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I've been studying topology and I am trying to find a relation between the compactification of a space $X$ and it's quotent space. For example, one-point compactification of the open interval $(0,1)$ can be thought of as bringing its end points to the point infinity (we consider the embedding $f(t) = (cos(2πt),sin(2πt)$ and then the closure of $f(X)$ is the unit circle which is a compact space). Same can be done with the same quotent map on the closed interval $[0,1]$ and we would get the unit circle as its quotent space. Is there any connection between these terms? (Since the identification of end points during this quotent map can be thought of as gluing them as well).

Mojo Top
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    Here is some related but somewhat technical discussion. – Izaak van Dongen Mar 11 '24 at 16:46
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    If $X$ is Tychonoff you can consider its Stone-Cech compactification, then if $\gamma X$ is any compactification of $X$, you have a map $\beta X\to \gamma X$ etending the inclusion $X\hookrightarrow \gamma X$, so $\gamma X$ is a quotient of $\beta X$. This quotient is obtained by identifying some points of $\beta X\setminus X$ with each other. – Jakobian Mar 11 '24 at 16:49
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    The one point compactification of $[0,1]$ is a disjoint union $[0,1]\sqcup{*}$, not the unit circle. $[0,1]$ is already compact. It's true that you can glue $0,1$ together to get a circle, but connecting that to the o.p.c. of $(0,1)$ is artificial. – Chad K Mar 11 '24 at 16:51
  • Sure, from category theory perspective we need to add a point, but its not an actual compactification then. Note that we're talking about two compactifications of $(0, 1)$, the interval $[0, 1]$ and $S^1$ @ChadK – Jakobian Mar 11 '24 at 16:54
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    @Jakobian: The one-point-compactification of a locally-compact Hausdorff space is a precisely defined concept, both in construction and universal property. The o.p.c. of $(0,1)$ is $S^1$, the o.p.c. of $(0,1]$ is $[0,1]$ and the o.p.c. of $[0,1]$ is $[0,1]\sqcup{*}$. – Chad K Mar 11 '24 at 16:56
  • I disagree, thats just a naming convention. Some authors can as well talk only about one point compactifications of locally compact Hausdorff, but non-compact spaces. What is more precise is talking about the Alexandroff extension of a space, that causes no confusion in nomenclature – Jakobian Mar 11 '24 at 16:57
  • Note that a compactification of a space $X$ is a compact Hausdorff space $Z$ and a dense embedding $X\hookrightarrow Z$. By taking disjoint union with a point, we lose density of $X$. So its not a compactification. The naming convention, for this reason, makes no sense, i.e. is a red herring, and in my opinion should be avoided. – Jakobian Mar 11 '24 at 17:10
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    I don't understand why we're talking about the one-point compactification of $[0, 1]$. I don't think the question mentions it? As I understand it, the question says "a way to think about the one-point compactification of $(0, 1)$ is that the endpoints are "brought together" to form a circle. This is also a way to think about the fact that $[0, 1]/{0, 1}$ is a circle. Is there a connection here?" I think that's an interesting question, and it's fair to say the answer is "yes", closely related to $[0, 1]$ being a compactification of $(0, 1)$. – Izaak van Dongen Mar 11 '24 at 17:12
  • @IzaakvanDongen Yes... sorry... It was a bit off-topic. Its a simple misunderstanding. – Jakobian Mar 11 '24 at 17:19

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Let $X$ be a Tychonoff space (or $T_{3.5}$ space), that is, a Hausdorff space which has a compactification.

There is two most important compactifications of $X$, first one always exists and its the Stone-Cech compactification $\beta X$ of $X$. It can be defined in multiple ways, and it has the property (which determines it up to homeomorphism that fixes $X$) that its the largest compactification of $X$, in the sense that if $\gamma X$ is another compactification, then there exists a (necessarily surjective) continuous function $\beta X\to \gamma X$ which extends the dense embedding $X\hookrightarrow \gamma X$. Then $\gamma X$ is a quotient of $\beta X$, since the map $\beta X\to \gamma X$ is closed, continuous and surjective, so a quotient map.

The second most important compactification of $X$ is the one point compactification $\alpha X$ (here $\alpha X$ is non-standard notation, people often denote it by $X^*$). It doesn't always exist, but when it does, and that is when $X$ is locally compact Hausdorff, its the smallest compactification of $X$ in the sense that if $\gamma X$ is a compactification of $X$, then there exists a continuous map $\gamma X\to \alpha X$ extending the dense embedding $X\hookrightarrow \alpha X$. Here the curious thing is that this property characterizes one point compactification, that is, if a smallest compactification exists, then its unique up to homeomorphism fixing $X$, $X$ is locally compact Hausdorff, and its the one point compactification $\alpha X$ of $X$. Note that here I assume the convention that $\alpha X = X$ when $X$ is already compact.

In your example, $S^1$ is a quotient of $[0, 1]$ by identifying the points of remainder (as its often called) $\{0, 1\} = [0, 1]\setminus (0, 1)$ of the compactification $[0, 1]$ of $(0,1)$. And indeed, since $S^1$ is the smallest compactification of $(0, 1)$, any compactification of $(0, 1)$ has the property that $S^1$ is its quotient.

Jakobian
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    @MojoTop Yes thats true, I suppose its an illustrative example. Here we have two dense embeddings $(x, y)\mapsto [x, y, 1]$ into $\mathbb{RP}^2$, and say, stereographic projection into $S^2$. Its true that $\mathbb{RP}^2$ is a quotient of the sphere via the map $(x, y, z)\mapsto [x, y, z]$. However, note that we are identifying points of $S^2$ in this construction without respecting the copy of $\mathbb{R}^2$ inside of it. – Jakobian Mar 11 '24 at 23:31
  • Sorry I wanted to edit my question and I mistakenly deleted it. I am going to rephrase it. My confusion arised while I was considering the compactification of the real plane. The one point compactification of the real plane is the sphere, but we can also consider the real projective plane as its compactification. We know that the sphere is the smallest one, but we can also notice that the real projective plane is the quotient space of the sphere. Considering the previous example, would this mean that there is a quotent map from the projective plane to the sphere? – Mojo Top Mar 12 '24 at 00:03
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    @MojoTop Yes, thats right. All you need to do is identify all points of the remainder, that is ${[x, y, 0]\in \mathbb{RP}^2 : (x, y, 0)\neq (0, 0, 0)}$ (sometimes you call this the line at infinity). This will be, topologically, a sphere. – Jakobian Mar 12 '24 at 00:18