Why is the following subset of $\mathbb{C}$ simply connected. Let $E=\mathbb{C} \setminus \lbrace te^{it} \mid t\in \mathbb{R}_{\geq{0}} \rbrace$. Then we can define a branch of logarithm on $E$. My complex analysis instructor said that we can do it for any simply connected domain of $\mathbb{C}$. 
How do we show using basic algebraic topology that $E$ is simply connected.
 A: The map $z\mapsto ze^{-i|z|}$ is a homeomorphism of $E$ onto $E':=\mathbb{C}\,\backslash\,\mathbb{R}_{\geq 0}$, which is star convex (for every $z\in E'$, the line segment connecting $z$ and $-1$ is contained in $E'$). This shows $E'$ is contractible (and in particular simply connected), hence so is $E$.
A: For a careful proof, you want to use Seifert-van Kampen. Fatten the removed line up to $L=\{te^{it}\}+\{z:|z|<\epsilon\}$ to get $\mathbb{C}=E\cup L$ as a union of open connected subsets. For sufficiently small $\epsilon$, the inclusion of $\{te^{it}\}$ into $L$ is a homotopy equivalence, and $\{te^{it}\}$ is homeomorphic to $\mathbb{R}$, for instance because the map $\mathbb{R}\to \{te^{it}\}$ is locally a continuous bijection of compact Hausdorff spaces. The intersection $E\cap L$ is also homeomorphic to a line, so $0=\pi_1(\mathbb{C})$ is the free product with amalgamation of $\pi_1(L)=0$ and $\pi_1(E)$ over $\pi_1(L\cap E)=0$, that is, it's just $\pi_1(E)$. So $E$ is simply connected.
