Can we determine an explicit construction of the functions $\beta(f)$, where $\beta$ is the Stone-Cech adjoint functor?

Question

$\newcommand{\top}{\mathsf{Top}}\newcommand{\ch}{\mathsf{CptHaus}}\require{AMScd}$This is a question about the Stone-Cech compactification, from the perspective of category theory.

The TLDR is that: we can explicitly construct $\beta X$, up to a certain standard of "explicit". Can we explicitly describe $\beta(f)$ for continuous $f$? The categorical algorithm for doing so is given below, but I am finding it rather difficult to get a clear construction from it.

In the pages $126$ onwards of MacLane's Categories for the Working Mathematician one may find the pieces of the proof of the "Special Adjoint Functor Theorem", which, together with some famous theorems of topology, guarantees the existence of a left adjoint to the inclusion functor: $$I:\ch\to\top$$Where $\top$ is the category of topological spaces and continuous functions between them, and $\ch$ is the full subcategory of $\top$ given by objects compact Hausdorff spaces. We can call this left adjoint $\beta\dashv I$, $\beta:\top\to\ch$.

I've carefully stepped through all the pieces of the proof - forming weird pullbacks and products in comma categories, etc. - but eventually we arrive constructively at: $$\beta X\cong\overline{\left\{(x_f)\in[0,1]^{\top(X,[0,1])}:\exists y\in X,\,x_f=f(y)\,\forall f\in\top(X,[0,1])\right\}}$$For any topological space $X$, where the closure is considered in the product topology on $[0,1]^{\top(X,[0,1])}$. We find that the unit of the adjunction (the continuous map $\eta_X:X\hookrightarrow\beta X$ with the universal property) can be given as: $$\eta_X(x)(F)=F(x),\,\forall F:\top(X,[0,1])\to[0,1]$$And we can more succinctly describe $\beta X$ as $\beta X=\overline{\eta_X(X)}$.

I'd like to follow the algorithm of the proof to construct the continuous functions $\beta f:\beta X\to\beta Y$ for any continuous $f:X\to Y$. To do so, I need to identify the unique intermediary arrows given by the universal property. To be clear, I refer to the property:

Fix a topological space $X$. For any compact Hausdorff space $K$ and any continuous map $\varphi:X\to K$, there exists a unique continuous $\psi:\beta X\to K$ such that $\psi\circ\eta_X=\varphi$.

The algorithm for this stage is a bit... complicated. Luckily the cogenerating set of $\ch$ is a one-object set, which enormously simplifies the calculations, but they are still a bit of a mess. We need to take the products, in the comma category $(X\downarrow I)$: $$Q:=\prod_{q\in\top(X,[0,1])}([0,1],q),\,H:=\prod_{h\in\ch(K,[0,1])}([0,1],h\circ\varphi)$$With projection arrows $\pi^{(Q)}_{\bullet}$ and $\pi^{(H)}_{\bullet}$ respectively, and we need to set $j:=\langle h\rangle_{h\in\ch(K,[0,1])}:(K,\varphi)\to H$ and $k:=\langle\pi^{(Q)}_{h\circ\varphi}\rangle_{h\in\ch(K,[0,1])}:Q\to H$ as arrows in the comma category. We then need to construct a pullback square (in the comma category!): $$\begin{CD}G@>j'>>Q\\@Vk'VV@VVkV\\K@>>j>H\end{CD}$$ And then trust in the process of the proof that $G$ is a subobject of $Q$ and thus there exists a continuous injection $\iota:\beta X\hookrightarrow G$ since $\beta X$ is formally constructible as the intersection of all subobjects of $Q$. Then we are done and we set $\psi=k'\circ\iota:\beta X\to K$, again trusting in the process of the proof that this is the unique intermediary arrow.

I've done my best to work through all that, and I've arrived at the following construction for the object $G$.

Let: $$G:=\{(a,b)\in K\times[0,1]^{\top(X,[0,1])}:h(a)=b(h\circ\varphi),\,\forall h\in\ch(K,[0,1])\}$$Endowed with the subspace topology of the product topology, and $k':=\pi$ the projection in the first coordinate.

So! I am almost done. We may let $\psi:=\pi\circ\iota$. But there is wrinkle - $\iota$ is very unclear! Note that in the construction of $G$ if $b=\eta_X(x)$ for some $x\in X$, then $h(a)=h(\varphi(x))$ holds for all $h$ if and only if $a=\varphi(x)$ due to the Urysohn lemma. Then, if we cross our fingers and hope that $\iota$ does the obvious thing, namely $\iota(\eta_X(x))=(\varphi(x),\eta_X(x))$, we can say $\psi(\eta_X(x))=\varphi(x)$ (which is exactly what we expect from the universal property).

However, determining $\iota$ on $\beta X\setminus\eta_X(X)$ is difficult. I would love to say: "if we set $\iota(\eta_X(x))=(\varphi(x),\eta_X(x))$ then there is a unique continuous extension of $\iota$ on $\beta X$ as $\eta_X(X)$ is dense in $\beta X$". I don't know how to demonstrate that this is true. I don't even know if it even possible to explicitly describe $\iota$, since the Wikipedia article on the Stone-Cech compactification claims that, in general (and with a small abuse of notation) the points in $\beta X\setminus X$ are extremely difficult to determine.

• 1i): Can we prove $\iota(\eta_X(x))\equiv(\varphi(x),\eta_X(x))$?

• 1ii): Can we prove this extends continuously on all of $\beta X$ in a unique way?

• 2: Are we lucky, in that we can go further to get a fully explicit characterisation of $\iota$ and thus of $\psi$ and of the value $\beta(f)$ for any function $f$?

Answer 1

$\newcommand{\top}{\mathsf{Top}}\newcommand{\ch}{\mathsf{CptHaus}}$I think I've figured out the final detail, but I'd really appreciate feedback confirming whether or not this is a correct solution. I'm confident but not totally so.

The nature of $G$ as a subobject of $Q$ is actually not so mysterious. Let $\pi_1:G\to K$ and $\pi_2:G\to Q$ be the first and second projections. Categorically, $\pi_2$ is monic as a pullback of a monic (which is essential to the proof, and I somehow completely forgot this when writing the original question), but we can topologically verify this as follows:

Suppose $(a,b)$ and $(a',b)$ are in $G$. Suppose $a\neq a'$: there exists $h\in\ch(K,[0,1])$ with $h(a)\neq h(a')$ (this is a reflection of $[0,1]$ as a cogenerator) by the Urysohn lemma. Then the condition: $h(a')=b(h\varphi)=h(a)$ for all $h$ fails, so one or both of $(a,b),(a',b)$ are not in $G$, a contradiction. Thus $\pi_2$ injects since there is at most one $a$, for every $b$, with $(a,b)\in G$.

This one-to-one correspondence implies that $G$ is, when viewed as a subobject of $Q$, merely a subset of $Q$ with the natural inclusion. Then $\iota$ is just the natural inclusion $\beta X\hookrightarrow G$, guaranteed to exist by the process of the proof, so $\iota(b)=(a,b)$ for the unique $a$ that makes this an element of $G$. In particular, problem 1i) disappears since $a=\varphi(x)$ when $b=\eta_X(x)$, so $\iota(\eta_X(x))=(\varphi(x),\eta_X(x))$ which is the obvious and natural thing.

This suffices to characterise $\psi:=\pi_1\circ\iota$ in a "more explicit" manner, now that $\iota$ is better understood as a simple inclusion. For any $b\in\beta X$, we claim (by categorical trickery) that there really does exist a unique $a\in K$ such that $h(a_b)=b(h\circ\varphi)$ for all continuous $h:K\to[0,1]$. We set $\psi(b)=a_b$ and are done.

On a topological level, it is interesting to see why such $a_b$ should exist. I have a proposed topological proof, but I'd appreciate feedback on the correctness of this proof.

Fix $b\in\beta X=\overline{\eta_X(X)}$, and take any continuous $h:K\to[0,1]$. There are different ways to go about the following, but we may as well use a sequence as that is the "nice" way. For all $n\in\Bbb N$ let $F_n=[b(h\circ\varphi)-1/n,b(h\circ\varphi)+1/n]\cap[0,1]$. As $h$ is continuous, $h^{-1}\{F_n\}$ is closed and thus compact in $K$; moreover, $h^{-1}\{F_{n+1}\}\subseteq h_0^{-1}\{F_n\}$ always.

By the product topology, we can take a basic, cylinder-set neighbourhood of $b$ on the coordinate $h\circ\varphi$ and the nonempty open set $F_n^{\circ}$ for each $n$. As $\eta_X(X)$ is dense in $\beta X$, this neighbourhood of $b$ will contain some $\eta_X(x_n)$ for some $x_n\in X$, thus $\eta_X(x_n)(h\circ\varphi)=h(\varphi(x_n))\in F_n^{\circ}$.

That guarantees $h^{-1}\{F_n\}$ is never empty, as it contains $\varphi(x_n)$. By the finite intersection property and compactness, $A_h:=\bigcap_{n\in\Bbb N}h^{-1}\{F_n\}$ is a nonempty compact subset of $K$. Each element $a\in A_h$ is such that $h(a)$ is in all the $F_n$; by basic properties of $[0,1]$ we must then have that $h(a)=b(h\circ\varphi)$ for all $a\in A_h$. We can also note that $a\in\overline{\varphi(X)}$.

Take these nonempty compact $A_h$ for all continuous $h:K\to[0,1]$. We want to show that their intersection is nonempty, since the intersection is necessarily a one-point set (if it is nonempty) by the Urysohn lemma. We can assert a nonempty intersection if we demonstrate the finite intersection property. Let $\{h_i\}_{i=1}^k\subset\ch(K,[0,1])$ be arbitrary - we want to show $\bigcap_{i=1}^kA_{h_i}\neq\emptyset$. Let, for $1\le i\le k$ and $n\in\Bbb N$, $F^{(i)}_{n}:=[b(h_i\circ\varphi)-1/n,b(h_i\circ\varphi)+1/n]\cap[0,1]$. Similarly can we construct basic cylinder neighbourhoods of $b$, for fixed $n$, by asserting the neighbourhood $(F^{(i)}_n)^{\circ}$ on the coordinates $h_i\circ\varphi$. By density will this contain an $\eta_X(x_n)$ for some $x_n\in X$, with $h_i(\varphi(x_n))\in(F^{(i)}_n)^{\circ}$ for all $i$. Then the intersected preimage $B_n:=\bigcap_{i=1}^kh_i^{-1}\{F^{(i)}_n\}$ is nonempty (and compact by continuity of each $h_i$) for each $n$, with the same property $B_{n+1}\subseteq B_n$ for all $n$. We have by the finite intersection property: $$\emptyset\neq\bigcap_{n\in\Bbb N}B_n\subseteq\bigcap_{i=1}^kA_{h_i}$$Which concludes the proof: $$\exists!a_b\in\bigcap_{h\in\ch(K,[0,1])}h^{-1}b(h\circ\varphi),\,\forall b\in\beta X$$It is pleasing to observe that this $a_b$ is in the closure $\overline{\varphi(X)}$ always, and can be intuitively thought of as a limit point of the $\eta_X(x)$ approaching $b$.

This is now as explicit a characterisation of $\psi,\iota$ as I think we can possibly get. Again, feel free to give feedback on this topological proof. I reiterate that there are no problems in the categorical sense, I fully accept the SAFT's proof and its consequences.