An open set of $ \mathbb{R}^2$ without a point is not simply connected

Question

An open set of $ \mathbb{R}^2$ without a point is not simply connected

I need a rigorous proof of this because I only have the intuitive idea that a loop around the point can not be deformed into a point.

Answer 1

Let $U$ be an open set and $x\in U$. Assume without loss of generality that $x=0$, the origin. We can define the continuous function

$$f:U\setminus\{0\}\to S^1, x\mapsto \frac x{\|x\|}.$$

Because $U$ is open, there is some $\epsilon>0$ so that $B(0,\epsilon)\subseteq U$. So take any closed curve $\gamma:[0,1]\to S^1$. You can embed this curve in $U\setminus\{0\}$ via $\gamma\mapsto \epsilon\gamma=:\gamma'$. If $U\setminus\{0\}$ is simply connected, there is a homotopy contracting $\gamma'$ to a point. Via $f$, this maps to a homotpy of $\gamma$ to a point in $S^1$. We know that $S^1$ is not simply connected, hence not every curve should be contractible. Contradiction. $\square$

Answer 2

By normalizing each nonzero vector you get a retraction of the plane minus the origin to a circle. This is a homotopy equivalence and therefore your question is equivalent to the nonsimplyconnectedness of the circle, which you say you are familiar with.

Answer 3

Maybe I can give another explanation of this problem, but this requires some previous knowledge about a certain kind of map called a retraction, which is defined as:

Definition: Suppose $X$ is a topological space, and $A$ is a subspace of $X$, and the map $r : X\rightarrow A$ is a continuous function. Then $r$ is a retraction if $r\circ \iota_{A} = Id_A$, where $\iota_{A} : A \rightarrow X$ is the identity map on $A$. If such a retraction function $r$ from $X$ to $A$ exists, $A$ is called a retract of $X$.

Then we deal with the question.

Step 1: We can denote this open set as $U$, the open set without a point is $U\setminus \{x\}$. Then by a translation and a dilation, which are all homeomorphisms, we can assume that $S^1$, the unit circle is in $U\setminus \{x\}$ and $x$ is the origin.

Step 2: It's easy to see that for any $p\in A$, if $r : X\rightarrow A$ is any contraction, then we can use $r : X\rightarrow A$ and $\iota_{A} : A \rightarrow X$ to induce their group homomorphisms $\iota_{A\star} : \pi_1(A,p)\rightarrow \pi_1(x,p)$, which is injective and $r_{\star}:\pi_1(x,p)\rightarrow \pi_1(A,p) $, which is surjective. This is because $(r\circ \iota_{A})_{\star} =r_{\star} \circ \iota_{A\star} = (Id_A)_{\star}$, which is the identity on $\pi_1(A,p)$.

Step 3: We can easily see from step 2 that a retract of a simply connected space is simply connected since $r_{\star}:\pi_1(x,p)\rightarrow \pi_1(A,p) $ is surjective and $\pi_1(x,p)$ is trivial. This proposition can also be seen by observing that if the retract of a space is not simply connected then the original space is not simply connected.

Step 4: Define $r: U\setminus \{x\} \rightarrow S^1$ as $r(x) = \frac{x}{\|x\Vert}$, which can be easily checked to be a retraction so that $S^1 $ is a retraction of $U\setminus \{x\}$. Then, since $S^1$ is not simply connected (you know its fundamental gruop is isomorphic to $\mathbb{Z}$), then $U\setminus \{x\}$ is not simply connected by step 3.

Answer 4

Probably the easiest proof is noting that $$\int_\limits{|z-\star| = \varepsilon} \frac{dz}{z-\star} \neq 0.$$ If $U \setminus \star$ would be simply connected, the integral would be zero by Cauchy's integral theorem.