(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.

  • $\begingroup$ If the space is a connected CW complex it should be true. Since it’s simply connected, it’s weakly contractible. If it’s a CW complex then it’s contractible. By your assumption that it contains an open ball, the space has a $2$-cell, so the inclusion of the $2$-cell into your space is a homotopy equivalence. Removing the point in the interior of the $2$-cell should induce a homotopy equivalence of the punctured space with $S^1$ $\endgroup$
    – Exit path
    Dec 16, 2019 at 19:13
  • $\begingroup$ You might also mentioned the significant feedback you had on MathSE, notably that the result holds when $S$ is compact (and hence, more generally when every compact subset of $S$ is contained in a simply connected compact subset). $\endgroup$
    – YCor
    Dec 17, 2019 at 0:20

2 Answers 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.

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].


Clearly, $B(x,r)\backslash \{x\}$ is a retract of $S\backslash\{x\}$ and so the inclusion $B(x,r)\backslash\{x\}\to S\backslash \{x\}$ induces an injection on $\pi_1$. I will argue below that the induced homomorphism is also surjective, giving $\pi_1(S\backslash\{x\})\cong\mathbb{Z}$.

Claim: The inclusion $B(x,r)\backslash\{x\}\to S\backslash \{x\}$ induces a surjection on fundamental groups.

Proof. Suppose $S\neq \mathbb{R}^2$ and take $r$ to be the maximum radius such that $B(x,r)\subseteq S$.

Case 1: there exists a point $y\in \overline{B}(x,r)\backslash S$.

Suppose $\alpha:[0,1]\to S\backslash \{x\}$ is a path such that

  • $\alpha(0)=a_0$ and $\alpha(1)=a_1$ lie in $B(x,r/2)\backslash B(x,r/3)$,
  • $Im(\alpha)\cap B(x,r)$ is the union of two half-open arcs which lies in $B(x,r)\backslash B(x,r/3)$ and does not separate $x$ and $y$ in $B(x,r)\cup\{y\}$.

The Peano continuum $P=Im(\alpha)\cup \overline{B}(x,r/2)$ lies in $S$. If $y$ was in a bounded component of $\mathbb{R}^2\backslash P$, then $S$ would not be simply connected. Therefore, $y$ must lie in the unbounded component of $\mathbb{R}^2\backslash (Im(\alpha)\cup B(x,r/2))$. Let $R$ be an infinite ray that consists of arc arc in $\overline{B}(x,r)$ from $x$ to $y$ that does not meet $Im(\alpha)\cap B(x,r)$ and an infinite curve from $y$ off to infinity which does not meet $Im(\alpha)$. Now choose an arc $\beta$ from $a_1$ to $a_0$ in $B(x,r/2)\backslash R$. Since the loop $\alpha\cdot\beta$ does not meet the ray $R$, $R$ lies in the unbounded component $O$ of $\mathbb{R}^2\backslash Im(\alpha\cdot\beta)$. Fischer and Zastrow define $I(\alpha\cdot\beta)=\mathbb{R}^2\backslash O$ to be the "inner region" of $\alpha\cdot\beta$ in Definition 11 of the paper [FZ] referenced below. Since $\alpha\cdot\beta$ is a loop that contracts in $S$, Lemma 13 in [FZ] guarantees that $\alpha\cdot\beta$ contracts in the set $S\cap I(\alpha\cdot\beta)$, which does not contain $x$. Therefore, $\alpha$ is homotopic to a path in $B(x,r/2)$ by a homotopy in $S$ that does not meet $x$.

Now given any loop $\ell$ in $S\backslash \{x\}$ based at a point in $B(x,r/2)\backslash B(x,r/3)$, one can finitely subdivide $\ell$ into segments that lie in $B(x,r/2)$ or lie in $S\backslash B(x,r/3)$. One can easily modify the subdivision and perform finitely many homotopies in the annular region to ensure that each subpath in $S\backslash B(x,r/3)$ is of the form $\alpha$ from above, i.e. satisfying the two bullet points. Applying the previous paragraph to these subpaths, we see that $\ell$ is path-homotopic in $S\backslash \{x\}$ to a loop in $B(x,r)\backslash \{x\}$.

Case 2: $\overline{B}(x,r)\subseteq S$. Notice that there exists points $y_n\notin S$ that converge to a point in the boundary circle $\partial\overline{B}(x,r)$. Suppose $\alpha:[0,1]\to S\backslash\{x\}$ is a path such that $\alpha((0,1))\cap \overline{B}(x,r)=\emptyset$ and $\alpha(0),\alpha(1)\in \partial\overline{B}(x,r)$. Let $\beta:[0,1]\to\partial\overline{B}(x,r)$ be the shortest arc on the boundary circle from $\alpha(1)$ to $\alpha(0)$. Consider the loop $\alpha\cdot\beta$ and it's inner region $I(\alpha\cdot\beta)$. If there exists a ray $R$ containing the line segment from $x$ to a point $y\in\partial\overline{B}(x,r)$ and an infinite curve from $y$ to $\infty$ that lies in the unbounded component of $\mathbb{R}^2\backslash Im (\alpha\cdot\beta))$, then we may proceed as before (using Lemma 13) to contract $\alpha\cdot\beta$ in $S\backslash\{x\}$. If there is no such ray, then a neighborhood of $\overline{B}(x,r)$ must lie in $I(\alpha\cdot\beta)$. However, this forces a bounded component of $\mathbb{R}^2\backslash Im(\alpha\cdot\beta)$ to contain a point $y_n$ and contradicts that $S$ is simply connected. Thus, we can always contract $\alpha\cdot\beta$ within $S\cap I(\alpha\cdot\beta)\backslash \{x\}$ as desired.

Now if $\ell$ is any loop in $S\backslash\{x\}$ based at a point in $B(x,r)$, then we consider the open set $\ell^{-1}(S\backslash \overline{B}(x,r))$, which might have countably many components. If $(a,b)$ is such a component, then $\alpha=\ell|_{[a,b]}$ is path-homotopic to the shortest arc $\beta$ along $\partial \overline{B}(x,r)$ from $\alpha(a)$ to $\alpha(b)$. In particular, the homotopy used may be chosen to be of the form $[a,b]\times [0,1]\to I(\alpha\cdot\beta^{-1})$. Since we chose this homotopy to have size relative to that of $\alpha$, the diameters of the homotopies used goes to $0$ and we may glue them together to form a homotopy of $\ell$ to a loop entirely in $\overline{B}(x,r)\backslash \{x\}$. Of course, such a loop can then be deformed into $B(x,r)\backslash \{x\}$ $\square$

Patching up this proof felt like whack-a-mole, and I fully admit it's probably not the most elegant approach, but I think the core idea is sound. At the end of the day, the results in [FZ] should take care of this problem.

[FZ] 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.

  • 1
    $\begingroup$ My impression is that this argument only proves that the quotient of $\pi_1(U)$ by the normal closure of some cyclic (retract) subgroup is trivial. $\endgroup$
    – YCor
    Dec 16, 2019 at 21:22
  • $\begingroup$ More precisely it is not true that given a path-connected space $X$ with a collared circle $C$ (i.e., with a closed subset homeomorphic to $C\times [0,1]$ with $C\times [0,1[$ being open in $X$, the "collared circle" being the copy of $C\times\{0\}$, assuming that gluing a disc along $C$ yields a simply connected space, does not imply that $X$ has cyclic fundamental group, and even if one assumes that $\pi_1(X)$ is torsion-free and that $\pi_1(C)$ is a retract subgroup of $\pi_1(X)$. $\endgroup$
    – YCor
    Dec 16, 2019 at 21:52
  • $\begingroup$ You are right, a more geometric argument is required. $\endgroup$ Dec 16, 2019 at 21:53
  • $\begingroup$ To elaborate on @YCor's first comment: if $G=\langle x,y|yx=xy^2\rangle$ then the pushout of $1\xleftarrow{}\langle x\rangle\to G$ is $1$, but $\langle x\rangle$ is an infinite cyclic group and a retract of $G$ as we see by sending $y$ to $1$. $\endgroup$ Dec 16, 2019 at 22:01
  • 1
    $\begingroup$ Yes, something here will need to be fixed for the case when $S$ includes $\overline{B}(x,r)$ with other things attached. If $S$ is just a closed ball, the conclusion is pretty obvious. $\endgroup$ Dec 17, 2019 at 3:10

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