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$\mathbb{R}$ is an example of a simply-connected space with non-simply connected compactification (one-point compactification of $\mathbb{R}$ is $\mathbb{S}^1$ so not simply connected.) but 2-point compactification of $\mathbb{R}$ is simply-connected.

So a simply connected space might or might not have simply connected compactifications.

Let $X$ be a locally compact Hausdorff simply connected space. Are the following true?

  1. There exists Hausdorff compactification $C_1$ of $X$ such that $C_1$ is not simply connected.

  2. There exists compactification $C_2$ of $X$ such that $C_2$ is simply connected. (but $C_2$ is not necessarily Hausdorff)

talekhine
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    Those voting to close should provide feedback in the comments. – Steven Clontz Dec 24 '24 at 03:51
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    Given any space $X$ you can define its compactification $Y:= X\sqcup {p}$, where the only open subset of $Y$ containing $p$ is $Y$ itself. Then $Y$ is simply connected. But this $Y$ is usually non-Hausdorff. Would you be satisfied with such compactifications? – Moishe Kohan Dec 24 '24 at 05:04
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    I’m pretty sure every (Hausdorff) compactification of $X=[0,\infty)$ in which $X$ is open is simply connected, in fact contractible. – David Gao Dec 24 '24 at 05:06
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    According to pi-base, the long ray is simply connected, but its only Hausdorff compactification, the closed long ray, is not. – Ulli Dec 24 '24 at 07:20
  • @Moishe Kohan: is $X \sqcup {p}$ the one-point compactification of $X$? Then, in fact, this is an example of a possibly not simply connected space, as pointed out by the OP. On the other hand, it might also be simply connected, see the comment of David Gao. – Ulli Dec 24 '24 at 07:26
  • Anyway, @ talekhine, please specify, whether you want to consider Hausdorff compactifications, only. It certainly makes up a very big difference. – Ulli Dec 24 '24 at 07:29
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    @Ulli The example of Moishe Kohan is not the usual one-point compactification of $X$. Instead, the only open neighborhood of the added point $p$ is $X \sqcup {p}$. An example of such a construction can be found in, for example, here. It is highly non-Hausdorff (in fact, even highly non-$T_1$). But a space that has a point whose only neighborhood is the entire space will be contractible and thus also simply connected. See here. – David Gao Dec 24 '24 at 08:06
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    @David Gao: Oh yes, I should have read more carefully. Thank you for clarification! – Ulli Dec 24 '24 at 08:58
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    @DavidGao $[0,\infty)$ has many Hausdorff compactifcations which are not even path connected. See https://math.stackexchange.com/q/4610897/349785. – Paul Frost Dec 24 '24 at 09:31
  • @PaulFrost Ah, you’re right. I thought I had a proof that they are all contractible, but I was badly mistaken. Hmm, I’m now wondering if the Stone-Čech compactification of a non-compact LCH space is ever path-connected. Your example shows it is not for $[0, \infty)$ (and easily generalizes to $\mathbb{R}$ as well - I think it also generalizes to $\mathbb{R}^n$ and the half space of $\mathbb{R}^n$ for any $n \geq 1$, maybe even to all non-compact manifolds with/without boundary?). Ulli’s example shows it is not for the long ray. – David Gao Dec 25 '24 at 01:10
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    The Stone-Chech compactification of a normal noncompact (locally compact, Haussdorff) space is never path-connected. I am not sure if normality can be dropped. – Moishe Kohan Dec 25 '24 at 04:43
  • In Question 1 you want the further assumption that $X$ is not compact. – user14111 Dec 25 '24 at 05:48
  • @MoisheKohan are you sure this holds without assuming that $X$ is realcompact (or that $\nu X$ is normal)? – Jakobian Dec 27 '24 at 03:58
  • @Jakobian: Yes, I am sure. – Moishe Kohan Dec 27 '24 at 03:59
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    @MoisheKohan I see this now, and I wrote this and some of my other takes on $\beta X$ not being path-connected in here. For example, theorem 3 implies that Moore plane is a path-connected non-normal Tychonoff space with its Stone-Cech compactification not path-connected. – Jakobian Dec 27 '24 at 05:51
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    @Jakobian: My proof seems to be different, but maybe the difference is superficial. See below. – Moishe Kohan Dec 27 '24 at 18:44
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    @MoisheKohan Its not actually different, its more verbose and uses the weaker result (which my proof in the link also used before I replaced normality with pseudonormality). But I believe the one for pseudonormality would look pretty much the same when the details are made explicit. – Jakobian Dec 27 '24 at 18:52
  • Even more generally normality can be replaced by "property $D$": that for any disjoint closed $A, B\subseteq X$ with $A$ countable and discrete, there exists disjoint open $U, V$ with $A\subseteq U$ and $B\subseteq V$. This includes normal, pseudonormal and countably compact spaces. A locally compact non-compact space with property $D$ is such that $\beta X$ is not path-connected. – Jakobian Dec 28 '24 at 13:17

1 Answers1

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  1. Suppose that $X$ is a locally compact normal noncompact Hausdorff space. In particular, $X$ is completely regular and, hence, the canonical embedding $\iota: X\to \beta X$ from $X$ to its Stone–Čech compactification, is a topological embedding with open image (which I will identify with $X$ itself). I claim that $\beta X$ cannot be path-connected, moreover, there is no continuous path connecting any point $x\in X$ to any point $y\in Y:=\beta X\setminus X$. Indeed, suppose such path exist. Then, since $X$ is open in $\beta X$, without loss of generality, we can assume that such a path has the form $p: [0,1]\to \beta X$, $p([0,1))\subset X$, $p(1)=y\in Y$. Then $p: [0,1)\to X$ is a proper map. Consider two sequences $s_n, t_n\in [0,1)$ converging to $1$ and having disjoint images $A$ and $B$ respectively. After passing to subsequences, we can assume that $p$ is injective on the images of $(t_n)$ and $(s_n)$. By the construction, $A\cup B$ is closed in $X$. Define a continuous map $g: A\cup B\to [0,1]$ by $$ g(p(s_n))= s_n, g(p(t_n))=1-t_n. $$ Since $X$ was assumed to be normal, by Tietze–Urysohn extension theorem, $g$ extends to a continuous map $f: X\to [0,1]$. By the universal property of the Stone–Čech compactification, the map $f$ extends to a continuous map $F: \beta X\to [0,1]$. Since $p$ was a continuous map and $$ \lim_{n\to\infty} p(s_n)= \lim_{n\to\infty} p(t_n)=y, $$ we have $$ \lim_{n\to\infty} F(p(s_n))= \lim_{n\to\infty} F(p(t_n))= F(y). $$ However, by the construction, $F(p(s_n))=s_n, F(p(t_n))=1-t_n$ and, thus, these sequences have distinct limits in $[0,1]$ (respectively, $1$ and $0$). This is a contradiction. Hence, $\beta X$ is not path-connected. This answers half of your question.

  2. The answer to the second half is in my earlier comment, here it is for the sake of completeness.

Let $X$ be any topological space. Define a new topological space $Y=X\sqcup \{a\}$ where open subsets in $Y$ are either open subsets of $X$ or (if they contain $a$) are equal to the entire $Y$. Of course, $Y$ is only Hausdorff if $X=\emptyset$. Nevertheless, $Y$ is a compactification of $X$: Clearly every open cover of $Y$ contains $Y$ as one of its elements, hence, admits a finite subcover. Furthermore, the inclusion $X\to Y$ is a homeomorphism to its image and the image is dense in $Y$. I claim that $Y$ is contractible, hence, simply-connected. Indeed, define a map $H: Y\times [0,1]\to Y$ by $H(y,t)=y$, $t<1$ and $H(y,1)=a$ for all $y\in Y$. Let's check that this map is continuous: If $U\subset X$ is open, then $H^{-1}(U)=U\times [0,1)$. If $U=Y$, then $H^{-1}(U)=Y\times [0,1]$. And there are no other open subsets of $Y$. Hence, $H$ is continuous.

Edit. The true question, I think, is about existence of a compactification with is path-connected but not simply-connected. However, one can prove that $[0,1)$ does not have one.

Moishe Kohan
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