CW-complex, Hausdorff space, second-countable space, sober space
connected space, locally connected space, contractible space, locally contractible space
A compactum, or compact Hausdorff space, is a space in which every limit that should exist does exist and does so uniquely.
One can define this literally for topological spaces, or in terms of convergence, or for locales; although these are all different contexts, the resulting notion of compactum is (at least assuming the axiom of choice) always the same. Interestingly, there is even an algebraic definition, not one that uses only finitary operations, but one which uses a monad.
If you know what a compact space is and what a Hausdorff space is, then you know what a compact Hausdorff space is, so let's be fancy. (Full justifications will be provided in section on compacta as algebras.)
Given a set $S$, let $\beta S$ be the set of ultrafilters on $S$. Note that $\beta$ is an endofunctor on Set; every function $f: S \to T$ induces a function $\beta f: \beta S \to \beta T$ using the usual application of functions to filters. In fact, $\beta$ is a monad; it comes with a natural (in $S$) unit $\eta_S: S \to \beta S$, which maps a point $x$ to the principal ultrafilter that $x$ generates, and multiplication $\mu_S: \beta \beta S \to \beta S$, which maps an ultrafilter $U$ on ultrafilters to the ultrafilter of sets whose principal ultrafilters of ultrafilters belong to $U$. That is,
Then a compactum is simply an algebra for this monad; that is, a set $X$ together with a function $\lim: \beta X \to X$, such that
It is then a theorem that this $\lim$ generates a convergence on $S$ that is compact, Hausdorff, and topological. The converse, that every compact Hausdorff topological convergence is of this form, is equivalent to the ultrafilter principle.
Every compact Hausdorff space is regular and sober and so defines a compact regular locale. Again, the axiom of choice gives us a converse: every compact regular locale is spatial and so comes from a compactum.
Probably this is also equivalent to the ultrafilter principle, but I need to check.
Note that every compact Hausdorff space (topological or localic) is not only regular but also normal.
Throughout this section, $CH$ will be used to denote the category of compact Hausdorff spaces (compacta).
Let $Bool$ be the category of Boolean algebras. The functor $\hom(-, \mathbf{2}): Bool^{op} \to Set$ has a left adjoint $P: Set \to Bool^{op}$ given by power sets, and we define the ultrafilter monad to be the composite $\beta \coloneqq \hom(P -, \mathbf{2})$.
For a set $S$, topologize $\beta S$ by declaring a basic open set to be one of the form
for $A$ a subset of $S$. Notice $\hat{\emptyset}$ is empty. Indeed, $\widehat{(-)}$ defines a Boolean algebra map
so that in particular $\widehat{A \cap B} = \hat{A} \cap \hat{B}$, which immediately implies that the $\hat{A}$ form a basis of a topology.
In fact, $\widehat{(-)}$ is the component at $P(S)$ of the counit of the adjunction $P \dashv \hom(-, \mathbf{2})$, as a morphism in $Bool^{op}$. If $f: S \to T$ is a function, the commutativity of the naturality square
implies that if $U = \hat{B} \in P(\beta T)$ is a basic (cl)open, then so is $(\beta f)^{-1}(U) = P(\beta f)(\hat{B}) = \widehat{f^{-1}(B)}$. It then follows that $\beta f$ is continuous.
These results show that the monad $\beta: Set \to Set$ lifts through the forgetful functor $U: Top \to Set$.
The unit of the monad $\beta$ is given componentwise by functions
where $prin_X$ takes $x \in X$ to the principal ultrafilter
It is evident that $prin_X$ is injective.
The injection $prin_X: X \to \beta X$ exhibits $X$ as a dense subset of $\beta X$.
If $\hat{A}$ is a basic open neighborhood containing an ultrafilter $F$, then $A$ is nonempty and hence contains some $x \in X$, which is to say $A \in prin_X(x)$ or that $prin_X(x) \in \hat{A}$.
$\beta S$ is Hausdorff.
Let $F, G$ be distinct ultrafilters, so there is $A \subseteq S$ with $A \in F$ and $\neg A \in G$. Then $\hat{A}$ and $\widehat{\neg A}$ are disjoint neighborhoods which contain $F$ and $G$ respectively.
$\beta S$ is compact.
It is enough to show that if $\mathcal{O}$ is a collection of opens such that the union of any finite subcollection is a proper subset, then the union of $\mathcal{O}$ is also proper.
If $\mathcal{O}$ covers $U$, it admits a refinement by basic clopens also covering $U$, and thus we may assume WLOG that $\mathcal{O}$ consists of basic clopens $\hat{A}$. If every finite union of elements of $\mathcal{O}$ is a proper subset of $\beta S$, then every finite intersection $\widehat{\neg A_1} \cap \ldots \cap \widehat{\neg A_n}$ is nonempty, so that the $\neg A$ generate a filter, which is contained in some ultrafilter $F$. This $F$ lies outside the union of all the $\hat{A}$’s.
Let $X$ be a topological space, and $F$ an ultrafilter on the underlying set $U X$. We say $F$ converges to a point $x$ (in symbols, $F \rightsquigarrow x$) if the neighborhood filter $N_x$ of $x$ is contained in $F$.
Convergence defines a relation $\xi$ from $\beta(U X)$ to $U X$.
If $X$ is Hausdorff, then the relation $\xi$ is well-defined, or functional (i.e., there is at most one point to which a given ultrafilter $F$ converges).
If $x \neq y$, then there are disjoint neighborhoods $U$, $V$ of $x$ and $y$. We cannot have both $U \in F$ and $V \in F$ (otherwise $\emptyset = U \cap V$ would be an element of $F$), so at most one of the neighborhood filters $N_x, N_y$ can be contained in $F$.
If $X$ is compact, then the relation $\xi$ is total (i.e., there exists a point to which a given ultrafilter $F$ converges).
If not, then for each $x \in X$ there is an open neighborhood $U_x$ that does not belong to $F$. Then $\neg U_x \in F$. Some finite number of neighborhoods $U_{x_1}, \ldots, U_{x_n}$ covers $X$. Then $\neg U_{x_1} \cap \ldots \cap \neg U_{x_n} = \emptyset \in F$, which is a contradiction.
If $X$ is compact Hausdorff, then the function $\xi: \beta(U X) \to X$ is continuous.
Let $U$ be an open neighborhood of $x \in X$; we must show that $\xi^{-1}(U)$ contains an open neighborhood of any of its points (i.e., ultrafilters $F$ such that $F \rightsquigarrow x$). Since $X$ is $T_3$ (Hausdorff regular), we may choose a neighborhood $V \in N_x$ whose closure $\bar{V}$ is contained in $U$. Then $\hat{V}$ is an open neighborhood of $F$ in $\beta(U X)$, and we claim $\hat{V} \subseteq \xi^{-1}(U)$.
For this, we must check that if $G \in \hat{V}$ and $G \rightsquigarrow y$, then $y \in U$. But if $y \in \neg U \subseteq \neg \bar{V}$, then $\neg \bar{V} \in N_y$, whence $G \rightsquigarrow y$ implies $\neg \bar{V} \in G$. This contradicts $G \in \hat{V}$, i.e., contradicts $V \in G$, since $V \cap \neg \bar{V} = \emptyset$.
If $S$ is a set and $X$ is a compact Hausdorff space, then any function $f: S \to X$ can be extended (along $prin_S: S \to \beta S$) to a continuous function $\hat{f}: \beta S \to X$.
We define $\hat{f}$ to be the composite
where $\beta(f)$ is continuous by Remark 1 and $\xi$ is continuous by Proposition 6. It remains to check that the following diagram is commutative:
The square commutes by naturality of $prin$, and commutativity of the triangle simply says that the ultrafilter $prin_{U X}(x)$ converges to $x$, or that $N_x \subseteq prin(x)$, which reduces to the tautology that $x \in V$ for every neighborhood $V \in N_x$.
For any set $S$, the function $prin_S: S \to \beta S$ is universal among functions from $S$ to compact Hausdorff spaces. Hence the functor $F: Set \to CH$ that takes $S$ to the compact Hausdorff space $\beta S$ is left adjoint to the forgetful functor $CH \to Set$.
Proposition 7 shows that for any function $f: S \to U X$ to a compact Hausdorff space, there exists continuous $g: \beta S \to X$ such that $g \circ prin_S = f$. All that remains is to establish uniqueness of such $g$. But if two maps $g, g': \beta S \to X$ to a Hausdorff space $X$ agree on a dense subspace, in this case the subspace $prin_S : S \hookrightarrow \beta S$ by Proposition 1, then they must be equal. Indeed, the pullback of the closed diagonal defines a closed subspace $D$ of $\beta S$,
and $D$ contains a dense subspace $S$, therefore $D = \beta S$; i.e., the equalizer of $g$ and $g'$ is all of $\beta S$, hence these two maps are equal.
We recall hypotheses of Beck’s precise monadicity theorem: a functor $U: C \to D$ is monadic if and only if
$U$ has a left adjoint,
$U$ reflects isomorphisms: a morphism $f: X \to Y$ of $C$ is an isomorphism if $U f: U X \to U Y$ is an isomorphism in $D$,
$D$ has, and $U$ preserves, coequalizers of parallel pairs that are $U$-split. (We say
is $U$-split if there is a coequalizer
that is split in $D$: there exists $i: Z \to U Y$ and $j: U Y \to U X$ such that $U h \circ i = 1_Z$, $U g \circ j = i \circ h$, and $U f \circ j = 1_{U Y}$.)
In the case where $D = Set$, we have the following useful lemma:
Suppose given a coequalizer in $Set$
split by $i: Z \to Y$, $j: Y \to X$ (so that $f j = 1_Y$, $h i = 1_Z$, $g j = i h$). Let $R \hookrightarrow Y \times Y$ be the image of $\langle f, g \rangle : X \to Y \times Y$, and let $\langle p_1, p_2 \rangle : E \to Y \times Y$ be the equivalence relation given by the kernel pair $(p_1, p_2)$ of $h$. Then $E = R \cdot R^{op}$, the relational composite given by taking the image of the span composite $R \times_Y R^{op} \to Y \times Y$.
Clearly $R \subseteq E$ since $h f = h g$, and we have $R^{op} \subseteq E^{op} = E$ and $R \cdot R^{op} \subseteq E \cdot E \subseteq E$ by symmetry and transitivity of $E$. In the other direction, suppose $(y_1, y_3) \in E$, so that $h(y_1) = h(y_3)$. Put $x = j(y_1)$ and $x' = j(y_3)$ (so that $f(x) = y_1$ and $f(x') = y_3$), and put $y_2 = g(x)$. Clearly then $(y_1, y_2) \in R$. Moreover,
so that $(y_3, y_2) \in R$, or $(y_2, y_3) \in R^{op}$. Hence $(y_1, y_3) \in R \cdot R^{op}$, and we have shown $E \subseteq R \cdot R^{op}$.
The forgetful functor $U: CH \to Set$ is monadic.
By theorem 1, $U$ has a left adjoint. Since bijective continuous maps between compact Hausdorff spaces are homeomorphisms, we have that $U$ reflects isomorphisms. Finally, suppose $(f, g)$ is a $U$-split pair of morphisms $X \to Y$ in $CH$; let $h: Y \to Z$ be their coequalizer in $Top$, given by a suitable quotient space. Being a quotient of a compact space, $Z$ is compact. Since $CH$ is a full subcategory of $Top$, the map $h$ is a coequalizer in $CH$ once we prove the following claim:
Furthermore, since the forgetful functor $Top \to Set$ has a right adjoint (given by taking indiscrete topologies on sets), the underlying function of $h$ (again denoted $h$) is the coequalizer of $(U f, U g)$ in $Set$, so that $U$ would preserve the claimed coequalizer.
In other words, to complete the proof, it suffices to verify the claim. Letting $p_0, p_1: E \stackrel{\to}{\to} Y$ be the kernel pair of $h$ in $Top$, to show $Z = Y/E$ is Hausdorff, it suffices to prove that the equivalence relation $\langle p_0, p_1 \rangle : E \to Y \times Y$ in $Top$ is closed. Let $R \hookrightarrow Y \times Y$ be the image of $\langle f, g \rangle: X \to Y \times Y$. By lemma 1, the subset $E$ of $U Y \times U Y$ coincides with the subset $R \cdot R^{op} \subseteq U Y \times U Y$. Now $R$ is the image of the compact space $X$ under the continuous map $\langle f, g \rangle$, so $R$ is a closed subset of $Y \times Y$. Similarly $R^{op}$ is a closed subset of $Y \times Y$. Under their subspace topologies, their fiber product $R \times_Y R^{op}$ is compact, and so its image $R \cdot R^{op}$ under the (continuous) span composite $R \times_Y R^{op} \to Y \times Y$ is also closed. This completes the proof.
In the absence of the axiom of choice, and especially in constructive mathematics, the best definition of compactum seems to be a compact regular locale. That is, it is the category of compact regular locales that has all of the nice properties, forming a nice category of spaces, and that has the desired examples, such as the unit interval. (See the discussion at Tychonoff theorem for an example of how the category of compact Hausdorff topological spaces might fail to be nice; see Frank Waaldijk’s PhD thesis (pdf) for a thorough discussion of what is needed to make the unit interval a compact Hausdorff topological space.)
The monadic definition, in particular, falls quite flat without some form of the axiom of choice; even excluded middle and COSHEP are powerless here. In fact, it is quite consistent to assume that every ultrafilter is principal (a strong denial of the ultrafilter principle), in which case $\beta$ is the identity monad. Then a compactum would be just a set if that were the definition used.
On the other hand, it is the monadic definition that gives an algebraic category with a nice relationship to Set. Without the ultrafilter principle, there is no reason to think that the set-of-points functor from compact regular locales to sets is even continuous.
By general nonsense, every $\beta S$, regarded as a free $\beta$-algebra, is a compactum, and the functor
is left adjoint to the forgetful functor $Comp \to Set$. Assuming the ultrafilter principle, this functor extends to a functor $\beta: Top \to Comp$ (identifying a set with its discrete space) that is left adjoint to the forgetful functor $Comp \to Top$. This is the Stone–Čech compactification functor (N.B.: for many authors, Stone–Čech compactification refers to the restriction of this functor to Tychonoff spaces $X$, which are precisely those spaces where the unit $X \to \beta X$ is an embedding so that we have a compactification in the technical sense).
A classical construction of the Stone–Čech compactification starts with the unit interval $I =[0, 1]$ and proceeds to the codensity monad induced from the functor
The monad is given on objects by $X \mapsto I^{\hom(X, I)}$; this lands in compact Hausdorff spaces. Let $\bar{X}$ be the closure of the image of the unit $u_X: X \to I^{\hom(X, I)}$; under the ultrafilter principle, $\bar{X}$ is compact Hausdorff.
If $X$ is a Tychonoff space, then the unit $u_X: X \to I^{\hom(X, I)}$ is a subspace embedding, so that $I$ is a cogenerator in the category of Tychonoff spaces. (In particular, $u_C$ is an embedding if $C$ is compact Hausdorff, so $I$ is also a cogenerator in the category of compact Hausdorff spaces.)
The proof is essentially Urysohn's lemma; see also related discussion at Tychonoff space and at uniform space (noting that compact Hausdorff spaces are uniform spaces for a unique uniformity).
The natural map $i_X: X \to \bar{X}$ is universal among maps from $X$ to compact Hausdorff spaces, thus giving a left adjoint $Top \to Comp$ to the (fully faithful) forgetful functor $U: Comp \to Top$.
Let $f: X \to C$ be a map, where $C$ is a compact Hausdorff space. Since $I$ is a cogenerator in the category of compact Hausdorff spaces, the unit for the codensity monad $M_I$,
is a continuous injection (and hence a closed subspace embedding, since $C$ is compact Hausdorff). Let $\hat{f} = M_I(f)$, and consider the pullback square
From an evident naturality square for the unit $u$, we have a map $h: X \to \hat{f}^{-1}(C)$, i.e., the map $u_X: X \to I^{\hom(X, I)}$ factors through the closed subspace $\hat{f}^{-1}(C) \hookrightarrow I^{\hom(X, I)}$. Therefore $h$ factors as
and since $\pi \circ h = f$, we conclude that $f$ factors through $i_X$. And moreover, there is at most one $k: \bar{X} \to C$ such that $k \circ i_X = f$, because $i_X$ maps $X$ onto a dense subspace of $\bar{X}$, and dense subspaces are epic in the category of Hausdorff spaces. This completes the proof.
We have a similar Stone–Čech compactification functor $Loc \to Comp$; we do not need the ultrafilter principle here if $Comp$ is defined in terms of locales.
The category $Comp$ of compact Hausdorff spaces and continuous maps is
Barr exact, since $U: Comp \to Set$ is monadic,
an extensive category, and
a total category (by monadicity over $Set$), and
a cototal category (because it is complete, well-powered, and has a cogenerator given by the unit interval $I = [0, 1]$).
From the first two properties, it follows that $Comp$ is a pretopos, meaning that $Comp$ enjoys the same finitary exactness properties that hold in a topos; in particular, first-order intuitionistic logic may be enacted within $Comp$.