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Given the following definitions:

Def (Compactification): $(\iota, Y)$ is a compactification of a topological space $X$, if $Y$ is compact and $\iota : X \to Y$ is an embedding such that $\iota(X)$ is dense in $Y$.

Def (Stone-Čech compactification): A compactification $(\beta, \beta X)$ of topological space $X$ is a Stone-Čech compactification of $X$, if for every compact Hausdorff space $Y$ and continuous function $f : X \to Y$ there exists a unique continuous function $g : \beta X \to Y$ such that $f = \beta \circ g$.

Often the Stone-Čech compactification is defined with $\beta X$ being necessarily Hausdorff. Can we deduct that $\beta X$ is Hausdorff from the above definition? In other words: is the above definition equivalent to the common definition?

My thoughts on this:

  1. If for every distinct $x, y \in \beta X$ there is a continuous function $g : \beta X \to Y$ with $g(x) \neq g(y)$ and $Y$ being a compact Hausdorff space, then $\beta X$ is Hausdorff: Since $Y$ is Hausdorff, we can find open disjoint neighborhoods $U, V$ of $g(x)$ and $g(y)$, thus $g^{-1}(U)$ and $g^{-1}(V)$ are open disjoint neighborhoods of $x$ and $y$.

  2. $X$ must be completely regular for the Stone-Čech compactification to exists. Can we use this to show existence of $g$ with $g(x) \neq g(y)$ in (1)?

  3. Kown constructions of the Stone-Čech compactification turn out to be Hausdorff. We can show this is unique up to homeomorphism via the universal property of the Stone-Čech compactification, but only if all other possible Stone-Čech compactifications are necessarily Hausdorff as well.

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    $\begingroup$ Compactifications are normally defined to be Hausdorff compactifications. If $X$ is a $T_1$ space, then its Wallman compactiification is a non-Hausdorff compactification with the required extension property. $\endgroup$
    – Tyrone
    Commented Feb 23, 2022 at 12:45
  • $\begingroup$ You have to add Hausdorff compactification to def. 2 in order to really get the definition of Stone-Čech compactification. Just compactification is not enough, see my answer. $\endgroup$ Commented Feb 23, 2022 at 13:47

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The short answer is no, we cannot deduce $T_2$ from the above definition. There is a classic construction, the Wallman compactification $\omega X$, based on the lattice of closed sets of $X$, which is defined for $X$ that are $T_1$. $\omega X$ is compact (not necessarily Hausdorff), $X$ embeds densely into it and every continuous map from $X$ to a compact Hausdorff space $Z$ can be extended to $\omega X$. For full details and construction, see
Engelking's book General Topology (revised edition, 1989). It's also known that $\omega X \text{ is } T_2 \iff X \text {is } T_4$ (so normal and $T_1$), and in that latter case $\omega X$ and the classically constructed $\beta X$ (as a Hausdorff compactification for any Tychonoff space $X$) are equivalent (and in particular homeomorphic).

So $\omega X$ obeys 2 (and 1), while not always being $T_2$ (take your favourite Tychonoff non-normal space as $X$ etc.)

As to your thoughts at the end, yes, on $\beta X$, like on any Tychonoff (includes $T_1$) space we can separate points by closed functions, yes. But if $X$ is Tychonoff, we have a function $f: X \to [0,1]$ separating $x_1 \neq x_2$ in $X$ in the sense that $f(x_1)=0$ and $f(x_2)=1$ and we can extend it to $\omega X$. But that doesn't help to prove Hausdorffness of $\omega X$! For that we have to separate the extra points in $\omega X\setminus X$ as well.. The remarks above show that we cannot expect anything from this route in general.

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A really easy way to see this doesn't work is to consider what happens if $X$ is already compact. Then $(1_X,X)$ trivially satisfies your definition of a Stone-Cech compactification of $X$, but $X$ may not be Hausdorff.

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Another counter-example, in which $X$ is a $T_{3\frac 1 2}$ space: Let $p=\langle \omega, \omega_1\rangle$ and $X^*=(\omega +1)\times (\omega_1+1)$ and $X=X^*\setminus \{p\},$ where $\omega +1$ and $\omega_1 +1$ each have the $\in$-order topology.

Then id$_X:X\to X^*$ is the Cech-Stone compactification of $X.$

Let $T_{X^*}$ be the topology on $X^*.$ Take $q\not\in X^*$ and let $Y=X^*\cup \{q\}.$ Let the topology on $Y$ be $T_{X^*}\cup \{(U\setminus \{p\})\cup \{q\}: p\in U\in T_{X^*}\}\cup \{Y\}.$ Then $Y$ is compact and id$_X:X\to Y$ is a dense homeomorphic embedding.

If $W$ is a compact Hausdorff space and if $f:X\to W$ is continuous then $f$ extends uniquely to a continuous $f:Y\to W$ (with $f(p)=f(q))$. But $Y$ is not Hausdorff.

I can add proofs upon request.

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