The Tychonoff Embedding and the Stone–Čech Compactification

Each coordinate ${\pi }_f\circ e$ equals $f$ and is continuous, so $e$ is continuous by [L1].

If $x\mathrm{\ne }y$, point separation supplies $f\in \mathcal{F}$ with $f(x)\mathrm{\ne }f(y)$, and then $e(x)(f)\mathrm{\ne }e(y)(f)$. Thus $e$ is injective.

Let $U$ be open in $X$ and $x\in U$. The complement $C=X\setminus U$ is closed, so choose $f\in \mathcal{F}$ with $f(x)=1$ and $f[C]=\{0\}$. Then $e(x)$ belongs to $e[X]\cap {\pi }_f^{-1}((1\mathrm{/}2,1])$, and this subspace-open set is contained in $e[U]$.

Step 1.3 shows that $e[U]$ is open in $e[X]$ for every open $U$, so $e^{-1}:e[X]\to X$ is continuous. Together with step 1.1 and injectivity from step 1.2, [L2] proves the assertion.

If $X$ is Tychonoff, let $J=C(X,[0,1])$. By [L1], complete regularity separates a point from a closed set and $T_1$ makes singletons closed, so this full family separates both points and points from closed sets. Hence [L4] embeds $X$ in $[0,1{]}^J$.

Conversely, suppose $X$ embeds in $[0,1{]}^J$. The interval is a metric space and hence Tychonoff by [L3], so [L2] makes the cube completely regular and $T_1$, and then makes its subspace $X$ completely regular and $T_1$.

The two implications are steps 1.1 and 1.2, so the equivalence holds.

By [L3], the full evaluation map embeds $X$ as $E=e[X]$ in the cube $Q=[0,1{]}^{C(X,[0,1])}$. By [L5] the interval is compact Hausdorff, so [L1] makes $Q$ compact and [L4] makes it Hausdorff.

Put $K=\bar{E}\subseteq Q$. It is closed, hence compact by [L2], and it is Hausdorff as a subspace of the Hausdorff space $Q$.

The evaluation embedding $e:X\to K$ has image $E$, which is dense in $K$ by the definition of closure and [L2]. Thus $(K,e)$ is a Hausdorff compactification in the sense of A Hausdorff compactification as a dense embedding into a compact Hausdorff space.

The coordinate projection ${\pi }_g$ is continuous by [L1]. Its restriction $\bar{g}={\pi }_g{\mathrm{\mid }}_B$ is continuous and satisfies $\bar{g}(e(x))=e(x)(g)=g(x)$.

If $h:B\to [0,1]$ is another such extension, then $h$ and $\bar{g}$ agree on $e[X]$, which is dense in $B$. Since $[0,1]$ is Hausdorff, [L2] gives $h=\bar{g}$.

By [L1], $K$ is Tychonoff. The forward implication of A space is Tychonoff if and only if it embeds in a cube $[0,1{]}^J$ then supplies an embedding of $K$ into a unit cube.

This is the asserted embedding.

By [L1], $B$ is compact Hausdorff and $e[X]$ is dense in it.

Use [L2] to fix an embedding $j:K\to [0,1{]}^J$. For each $a\in J$, the map ${\pi }_a\circ j\circ f:X\to [0,1]$ extends uniquely to a continuous $h_a:B\to [0,1]$ by [L5].

The family $(h_a{)}_{a\in J}$ assembles to a continuous map $h:B\to [0,1{]}^J$ by [L4], and $h\circ e=j\circ f$ coordinatewise.

The subset $j[K]$ is compact, hence closed in the Hausdorff cube by [L3]. It contains $h[e[X]]$, so it contains $h[B]$: the inverse image $h^{-1}[j[K]]$ is closed in $B$ and contains the dense subset $e[X]$.

Thus $\bar{f}=j^{-1}\circ h:B\to K$ is continuous and satisfies $\bar{f}\circ e=f$. If $q:B\to K$ is another extension, then $j\circ q$ and $h$ agree on dense $e[X]$, so [L3] gives equality and injectivity of $j$ gives $q=\bar{f}$.

Step 1.1 gives a Hausdorff compactification and step 4.1 gives the required unique extension for every compact Hausdorff target. This is exactly [L6].

Apply [L1] twice to obtain continuous $u:B\to B^{\mathrm{\prime }}$ and $v:B^{\mathrm{\prime }}\to B$ with $u\circ i=i^{\mathrm{\prime }}$ and $v\circ i^{\mathrm{\prime }}=i$.

The maps $v\circ u$ and ${\mathrm{id}}_B$ agree on $i[X]$, which is dense in $B$, so [L2] gives $v\circ u={\mathrm{id}}_B$. Similarly $u\circ v={\mathrm{id}}_{B^{\mathrm{\prime }}}$.

Hence $u$ is a homeomorphism over $X$; its uniqueness is the uniqueness clause in [L1].

The image $i[X]$ is compact by [L1], hence closed in the Hausdorff space $B$ by [L1].

By the compactification condition in The Stone–Čech compactification by its compact-Hausdorff extension property, $i[X]$ is dense in $B$. A closed dense subset equals $B$, so $i[X]=B$.