([This question](https://math.stackexchange.com/questions/3258252) 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.



Fundamental group of punctured simply connected subset of $\mathbb{R}^2$
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*](https://arxiv.org/abs/math/0512343v1https://).



> 
> **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].


