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If $X\subset\mathbb{R}^2$ is a connected open subset which is not simply connected, then there exists some point $p\in\mathbb{R}^2\setminus X$ such that the inclusion map $X\to\mathbb{R}^2\setminus \{p\}$ induces a nontrivial map on $\pi_1$. In other words, if a connected open subset of the plane is not simply connected, this is always because it contains a loop which winds around some point in its complement. More strongly, in fact, there is some point $p\in\mathbb{R}^2\setminus X$ such that $X$ contains a simple closed curve that has $p$ in its interior. (See here for one way to prove this.)

My question is whether this is still true for arbitrary (not necessarily open) subsets of $\mathbb{R}^2$. More precisely:

Suppose $X\subset\mathbb{R}^2$ is path-connected but not simply connected. Must there exist some point $p\in\mathbb{R}^2\setminus X$ such that the inclusion map $X\to\mathbb{R}^2\setminus \{p\}$ induces a nontrivial map on $\pi_1$? More strongly, must there exist such $p$ such that $X$ contains a simple closed curve that has $p$ in its interior, or at least a curve with winding number $1$ around $p$ (so the map on $\pi_1$ is surjective)?

I suspect the answer is yes, since I've heard of various results saying that arbitrary subsets of $\mathbb{R}^2$ are homotopically "tame" in various ways (for instance, unlike in higher dimensions, they cannot have nontrivial $H_k$ for $k\geq 2$). The closest result to my question that I know of is the result from this answer that if $X\subseteq \mathbb{R}^2$ then any nontrivial element in $\pi_1(X)$ remains nontrivial in $\pi_1(\mathbb{R}^2\setminus F)$ for some finite set $F\subseteq \mathbb{R}^2\setminus X$. This is very close to my question: it is weaker since it involves a finite set $F$ rather than a single point, but stronger since it applies to an arbitrary nontrivial element of $\pi_1(X)$ rather than just saying there exists an element of $\pi_1(X)$ that remains nontrivial. I would guess you can prove an affirmative answer to my question either from that result or using ideas similar to its proof, but I don't see an obvious way to do so. (Note that the finite set $F$ in that result cannot be reduced to a single point, since your element of $\pi_1(X)$ could be a commutator of loops around different points.)

(Alternatively, by the argument I linked above in the case of open subsets, it would suffice to show that there exists a nontrivial loop in $X$ with only finitely many self-intersections, since then you can use that to reduce to a nontrivial simple loop and then invoke the Jordan curve theorem.)

Eric Wofsey
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  • Is it known whether the fundamental group of a subset of the plane can be perfect? Commutators can't be detected by winding number. – Cheerful Parsnip Aug 07 '23 at 02:35
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    @CheerfulParsnip: That is impossible by the result I mentioned in the question: there exists a nontrivial homomorphism from $\pi_1(X)$ to a free group $\pi_1(\mathbb{R}^2\setminus F)$ whose image is free and therefore not perfect. – Eric Wofsey Aug 07 '23 at 05:05
  • Ah yes, I see.$,$ – Cheerful Parsnip Aug 07 '23 at 05:08
  • "such that contains a simple closed curve that has in its interior" — I would avoid using the word "interior" when discussing open sets but meaning "a bounded component of its complement in the plane". – Dan Asimov Aug 07 '23 at 06:15
  • I was thinking about the stronger conjecture of there being a simple closed curve in $X$ not contractible in $X$, and came across https://mathoverflow.net/questions/220709/existence-of-a-non-null-homotopic-simple-closed-curve which discusses a compact 2-dimensional counterexample embedded in $\mathbb R^3$, namely the Harmonic Archipelago. I'm thinking about asking about this stronger conjecture on MO. – D.R. Aug 08 '23 at 02:18
  • "Not simply-connected". I.e., there is a loop in it that is not homotopic to a point. Isn't that enough to answer your question? – Paul Sinclair Aug 08 '23 at 11:55
  • @PaulSinclair: I'm not sure what you mean. – Eric Wofsey Aug 08 '23 at 13:49
  • You have a loop which is not homotopic to a point. If the interior of that loop was contained in $X$, it would be contractable, so there is at least one point $p \notin X$ interior to the loop. The winding number about $p$ is not $0$. By tracing the branches of the loop closest to $p$, you get a curve where the winding number is $1$. – Paul Sinclair Aug 08 '23 at 14:05
  • @PaulSinclair: The loop may not be simple. As I mentioned in the question, it would suffice to show there is a nontrivial loop with only finitely many self-intersections, so you could reduce to the case of a simple loop. – Eric Wofsey Aug 08 '23 at 14:38
  • I wasn't assuming the loop was simple. But it is compact, and as $p$ is disjoint from it, in each direction from $p$ there is a unique point on the loop that is closest to $p$. Those closest points define a simple closed curve about $p$ which is contained in $X$. – Paul Sinclair Aug 08 '23 at 14:58
  • @PaulSinclair: The closest point is not necessarily a continuous function of the direction. – Eric Wofsey Aug 08 '23 at 15:09
  • It's also not clear how you are getting the point $p$ in the first place--what do you mean by the "interior" of the loop if it is not simple so you don't have the Jordan curve theorem? It is not obvious that the complement of the loop in the plane must be disconnected (or that $X$ couldn't contain all of the bounded components without the loop becoming trivial in $X$). – Eric Wofsey Aug 08 '23 at 15:16
  • I admit my conception here is simple-minded. But what stops a loop in a subset of the plane from being homotopic to a point is that it encircles some point not in the set. i.e., the winding number around the point is not $0$. I had realized my naive conception about finding the "inner loop" was flawed not long after giving it, but had other things to attend to. Perhaps the boundary of the component of the complement of the curve containing $p$. Anyway, I admit my knowledge is far inferior to yours here. It just seemed an obvious approach to consider. – Paul Sinclair Aug 08 '23 at 16:00
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    Well, my question is exactly about how to turn that obvious-seeming geometric intuition into a rigorous proof. My impression is that as a general rule, these kinds of geometric intuitions tend to not actually be correct for arbitrary subspaces of $\mathbb{R}^n$ for $n>2$, but for $\mathbb{R}^2$ they tend to be correct (but the proofs are difficult). I haven't found any reference that verifies this particular statement, though. – Eric Wofsey Aug 08 '23 at 16:12
  • Does https://mathoverflow.net/questions/338073/in-a-subset-of-mathbbr2-which-is-not-simply-connected-does-there-exist-a-s answer your strongest conjecture in the affirmative? – D.R. Aug 09 '23 at 06:27
  • @D.R.: Yes, thank you! – Eric Wofsey Aug 09 '23 at 14:08

1 Answers1

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The strong version of my question has been answered affirmatively by Jeremy Brazas on MathOverflow:

All the ingredients to prove what you want (and more) can be found in

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.

Proof Sketch. Suppose $X\subseteq \mathbb{R}^2$ is path connected and $\alpha:S^1\to X $ is not null-homotopic. Let $Y=\alpha(S^1)\cup\bigcup\{U\mid U\text{ is a bounded path component of }\mathbb{R}^2\backslash \alpha(S^1)\text{ with }U\subseteq X\}$. Fischer and Zastrow show $Y$ is a Peano continuum homotopy equivalent to a “Sierpinski-like” space, i.e. a copy of the Sierpinski Carpet with some, possibly empty, set of (originally deleted) squares filled back in. Since all we want to do is find some simple closed curve that doesn’t contract, pick any bounded path component $C$ of $\mathbb{R}^2\backslash Y$ and a point $z\in C\backslash X$. While $C$ is homeomorphic to an open disk, its boundary need not be a simple closed curve. However, you basically have a planar Peano continuum $Y$ surrounding a “hole” at $z$. Since $Y\subseteq X$ and $z\notin X$, any simple closed curve in $Y$ with winding number $1$ around $z$ will do the trick. If you want to create one explicitly, use some planar geometry to cover $Y$ with enough arc-wise connected open (in $Y$) sets to create a finite loop of neighborhoods in $Y$ with winding number $1$ around $z$ and build your curve piece-wise.

Eric Wofsey
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