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

  • $\begingroup$ @freakish How do you know H is isomorphic to $\mathbb{Z}$? $\endgroup$ Jun 11, 2019 at 12:02
  • $\begingroup$ @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. $\endgroup$
    – freakish
    Jun 11, 2019 at 13:41
  • 2
    $\begingroup$ 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 $\endgroup$ Jun 11, 2019 at 15:23
  • $\begingroup$ It’s worth noting that there are noncontractible simply connected (closed) sets. $\endgroup$ Jun 11, 2019 at 16:42
  • 1
    $\begingroup$ You can try Mathoverflow (the question is at the right level). Make sure you link to the original MSE question. $\endgroup$ Dec 16, 2019 at 18:31

1 Answer 1


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.


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 $<a>^H$ 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 $H/<a>^{H}$ 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 $G=<g>^G$, 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, $G=<g>^G$, 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

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

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