Maximal Non-Hausdorff Compactification

Question

I have recently started to read about compactifications of topological spaces, however I would like to clear my mind on a few things.

For starters, I am interested in generic topological spaces (not necessarily Tychonoff) although I imagine that not much can be said without imposing restrictions of some sort.

Given a topological space $X$, a $\textbf{compactification}$ (for me) is a $\textbf{compact}$ (not necessarily Hausdorff) space $X^*$ together with a continuous embedding $\phi:X\rightarrow X^*$ such that $\phi(X)$ is dense in $X^*$.

Now, I know a compactification always exists: either $X$ is already compact or we can consider the Alexandroff one-point extension, so that the Corona set $X^*\setminus X$ is, resp., empty or only contains one point. On the other side of the spectrum, there is the Stone-Čech compactification, which is in some sense the largest among Hausdorff compactifications (which exists if and only if the original space is Tychonoff), and this sense is made precise by its universal property.

My question is now as follows: If $X$ is a generic topological space, is there any compactification which satisfies a similar universal property* (so that it can be thought as the biggest compactification and not just the biggest among the Hausdorff ones) ? Related to this, if $X$ is indeed Tychonoff and this "biggest" compactification exists (at least for specific $X$), does it coincide with the Stone-Čech one?

Notes:

1. As I understand the matter so far, I'm aware that one can carry out the Stone-Čech construction in general, but for non-Tychonoff spaces the map $\phi$ will not be an embedding. This is not what I'm looking for (at least for now) since I'm interested in not dropping the embedding requirement, and the same for the density one.
2. I am not expecting any kind of constructional result (at least in general), the only issues of the sort that I'm usually interested in are foundational (e.g., does this need some version of the Axiom of Choice?).

Any reference or direction is highly appreciated.

*Edit for clarity: Honestly I'm not completely sure what the universal property would look like in general, but I mean something on the line of: for any topological space $Y$ and continuous map $f:X\rightarrow Y$ there is a unique (perhaps modulo some equivalence relation) continuous map $f^*:X^*\rightarrow Y$ such that $f^*\circ\phi=f$.

Answer 1

A possible candidate for your question is the Wallman compactification, which works for every $T_1$ space. It coincides with the Stone-Cech compactification iff $X$ is normal (more generally, $wX$ is $T_2$ iff $X$ is $T_4$).

The universal property it satisfies is:

There exists an embedding $\iota:X\to wX$ such that $\iota(X)$ is dense in $wX$. Moreover, given any continuous map $f:X\to K$ with $K$ a $T_2$ compact space, there exists $\tilde f:wX\to K$ such that $\tilde f\circ \iota=f$.

The construction of $wX$ is as follows: denote by $\mathcal{C}(X)$ the family of closed subsets of $X$, by $\mathcal{F}(X)$ the set of ultrafilters in that family and by $\mathcal{F}_0(X)$ the set of nonprincipal ultrafilters. Then $wX:=X\cup \mathcal{F}_0(X)$, with the topology induced by the following base $\mathcal{B}:=\{U\cup \{f\in \mathcal{F}_0(X):\exists A\in f: U\subset A\}:U\in \tau_X\}$, where $\tau_X$ is the topology of $X$.

For more informations on the Wallman compactification, check Engelking's "General topology", chap. 3 (in particular section 3.6)

Concerning your second point, i.e. the dependence on AC: this construction needs some version of AC (the ultrafilter lemma should suffice). I don't know off the top of my head if the existence of such a compactification for general (i.e. $T_1$) spaces also necessarily requires some form of AC, I'll think about it.

EDIT: Actually, the existence, for every $T_{2}$ space, of a compactification satisfying the universal property (from now on, E.C.) above implies the UL, so there's a full circle of implications (and so of equivalences). To prove it, it suffices to prove that E.C. implies Banach-Alaoglu, since that is known to be equivalent to UL (see here for a proof). To do this, first notice that $(X',\sigma(X',X))$ is necessarily $T_2$ (some care is needed here$^1$). Thus $B_{X'}$ with the subspace topology is also $T_2$. Now, given $x\in X$, we can see it as a (continuous) function $\hat x:B_{X'}\to \mathbb [-1,1]$, which must then by the E.C. extend to the compactification of $B$, call it $\beta(B)$. Since $B$ is dense in $\beta(B)$, $\|\varphi(x)\|_\infty\le \|x\|$. Given $g\in\beta(B)$, define $\varphi(g)\in X'$ as $\varphi(g)(x)=\hat x(g)$. By the previous observation, $\varphi:\beta(B)\to B_{X'}$ is a well defined continuous map. Since $B$ is the image of a compact set, it is compact.

$^1:$indeed, the same claim can fail miserably in $ZF$ if we are talking about the weak topology: in $ZF+\neg HB$ ($HB=$Hahn-Banach), there exists a Banach space without nontrivial continous functionals, so the weak topology is just the trivial topology. For the $w*$ topology, however, this is not an issue and the Hausdorfness can be proven as follows: given $y_1,y_2\in X':y_1\neq y_2$, there exists $x\in X:y_1(x)\neq y_2(x)$. Wlog, suppose $y_1(x)>y_2(x)$. Then $A_1:=\{y\in X': y(x)>y_1(x)+\frac{y_1(x)-y_2(x)}3\}$, $A_2:=\{y\in X': y(x)<y_2(x)+\frac{y_1(x)-y_2(x)}3\}$ are disjoint open sets containing respectively $y_1$ and $y_2$.