us radially project $\gamma''$ onto $\mathbb{S}^n$ to get a loop $\gamma'$ of $\mathbb{S}^n$: $$\gamma':=\frac{\gamma''}{\|\gamma''\|}.$$ $\gamma'$ is a loop composed with a finite number of arcs of circles, hence $\gamma'([0,1])$ has empty interior. Moreover, one can see that $\gamma'$ is homotopic to $\gamma$.

Answer 2

$\mathbb{R}$ is simply connected because it is contractible.

$S^n$ is the union of two open sets, $U = S^n$ minus one point, and $V = S^n$ minus a different point. Both are simply connected, since both are homeomorphic to $\mathbb{R}^n$ (One can see this by using a stereographic projection.), and $\mathbb{R}^n$ is contractible. Also, $U \cap V$ is path-connected, so we can use the Seifert-van Kampen theorem, which tells us that the fundamental group of $U \cup V$ is the amalgamated product of two trivial groups, which must be the trivial group. So $U \cup V$ is simply connected.