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Let $X = \mathbb{R}^2$ have a topology which is strictly finer than the topology of the Euclidean plane $\mathbb{E}^2$. Question: Is it true that $X$ is not simply connected? Note that a simply connected space is path connected.

Though this intuition is not precise in general, in simple cases passing to a finer topology cuts connections or adds slits between points. In the very simplest instance, declaring a nonempty closed proper subset $A \subsetneq \mathbb{E}^2$ to be open amounts to cutting $A$ out with an $A$-shaped cookie cutter. This results in a new space $X \cong (\mathbb{E}^2 \backslash A) \sqcup A$, which isn't even connected. If one instead obtains $X$ by declaring an arbitrary subset $S \subset \mathbb{E}^2$ to be open, a qualitative change is that now a path $[0, 1] \to X$ mapping $0$ into $S \cap \partial_{\mathbb{E}^2} S = S \setminus \mathrm{int}_{\mathbb{E}^2} S$ cannot instantaneously leave $S$; non-instantaneous means there's an $\epsilon > 0$ such that the path maps to $S$ until at least $\epsilon$. For example, declaring $[0, 1) \times \{0\}$ to be an open set results in an $X$ which looks like $\mathbb{R}^2$ with the segment $[0, 1) \times \{0\}$ hanging off from its only attachment at $(1, 0)$. This $X$ and other ad hoc examples I've tried are not simply connected, but in general the question seems complex.

This page demonstrates the existence of a strictly finer connected topology on the Euclidean unit interval, which shows the intuition of adding slits between points is not fully correct. Note however that no strictly finer path connected topology on the unit interval is possible, because a path connected Hausdorff space is arc connected, which immediately implies the interval with the finer topology is homeomorphic to the Euclidean unit interval. Applying the same fact to $\mathbb{E}^1$ shows for the $1$-dimensional case there is no strictly finer simply connected topology.

Geoffrey Sangston
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    If there was no strictly finer simply connected topology, then, of course, there would be no strictly finer topology, which is homeomorphic to the Euclidean topology. Or, equivalently, each continuous bijective map $\mathbb E^2 \rightarrow \mathbb E^2$ is a homeomorphism. Although this is true, I guess all proofs are non-trivial, for instance use invariance of domain. Hence, a proof that no strictly finer simply connected topology exists, would probably also be quite complicated. – Ulli Dec 21 '24 at 17:30
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    I'm not sure whether your argumentation that there is no finer arc connected topology on $\mathbb E^1$ is correct. Doesn't this show that such a topology coincides with the original one on each compact interval, only ? – Ulli Dec 21 '24 at 17:48
  • @Ulli You are correct. That was wishful thinking and the argument I had in mind does not work. I'm trying to figure out how to edit it now. – Geoffrey Sangston Dec 21 '24 at 18:05
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    @Ulli I think there's a more basic argument, but here's an argument just to affirm the claim because I have to run. I'll have to edit it later. A space $X$ with topology containing the Euclidean topology which is path connected has the same open sets on compact intervals, by the fact mentioned. It follows that the identity function $\mathbb{E}^1 \to X$ is continuous at every point. This is true for the inverse identity function, so the identity is a homeomorphism, hence the open sets coincide. – Geoffrey Sangston Dec 21 '24 at 19:05
  • yes, I think your last argument is valid. However, I think I was too fast with my assertion that the two topologies coincide on compact intervals. This would be true only, if the finer topology is also path connected on the compact intervals, isn't it? – Ulli Dec 21 '24 at 20:11
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    @Ulli I think so. A path of $X$ is also a path of $\mathbb{E}^1$. Let $a < b$ in $X$, and let $t_0 = \sup{t \in [0, 1] : \sigma(t) = a}$ and $t_1 = \inf{t \in (t_0, 1] : \sigma(t) = b}$. By construction, $t_0 < t_1$. Since $a < b$, the intermediate value theorem implies $\sigma([t_0, t_1]) \supset [a, b]$, and the construction of $t_0$ and $t_1$ implies the reverse inclusion. Reparameterizing $\sigma|_{[t_0, t_1]}$ results in a path in $[a, b] \subset X$ from $a$ to $b$. So every pair of points in any compact interval is connected by a path within the interval. – Geoffrey Sangston Dec 21 '24 at 21:12
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    Ah yes, that $\sigma$ is also a path to $\mathbb E^1$, might be helpful. And, starting with an injective path, probably shows immediately that it cannot leave $[x, y]$. But I have to go to bed now. Unfortunately, tomorrow, I probably will not have time to think about it. Summarizing our discussion, I believe more and more that your statement about $\mathbb E^1$ is true. But this was not your question, anyway, although a first step to prove or disprove. Good luck with your question! – Ulli Dec 21 '24 at 21:44

2 Answers2

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Sure. For instance, let $X$ be a wedge sum of $2^{\aleph_0}$ copies of $\mathbb{R}$. There is a continuous bijection $X\to\mathbb{R}^2$ taking the copies of $\mathbb{R}$ to the lines through the origin. Pushing the topology on $X$ forward to $\mathbb{R}^2$, this gives a topology strictly finer than the usual topology which is contractible.

Eric Wofsey
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    Agh! Very nice. A simplification of your example which modifies one from my question also works: Declare the ray $(0, \infty) \times {0}$ to be open. As with your space, this is contractible with a straight line homotopy to the origin. And there's lots of arbitrarily complicated variations on your idea, like letting the fibers/hairs extend from a line segment instead of the origin. – Geoffrey Sangston Dec 22 '24 at 01:56
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This answer establishes a restricted claim which coincides with the intuition that a loop which surrounds some "cuts" in the plane cannot contract over them. It ended up only using basic facts, so I tried to be detailed.

Claim: Equip $X = \mathbb{R}^2$ with a topology that contains the (Euclidean) topology of $\mathbb{E}^2$. Further suppose that the subspace topologies on $X \setminus D$ and $\mathbb{E}^2 \setminus D$ are equal, where $D \subset \mathbb{R}^2$ denotes a given bounded set. If $X$ is simply connected, then the topologies of $X$ and $\mathbb{E}^2$ coincide.

By rescaling, we may assume without loss of generality that the subspace topologies on $X \backslash (D^2)^\circ$ and $\mathbb{E}^2 \backslash (D^2)^\circ$ are equal, where $D^2$ denotes the closed unit disk and $(D^2)^\circ$ denotes the open unit disk. In particular the inclusion map $S^1 \hookrightarrow X$ is continuous, hence extends to a continuous map $D^2 \to X$ by the assumption that $X$ is simply connected. This further extends to a map $F : \mathbb{E}^2 \to X$ which restricts to the identity on $\mathbb{E}^2 \backslash (D^2)^\circ$, and which is continuous by the pasting lemma, using the assumption that the subspace topologies on $X \backslash (D^2)^\circ$ and $\mathbb{E}^2 \backslash (D^2)^\circ$ are equal. Since $F$ is also continuous when viewed as a map $\mathbb{E}^2 \to \mathbb{E}^2$, it follows that $F(D^2)$ is a compact, hence bounded, subset of $\mathbb{E}^2$. So there exists an origin-centered closed disk $D'$ which is possibly larger than $D^2$ such that $F(D') = D'$ and $F|_{\partial D'} = \mathrm{Id}|_{\partial D'}$. Now the claim clearly follows from the following lemma, which I thought might be nice to single out.

Lemma: Let $Y = D^2$ have a topology which contains the Euclidean subspace topology on $D^2$. Suppose there exists a continuous map $F : D^2 \to Y$ that restricts to the identity map on $S^1$. Then the topologies of $Y$ and $D^2$ coincide.

Proof: Note that $F$ is in particular continuous as a map $D^2 \to D^2$. It follows from the non-existence of a continuous retraction $D^2 \to S^1$ that $F$ surjects; see Continuous function from the closed unit disk to itself being identity on the boundary must be surjective?, and for the non-existence of a continuous retraction $D^2 \to S^1$ see the proof of Theorem 1.9 from Hatcher. Since $D^2$ is compact and Hausdorff, it follows by the closed map lemma that $F$ is closed as a map $D^2 \to D^2$. So if $S \subset Y$ is a closed set, then $S = F(F^{-1}(S))$ is closed as a subset of $D^2$, which means the two topologies coincide.

Geoffrey Sangston
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