# Fundamental group of punctured simply connected subset of $\mathbb{R}^2$

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 $$\pi_1(S\setminus\{x\})\cong\mathbb{Z}$$?

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 $$\newcommand{\ra}{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c}B\setminus\{x\}&\ra{}&S\setminus\{x\}&\ra{}&\mathbb{R}^2\setminus\{x\}\\\da{}&&\da{}\\B&\ra{}&S\end{array}$$ 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 homomorhpisms $$G\to A$$ 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.

• @freakish How do you know H is isomorphic to $\mathbb{Z}$? Jun 11, 2019 at 12:02
• @ConnorMalin Ok, after diving into details I can only prove that $H$ is a normal closure of some image of $\mathbb{Z}$ in $\pi_1(V)$. That's not enough. Jun 11, 2019 at 13:41
• If $S$ is open, then complex analysis tells us that it's either $\mathbb{R}^2$, in which case the answer is clear, or it's conformally equivalent, so in particular homeomorphic to $D^2$, in which case the answer is clear too. So we're looking at widely non-open sets. I think the best way to go would then be to somehow manage to get back to the open case Jun 11, 2019 at 15:23
• It’s worth noting that there are noncontractible simply connected (closed) sets. Jun 11, 2019 at 16:42
• You can try Mathoverflow (the question is at the right level). Make sure you link to the original MSE question. Dec 16, 2019 at 18:31

This is a complete revision of an earlier answer.

I am not sure about the question in full generality, but here is what I can prove:

Theorem 1. Suppose that $$S\subset R^2$$ is a simply-connected compact subset. Then for every $$x\in int(S)$$, the group $$G=\pi_1(S-\{x\})$$ is infinite cyclic.

Proof.

Lemma 1. Let $$S\subset R^2$$ be a subset (not necessarily compact) with $$H_0(S)=H_1(S)=0$$. Then for every $$x\in int(S)$$, $$H_1(S-\{x\})\cong {\mathbb Z}$$.

Proof. Let $$B\subset int(S)$$ be an open ball centered at $$x$$. The set $$S$$ is the union of $$X=S-\{x\}$$ and $$B$$, with $$X\cap B= B-\{x\}$$.
We have the (exact) Mayer-Vietoris sequence $$0= H_1(S) \leftarrow H_1(B)\oplus H_1(X) = H_1(X) \leftarrow H_1(B-\{x\})={\mathbb Z}\leftarrow H_2(S)=0$$ (the equality $$H_2(S)=0$$ is not completely trivial). From this, we obtain $$H_1(X)= {\mathbb Z}$$. qed

In particular, the abelianization of $$G=\pi_1(S-\{x\})$$ is infinite cyclic.

In order to proceed, I will need a bit of group theory.

Definition. A group $$F$$ is called fully residually free if for every finite subset $$E\subset F$$, there exists a homomorphism $$f: F\to F_n$$, a free group of some rank (depending on $$E$$), such that $$f|_E$$ is one-to-one.

Notation. Given a group $$H$$ and an element subset $$a\in H$$, I will denote $$^H$$ the normal closure of $$a$$ in $$H$$, i.e. the smallest normal subgroup of $$H$$ containing $$a$$.

Lemma 2. Suppose that $$H= F_\alpha$$ is a free group of (possibly infinite) rank $$\alpha\ge 1$$ and assume that $$H/^{H}$$ is trivial. Then $$n=1$$, i.e. $$H$$ is infinite cyclic.

Proof. Use the fact that the abelianization of $$F$$ is $$\cong {\mathbb Z}^\alpha$$ and if the quotient of $${\mathbb Z}^\alpha$$ by a rank $$\le 1$$ subgroup is trivial then $$\alpha=1$$. qed

Lemma 3. Suppose that $$G$$ is a fully residually free group whose abelianization is isomorphic to $${\mathbb Z}$$, and $$g\in G$$ is such that $$G=^G$$, i.e. $$G$$ is normally generated by $$g$$. Then $$G\cong {\mathbb Z}$$.

Proof. First of all, every quotient group $$Q$$ of $$G$$ is normally generated by the projection of $$g$$. Therefore, if such quotient is a free group, by Lemma 2, the group $$Q$$ is cyclic, in particular, abelian. I now claim that $$G$$ is abelian. Indeed, take $$g_1, g_2\in G$$ and let $$E=\{1, [g_1, g_2]\}$$. Since $$G$$ is fully residually free, there is a free quotient group of $$G$$ to which the set $$E$$ projects injectively. As noted above, such quotient is abelian, forcing $$[g_1, g_2]=1$$. Hence, $$G$$ is abelian. But $$G$$ is assumed to have infinite cyclic abelianization, which implies that $$G\cong {\mathbb Z}$$. qed

I will also need the following theorem proven in Corollary 7 of

H. Fischer, A. Zastrow, The fundamental groups of subsets of closed surfaces inject into their first shape groups, Algebraic and Geometric Topology 5 (2005) 1655-1676.

Theorem 2. For every path-connected compact subset $$X\subset S^2$$, the fundamental group $$\pi_1(X)$$ is fully residually free.

We can now finish the proof of Theorem 1. Recall that $$S\subset R^2\subset S^2$$ is compact, $$G=\pi_1(X)$$, where $$X=S-\{x\}$$. Since $$x$$ is in the interior of $$S$$ and $$S$$ is path-connected, it follows that $$X=S-\{x\}$$ is path-connected. Let $$g\in \pi_1(X)$$ be an element represented by a small circle centered at $$x$$. Since $$\pi_1(S)=1$$, by the van Kampen theorem, $$G=^G$$, i.e. $$G$$ is normally .generated by $$G$$. By Lemma 1, $$G$$ has infinite cyclic abelianization. Thus, by combining Theorem 2 with Lemma 3, we see that $$G\cong {\mathbb Z}$$. qed

• Doesn't Corollary 7 in the paper require that $X$ is proper and compact? Dec 14, 2019 at 16:58
• @ThomasBrowning: You are right, I was careless. See the edit. Dec 14, 2019 at 23:21