Fundamental group of $\mathbb R^2$ without a closed semi-line

Question

Intuitively the fundamental group of the $\mathbb R^2$ without a closed semi line is the trivial group, but I don't know how to prove it.

Any ideas? thanks

Answer 1

Here's an idea (which might use unnecessarily strong results): Let $$A=\{(x,y)\mid y>0\},\quad B=\{(x,y)\mid x<0\}\quad\text{and}\quad C=\{(x,y)\mid y<0\}.$$ Then $\Bbb R^2\setminus ([0,\infty)\times\{0\})=A\cup B\cup C$. Use Seifert-van Kampen for $A$ and $B$ and then for $A\cup B$ and $C$.

Answer 2

$D=\mathbb{R}^2-[0,\infty)$ is star-shaped. Take for instance $P$ to be the point with coordinates $(-1,0)$. Then for any point $Q\in D$, the segment joining $P$ and $Q$ is completely contained in $D$. It is easy to show that any star-shaped domain is simply connected.

Another possibility is to show that $D$ is homeomorphic to a domain that you already know to be simply connected. using complex variables, the map $\sqrt z$ (the branch defined on $D$ such that $\sqrt{-1}=i$) is an homeomorphism between $D$ and the upper half-plane. And the map $$ \frac{\sqrt z-i}{\sqrt z+i} $$ is an homeomorphism between $D$ and the unit disk.

Answer 3

HINT:

The function $$F:(\mathbb{R}^2-[0,\infty))\times [0,1] \longrightarrow \mathbb{R}^2-[0,\infty)$$ with $(x,y,t) \mapsto (x-t(x+1),(1-t)y)$ is a deformation retraction of $\mathbb{R}^2-[0,\infty)$ to a point.