Let me first make sure I have the correct definitions because my question will be about the difference about the two and there may be some massive confusion on my part.


A topological space $X$ is said to be **completely regular** or **Tychonoff** when it is Hausdorff and satisfies the following equivalent conditions:


* For every $x \in X$ and closed $F \subseteq X$ such that $x\not\in F$, there exists a continuous $f\colon X\to\mathbb{R}$, which we can assume to have values in $[0,1]$, such that $f(x) = 0$ and $f|\_F = 1$.
* The map $X \to [0,1]^{C(X,[0,1])}$ taking $x\in X$ to the family $(f(x))\_{f\in C(X,[0,1])}$ of its images under every continuous $f\colon X\to[0,1]$ defines a homeomorphism of $X$ to its image.
* The Stone-Čech compactification map $X \to \beta X$ defines a homeomorphism of $X$ to its image.
* There exists a compact [Hausdorff] space $K$ such that $X$ is homeomorphic to a subspace of $K$.


On the other hand, a (necessarily Hausdorff) topological space $X$ is said to be **functionally Hausdorff** (or **Urysohn**, but some people use this to mean something different, so it's probably best to avoid this terminology) when it satisfies the following equivalent conditions:


* For every $x,y \in X$ such that $x\neq y$, there exists a continuous $f\colon X\to\mathbb{R}$, which we can assume to have values in $[0,1]$, such that $f(x) = 0$ and $f(y) = 1$.
* The map $X \to [0,1]^{C(X,[0,1])}$ taking $x\in X$ to the family $(f(x))\_{f\in C(X,[0,1])}$ of its images under every continuous $f\colon X\to[0,1]$ is injective.
* The Stone-Čech compactification map $X \to \beta X$ is injective.
* There exists a continuous injective map $X \to K$ with $K$ a compact [Hausdorff] space.


I note that example 91 (the “deleted Tychonoff corkscrew”) in Steen & Seebach's *Counterexamples in Topology* gives an example of a functionally Hausdorff space which is not completely regular, showing that the two notions are not equivalent.


Since until recently I thought these two notions were equivalent (I somehow thought that $X \to \beta X$ was automatically an embedding when it is injective), my goal is essentially to dispel the confusion I had; I first have to ask:



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> **Question 0a:** Is the above account correct? (Are the properties I claim to be equivalent indeed equivalent, and equivalent to standard definitions for the terms they claim to define?)
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Every topological space $X$ has a **complete regularization** or **Tychonoff-ization**, namely a continuous map $X \to X'$ with $X'$ a completely regular space, such that every continuous map $X \to Y$ with $Y$ completely regular uniquely factors as $X \to X'\to Y$. (I.e., the functor $X \mapsto X'$ is left adjoint to the inclusion functor from the full subcategory of completely regular spaces to that of topological spaces.) This $X'$ can be defined as the image of the Stone-Čech compactification map $X \to \beta X$ with the subspace topology; in particular, $X \to X'$ is always surjective.



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> **Question 0b:** Is this still correct?
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**Edit** (2019-06-08): Essentially the above description of complete regularization functor is given, under the name “**Tychonoff functor**” in the paper “The Tychonoff Functor and Related Topics” by T. Ishii, chapter 6 (p.203–243) in K. Morita & J. Nagata (eds.), *Topics in General Topology* (1989). So it would appear that it is correct.


Now I thought $X'$ was a quotient space of $X$. This can't be the case because, if what I wrote above is correct, the equivalence relation (“having the same image in $X'$”) is simply “having the same image under every continuous function $X\to\mathbb{R}$ (or equivalently $X\to[0,1]$)”, which is trivial for a functionally Hausdorff space, yet the latter is not necessarily completely regular.


But this is problematic because in section d-2 (“Higher Separation Axioms”, p.158–159) of the *Encyclopedia of General Topology* (Hart, Nagata & Vaughan eds.) one reads:



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> “To every space $X$ one can associate a Tychonoff space $Y$ as follows. Two points $x$ and $y$ in $X$ are equivalent if $f(x) = f(y)$ for all continuous real-valued functions $f$ on $X$. The corresponding *quotient space* $Y$ is Tychonoff and the rings $C(X)$ and $C(Y)$ of real-valued continuous functions are isomorphic; the same holds for the rings $C^\*(X)$ and $C^\*(Y)$ of bounded real-valued continuous functions.”
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It would seem to me that this assertion is contradicted by the existence of the aforementioned counterexample in Steen & Seebach.



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> **Question 0c:** Am I correct in believing that the above quote is in error? (Or did I miss some fine print or hidden assumption?)
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Now assuming all of the above is correct, there are two natural questions which are left open:



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> **Question 1a:** Does every topological space $X$ have a “functional Hausdorffization” (or “Urysohnization”), namely, does the inclusion functor from the full subcategory of functionally Hausdorff spaces to that of topological spaces have a left adjoint? • **Question 1b:** If so, is it given by quotienting by the equivalence relation “$f(x) = f(y)$ for all continuous real-valued functions $f$ on $X$” or is there some subtlety?
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**Edit** (2019-06-08): The paper “A universal factorization theorem in topology” (*Canad. Math. Bull.* **9** (1966) 201–207), by R. Sharpe, M. Beattie & J. Marsen defines an equivalence relation $T\_{3/2}$ (it is not clear to me whether this is a typo for $T\_{3\frac{1}{2}}$, and whether they may be under the same confusion as mentioned above) on any topological space by $x\,\mathbin{T\_{3/2}}\,y$ when for every continuous $f\colon X\to [0,1]$ we have $f(x)=f(y)$, i.e., the relation defined above. Calling just $R$ this equivalence relation, according to this paper's main theorem, the crucial point in answering questions 1ab positively is to check that $R = \lim R$ in the notation of the paper, where $\lim R$ is the equivalence relation defined by $x\mathbin{(\lim R)}y$ when for every continuous $f\colon X\to Z$ with $Z$ such that $R$ is just equality (i.e., in our case, every functionally Hausdorff $Z$) we have $f(x)=f(y)$; but this is trivial: so it appears that questions 1ab have a positive answer.



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> **Question 2a:** Even if complete regularization is not given by a quotient, is there still a unique coarsest equivalence relation $R$ on any topological space such that $X/R$ is completely regular? • **Question 2b:** If so, can we describe $R$ concretely, and can we describe the natural continuous map $X' \to X/R$ (where $X'$ is complete regularization as defined above)?
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Tychonoff-ization and Urysohn (functionally Hausdorff) topological spaces
**0a** Correct, except for one point: is $X$ is not completely regular then there is no $\beta X$, so the third equivalence in `functionally Hausdorff' does not exist.


**0b** Not quite, the paper mentioned in your edit maps $X$ into a Tychonoff cube and lets $X'$ be the image, see 0a: there is no $\beta X$ available.


**0c** That is indeed erroneous; it is the quotient of the **set** $X$ endowed with the (weaker) topology generated by the continuous functions.


**1a** This is correct, as you point out below the question.


**1b** There is no subtlety, every continuous $f:X\to[0,1]$ will factor through that equivalence relation.


**2a** Assuming [this definition of coarse](https://en.wikipedia.org/wiki/Partition_of_a_set#Refinement_of_partitions): the coarsest equivalence relation on the underlying set (all points equivalent) gives a one-point quotient, which is certainly completely regular. In that case the answer to **2b** is an obvious yes.


**2b** Let $X$ be the deleted corkscrew; its complete regularization $X'$ is just the set underlying set $X$ with a weaker topology: that generated by the continuous functions. It has many completely regular quotients: for every $\alpha<\omega\_1$ and $n<\omega$ let $C(\alpha,n)$ be the set of points $(i,\beta,m)$ with $i\in\mathbb{Z}$, $\beta<\alpha$ and $m>n$. The set $C(\alpha,n)$ is clopen in the corkscrew-without-$\{a^+,a^-\}$, hence so is $\{a^+\}\cup C(\alpha,n)$ in $X$. Identifying that set to a point yields a completely regular quotient. There is no finest equivalence relation among these and their common refinement of all these yields the identity relation, which does not yield a completely regular quotient.


