TITLE: Fundamental group of punctured simply connected subset of $\mathbb{R}^2$
QUESTION [18 upvotes]: (This question is originally from Math.SE where it was suggested that I ask the question here)
Let $S$ be a simply connected subset of $\mathbb{R}^2$ and let $x$ be an interior point of $S$, meaning that $B_r(x)\subseteq S$ for some $r>0$.
Is it necessarily the case that $\pi_1(S\setminus\{x\})\cong\mathbb{Z}$?
This is true if $S$ is open by the Riemann mapping theorem.
Moishe Kohan's answer on Math.SE shows that the result holds if $S$ is compact.

Let $B=B_r(x)$ and let $G=\pi_1(S\setminus\{x\})$.
I can show that $G^{ab}\cong\mathbb{Z}$.
Consider the commutative diagram of topological spaces
$$
\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}
\begin{array}{c}B\setminus\{x\}&\ra{}&S\setminus\{x\}&\ra{}&\mathbb{R}^2\setminus\{x\}\\\da{}&&\da{}\\B&\ra{}&S\end{array}
$$
Applying the fundamental group functor gives a commutative diagram of groups
$$
\begin{array}{c}\mathbb{Z}&\ra{}&G&\ra{}&\mathbb{Z}\\\da{}&&\da{}\\1&\ra{}&1\end{array}
$$
where the composition of the top two maps is the identity homomorphism on $\mathbb{Z}$.
By the Seifert-van Kampen theorem, the square is a pushout diagram of groups, meaning that the normal closure of the image of $\mathbb{Z}\to G$ is all of $G$.
In particular, if $A$ is an abelian group then any two homomorphisms $G\to A$ that agree on the image of $\mathbb{Z}\to G$ must agree on all of $G$.
Another way to put this is that if we have two homomorphisms $G\to A$ such that the two compositions $\mathbb{Z}\to G\to A$ are equal then the two homomorphisms $G\to A$ are equal.
I claim that the map $G\to\mathbb{Z}$ is an abelianization map.
To see this, let $A$ be an abelian group and let $G\to A$ be a homomorphism.
Now recall that the composition $\mathbb{Z}\to G\to\mathbb{Z}$ is the identity.
Then the composition $\mathbb{Z}\to G\to\mathbb{Z}\to G\to A$ agrees with the composition $\mathbb{Z}\to G\to A$.
By the remark at the end of the previous paragraph, this means that the composition $G\to\mathbb{Z}\to G\to A$ agrees with the map $G\to A$.
In other words, the composition $\mathbb{Z}\to G\to A$ makes the abelianization diagram commute.
To show uniqueness, let $\mathbb{Z}\to A$ be a map making the abelianization diagram commute.
Then the composition $G\to\mathbb{Z}\to A$ agrees with the map $G\to A$.
Then the composition $\mathbb{Z}\to G\to\mathbb{Z}\to A$ agrees with the composition $\mathbb{Z}\to G\to A$.
Since the composition $\mathbb{Z}\to G\to\mathbb{Z}$ is the identity, this shows that the map $\mathbb{Z}\to A$ is given by the composition $\mathbb{Z}\to G\to A$.
This shows that the map $G\to\mathbb{Z}$ is an abelianization map.

REPLY [6 votes]: Here is a proof that $\pi_1(S\setminus\{x\})\cong\mathbb Z$. It is along similar lines to Jeremy Brazas' answer, and also makes use of Lemma 13 from the paper The fundamental groups of subsets of closed surfaces inject into their first shape groups.

Lemma 13: Let any set $A\subseteq\mathbb R^2$, and map $\alpha\colon S^1\to A$ be given. Let $U$ be the unbounded connected component of $\mathbb R^2\setminus{\rm Im}(\alpha)$. If $\alpha\colon S^1\to A$ is null-homotopic, then so is $\alpha\colon S^1\to A\setminus U$.

As we know, the embedding $B_r(x)\setminus\{x\}\to\mathbb R^2\setminus\{x\}$ gives an isomorphism of fundamental groups, which are isomorphic to $\mathbb Z$. Hence, the embeddings $B_r(x)\setminus\{x\}\to S\setminus\{x\}\to\mathbb R^2\setminus\{x\}$ gives homomorphisms
$$
\mathbb Z\stackrel{f}{\to}\pi_1(S\setminus\{x\})\stackrel{g}{\to}\mathbb Z
$$
with $gf=\iota$. To show that $f,g$ are isomorphisms, it is enough to show that $g$ has trivial kernel. So, suppose that $g([\gamma])=0$ for some closed curve $\gamma\colon S^1\to S\setminus\{x\}$. This means that $\gamma$ is null-homotopic in $\mathbb R^2\setminus\{x\}$.
Letting $U$ be the unbounded connected component of $\mathbb R^2\setminus{\rm Im}(\gamma)$, Lemma 13 states that $\gamma$ is null-homotopic in $\mathbb R^2\setminus(U\cup\{x\})$. If we can show that $\mathbb R^2\setminus U\subseteq S$ then we are done. It is enough to show that each bounded connected component $V$ of $\mathbb R^2\setminus{\rm Im}(\gamma)$ is contained in $S$.
From the proof of Lemma 13 in the linked paper, $V$ is simply connected and, by the Riemann mapping theorem, there is a conformal $h\colon\mathbb D\to V$ with boundary map $h\vert_{\partial\mathbb D}$ mapping to $\partial V\subseteq {\rm Im}(\gamma)\subseteq S$. By shrinking this down to a circle in $\mathbb D$, we see that it has winding number $1$ about every point in $V$, so cannot be null-homotopic in $S$ unless $V\subseteq S$.
[I was hoping to complete this proof only using the statement of Lemma 13, but it is a bit trickier than expected, so had to refer to the proof of the lemma from the linked paper].