nLab separation axioms in terms of lifting properties

Idea

In point-set topology, most of the separation axioms that are traditionally considered on topological spaces turn out (Gavrilovich 2014) to have an equivalent reformulation in terms of lifting properties, namely of the given space against, typically, a map of finite topological spaces which reflects the “opposite property” or the “archetypical counterexample” to the given separation condition, in a sense (“Quillen negation”).

To see this, first notice/recall the following two basic examples of lifting properties in diagrams in TopSp.

The property

• $X \xrightarrow{f} Y$ is a surjective function.

means equivalently that $f$ has the right lifting property against the unique map from the empty space to the point space, the simplest map which is not surjective:

Namely, a commuting diagram of the outer form is equivalently just the choice of a single point in $Y$ (this being the image of the bottom map), and the existence of the dashed lift means that any such point has a preimage through $f$. This is the very definition of surjectivity.

Note that we have defined surjectivity with help of a simplest counterexample: the map on the left is perhaps the simplest example of a non-surjective map. We shall see that this pattern holds in other examples as well. Part of the reason is that a map never has the left lifting property with respect to itself unless it is an isomorphism, and thus taking the lifting property is a simplest way to define a class of morphism not containing a given counterexample or without a given property in a matter useful in a diagram chasing computation.

Similarly, the property

• $X \xrightarrow{f} Y$ is an injective function.

means equivalently that $f$ has the right lifting property against the unique map from the discrete space with two elements $Dsc(\{0,1\})$

$$Opens \Big( Dsc \big( \{0,1\} \big) \Big) \;\; \coloneqq \;\; \Big\{ \varnothing ,\, \{0\} ,\, \{1\} ,\, \{0,1\} \Big\}$$

to the point space, one of the simplest non-injective maps:

Namely, that such a lift exists means that the two points in the image of the top map – which may be any two points $x_1, x_2 \,\in\, X$ (by the nature of the discrete topology) such that (by commutativity of the diagram) their images are equal, $f(x_1) \,=\, f(x_1) \;\in\; Y$ – must already have been equal themselves. This is the very definition of injectivity.

Now consider the variant of the previous example where the 2-element set is equipped instead with the codiscrete topology, whose only open subsets are the empty set and the set containing both elements

(1)$$Opens \Big( CoDsc \big( \{0,1\} \big) \Big) \;\coloneqq\; \big\{ \begin{array}{l} \varnothing, \{0,1\} \end{array} \big\} \,.$$

In terms of this, the property

• $X$ is a $T_0$-space.

means equivalently that its unique map $X \to \ast$ to the point space has the right lifting property against the unique map from the codiscrete space with two elements to the point space:

Namely, now continuity restricts the top map to be such that neither $x_i \in\; X$ is contained in an open subset that does not contain the other (this is the usual way to state the axiom): For if it were, then the pre-image of that open subset would be $\{x_i\} \,\subset\, CoDsc\big( \{0,1\} \big)$, which however is not open (1). This means that if $X_1$ is a $T_0$-space then any such top map must be the constant function, and this is equivalent to the existence of a lift in the diagram.

Note that codiscrete space with two points is a simple example of not a $T_0$-space.

Proceeding in this manner, one sees that the property

• $X$ is a $T_1$-space.

means equivalently that its unique map $X \to \ast$ to the point space has the right lifting property against the unique map from the Sierpinski space $Sierp$, again the simplest not a $T_1$-space.

$$Set \big( Sierp \big) \;\coloneqq\; \{0,1\} \,, \;\;\;\;\;\; Opens \big( Sierp \big) \;\coloneqq\; \big\{ \varnothing ,\, \{0\} ,\, \{0,1\} \big\}$$

to the point space:

Namely, the previous argument applies, but only to the point $1 \in Sierp$, while now $\{0\} \,\subset\, Sierp$ is open. Therefore the existence of lifts now means that any two points must both have an open neighbourhood not containing the other, which is the definition of a $T_1$-space.

Next, a space is Hausdorff (axiom $T_2$) iff every two distinct points have disjoint open neighbourhoods. To give two disjoint open subsets is to give a map to the space with two open points and one closed point. To give two distinct points is to give an injective map. Hence, every two distinct points have disjoint open neighbourhoods iff any such injective map extends to a map to that space with three points.

• $X$ is a $T_2$-space.

This is represented by the following lifting property diagram.

Axioms $T_3-T_5$ and others require finite topological spaces with 4 to 7 points, and we need to introduce appropriate notation.

We give one more reformulation which does not require special notation. A space is extremally disconnected iff the closure of an open subset is open, or, equivalently, the closures of disjoint open subsets are disjoint.

• $X$ is extremally disconnected.

This is represented by the following lifting property diagram.

Background and notation

In order to economically define and denote the finite topological spaces which will appear in the lifting problems discussed below, we may encode them through their specialization order.

Recall that the specialisation preorder on the underlying set of points of a topological space $X$ is the preorder whose order relation, for any $x, y \in X$, is

(2)$$x \,\leq\, y \;\;\;\;\;\;\text{ iff }\;\;\;\;\;\; x \,\in\, cl(y) \,,$$

where the right hand side means that the following two equivalent conditions hold:

1. $x$ is in the topological closure of $y$;

2. every open subset which contains $x$ also contains $y$.

We may regard these preordered sets equivalently as (thin and strict) categories, whose

• objects are the points of $X$,

• morphisms reflect the order relation:

  for $x, y \,\in\, X$ there exists a unique morphism

  $$y \,\leftarrow\, x \;\;\;\text{ iff }\;\;\; y \,\leq\, x\;$$

  (the mnemonic: view the arrow head of “$\leftarrow$” as a “$\lt$”-symbol)

For a finite topological space $X$, this specialisation preorder $Spec(X)$ – or equivalently the corresponding category, which we shall conceptually conflate with the pre-ordering – uniquely determines the topology:

A subset $C \,\subset X\,$ is closed iff the following equivalent conditions hold:

1. $C$ downward closed in the specialization order;

2. there are no morphisms going out of $C$ in the corresponding category.

Accordingly, we may and will denote finite topological spaces by showing the graph containing their points with a system of arrows indicating the generating morphisms in their corresponding specialization preorder-category.

In doing so, often it will be convenient to show multiple copies of the same object, i.e. the same point. Noticing that in the strict category corresponding to a preorder, an isomorphism between two objects does not imply their equality, we have and distinguish the following two notations:

• ‘$=$’ denotes an identity morphism,

• ‘$\leftrightarrow$’ denotes an isomorphism

  (hence, because the category is thin, the existence of arrows back and forth $\leftrightarrows$).

For example:

finite topological space	open subsets	specialization order	as picture
discrete space $Dsc\big(\{ 0,1 \}\big)$	$\Big\{\; \varnothing,\, \{0\},\, \{1\},\, \{0,1\} \;\Big\}$	$\Big\{\; 0 \phantom{\leftarrow} 1 \;\Big\}$	$\boxed{\{\boxed{0},\boxed{1}\}}$
Sierpinski space $Sierp$	$\Big\{\; \varnothing,\, \{0\},\, \{0,1\} \;\Big\}$	$\Big\{\; 0 \rightarrow 1 \;\Big\}$	$\boxed{\{\boxed{0}\rightarrow 1\}}$
codiscrete space $CoDsc\big( \{0,1\} \big)$	$\Big\{\; \varnothing,\, \{0,1\} \;\Big\}$	$\Big\{\; 0 \leftrightarrow 1 \;\Big\}$	$\boxed{\{0\leftrightarrow 1\}}$
point space $\ast$	$\Big\{ \varnothing,\, \{0\} = \{1\} \;\Big\}$	$\Big\{\; 0 = 1 \;\Big\}$	$\boxed{*}$

Notice here how in $\big\{\; 0 \rightarrow 1 \;\}$ the point $0$ is not closed (as there do emanate arrows form it) while the point ${1}$ is closed (as no arrows emanate from it).

Under this identitification of finite topological spaces $X$ with preordered sets regarded as thin categories $Spec(X)$, the continuous maps between topological spaces correspond to functors between their specialization preorders:

$$\overset{ \color{blue} { \text{continuous function betw.} \atop \text{finite topological spaces} } \mathclap{\phantom{\vert_{\vert_{\vert_{\vert_{\vert_{\vert}}}}}}} }{ X \xrightarrow{f} Y } \;\;\;\;\;\;\;\;\; \text{corresponds to} \;\;\;\;\;\;\;\;\; \overset{ \color{blue} { \text{functor between} \atop \text{ specialization orders } } \mathclap{\phantom{\vert_{\vert_{\vert_{\vert_{\vert_{\vert}}}}}}} }{ Spec(X) \xrightarrow{Spec(f)} Spec(Y) } \mathrlap{\,.}$$

With specialization orders denoted by their generating graphs as before, and using that there is at most one morphism for every ordered pair of objects, we may specify such functors $Spec(f)$ simply by labeling each object in their codomain by the same symbol as its preimage.

For example

$$\big\{\, 0 \,\big\} \;\;\; \xrightarrow{\phantom{---}} \;\;\; \big\{\, 0,\, 1 \,\big\}$$

is to denote the functor between specialization orders of discrete spaces with a single and with two elements, respectively, which takes the point denoted “$0$” on the left to the point denoted by the same symbol “$0$” on the right.

In this notation, the following shows the canonical functors between the four examples of specialization orders from the above list:

$$\overset{ \color{blue} { \text{discrete space} \atop \phantom{-} } }{ \Big\{\; 0 \phantom{\leftarrow} 1 \;\Big\} } \;\;\xrightarrow{\phantom{---}}\;\; \overset{ \color{blue} { \text{Sierpinski space} \atop \phantom{-} } }{ \Big\{\; 0 \rightarrow 1 \;\Big\} } \;\;\xrightarrow{\phantom{---}}\;\; \overset{ \color{blue} { \text{codiscrete space} \atop \phantom{-} } }{ \Big\{\; 0 \leftrightarrow 1 \;\Big\} } \;\;\xrightarrow{\phantom{---}}\;\; \overset{ \color{blue} { \text{point space} \atop \phantom{-} } }{ \Big\{\; 0 = 1 \;\Big\} } \mathrlap{\,.}$$

Notice here the role of the equality sign: In the denotation of a functor as above, arrows may be sent to equality signs (but not the other way around): This corresponds to the corresponding continuous function “gluing” these points, in that it is (at least locally) the coprojection onto the quotient by the subset of points that are being identified.

For example, the following denotes the functor corresponding to a map that “glues” three points to each other:

$$\left\{ \;\; \array{ & & U && && V \\ & \swarrow && \searrow && \swarrow && \searrow \\ a && && x && && b } \;\; \right\} \;\;\;\;\;\;\;\; \xrightarrow{\phantom{------}} \;\;\;\;\;\;\;\; \left\{ \;\; \array{ && U &\color{red}=& x &\color{red}=& V \\ & \swarrow & && && & \searrow \\ a & & && && && b } \;\; \right\}$$

In this notation, the open subsets of the domain are

$$\{\{U\}, \{V\}, \{a,U\}, \{V,b\}, \{U,x,V\}\},\{a,U,x,V,b\}\}$$

In the codomain, points $a$ and $b$ are closed, and the only point left is open.

Separation of subsets

Here we describe how topological separation of subsets of a topological space may be expressed in terms of factorizations of their joint characteristic function.

In all of the following, $S$ denotes a topological space and

$$F,\,G \;\subset\; S$$

a pair of subsets.

Disjoint

We say that a pair of subsets is disjoint if their intersection is empty:

(3)$$F,\, G \,\subset\, S \,, \;\;\;\; F \cap G \,=\, \varnothing$$

This situation (3) may equivalently be expressed by a characteristic continuous function from $S$ to the codiscrete topological space of 3 elements. We will suggestively denote these three elements by $e_F$, $e_G$ and $e_\varnothing$, respectively, so that in the above notation the codiscrete topology on them reads like this:

$$CoDisc \big( \{ e_F ,\, e_F ,\, e_{\varnothing} \} \big) \;\; = \;\; \big\{ e_F \leftrightarrow e_G \leftrightarrow e_\varnothing \big\} \,.$$

Namely, any function to such a codiscrete space is continuous, and any function to the set with 3 elements partitions its domain into the disjoint preimages of these three elements, which we may regard as a pair $F, G$ of disjoint subsets and their complement $S \setminus \{F \cup G\}$:

(4)$$\array{ S_{F,G} &\colon& S &\longrightarrow& \big\{ e_F \leftrightarrow e_G \leftrightarrow e_{\varnothing} \big\} \\ && x &\mapsto& \left\{ \array{ e_F &\vert& x \in F \\ e_G &\vert& x \in G \\ e_\varnothing &\vert& otherwise } \right. }$$

We may now encode topological separation properties of the two subsets in terms of factorizations (hence liftings) of this their characteristic function:

Often $F$ and $G$ will be points (identified with their singleton subsets); in that case, one usually says distinct in place of disjoint.

Often $F$ or $G$ will be closed sets; notice that disjoint closed sets are automatically separated, while a closed set and a point, if disjoint, are automatically topologically disjoint.

To express the assumption that $F$ or $G$ is closed, we may modify the topology on $\{e_F,e_{\varnothing}, e_G$ and make the points $e_F$ and/or $e_G$ closed as appropriate.

Topologically disjoint

We say that a disjoint pair of subsets (3) is topologically disjoint if there exists a neighbourhood of one set that is disjoint from the other set:

(5)$$F, G \;\text{are topologically disjoint} \;\;\;\;\;\;\;\;\; \Leftrightarrow \;\;\;\;\;\;\;\;\; \Big( \underset { U \underset{\mathclap{nbhd}}{\supseteq} F } {\exists} \;\; U \cap G = \varnothing \Big) \;\;or\;\; \Big( \underset { V \underset{\mathclap{nbhd}}{\supseteq} G } {\exists} \;\; F \cap V = \varnothing \Big) \,.$$

(Notice that topologically disjoint sets must be disjoint.)

The topological separation condition (5) on a pair of disjoint subsets means equivalently that their characteristic function (4) factors as

or as

The required neighbourhoods $U\supset F$ or $V\supset B$ are the preimages under the diagonal arrow of the open neighbourhoods $\big\{ e_F \leftrightarrow e_{\varnothing} \big\}$ of point $e_F$, or $\big\{ e_{\varnothing} \leftrightarrow e_G \big\}$ of point $e_G$, respectively.

Topologically separated

We say that a disjoint pair of subsets (3) is separated if each set has a neighbourhood that is disjoint from the other set:

(6)$$F, G \;\text{are topologically disjoint} \;\;\;\;\;\;\;\;\; \Leftrightarrow \;\;\;\;\;\;\;\;\; \Big( \underset { U \underset{\mathclap{nbhd}}{\supseteq} F } {\exists} ,\; U \cap G = \varnothing \Big) \;\;and\;\; \Big( \underset { V \underset{\mathclap{nbhd}}{\supseteq} G } {\exists} ,\; F \cap V = \varnothing \Big) \,.$$

(Notice that separated sets must be topologically disjoint and disjoint.)

The separation condition (6) on a pair of disjoint subsets means equivalently that their characteristic function (4) factors as

and as

The two diagrams above can equivalently be combined into a single diagram; the separating neighbourhoods $U\supset F$ and $V\supset B$ are the preimages under the diagonal arrow of the open subsets $\big\{ e_F \leftrightarrow e_U \leftarrow e_{\varnothing} \big\}$ and $\big\{ e_{\varnothing} \rightarrow e_V \leftrightarrow e_G \big\}$.

Separated by neighbourhoods

We say that a disjoint pair of subsets (3) is separated by neighbourhoods if the sets have disjoint neighbourhoods

(7)$$U \subseteq V \;\text{and}\; V \subseteq G \;\;\;\text{such that}\;\;\; U \cap V = \varnothing \,.$$

The separation condition (7) on a pair of disjoint subsets means equivalently that their characteristic function (4) factors as

The separating neighbourhoods $U\supset F$ and $V\supset B$ are the preimages under the diagonal arrow of the open subsets $\big\{ e_F \leftrightarrow e_U \big\}$ and $\big\{e_V \leftrightarrow e_G \big\}$.

Separated by closed neighbourhoods

We say that a disjoint pair of subsets (3) is separated by neighbourhoods if the sets have disjoint closed neighbourhoods, i.e. there exist $U\subseteq V$ and $V\subseteq G$ such that their closures $\bar U$ and $\bar V$ do not intersect

(8)$$U \subseteq V \;\text{and}\; V \subseteq G \;\;\;\;\;\; \text{such that} \;\;\;\;\;\; \bar U \cap \bar V \;=\; \varnothing$$

The separation condition (8) on a pair of disjoint subsets means equivalently that their characteristic function (4) factors as

The separating closed neighbourhoods $\bar U\supset F$ and $\bar V\supset B$ are the preimages under the diagonal arrow of the closed neighbourhoods $\big\{ e_F \leftrightarrow e_U \rightarrow e_{\bar U} \big\}$ of point $e_F$, and $\big\{ e_{\bar V} \leftarrow e_V \leftrightarrow e_G \big\}$ of point $e_G$.

Separated by a function

We say that two disjoint subsets $F$ and $G$ are separated by a function if there exists a continuous real-valued function on the space that maps $F$ to $0$ and $G$ to $1$:

(9)$$\underset {f \,\colon\, S \to \mathbb{R}} {\exists} \left( F \subseteq f^{-1}(\{0\}) \;\wedge\; G \subseteq f^{-1}(\{1\}) \right)$$

Equivalently, we may assume that $f$ takes values in $[0,1]\subseteq \mathbb{R}$.

This separation condition (9) on a pair of disjoint subsets means equivalently that their characteristic function (4) factors as

If $f:S\to [0,1]$ in a separating function as above, we may take the diagonal arrow defined as

(10)$$\array{ f' &\colon& S &\longrightarrow& [0,1]\vee_{\{0,1\}} \{0'\leftrightarrow 0, 1\leftrightarrow 1\} \\ && x &\mapsto& \left\{ \array{ e_F &\vert& x \in F \\ e_G &\vert& x \in G \\ f(x) &\vert& otherwise } \right. }$$

Precisely separated by a function

Finally, we say that two disjoint subsets $F$ and $G$ are separated by a function if there exists a continuous real-valued function on the space that maps precisely $F$ to $0$ and $G$ to $1$:

(11)$$\underset {f \,\colon\, S \to \mathbf{R}} {\exists} \left( F \,=\, f^{-1}(\{0\}) \;\;\wedge\;\; G \,=\, f^{-1}(\{1\}) \right)$$

Equivalently, we may assume that $f$ takes values in $[0,1]\subseteq \mathbb{R}$.

This separation condition (11) on a pair of disjoint subsets means equivalently that their characteristic function (4) factors as

If $f:S\to [0,1]$ in a separating function as above, we may take the diagonal arrow defined as

(12)$$\array{ f' &\colon& S &\longrightarrow& {[0,1]} \\ && x &\mapsto& \left\{ \array{ 0 &\vert& x \in F \\ 1 &\vert& x \in G \\ f(x) &\vert& otherwise } \right. }$$

Notice that sets separated by a function must be separated by closed neighbourhoods (the preimages of $[-\epsilon, \epsilon]$ and $[1-\epsilon, 1+\epsilon]$). Notice that sets precisely separated by a function must be separated by a function.

Separation axioms as lifting properties

In all of the following definitions, ${X}$ is a topological space.

We shall use the symbol “$⧄$” to denote the lifting property of one map against another (compare Joyal-Tierney calculus):

So for $S \xrightarrow{\phi} T$ and $X \xrightarrow{f} Y$ a pair of continuous functions between topological spaces, we write

$$\phi \;⧄\; f \;\;\;\;\;\; \Leftrightarrow \;\;\;\;\;\; \phi \;\text{has the left lifting property against}\; f$$

With this notation, we have the following dictionary between lifting properties and separation axioms.

Kolmogorov spaces ($T_0$)

A topological space ${X}$ is $T_0$ (or Kolmogorov) if any two distinct points in ${X}$ are topologically disjoint.

(It will be a common theme among the following separation axioms to have one version of an axiom that requires $T_0$ and one version that doesn’t.)

This condition is equivalent to the lifting property

$$\big( \{x\leftrightarrow y\} \longrightarrow \{x=y\} \big) \;\;\,⧄\,\;\; \big( {X} \longrightarrow \{*\} \big)$$

hence:

The same property is also defined by the following diagram:

Symmetric spaces ($R_0$)

A topological space ${X}$ is an $R_0$-space (also: symmetric space), if any two topologically distinguishable points in $X$ are topologically separated.

This condition is equivalent to the following lifting property:

$$\big( \{x{\searrow}y\} \longrightarrow \{x\leftrightarrow y\} \big) \;\;\;\,⧄\,\;\;\; \big( {X} \longrightarrow \{*\} \big) \,.$$

Accessible spaces ($T_1$)

$X$ is $T_1$ (or accessible or Fréchet) if any two distinct points in $X$ are separated, i.e.

$$\{x{\searrow}y\} \longrightarrow \{x=y\} \;\;\;\,⧄\,\;\;\; {X} \longrightarrow \{*\}$$

Thus, $X$ is T1 if and only if it is both T0 and R0: indeed, a morphism lifts wrt the composition $\{x{\searrow}y\} \longrightarrow \{x\leftrightarrow y\} \longrightarrow \{x=y\}$ iff it lifts wrt either one of the two morphisms. (Although you may say such things as “T1 space”, “Frechet topology”, and “Suppose that the topological space $X$ is Frechet”, avoid saying “Frechet space” in this context, since there is another entirely different notion of Frechet space in functional analysis.)

The same property is also defined by the following diagrams:

Reregular spaces ($R_1$)

$X$ is R1, or preregular, if any two topologically distinguishable points in X are separated by neighbourhoods. Every R1 space is also R0.

Weakly Hausdorff spaces

$X$ is weakly Hausdorff, if the image of every continuous map from a compact Hausdorff space into $X$ is closed. All weak Hausdorff spaces are T1, and all Hausdorff spaces are weak Hausdorff.

Hausdorff spaces ($T_2$)

$X$ is Hausdorff, or T2 or separated, if any two distinct points in $X$ are separated by neighbourhoods, i.e.

$$\{x,y\} \hookrightarrow \underset{X}{} \;\;\;\,⧄\,\;\;\; \{x{\searrow}\underset{X}{}{\swarrow}y\} \longrightarrow \{x=X=y\}$$

As a lifting diagram this is The interesting case is when the top horizontal arrow $\{x,y\} \to \{x{\searrow}\underset{X}{}{\swarrow}y\}$ maps $x$ and $y$ to the open points $x$ and $y$; in that case the separating neighbourhoods are the preimages of the open points $x$ and $y$ under the diagonal arrow. In the other cases both points map to a copy of $\{0\to 1\}$, and the lifting property reduces to $T_1$. In particular, we see that every Hausdorff space is also T1.

Also, $X$ is Hausdorff if and only if it is both T0 and R1.

Urysohn spaces ($T_{2 \tfrac{1}{2}}$)

$X$ is $T2\frac{1}{2}$, or Urysohn, if any two distinct points in $X$ are separated by closed neighbourhoods. As a lifting property this is

$$\{x,y\} \hookrightarrow {X} \;\;\;\,⧄\,\;\;\; \boxed{\{\overset{\boxed{x}}{}{\searrow}\underset{U}{}{\swarrow}\overset{\boxed{X}}{} }\!\!\!\!\!\!\! \boxed{ {\,\,\,\,\,\,\searrow}\underset{V}{}{\swarrow}\overset{\boxed{y}}{} } \} \longrightarrow \{\boxed{x=x'=X=y'=y}\}$$

As a lifting diagram this is The interesting case is when the top horizontal arrow maps $x$ and $y$ to the open points $x$ and $y$; in that case the separating closed neighbourhoods are of closed subsets $\big\{\overset{\boxed{x}}{}{\searrow}\underset{U}{}\big\}$ and $\big\{\underset{V}{}{\swarrow}\overset{\boxed{y}}{}\big\}$ under the diagonal arrow. The other cases are similar but easier.

Note the 5-point space here contains a copy of the 3-point space used in $T_2$. Hence, every $T2\frac{1}{2}$ space is also Hausdorff.

Completely Hausdorff spaces

$X$ is completely Hausdorff, or completely T2, if any two distinct points in X are separated by a continuous function, i.e.

$$\{x,y\} \hookrightarrow {X} \;\;\;\,⧄\,\;\;\; [0,1]\longrightarrow \{*\}$$

where $\{x,y\} \hookrightarrow X$ runs through all injective maps from the discrete two point space $\{x,y\}$. As a lifting diagram this is

Every completely Hausdorff space is also $T2\frac{1}{2}$ because there is a surjection ${[0,1]}\vee_{\{0,1\}} \{x\leftrightarrow 0, 1\leftrightarrow y\} \to \{\boxed{\overset{\boxed{x}}{}{\searrow}\underset{U}{}{\swarrow}\overset{\boxed{X}}{} }\!\!\!\!\!\!\! \boxed{ {\,\,\,\,\,\,\searrow}\underset{V}{}{\swarrow}\overset{\boxed{y}}{} } \}$.

Regular spaces

$X$ is regular if, given any point ${x}$ and closed subset $F$ in $X$ such that ${x}$ does not belong to $F$, they are separated by neighbourhoods, i.e.

$$\{x\} \longrightarrow {X} \;\;\;\,⧄\,\;\;\; \{ \boxed{ \boxed{ \overset{\boxed{x}}{} \searrow \underset{X}{} \swarrow \overset{\boxed{U}}{} }\!\!\!\!\!\!\! {\,\,\,\,\,\,\searrow\underset{F}{} } } \} \longrightarrow \{\overset{\boxed{x=X=U}}{}\searrow\underset{F}{}\}$$

As a lifting diagram

Indeed, in the interesting case that $x$ maps to $x$ by the top horizontal arrow, the separating neighbourhoods would be the preimage under the diagonal arrow of $\{\boxed{x}\}$ and $\{\overset{\boxed{U}}{}\searrow\underset{F}{}\}$.

In fact, in a regular space, any such ${x}$ and ${F}$ will also be separated by closed neighbourhoods obtained by “applying” the lifting property again to $\{\overset{\boxed{x}}{}\searrow\underset{X=U=F}{}\}$. Every regular space is also R1.

Regular Hausdorff space ($T_3$)

$X$ is a regular Hausdorff space, or T3, if it is both T0 and regular.[1] Every regular Hausdorff space is also $T2\frac{1}{2}$.

Completely regular spaces

$X$ is completely regular if, given any point ${x}$ and closed set $F$ in $X$ such that ${x}$ does not belong to $F$, they are separated by a continuous function, i.e.

$$\{x\} \longrightarrow {X} \;\;\;\,⧄\,\;\;\; [0,1] \vee_{\{1\}} {\{1\leftrightarrow F\}} \longrightarrow \{ \overset{e_{[0,1]}}{} \searrow F\}$$

Here in $[0,1] \vee_{\{1\}} {\{1\leftrightarrow F\}}$ the points $F$ and $1$ are topologically indistinguishable, $[0,1]$ goes to $x$, and $F$ goes to $F$.

Every completely regular space is also regular.

Tychonoff spaces ($T_{3 \tfrac{1}{2}}$)

$X$ is Tychonoff, or T3$\frac{1}{2}$, completely T3, or completely regular Hausdorff, if it is both T0 and completely regular.[2] Every Tychonoff space is both regular Hausdorff and completely Hausdorff.

Normal spaces

$X$ is normal if any two disjoint closed subsets of $X$ are separated by neighbourhoods, i.e.

$$\emptyset \longrightarrow {X} \;\;\;\,⧄\,\;\;\; \{\underset{x}{}{\swarrow} \overset{U}{}{\searrow} \underset{X}{}{\swarrow}\overset{V}{}{\searrow}\underset{y}{}\} \longrightarrow \{\underset{x}{}{\swarrow}\overset{\boxed{U=X=V}}{}{\searrow} \underset{y}{}\}$$

As a lifting diagram

Indeed, the disjoint closed subsets are the preimges of the closed points $x$ and $y$ under the bottom horizontl arrow, and their separating neighbourhoods are the preimages of open subsets $\big\{\underset{x}{}{\swarrow} \overset{U}{}\big\}$ and $\big\{\overset{V}{}{\searrow}\underset{y}{}\big\}$.

In fact, by Urysohn's lemma a space is normal if and only if any two disjoint closed sets can be separated by a continuous function, i.e.

$$\emptyset \longrightarrow {X} \;\;\;\,⧄\,\;\;\; [0,1]\vee_{\{0,1\}} \{0'\leftrightarrow 0, 1\leftrightarrow 1'\} \longrightarrow \{0=0'{\searrow}e_{[0,1]}{\swarrow}1=1'\}$$

Here in $[0,1]\vee_{\{0,1\}} \{0'\leftrightarrow 0, 1\leftrightarrow 1'\}$ the points $0',0$ and $1,1'$ are topologically indistinguishable, $[0,1]$ goes to $e_{[0,1]}$, and both $0,0'$ map to point $0=0'$, and both $1,1'$ map to point $1=1'$.

Normal Hausdorff spaces ($T_4$)

$X$ is normal Hausdorff, or T4, if it is both T1 and normal. Every normal Hausdorff space is both Tychonoff and normal regular.

Completely normal spaces

$X$ is completely or heriditarily normal if any two separated sets $A$ and $B$ are separated by neighbourhoods $U\supset A$ and $V\supset B$ such that $U$ and $V$ do not intersect, i.e.

$$\emptyset \longrightarrow {X} \;\;\;\,⧄\,\;\;\; \{ \underset{X}{}{\swarrow} \overset{A\leftrightarrow U}{} {\searrow} \underset{U'}{}{\swarrow} \overset{W}{} {\searrow} \underset{V'}{}{\swarrow} \overset{V\leftrightarrow B}{}{\searrow} \underset{X}{} \} \longrightarrow \{U=U',V'=V\}$$

Every completely normal space is also normal.

It is easy to check that a closed inclusion into a heridarily normal space has the same lifting property. As left lifting property is stable under colimits, this implies that a filtered colimit (transfinite composition) of heriditarily normal spaces is heriditarily normal. Compare this argument to colimits of normal spaces.

Perfectly normal spaces

$X$ is perfectly normal if any two disjoint closed sets are precisely separated by a continuous function, i.e.

$$\emptyset\longrightarrow {X} \;\;\;\,⧄\,\;\;\; [0,1]\longrightarrow \{0{\swarrow}X{\searrow}1\}$$

where $(0,1)$ goes to the open point $X$, and $0$ goes to $0$, and $1$ goes to $1$.

Every perfectly normal space is also completely normal.

Applications.

The following notation and terminology helps to discuss applications.

Quillen negation

For a class $P$ of morphisms in a category, its left Quillen negation $P^{⧄ l}$ with respect to the lifting property, respectively its right Quillen negation $P^{⧄ r}$, is the class of all morphisms which have the left, respectively right, lifting property with respect to each morphism in the class $P$. In notation,

$$P^{⧄ l} := \{ i \,\,:\,\, \forall p\in P\,\, i\;\;\;\,⧄\,\;\;\; p\}, P^{⧄r} := \{ p \,\,:\,\, \forall i\in P\,\, i\;\;\;\,⧄\,\;\;\; p\}, P^\lr:=(P^\lrl)^\rlr,..$$

As the terminology might suggest, taking the Quillen negation of a class/property $P$ is a simple way to define a class of morphisms excluding non-isomorphisms from $P$, in a way which is useful in a diagram chasing computation, and is often used to define properties of morphisms starting from an explicitly given class of (counter)examples. A number of elementary notions may also be expressed using the lifting property starting from a list of (counter)examples.

Good behaviour under (co)limits

Reformulations in terms of lifting properties clarify behaviour with respect to (co)limits, as notions defined by lifting properties are closed under (co)limits.

For example, as remarked above, it is easy to check that a closed inclusion into a heridarily normal space has the same lifting property; this is not true for the lifting property defining normality. Hence, filtered colimit (transfinite composition) of closed inclusions of heriditarily normal spaces is heriditarily normal. Compare this argument to what is given in colimits of normal spaces.

The same argument proves the Tychonoff theorem that a product of compact spaces is compact.

Reflection and weak factorisation systems

Lifting properties give rise to reflection and weak factorisation systems. For example, each map $X\to\{o\}$ decomposes as

$$X \xrightarrow{(T_i)^{⧄rl}} X'\xrightarrow{(T_i)^{⧄l}}\{o\}$$

where $(T_i)$ denotes the morphism appearing in the definition of axiom $T_i$, for $i0,1$. This gives a statement about reflection close to separation axioms#Reflection.

In terms of Quillen negation we can understand the $k$-coreflection $kTop\to Top$ from the category of compactly generated spaces as follows.

Let $(CHaus)=\big(\{0,1\}\to\{0=1\}\cup\{\varnothing\to K\,\,:\,\,K\,\,\text{is compact Hausdorff}\,\}\big)$.

First notice it holds $k(X) \xrightarrow{(CHaus)^{⧄r}}X$.

Now consider a decomposition of a map $\varnothing\to X$ as

$$\varnothing \xrightarrow{(CHaus)^{⧄rl}} X_{c.g.}\xrightarrow{(CHaus)^{⧄r}}X$$

such that the latter arrow is injective.

It is easy to see that $X_{c.g.}\xrightarrow{(CHaus)^{⧄r}}X$ means that a subspace of $X_{c.g}$ is closed whenever it is closed in $X$ and its intersection with every compact Hausdorff subsets of (the original topology on) $X$ is closed (in the original topology on $X$). Then $X_{c.g.}$ has all the same closed sets and possibly more, hence all the same open sets and possibly more.

On the other hand,

$$\varnothing \xrightarrow{(CHaus)^{⧄rl}} X_{c.g.} \,\,\,⧄\,\,\,\, k(X) \xrightarrow{(CHaus)^{⧄r}}X$$

hence the obvious map $X_{c.g.}\to k(X)$ is continuous.

Hence, $X_{c.g.}$ is isomorphic to $k(X)$.

Compare our proof of the following lemmas with the one given in subspace topology#pushout.

Lemma

The pushout in Top of any (closed/open) subspace $i \colon A \hookrightarrow B$ along any continuous function $f \colon A \to C$, is a (closed/open) subspace $j: C \hookrightarrow D$.

Lemma

Let $\kappa$ be an ordinal, viewed as a preorder category, and let $F: \kappa \to Top$ be a functor that preserves directed colimits. Then if $F(i \leq j)$ is a (closed/open) subspace inclusion for each morphism $i \leq j$ of $\kappa$, then the canonical map $F(0) \to colim_{i \in \kappa} F(i)$ is also a (closed/open) inclusion.

Proof

This follows from the fact that being a (closed/open) inclusion is a left lifting property:

• $(\emptyset\longrightarrow \{o\})^{rr}=\{\{x\leftrightarrow y\rightarrow c\}\longrightarrow\{x=y=c\}\}^l=\{\{x\leftrightarrow y\leftarrow c\}\longrightarrow\{x=y=c\}\}^l$ is the class of subsets, i.e. injective maps $A\hookrightarrow B$ where the topology on $A$ is induced from $B$

• $\{ \{z\leftrightarrow x\leftrightarrow y\rightarrow c\}\longrightarrow\{z=x\leftrightarrow y=c\} \}^\lrl = \{\{c\}\longrightarrow \{o\rightarrow c\}\}^\lr$ is the class of closed inclusions $A\subset B$ where $A$ is closed

• $\{ \{z\leftrightarrow x\leftrightarrow y\leftarrow c\}\longrightarrow\{z=x\leftrightarrow y=c\} \}^\lrl$ is the class of open inclusions $A\subset B$ where $A$ is open

Urysohn lemma and Tietze extension theorem.

We now explain how to view the standard proof of the Urysohn lemma in terms of lifting properties.

Let $\Lambda_n=\big\{ \underset{c_0}{} \swarrow \overset{o_1}{} \searrow \cdots \swarrow \overset{o_n}{} \searrow \underset{c_n}{} \big\}$ for $n\geq 1$. Pick a sequence of “subdivision” maps, e.g.

$$\Lambda_{2({n+1})}\to \Lambda_{2n}$$

$$c_{2i}\mapsto c_i,\,\,\, o_{2i+1},c_{2i+1},o_{2i+2}\mapsto o_i,\,\,\, 0\leq i\leq n-1,\,\,\,c_{2n}\mapsto c_n,$$

and let $\Lambda_\infty=\lim_{n\to\infty} \Lambda_{2^n},$ be the limit in the category of topological spaces.

In this notation normality (T4) is defined by $\varnothing \to X \;\;\;\,⧄\,\;\;\; \Lambda_2\to \Lambda_1$. Iterating the lifting property shows that $f \;\;\;\,⧄\,\;\;\; \Lambda_2\to \Lambda_1$ implies that $f \;\;\;\,⧄\,\;\;\; \Lambda_{2n}\to \Lambda_n$ and, passing to the limit, $f \;\;\;\,⧄\,\;\;\; \Lambda_\infty\to \Lambda_1$.

Iterating the lifting property $T_4$ implies that

$$\Lambda_\infty\to \Lambda_1 \in \{ \Lambda_2 \to \Lambda_1\}^{lr}.$$

The standard arguments from the proof of Urysohn lemma give the following relation between $\Lambda_\infty$ and $\mathbf{R}$.

1. $[0,1]$ and ${[0,1]}\vee_{\{0,1\}} \{e_F\leftrightarrow 0, 1\leftrightarrow e_G\}$ are retracts of $\Lambda_\infty$.

2. The map $\Lambda_\infty\to \Lambda_1$ factors as

   $$\Lambda_\infty\to {[0,1]}\vee_{\{0,1\}} \{e_F\leftrightarrow 0, 1\leftrightarrow e_G\}\to \Lambda_2 \to \Lambda_1$$

Item (1) implies that $A \to X \;\;\;\,⧄\,\;\;\; \Lambda_\infty\to \{o\}$ implies $A \to X \;\;\;\,⧄\,\;\;\; [0,1]\to \{o\}$ (and $A \to X \;\;\;\,⧄\,\;\;\; [0,1]\vee_{\{0,1\}} \{e_F\leftrightarrow 0, 1\leftrightarrow e_G\}\to \{o\}$).

This reminds one of the Tietze extension theorem but is different: it is not true that any closed inclusion in a normal space lifts wrt to $\Lambda_2 \to \Lambda_1$, although it is true for closed inclusions into a heriditarily normal space.

Item (2) (see the diagram above) shows that $A\to B \;\;\;\,⧄\,\;\;\; \Lambda_\infty\to \Lambda_1$ implies that $A\to B \;\;\;\,⧄\,\;\;\; {[0,1]}\vee_{\{0,1\}} \{e_F\leftrightarrow 0, 1\leftrightarrow e_G\}\to \Lambda_1$. Finally, for $A=\varnothing$ the latter implies that $\varnothing \to B \;\;\;\,⧄\,\;\;\; \Lambda_2 \to \Lambda_1$. This implies the Urysohn lemma that

$$\varnothing \to B \;\;\;\,⧄\,\;\;\; \Lambda_2 \to \Lambda_1\text{ iff }\varnothing \to B \;\;\;\,⧄\,\;\;\; {[0,1]}\vee_{\{0,1\}} \{e_F\leftrightarrow 0, 1\leftrightarrow e_G\} \to \Lambda_1$$

The following theorem is a summary of considerations above.

Theorem

1. $\{\Lambda_2\to \Lambda_1\} ^{⧄ l} \subset \{\Lambda_\infty\to \Lambda_1\}^{⧄ l} \subset \{{[0,1]}\vee_{\{0,1\}} \{e_F\leftrightarrow 0, 1\leftrightarrow e_G\} \to \Lambda_1\}^{⧄ l}$

2. $\{\Lambda_2\to \Lambda_1\} ^{⧄ l} \subset \{\Lambda_\infty\to \{o\}\}^{⧄ l} \subset \{{[0,1]} \to \{o\}\}^{⧄ l}$

3. for arbitrary space $B$ it holds

   $$\varnothing \to B \;\;\;\,⧄\,\;\;\; \Lambda_2 \to \Lambda_1\,\,\,\text{ iff }\,\,\,\varnothing \to B \;\;\;\,⧄\,\;\;\; {[0,1]}\vee_{\{0,1\}} \{e_F\leftrightarrow 0, 1\leftrightarrow e_G\} \to \Lambda_1$$

A noted above, item (3) is the usual statement of Urysohn lemma, and item (2) is similar but not equivalent to the Tietze extension theorem.

Extremally disconnected spaces being projective

The morphism $f_{e.d.}:= \boxed{ \boxed{{}^{\boxed{u}}\!\! \searrow_{\,c_1}} \,\,\, \boxed{{}_{c_2}\swarrow^{\boxed{v}}} } \to \boxed{ \overset{ \boxed{ \boxed{u} \;\; \, \;\; \boxed{v}} }{ \underset{ c } { \searrow \;\, \swarrow } } }$ is surjective, proper, and has the right lifting property $T_1$.

Both being surjective and being proper are the right lifting properties. This means that each morphism $S\xrightarrow { (f_{e.d.})^{lr} } X$ is surjective and proper, and if $X$ has $T_1$, so does $S$. In particular, if $X$ is compact Hausdorff, so is $S$.

Each morphism $\varnothing \to X$ can be decomposed as

$$\varnothing \xrightarrow{ (f_{e.d.})^l } S \xrightarrow { (f_{e.d.})^{lr} } X$$

This means there is a surjective proper map onto each space from an extremally disconnected space, and that in this both spaces can be assumed compact Hausdorff.

Gleasons theorem (here) which says that extremally disconnected spaces are projective in the category of topological spaces and proper maps, can be expressed by saying that $\varnothing \to S \in (proper)^{l}$ whenever $\varnothing \xrightarrow{ (f_{e.d.})^l } S$.

In fact there is a proper morphism $f_{proper}$ of finite spaces such that for compact Hausdorff spaces it holds $S \xrightarrow { (f_{proper})^{lr} } X$, i.e. $(f_{proper})^{lr}$ is a class of proper maps containing (necessarily proper) maps of compact Hausdorff spaces.

Related concepts

• Tychonoff theorem – Proof via lifting properties

• colimits of normal spaces

References

• Misha Gavrilovich, Point set topology as diagram chasing computations, The De Morgan Gazette. 2014. Vol. 5. No. 4. P. 23-32. (arXiv:1408.6710, pdf)

• Misha Gavrilovich, The unreasonable power of the lifting property in elementary mathematics, 2017 (arXiv:1707.06615, pdf)

• Misha Gavrilovich, Konstantin Pimenov. A naive diagram chasing approach to formalisation of tame topology., 2018 (pdf)

• Misha Gavrilovich, Extremally disconnected spaces as $\{\{u\to a,b\leftarrow v\}\to\{u\to a=b\leftarrow v\}\}^{lr}$, and being proper as $(\{\{o\}\to\{o\to c\}\}^r_{\le 4})^{lr}$, 2021 (pdf)