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Show that $\mathbb{R}^{2} \setminus \{ (0, 0)\}$ is not simply connected. Does it also hold for $\mathbb{C}\setminus \{0\}$?

Suppose, for the sake of contradiction, it was simply connected.

Consider the functions \begin{align*} &\varphi \colon [0, \pi ] \to \mathbb{R}^{2}, &&\psi \colon [0, \pi ] \to \mathbb{R}^{2} \\ &\varphi( x) = \left(-\cos\left(x\right), \sin\left(x\right) \right), &&\psi ( x) = ( -\cos\left( x\right) , -\sin\left( x\right) ) \end{align*} which parametrise the upper and lower half of the unit circle respectively. Now there must exist some homotopy $H\colon [0, \pi ]\times [0, 1]\to \mathbb{R}^{2} $ which transforms $\varphi$ into $\psi$. Since $H$ is continuous by definition, in particular the function \begin{align*} g\colon [0, 1] \to \mathbb{R}^{2}, \quad g( t) = H( \pi /2, t) \end{align*} is continuous (so with this function one basically only considers what happens on the straight line joining $(0, 1)$ and $(0, -1)$, and the idea is that our homotopy must at some point pass through $(0, 0)$) . $g(0) = (0, 1)$ and $g( 1) =( 0, -1) $. Even though it is intuitively clear that the first component of $g( x)$ will always be $0$, I don't know how to show this. After this I could then apply the intermediate value theorem.

For the second question I would say yes, since the proof is probably analogous to the above.

Could someone help me to finish my proof?

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  • $\begingroup$ The proof of this fact is not trivial. There are many different ways to do that, e.g. Brouwer fixed point theorem or topological degree. But all of them require more sophisticated tools. $\endgroup$
    – freakish
    Mar 28, 2022 at 20:19
  • $\begingroup$ Imho, the easiest way to do it is using topological degree or the homotopy lifting property. Both still use some non elementary tools. $\endgroup$ Mar 28, 2022 at 20:20
  • $\begingroup$ @SolubleFish (and freakish) geometrically it looks so intuitive, the more interesting it is that the proof is rather involved. Unfortunately I just started learning about these concepts and am not familiar with the mentioned tools. $\endgroup$
    – Richard
    Mar 28, 2022 at 20:27
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    $\begingroup$ @Richard yes, it looks intuitive, but I doubt that you will be able to solve this with elementary approach. This is one of those intuitive and easy to understand problems that does not have an easy proof. $\endgroup$
    – freakish
    Mar 28, 2022 at 20:56
  • $\begingroup$ See also math.stackexchange.com/q/4394358/861687. $\endgroup$
    – user264745
    Apr 27, 2022 at 13:38

3 Answers 3

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For the second part, since $\mathbb C^*$ and $\mathbb R^2 \backslash \{(0,0)\}$ are homeomorphic, they have the same topological properties. In particular, one is simply connected iff the other one is. In the following, I use $\mathbb C^*$, as the complex notation is more convenient.

As we discussed in the comments, the full proof is non elementary, requiring some tools from algebraic topology. However, if we restrict ourselves to paths and homotopies which are $C^1$, then there is an elementary way to construct lifts and winding numbers.

$C^1$ path lifting property

Let $\gamma:[0,1]\to \mathbb C^*$ be a $C^1$ path. Let $\theta_0$ be such that $\gamma(0) = |\gamma(0)|e^{i\theta_0}$. Then, there are unique $C^1$ maps $\rho : [0,1]\to \mathbb R_{>0}$ and $\theta : [0,1]\to \mathbb R$ such that $\theta(0) = \theta_0$ and : $$\forall t\in[0,1], \gamma(t) = \rho(t)e^{i\theta(t)}$$

Proof : The existence and unicity of $\rho$ is trivial as $\rho = |\gamma|$. Then, we want to find $\theta$ such that $e^{i\theta} = \frac{\gamma}{|\gamma|}$.

If there is such a theta, then, taking derivatives, we get : $$i\theta'e^{i\theta} = \left(\frac{\gamma}{|\gamma|}\right)'$$ Therefore : $$\theta' = -i\frac{\gamma^*}{|\gamma|}\left(\frac{\gamma}{|\gamma|}\right)'$$ and $$\forall t\in[0,1], \theta(t) = \theta_0 -i \int_0^t \frac{\gamma^*}{|\gamma|}\left(\frac{\gamma}{|\gamma|}\right)' \text dt'$$ It is easy to check that this formula defines a $C^1$ function $\theta$ which solves our problem.

Winding number (or degree) of a closed path

Now, let $\gamma :[0,1]\to \mathbb C^*$ be a $C^1$ closed path : $\gamma(0) = \gamma(1)$. Let $\rho,\theta$ be maps as above. Then, because the path is closed, we have $\rho(0) = \rho(1)$ and $e^{i\theta(0)} = e^{i\theta(1)}$. This implies that there is an integer, which we write $\deg(\gamma)$ such that : $$\deg(\gamma) = \frac{1}{2i\pi}(\theta(1) - \theta(0) ) = -\frac{1}{2\pi} \int_0^1 \frac{\gamma^*}{|\gamma|}\left(\frac{\gamma}{|\gamma|}\right)' \text dt$$

Homotopy invariance of the winding number

If $H:[0,1]\times [0,1]\to \mathbb C^*$ is $C^1$ and $\gamma_s(t) = H(s,t)$, then the above formula makes it clear that $s\mapsto \deg(\gamma_s)$ is an integer value continuous function. Therefore it is constant : the winding number is $C^1$-homotopy invariant.

Conclusion : $\mathbb C^*$ is not $C^1$-simply connected

The constant path equal to $1$ has degree $1$, while the path $\gamma(t) =e^{2i\pi t}$ has degree $1$. (In general, $t\mapsto e^{2i\pi nt}$ has degree $n\in\mathbb Z$).

Because the degree is homotopy invariant, those paths belong to different $C^1$-homotopy classes.

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I was thinking that a topological space is simply connected if and only if is path-connected and the fundamental group of at each point is trivial. In this case via stereographic projection $\mathbb{R} \backslash \{(0,0)\}$ is homeomorphic to $\mathbb{S}^2 \backslash \{N,P\}$ (north pole and south pole). Homeomorphic spaces has same fundamental group so since the sphere without two point can be retracted to a circle (I suggest you to take a look also at this), we got that $\pi_1(\mathbb{R} \backslash \{(0,0)\}) = \mathbb{Z}$ and so it's not simply connected. For the second part we can reiterate the same reasoning with this homeomorphism and we obtain that $\mathbb{C}\backslash\{0\}$ is not simply connected.

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    $\begingroup$ thanks for the answer, I haven't learned about most of the concepts you mention so I'm not yet able to understand the answer. As soon as I have read up on these things I will be able to evaluate it. $\endgroup$
    – Richard
    Mar 28, 2022 at 20:32
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The punctured plane is homeomorphic to the cylinder of infinite height $\mathbb{S}^1\times \mathbb{R}$. Though such a cylinder is path connected, it must have $\mathbb{Z}$ as its first fundamental group.

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