Let $U$ be a subset of $\mathbb{R}^2\setminus\{0\}$, $i_\ast(\pi_1(U))=0$ or $\mathbb{Z}$ Let $U$ be a connected open subset of the punctured plane $\mathbb{R}^2\setminus\{0\}$ and $i\colon U\to \mathbb{R}^2\setminus\{0\}$ the inclusion mapping, $\pi_1(U)$ the fundamental group of $U$. Can we deduce that $i_\ast(\pi_1(U))$ is either $0$ or $\mathbb{Z}$? Can it happen that $\pi_1(U)=2\mathbb{Z},3\mathbb{Z},\dots$? 
Is it correct that we can define holomorphic branch of square root of $z$ on $U$ if and only if we can define holomorphic branch of $\log z$ on $U$?
 A: It suffices to show that the image is either $\mathbb{Z}$ or $0$ for any closed, bounded subset, since any loop is contained inside a closed, bounded subset.
Let $B$ be the bounded subset. We construct a map $P \rightarrow B$ that when composed with the inclusion of $B$ into the punctured plane $P$ is a retraction of $\pi_1(B)$ onto a $\mathbb{Z}$ retract. This implies that there is no $n \neq 0$ so that $na$ is in the summand when $a$ is not in the summand which implies that the image of the generator of this summand in $\pi_1(P)$ is either $0$ or a generator of $\pi_1(P)$.
Since $B$ is bounded, there is a circle $C$ containing it. Define a map $f:C \rightarrow B$ by sending $p$ to the element of $B$ that is closest to $p$ on the line from $p$ to the origin (if no such point exists the fundamental group of $B$ is trivial).
Now let $g: P \rightarrow C$ be the map radially deforming $P$ into $C$. Then $f \circ g \circ i$ is a retraction onto something homeomorphic to a circle, so its fundamental group is $\mathbb{Z}$.
