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 $$U$$?

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 $$\mathbb{Z}$$.
• I feel that $f\circ g\circ i$ maybe not continuous, since the set $B$ maybe not convex.