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Let $\mathbb{S}$ be the Sierpiński space, the two pointed space $\{ 0, 1 \}$ with open sets $\{0 \}$, $\emptyset$, $\{ 0, 1 \}$. We give $\{ 0, 1 \}$ a partial order where $0 < 1$.

Let $X$ be a topological space. Consider the space $Y = \prod_{f : X \rightarrow \mathbb{S}} \mathbb{S}$. This space is compact by Tychonoff's theorem.

There is a natural map $f : X \rightarrow Y$ and the closure of the image of $f$ in $Y$, $X_0$. This is a closed subspace of a compact space and so it is compact.

Is $X_0$ the Stone–Čech compactification of $X$ under certain conditions? I suspect that it always is. However, some have told me that this construction only works for $T_{ 3 \frac{1}{2}}$ spaces. I would find it quite helpful if someone could link me to a source explaining the limitations of this construction if there are some. Alternatively, I think there is an adjoint functor theorem in terms of cogenerators? See the special adjoint functor theorem at nLab. It seems like every space is a subspace of a product $\prod_{i \in I} \mathbb{S}$.

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    $\begingroup$ A subspace of $\prod_{i \in I} \Bbb S$ is always $T_0$. Not all spaces are $T_0$. If you define $\Bbb S'= \{0,1,2\}$ with topology with only non-trivial open set $\{0\}$, we can embed all spaces in $\prod_{i \in I} \Bbb S'$ products. $\endgroup$ Commented Jan 11, 2022 at 15:03

3 Answers 3

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No, the closure of the image of $f$ in $Y$ is never the Stone-Čech compactification of $X$ unless $X$ is empty. In particular, consider the element $a\in Y$ which is $1$ on every coordinate. Note that the only open subset of $Y$ that contains $a$ is $Y$ itself. So, if $X$ is nonempty, then $a$ will be in the closure $X_0$ of the image of $f$. This means that $X_0$ contains a point whose only neighborhood is the whole space $X_0$. Also, $a$ is not in the image of $f$ (no element of $X$ is in every closed subset of $X$), so $a$ is not the only point of $X_0$. This means $X_0$ is not $T_1$, so it cannot be the Stone-Čech compactification of $X$.

The main moral here is that the Stone-Čech compactification is not about making a space compact: it is about making a space compact Hausdorff. So if you have some sort of universal construction that does not enforce the Hausdorffness condition, there is no reason to expect to get the Stone-Čech compactification.

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  • $\begingroup$ I find interesting your remark "if you have some sort of universal construction that does not enforce the Hausdorffness condition, there is no reason to expect to get the Stone-Čech compactification.". Stone-Cech compactification fits into a weak factorisation system generated by the class of all proper (=closed) morphisms of finite topological spaces (in fact a few are enough), but it is not clear how this universal construction does enforce the Hausdorff condition. $\endgroup$
    – user420620
    Commented Jan 11, 2022 at 10:37
  • $\begingroup$ @user420620: I am not familiar with the construction you are describing and am not sure what the precise statement is, but it sounds like it cannot be correct. If you are starting from all proper morphisms of finite spaces, then it would seem that the "compactification" of a finite space you would get would be itself (and thus not necessarily Hausdorff). $\endgroup$ Commented Jan 11, 2022 at 16:36
  • $\begingroup$ Your counterexample is correct, but a weaker Hausdorff-kind condition does hold. The construction is as follows. Let $P$ be the class of all proper (=closed) morphisms of finite spaces, and let $P^l$, and $P^{lr}:=(P^l)^r$ be its weak orthogonals. Then each map in $P^{lr}$ is proper. A map $f:X\to\{o\}$ factors as $X\xrightarrow {f_l} X'\xrightarrow{f_{lr}} \{o\}$ where $f_l\in P^l$ and $f_{lr}\in P^{lr}$. Then $X'$ is quasi-compact, $\endgroup$
    – user420620
    Commented Jan 11, 2022 at 20:23
  • $\begingroup$ and by construction $X'$ is universal among compactifications $X\to K$ such that $K\to \{o\}$ is in $P^{lr}$, and the latter holds for any compact Hausdorff $K$. In pacticular, $X'$ maps to the Stone-Cech compactification of $X$, which arguably is a Hausdorff-kind property. And I think for Hausdorff normal $X$ you can take $X'$ to be its Stone-Cech compactification. $\endgroup$
    – user420620
    Commented Jan 11, 2022 at 21:14
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Correction (2022-01-01): In the first version of this answer, I had defined $E$-compact spaces as those $X$ such that the natural evaluation map $X \to E^{C(X,E)}$ is a homeomorphism on a closed subspace: this is because I had missed an assumption that $E$ is Hausdorff (in section IV of Mrówka's paper cited below): the correct definition of being $E$-compact (corrected below) is that $X$ is homeomorphic to a closed subspace of some power $E^I$ of $E$ (again, when $E$ is Hausdorff, this is equivalent to the definition I had initially written — see theorems 2.1(d) and 4.10 of Mrówka — but not in general). After fixing the answer below, it is less directly relevant to the question as it was asked (which has, in any case, been satisfactorily answered by Eric Wofsey), but I believe it and the references given are still very relevant to the spirit of the question, so I am leaving it (after correction) rather than deleting.

The question is related to the concept (introduced by Engelking and Mrówka) of $E$-compact spaces, where $E$ is an arbitrary topological space: a space $X$ is called $E$-compact when $X$ is homeomorphic to a closed subspace of some power $E^I$ of $E$; furthermore, the $E$-compactification of $X$ (which can be denoted $\beta_E X$) is the closure of the image of the natural evaluation map $X \to E^{C(X,E)}$ (where $C(X,E)$ denotes the set of continuous maps $X\to E$); see Mrówka, “Further results on $E$-compact spaces. I”, Acta Math. 120 (1968), 161–185 for more on $E$-compactness (note that in this paper, the Sierpiński space is called the “Alexandrov connected dyad” and denoted $\mathcal{F}$; but nothing is said about $\mathcal{F}$-compact spaces because $E$-compactness is studied only for $E$ Hausdorff).

The closest thing to the study of $\mathbb{S}$-compact spaces and $\mathbb{S}$-compactifications in the literature (where $\mathbb{S}$ is Sierpiński space) seems to be the paper by Horst Herrlich, “Compact $T_0$-spaces and $T_0$-compactifications“, Applied Categorical Structures 1 (1993) 111–132, especially remark 1.4(b) which gives necessary and sufficiant conditions for being $\mathbb{S}$-compact (but the previous theorem may be of interest to you as it gives a result of equivalence to compactness replacing “closed” by “nearly closed”), and various results on the $\mathbb{S}$-compactification (esp. in theorem 2.3, proposition 3.1 and section 4).

The preprint “Sierpinski object for affine systems” by Denniston &al. may also be of some relevance here.

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  • $\begingroup$ How is the notion of E-compactness related to the weak factorisation system genearetd by the maps $E\to \{o\}$ sending $E$ to a point, and the embedding $\{c\}\to\{o\to c\}$ of a closed point into the Sierpinski space ? For example, is the class of $E$-compact spaces closed under retracts ? A related notion, which coincides for Hausdorff spaces, of extension closed subspaces is indeed closed under retracts by (D.Harris, Compact spaces and products of finite spaces, Proc.AMS 35(1) 1972). $\endgroup$
    – user420620
    Commented Jan 13, 2022 at 18:36
  • $\begingroup$ Let me explain the remark on weak factorisation systems; in fact I am not sure the weak factorisation system as I describe is well-defined. Let $P^l$, $P^{lr}:=(P^l)^r$ denote the left, resp. left-right, weak orthogonal of a class $P$ of maps. The class $\{\{c\}\to\{o\to c\}\}^{lr}$ is precisely the class of closed subsets because $\{\{c\}\to\{o\to c\}\}^{l}$ is precisely the class of maps with dense image. Hence, $\{E\to\{o\}, \{c\}\to\{o\to c\}\}^{lr}$ does contain all $X\to \{o\}$ for all $X$ $E$-compact because an $lr$-orthogonal is closed under products and composition. $\endgroup$
    – user420620
    Commented Jan 13, 2022 at 18:39
  • $\begingroup$ But an $lr$-orthogonal is also closed under retracts, and I think for Hausdorff $E$ (Harris, above) implies a retract of an $E$-compact space is $E$-compact. But is a space $E$-compact iff $X\to \{o\}$ lies in $\{E\to\{o\}, \{c\}\to\{o\to c\}\}^{lr}$ ? Would this be true if we modify the definition of $E$-compact to extension closed subspaces of powers of $E$ ? $\endgroup$
    – user420620
    Commented Jan 13, 2022 at 18:43
  • $\begingroup$ ah, sorry, it seems a standard argument proves it is the same $\endgroup$
    – user420620
    Commented Jan 14, 2022 at 19:23
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A category-theoretic way to phrase this is to take the weak factorisation system cogenerated by $\mathbb{S}\to\{0\}$ and $\{1\}\to\mathbb{S}$ (where the latter denotes the inclusion of the closed point in $\mathbb{S}$). If you start with more of proper maps, you have a better chance of getting something like Stone-Cech compactification. If you start with cogenerators $E\to\{0\}$ and $\{1\}\to\mathbb{S}$, you would probably get the notion of $E$-compactification mentioned in an answer below.

To make notation more expressive, let me denote $\mathbb{S}$ via its closure preorder $\{o\to c\}$.

Let me explain this remark on weak factorisation systems. Let $P^l$, $P^{lr}:=(P^l)^r$ denote the left, resp. left-right, weak orthogonal of a class $P$ of maps. The class $\{\{c\}→\{o→c\}\}^{lr}$ is precisely the class of closed subsets because $\{\{c\}→\{o→c\}\}^{l}$ is precisely the class of maps with dense image. Hence, $\{E→\{o\},\{c\}→\{o→c\}\}^{lr}$ does contain all $X→\{o\}$ for all $X$ closed subsets of a power of $E$ (i.e. $E$-compact) because an lr-orthogonal is closed under products and composition. But an lr-orthogonal is also closed under retracts, and I think for Hausdorff $E$ a retract of an $E$-compact space is $E$-compact. As described in an answer below, for a space $X$ you may decompose a map $X\to\{o\}$ as $X\to X'\to \{o\}$ where $X'$ is

$E$-compactification of X (which can be denoted $βEX$) is the closure of the image of the natural evaluation map $X→E^{C(X,E)}$ (where $C(X,E)$ denotes the set of continuous maps $X→E$);

Diagram chasing considerations show that $X'\to \{o\}$ is in $\{E→\{o\},\{c\}→\{o→c\}\}^{lr}$, and by construction $X\to X'$ is in $\{E→\{o\},\{c\}→\{o→c\}\}^{l}$.

This is your construction.

Alternatively, I think there is an adjoint functor theorem in terms of cogenerators?

To relate to this, you should probably consider $\{E→\{o\},\{c\}→\{o→c\}\}^{lr}$ as a category, and take the obvious functor to topological spaces.

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