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The compact complement topology on $X = \mathbb{R}$ is the one where closed subsets are the Euclidean-compact subsets and $X$.

With this topology, any bounded subset of $X$ has Euclidean topology, and so it's clear path-connected.

The compact subsets of $X$ are the Euclidean-closed subsets, and so they can be unbounded. The compact Hausdorff subsets are the Euclidean-compact subsets.

In particular there exist loops $S^1\to X$ with unbounded image.

Partial result: If a loop $g:S^1\to X$ has image bounded from above/below, then it's homotopic to a constant.

Proof: Let $f:[0, 1)\to [0, \infty)$ be an increasing bijection and define $H:[0, 1]\times S^1\to X$ by $H(t, x) = \begin{cases} g(x)+f(t), & t < 1 \\ 0, & t = 1\end{cases}$ then if $g(S^1)$ is bounded from below, then $H$ is continuous. If $g(S^1)$ is bounded from above then replace $g(x)+f(t)$ by $g(x)-f(t)$. To see that it's continuous, consider any Euclidean-compact set $K\subseteq X$, then simply observe that there must be $t_0 < 1$ such that $g(x)+f(t)\notin K$ for $t> t_0$, and so $H^{-1}(K)$ is closed. $\square$

Is $X$ simply connected? As an example of a function which can possibly be not homotopic to a constant map, I propose the map $g:S^1\to X$ defined by $g(1, 0) = 0$ and $g$ maps $S^1\setminus \{(1, 0)\}\cong \mathbb{R}\to X$ where $\cong$ is some homeomorphism. If any maps aren't homotopic, it's the surjective ones.

Jakobian
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  • Isn't any loop $f$ the sum of a bounded-from-above loop $f_-$ and a bounded-from-below loop $f_+$? If that statement is true, my suggestion is to study how point-wise addition behaves with respect to homotopy. It's something I'd never considered before. – XavierO May 21 '25 at 14:11
  • @XavierO how do you want to write $f$ as a sum of such loops? If you write $f_+ = \max(f, 0), f_- = \min(f, 0)$ then $f_+, f_-$ don't have to be continuous anymore. And similarly, if $f, g$ are loops then $f+g$ is not necessarily continuous. – Jakobian May 21 '25 at 14:56
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    see https://math.stackexchange.com/questions/5009984 – PatrickR May 22 '25 at 22:23

1 Answers1

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$X$ is contractible and so simply connected:

Define a map $H:[0, 1]\times X\to X$ by $H(t, x) = \begin{cases} x, & t = 1 \\ f(t), & t <1\end{cases}$ where $f:[0, 1)\to [0, \infty)$ is an increasing bijection. If $K\subseteq X$ is a proper closed set in $X$, i.e., Euclidean-compact, then $$H^{-1}(K) = (\{1\}\times K) \cup (\{t\in [0, t_0] : f(t)\in K\}\times X)$$ for some $t_0 < 1$, which is closed in $[0,1]\times X$, and so $H$ is continuous.

PatrickR
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Jakobian
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    Nice answer, a lot simpler than the one at https://math.stackexchange.com/questions/5009984 – PatrickR May 22 '25 at 22:56
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    Amazing answer. Am I mistaken in thinking this $H$ is also continuous given the finite complement topology on $\mathbb{R}$? This would give a simpler argument than the one here for cardinality $\mathfrak{c}$, which is currently linked to by pi-base. – Geoffrey Sangston May 23 '25 at 00:50
  • @GeoffreySangston it definitely is, good observation! – Jakobian May 23 '25 at 00:55
  • Minor nitpick: Your formulation of the inverse image isn't quite correct, as it's just ${1} \times K$ if $K$ consists only of negative values. But the result is the same. – MartianInvader Jun 03 '25 at 16:31
  • @MartianInvader But thats the same as what I wrote then. If $K$ consists only of negative values then the second set of the union is empty. – Jakobian Jun 03 '25 at 16:41