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In general topology by Wiilard, He mentioned in the exercises to the Looped line topology defines as follows: At each point $x$ of the real line other than the origin, the basic neighbourhoods of $x$ will be the usual open intervals centred at $x$. Basic neighbourhoods of the origin will be the sets: $(-\epsilon, \epsilon) \cup (-\infty, -n) \cup (n, \infty)$, for all possible choices $\epsilon > 0$ and $n \in \mathbb{N}$. If there is another that it mentions to Looped line topology?

Also, in, The looped line topology (Willard #4D), the user was asked to verify that the Looped line is a topology.

Done so far. I was able to see that it is $T_2$, compact, by using the definition. Also, it is metrizable since it is regular and second countable.

Interested in. I would like to see the reason why it is homomorphic to extend topology on the real line, $[-\infty, \infty].$

Attempt. I was trying to send $-\infty$ and $\infty$ to $0$ and send other points to themself but I could not finish.

Any help?

00GB
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    The looped line is homeomorphic to a figure eight: in the extended reals $[-\infty,+\infty]$ we identify ${0,-\infty,+\infty}$ to a single point. – Henno Brandsma Oct 28 '21 at 21:00
  • @HennoBrandsma, How can see this? by defining function,for example. – 00GB Oct 28 '21 at 21:12
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    What I described is the function. – Henno Brandsma Oct 28 '21 at 21:13
  • the single point is zero. Right? – 00GB Oct 28 '21 at 21:14
  • Yes indeed. You can tell by the type of neighbourhoods it has. – Henno Brandsma Oct 28 '21 at 21:15
  • @HennoBrandsma, I am assuming this function must be a homomorphism. Right? But we will be sending three points to the same point. So, it is not injective. – 00GB Oct 28 '21 at 21:26
  • In the quotient space I described the three points are one class. So it is 1-1. – Henno Brandsma Oct 28 '21 at 21:28
  • @HennoBrandsma, Maybe I am missing something obvious but I did not see the picture yet. Could you explain if you do not mind? – 00GB Oct 28 '21 at 21:42
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    A picture appears here. (It is exact, if we push along $f(x) = 2 \arctan(x)$ first to map $(-\infty, \infty)$ to the $(-\pi,\pi)$ preimage of that diagram.) There is no sending three points to one point - the open ends at $\infty$ and $-\infty$ are abutted against the point $0$, but no points are identified. – Eric Towers Oct 28 '21 at 22:14
  • @EricTowers, I see the picture. Basically, we will have $f:(-\infty,\infty)\to (-\pi,\pi)$ and take the inverse image of the graph. Right? But endpoints $-\infty$ and $\infty$ will be sent it where? – 00GB Oct 29 '21 at 02:56
  • @HennoBrandsma, what quotient space is defined on $\mathbb R$ but $\mathbb R$ with extend topology. Right? – 00GB Oct 29 '21 at 03:38
  • Yes, but the picture @EricTowers linked to is the result of the extended line with $0$ and two endpoints identified. The looped line is homeomorphic to it. Both open loops correspond to the open rays left and right of $0$ etc. – Henno Brandsma Oct 29 '21 at 04:52
  • @HennoBrandsma : The linked map is from $(-\pi, \pi)$, so there is no need for the extended reals or non-injectivity. – Eric Towers Oct 29 '21 at 05:15
  • @EricTowers that map essentially maps from $\Bbb R$, and is an immersion. I’m talking about a homeomorphism of the looped line space with that image space as a quotient space. Different thing. – Henno Brandsma Oct 29 '21 at 05:17
  • A couple of things: homeomorphism, with an 'e', a topological isomorphism. If a function sends two different things ($-\infty$ and $+\infty$) to the same same thing ($0$), then it's not 1-to-1, isn't a bijection, can't be a homeomorphism. – BrianO Oct 29 '21 at 06:52
  • Brian. If this comment is about my attempt ? I know it would not be this why I was asking. – 00GB Oct 29 '21 at 11:44
  • @HennoBrandsma, I think, we can see the compactness by just suing the definition directly. Like, take open cover ${U_\alpha\colon \alpha\in\Delta}$ for $\Bbb R$ then there exists $\alpha_0$ such that $U_{\alpha_0}=(-\infty,n)\cup(-\epsilon,\epsilon)\cup(n,\infty)$ with $0\in U_{\alpha_0}$. Next, notice that $[-n,-\epsilon]$ and $[\epsilon,n]$ are closed and bounded, so they are compact, as needed. Is that right? – 00GB Oct 31 '21 at 16:13
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    Yes that works fine. – Henno Brandsma Oct 31 '21 at 16:52
  • @HennoBrandsma, Are you aware of any book that mentioned looped line topology rather than Willard. – 00GB Oct 31 '21 at 16:57
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    No I’m not. It’s just his way to define the lemniscate. – Henno Brandsma Oct 31 '21 at 17:00
  • @HennoBrandsma, Following looped line, it is not holomorphic to unite circle, $S^{1}$? Right? The way that I see, by connectedness. Like $\Bbb R\setminus{0}$ is connected but remove one point from $S^{1}$. Is that right? If there is another way to see they are holomorphic. – 00GB Nov 02 '21 at 03:05
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    They are not homeomorphic by connectedness arguments, yes. The looped line has a cut point while the circle has not. – Henno Brandsma Nov 02 '21 at 05:37
  • @HennoBrandsma, Is there another way to see they are not homeomorphic rather than using connectedness? Thank you much – 00GB Nov 03 '21 at 22:15

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Let $L$ be the image of the map $\beta: (-\pi,\pi) \to \Bbb R^2$, a "lemniscate", where $\beta(t)=(\sin 2t, \sin t)$, see this question for a picture. Define the map $f: \Bbb R \to L$ by $\beta(2\arctan(x))$ and note that this is continuous (reals in the usual topology), and 1-1.

Also let $Y = [-\infty,+\infty]{/}\{-\infty, +\infty,0\}$ in the quotient topology induced by the identification map $q: [-\infty,+\infty]$ that identifies the three points, and the extended reals have their standard topology with basic neighbourhoods $[-\infty, n), n \in \Bbb Z$ for the left end point and $(n,+\infty]$ for the right end point. We can extend $f$ to $\hat{f}: [-\infty,+\infty] \to L$ by defining $\hat{f}(\pm \infty) = 0$ as well, which is continuous and then $Y \simeq L$ as $q$ factorises through it.

There remains a simple verification that $L \simeq \Bbb R$ where the latter has this looped line topology. $f$ essentially becomes the homeomorphism.

Henno Brandsma
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  • Henno Brandsma, follow about looped line. I know looped line has the fixed point property. So, mu question if removed $1$ from $\Bbb R$, that is, $\Bbb R\setminus{1}$ with looped line . does still have the fixed point property? – 00GB Jan 15 '22 at 05:10
  • Henno Brandsma, did you see my last comment? – 00GB Jan 16 '22 at 02:37
  • @00GB Yes, but I don't know. – Henno Brandsma Jan 16 '22 at 06:19