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Show that $\mathbb{R}^n\setminus \{0\}$ is simply connected for $n\geq 3$.

To my knowledge I have to show two things:

  1. $\mathbb{R}^n\setminus \{0\}$ is path connected for $n\geq 3$.

  2. Every closed curve in $\mathbb{R}^n\setminus \{0\}, n\geq 3$ is null-homotop.

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

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    $\begingroup$ Path-connectedness is easy. For the second part, you can project each closed curve onto the unit sphere (it's not hard to find a homotopy if you explicitly want one). Then if the resulting curve is not surjective, it's easy because $S^{n-1} \setminus \{p\}$ is homeomorphic to $\mathbb{R}^{n-1}$. If the resulting curve is surjective, you can partition it into parts each leaving out one of two opposed caps, and use that partition to find a homotopy with a non-surjective closed path. $\endgroup$ – Daniel Fischer Jul 6 '13 at 17:20
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    $\begingroup$ Can you think of a way to deformation retract to a more familiar space, maybe to one with $\pi_1=0$? $\endgroup$ – Derek Allums Jul 6 '13 at 17:21
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    $\begingroup$ @DanielFischer I suggest you write it as an answer and improve it afterwards if you feel it's necessary. $\endgroup$ – Git Gud Jul 6 '13 at 17:22
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    $\begingroup$ How about this for 1? Your point $p$ lives on a sphere of radius $r>0$ about the origin. This sphere intersects the ray starting at the origin and going through $q$. Take any path along the sphere from $p$ to the intersection point, and then continue along the straightline to $q$. $\endgroup$ – Matt Jul 6 '13 at 18:17
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    $\begingroup$ Think about it in spherically, given $x,y$ we know $|x|,|y|>0$ so project $y$ onto the sphere of radius $|x|$ and then move on this sphere to $x$. $\endgroup$ – James Jul 6 '13 at 18:19
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The path connectedness is easy. Given $x,\, y \in \mathbb{R}^n\setminus\{0\}$, the straight line segment $t \mapsto (1-t)\cdot x + t\cdot y$ connects $x$ and $y$ in $\mathbb{R}^n\setminus\{0\}$ unless $y$ is a negative multiple of $x$. If $y = c\cdot x$ for a $c < 0$, then you can compose a path of two straight line segments, one going a little away from $x$ to $x + \varepsilon\cdot e_i$ 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

$$F\colon S^{n-1} \times (0,\,\infty) \to \mathbb{R}^n\setminus\{0\};\quad (\xi,\,r) \mapsto r\cdot \xi.$$

(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 $\gamma\colon [0,\,1] \to S^{n-1}$. 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 $E = \{x\in S^{n-1}\colon \lvert x_n\rvert \leqslant \frac13\}$.

By uniform continuity of $\gamma$, there is a $\varepsilon > 0$ such that $\gamma(t) \in B \land \lvert s-t\rvert < \varepsilon \Rightarrow \gamma(s) \notin E \cup T$, 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 $t_{k+1} = \min\bigl(\{s > t_k \colon \gamma(s) \in E\} \cup \{1\}\bigr)$.
  • If $\gamma(t_k) \in E$, let $t_{k+1} = \min \{s > t_k \colon \gamma(s) \in T\cup B\}$.
  • 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 $S^{n-1}$.

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  • $\begingroup$ I do not understand what the homeomorphism F does... $\endgroup$ – math12 Jul 6 '13 at 19:07
  • $\begingroup$ In what sense do you not understand? As a mapping, it maps a direction (given by a point on the sphere) and a radius to the corresponding point in the space. Polar coordinates, pretty much. If you wonder what role it plays here, not a big one, but it exposes $\mathbb{R}^n\setminus\{0\}$ as the product of two path-connected spaces, so shows the path-connectedness, and it reduces the homotopy question to homotopies on a sphere. $\endgroup$ – Daniel Fischer Jul 6 '13 at 19:17
  • $\begingroup$ Ah, you show that every point in $\mathbb{R}^n\setminus\left\{0\right\}$ can be identified with a point on the unit sphere and so two different points can be connected by walking on the unit sphere? $\endgroup$ – math12 Jul 6 '13 at 19:26
  • $\begingroup$ Not exactly, $F^{-1}(x) = (x/\lVert x\rVert,\, \lVert x\rVert)$, so a point in $\mathbb{R}^{n-1}\setminus\{0\}$ can be identified with the intersection of the ray (emanating from $0$) through this point, and the distance of the point to $0$. You can then connect any two points by 1) going to the sphere along the ray, 2) walking on the sphere, 3) going from the sphere to the endpoint on the ray. $\endgroup$ – Daniel Fischer Jul 6 '13 at 19:30
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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

connected

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_i\gamma_{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.

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  • $\begingroup$ @Daniel: Thanks for the comment. I'm busy preparing travel from N. Wales to Strasbourg tomorrow for the Loday Memorial Conference, so I hope to have time to think on it on the train. I know the next stage of deforming a subdivided homotopy does need the 3-fold intersection criterion, but thought we could get away with this argument. $\endgroup$ – Ronnie Brown Sep 2 '13 at 20:03
  • $\begingroup$ @Daniel: The point is that the intersection of the sets has to contain the base point! Thanks again. $\endgroup$ – Ronnie Brown Sep 3 '13 at 5:43
  • $\begingroup$ Ah, of course. That's a nice proof then. $\endgroup$ – Daniel Fischer Sep 3 '13 at 8:39
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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 $n\geq 3$. 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 $n \geq 3$.

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    $\begingroup$ Notice that the proof of the van Kampen theorem itself involves (a generalization of) the exact same argument given by Daniel in his answer (In fact, that is the easy part of the proof of vK's theorem!). Also, notice that (up to equivalence) the question of the OP is precisely «how to prove $S^n$ is simply connected?» which is, exactly, the one you waved away with an «it is easy» :-) $\endgroup$ – Mariano Suárez-Álvarez Jul 6 '13 at 19:32
  • $\begingroup$ @MarianoSuárez-Alvarez I added a proof that $S^{n-1}$ is simply connected for $n\geq 3$. $\endgroup$ – user38268 Jul 6 '13 at 19:47
  • $\begingroup$ Ben, in that special case the proof of the vK theorem is exactly that given by Daniel. $\endgroup$ – Mariano Suárez-Álvarez Jul 6 '13 at 19:51
  • $\begingroup$ @MarianoSuárez-Alvarez I don't remember the proof of van Kampen very much and I haven't looked at algebraic topology for a while, so that was my shot above :) $\endgroup$ – user38268 Jul 6 '13 at 19:56
  • $\begingroup$ May I ask why there is such a hyperplane H? Thanks. $\endgroup$ – awllower Apr 21 '14 at 14:03

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