Krdytkyu

Show that $mathbb{R}^nsetminus {0}$ is simply connected for $ngeq 3$.

To my knowledge I have to show two things:

1. $mathbb{R}^nsetminus {0}$ is path connected for $ngeq 3$.




2. Every closed curve in $mathbb{R}^nsetminus {0}, ngeq 3$ is null-homotop.

The problem is that I do not know exactly how to show that.

The path connectedness is easy. Given $x,, y in mathbb{R}^nsetminus{0}$, the straight line segment $t mapsto (1-t)cdot x + tcdot y$ connects $x$ and $y$ in $mathbb{R}^nsetminus{0}$ unless $y$ is a negative multiple of $x$. If $y = ccdot x$ for a $c < 0$, then you can compose a path of two straight line segments, one going a little away from $x$ to $x + varepsiloncdot e_i$ for a small enough $varepsilon > 0$ and $i$ such that $x$ is not a multiple of $e_i$.

It could also be easily seen from the homeomorphism

$$Fcolon S^{n-1} times (0,,infty) to mathbb{R}^nsetminus{0};quad (xi,,r) mapsto rcdot xi.$$

(A proof that $F$ is indeed a homeomorphism may be required, or that may be considered obvious, depending on what was treated previously.)

Given a closed path $gamma colon t mapsto (xi(t),, r(t))$ in $S^{n-1}times (0,,infty)$, the map $H(s,t) = (xi(t),, r(t)^{1-s})$ provides a homotopy to a closed path in $S^{n-1}$.

If you already know that $S^d$ is simply connected for $d geqslant 2$, you are done now.

Otherwise, to prove that fact, consider a closed path $gammacolon [0,,1] to S^{n-1}$. If $gamma([0,,1]) neq S^{n-1}$, assume without loss of generality that the north pole $N = (0,,ldots,,0,,1)$ is not on the trace of $gamma$. Stereographic projection from the north pole gives a homeomorphism $S^{n-1}setminus {N} to R^{n-1}$ (proof may be required), and that shows that $gamma$ is null-homotopic in $S^{n-1}setminus {N}$, hence a fortiori in $S^{n-1}$.

If $gamma$ is surjective, you can partition it into parts that each omit one of two opposing caps of the sphere.

Consider for example the two caps $T = {x in S^{n-1}colon x_n geqslant frac23}$, $B = {x in S^{n-1}colon x_n leqslant -frac23}$ and the belt $E = {xin S^{n-1}colon lvert x_nrvert leqslant frac13}$.

By uniform continuity of $gamma$, there is a $varepsilon > 0$ such that $gamma(t) in B land lvert s-trvert < varepsilon Rightarrow gamma(s) notin E cup T$, and similar for all other combinations of $B,,E,,T$.

Without loss of generality, suppose that $gamma$ starts (and ends) in the south pole.

Set $t_0 = 0$. While $t_k < 1$, find the next partition point $t_{k+1}$ in the following way:

• If $gamma(t_k) in B$, let $t_{k+1} = minbigl({s > t_k colon gamma(s) in E} cup {1}bigr)$.


• If $gamma(t_k) in E$, let $t_{k+1} = min {s > t_k colon gamma(s) in Tcup B}$.


• If $gamma(t_k) in T$, let $t_{k+1} = min {s > t_k colon gamma(s) in E}$.

By the uniform continuity mentioned above, $t_m = 1$ for an $m leqslant frac{1}{varepsilon}$.

Let $gamma_k$ be the restriction of $gamma$ to $[t_k,,t_{k+1}]$.

If $gamma(t_{k}) in T$, then the composition $gamma_{k-1}gamma_k$ is a path connecting two points $a,, b in E$ in $S^{n-1}setminus B$. The latter is homeomorphic to an open ball, so $gamma_{k-1}gamma_k$ is homotopic to a path connecting $a$ and $b$ in $E$.

Replacing all the $gamma_{k-1}gamma_k$ with homotopic paths in $E$, you obtain a path homotopic to $gamma$ whose trace omits $T$, hence by the above, it, and therefore also $gamma$ is null-homotopic in $S^{n-1}$.

The easy part of the Seifert-van Kampen theorem and which does not require an algebraic context (e.g. groups or groupoids) is that if $X$ is the union of a family $mathcal U$ of path connected and simply connected open sets $U_i$ such that any pairwise intersection $U_i cap U_j$ is path connected, and all containing a "base point" $a in X$, then $X$ is simply connected. The key part of the proof is conveyed by the following diagram

in which $alpha$ is a loop in $X$ at the base point $a$ subdivided so that each part $alpha_i$ lies in a set say $U_i$ of $mathcal U$. The start point $x_i$ of $alpha_i$ shown as $circ$ need not be at the base point. However $x_i$ lies in $U_{i-1} cap U_i$ which is path connected; so there is a path $gamma_i$ in $ U_{i-1} cap U_i $ joining $x_i$ to the base point, and we set $gamma_i$ to be constant for $i=0,n$. Now each path $zeta_i= gamma_i^{-1}alpha_igamma_{i+1} $lies in $U_i$ which is simply connected, and so is homotopic in $U_i$ rel end points to a constant path $beta_i$. Hence $alpha$ is null homotopic, using the "horizontal" composition of these homotopies.

Note that even for this result we need a " horizontal composition of squares", and such an intuition is the start of a $2$-dimensional theory.

This type of argument was generalised by J.F. Adams (in unpublished lecture notes) to prove any map $S^r to S^n$ is null homotopic for $r <n$, see 7.6.1 of Topology and Groupoids.

You can sledgehammer proof your problem by induction on the number of points that you remove and the Seifert-van Kampen theorem. Say $x_1,ldots,x_m$ are the points that you remove. We will induct on $m$. When $m = 1$ this is easy because $Bbb{R}^n$ minus a point is homotopy equivalent to $S^{n-1}$ that is simply connected for $ngeq 3$. For the general case of $x_1,ldots,x_m$ points removed, I claim we can find a hyperplane $H$ that separates the first $x_1,ldots,x_l$ points and last $x_{l+1},ldots,x_m$ points. Let $U$ be the open cube minus contains the first $l$ points and $V$ the open cube minus the rest of the points. Then $Bbb{R}^n - {x_1,ldots,x_m}) = U cup V$ and then Seifer-van Kampen says that

$$pi_1(Bbb{R}^n - {x_1,ldots,x_m}) cong pi_1(U)ast pi_1(V)/operatorname{some subgroup}.$$

We don't care about the subgroup at the moment because by induction $pi_1(U) = pi_1(V) = 0$ and so $pi_1(Bbb{R}^n - {x_1,ldots,x_m}) = 0$. Done.

Edit: For those not convinced $S^{n-1}$ is simply connected for $n geq 3$ here's a proof: Let $U$ be the upper hemisphere and $V$ the lower hemisphere, choose epsilon fattenings of these if you wish. For $n geq 3$, $U cap V$ is homotopy equivalent to the equator $S^{n-2}$ that is path - connected for $n geq 3$. $U,V$ are homotopy equivalent to $D^{n-1}$ which is simply connected for $n geq 3$. The hypotheses of the Seifert-van Kampen theorem are now satisfied and applying it to $U$ and $V$ shows $pi_1(S^{n-1}) = 0$ for $n geq 3$.