The Extension Problem
A compactification of \(X\) places \(X\) densely inside a compact space, so it is natural to ask whether maps defined on \(X\) can be continued to that larger space. For the Stone-Cech compactification, the answer is especially strong: the target may be any compact Hausdorff space, and the extension is unique. The proof works by extending scalar-valued coordinate functions and then showing that their combined values still describe a point of the target.
A space is completely regular Hausdorff if it is Hausdorff and, whenever \(x\) is a point outside a closed set \(F\), there is a continuous function \(u:X\to[0,1]\) such that \(u(x)=0\) and \(u(z)=1\) for every \(z\in F\). For spaces \(X\) of this kind, write \(C(X,[0,1])\) for the set of continuous functions from \(X\) to the usual unit interval.
Here and below, a product has the product topology. For \(g\in C(X,[0,1])\), let \(\pi_g\) be the projection of the cube onto its coordinate indexed by \(g\). Then \(\pi_g|_{\beta X}\) is a continuous extension of \(g\), because \(\pi_g(e_X(x))=g(x)\). It is the only such extension: two continuous maps from \(\beta X\) to the Hausdorff space \([0,1]\) that agree on the dense subset \(e_X(X)\) agree everywhere. For this last fact, their equalizer is closed: the diagonal in \([0,1]\times[0,1]\) is closed, and the equalizer is its preimage under the map whose coordinates are the two maps.
Representing the Target by Coordinates
To extend a map into \(Y\), we first represent \(Y\) inside a product of intervals. For each \(u\in C(Y,[0,1])\), let \(\pi_u\) be the corresponding coordinate projection from \([0,1]^{C(Y,[0,1])}\), and define \(j:Y\to[0,1]^{C(Y,[0,1])}\) by \(j(y)_u=u(y)\). In particular, the correct coordinate identity is \(\pi_u\circ j=u\). Each coordinate of \(j\) is continuous, so \(j\) is continuous: a map into a product is continuous if all its coordinate maps are continuous. This product criterion follows because a basic open set in the product restricts only finitely many coordinates.
The map \(j\) is an embedding. Indeed, if \(y\ne z\), the Hausdorff condition makes \(\{z\}\) closed, and complete regularity supplies \(u\) with \(u(y)=0\) and \(u(z)=1\). Thus \(j(y)\ne j(z)\), so \(j\) is injective. To check that its inverse on its image is continuous, take an open set \(V\subseteq Y\) and \(y\in V\). The set \(F=Y\setminus V\) is closed. Complete regularity gives \(u\) with \(u(y)=0\) and \(u(F)=\{1\}\). The coordinate neighborhood \(\{w:w_u<1/2\}\) contains \(j(y)\), and its intersection with \(j(Y)\) is contained in \(j(V)\). Such neighborhoods show that \(j(V)\) is open relative to \(j(Y)\). Hence \(j\) is a homeomorphism onto its image.
Because \(Y\) is compact and \(j\) is continuous, \(j(Y)\) is compact. A compact subset of a Hausdorff space is closed: for any point outside the compact set, separate it from each point of the set by disjoint open neighborhoods, and use compactness to retain finitely many neighborhoods covering the set; the intersection of the corresponding finitely many neighborhoods of the outside point misses the set. Therefore \(j(Y)\) is closed in the product cube.
Proof of the Universal Property
For each \(u\in C(Y,[0,1])\), the composite \(u\circ f:X\to[0,1]\) is continuous. Its extension to \(\beta X\) is the coordinate projection \(\pi_{u\circ f}|_{\beta X}\). Define a map \(H:\beta X\to[0,1]^{C(Y,[0,1])}\) by declaring its \(u\)-coordinate to be that extension. Each coordinate of \(H\) is continuous, so the product criterion makes \(H\) continuous.
Proof. For every \(x\in X\) and every \(u\in C(Y,[0,1])\), the \(u\)-coordinate of \(H(e_X(x))\) is \(\pi_{u\circ f}(e_X(x))=(u\circ f)(x)=u(f(x))\). The \(u\)-coordinate of \(j(f(x))\) is \(\pi_u(j(f(x)))=u(f(x))\). Thus \(H(e_X(x))=j(f(x))\), and in particular \(H(e_X(X))\subseteq j(Y)\). Since \(j(Y)\) is closed and \(H\) is continuous, \(H^{-1}(j(Y))\) is closed in \(\beta X\). It contains the dense set \(e_X(X)\), so it is all of \(\beta X\). Define \(\widehat f=j^{-1}\circ H\), with \(j^{-1}\) understood as the continuous inverse from \(j(Y)\) to \(Y\). This is continuous and satisfies \(\widehat f(e_X(x))=f(x)\) for every \(x\in X\).
For uniqueness, suppose \(g_1,g_2:\beta X\to Y\) are continuous and both extend \(f\). The set \(E=\{p\in\beta X:g_1(p)=g_2(p)\}\) is closed: it is the preimage of the diagonal in \(Y\times Y\), which is closed because \(Y\) is Hausdorff. Since both maps agree on \(e_X(X)\), the closed set \(E\) contains a dense subset of \(\beta X\), and hence \(E=\beta X\). Therefore \(g_1=g_2\). If \(X\) is empty, \(\beta X\) is empty and there is exactly one map from it to \(Y\), so the statement also holds in that case. \(\square\)
Worked Applications
Worked Example: A Map from a Finite Discrete Space
Let \(X=\{a,b\}\) have the discrete topology, let \(Y=[0,1]\) have its usual topology, and define \(f(a)=1/4\), \(f(b)=3/4\). The space \(X\) is compact, so its evaluation image \(e_X(X)\) is compact. The product cube containing it is Hausdorff, so that image is closed. Consequently its closure is itself: \(\beta X=e_X(X)\), which identifies \(\beta X\) with \(X\). The extension is therefore \(\widehat f(e_X(a))=1/4\) and \(\widehat f(e_X(b))=3/4\).
Check: for each \(u\in C([0,1],[0,1])\), the coordinate of \(H(e_X(a))\) is \((u\circ f)(a)=u(1/4)\), and the corresponding coordinate of \(j(f(a))\) is \(u(1/4)\). At \(e_X(b)\), both coordinates are \(u(3/4)\). Thus \(H=j\circ f\) on all of \(\beta X\), as required.
Worked Example: Extending an Inclusion into the Interval
Let \(X=(0,1)\) with the subspace topology from \(\mathbb R\), let \(Y=[0,1]\) with its usual topology, and take \(f(x)=x\). The coordinate indexed by \(f\) in the evaluation cube for \(\beta X\) gives the extension \(\widehat f=\pi_f|_{\beta X}:\beta X\to[0,1]\). For every \(x\in(0,1)\), \(\widehat f(e_X(x))=\pi_f(e_X(x))=f(x)=x\), so this really extends the inclusion.
Its image is exactly \([0,1]\). It is contained in \([0,1]\) because that is the coordinate space. It is compact because \(\beta X\) is compact and \(\widehat f\) is continuous, so it is closed in \([0,1]\). It contains \((0,1)\), since \(\widehat f(e_X(x))=x\) for every \(x\in(0,1)\). The only closed subset of \([0,1]\) containing \((0,1)\) is all of \([0,1]\): closedness forces it to contain the closure of \((0,1)\), namely \([0,1]\). Hence the extension attains both endpoint values as well. Uniqueness follows from the theorem.
Worked Example: An Alternating Map from the Natural Numbers
Give \(X=\mathbb N\) the discrete topology and \(Y=\{0,1\}\) the discrete topology. Define \(f(n)=0\) when \(n\) is even and \(f(n)=1\) when \(n\) is odd. The target \(Y\) is finite, hence compact Hausdorff. Since \(f\) is continuous and has values in \([0,1]\), it is one of the coordinates in the evaluation cube for \(\beta\mathbb N\). The extension is \(\widehat f=\pi_f|_{\beta\mathbb N}\).
For \(n=2k\), where \(k\in\mathbb N\), \(\widehat f(e_X(2k))=\pi_f(e_X(2k))=f(2k)=0\). For \(n=2k-1\), \(\widehat f(e_X(2k-1))=\pi_f(e_X(2k-1))=f(2k-1)=1\). Its image lies in \(\{0,1\}\): the image is compact and thus closed in \([0,1]\), and it is contained in the closure of \(\{0,1\}\) because the dense set \(e_X(\mathbb N)\) has coordinate values only in \(\{0,1\}\). Both values occur, as the even and odd calculations show. This gives a concrete extension even though \(\beta\mathbb N\) has many points beyond the embedded natural numbers.
What the Hypotheses Do—and a Common Pitfall
The compactness and Hausdorff conditions on the target are used at distinct points. Compactness makes \(j(Y)\) closed in the product cube, which forces the cube-valued extension \(H\) to land in \(j(Y)\). Hausdorffness makes the diagonal closed, which is what turns agreement on a dense subset into uniqueness. Complete regularity provides enough real-valued functions to embed the target in the cube; the Hausdorff part of the hypothesis on \(X\) is also part of the evaluation-embedding construction of \(\beta X\).
Common pitfall. The expression \(u\circ j\) is not defined when \(u:Y\to[0,1]\) and \(j:Y\to[0,1]^{C(Y,[0,1])}\): the output of \(j\) is a point of the product, not a point of \(Y\). The correct coordinate calculation is \(\pi_u\circ j=u\), where \(\pi_u\) projects the product onto the coordinate indexed by \(u\). This type check is essential when verifying that the constructed maps agree coordinate by coordinate.
Another tempting overreach is to remove compactness of the target and expect every map to extend. For example, give \(\mathbb N\) the discrete topology and consider its identity map into \(\mathbb N\), also discrete. If a continuous extension \(\beta\mathbb N\to\mathbb N\) existed, its image would be compact, since \(\beta\mathbb N\) is compact. The image would contain every natural number because the extension agrees with the identity on the embedded \(\mathbb N\). But \(\mathbb N\) is not compact in the discrete topology: the cover by all singleton sets has no finite subcover. Thus no such extension exists. The compact Hausdorff target is not a technical ornament; it is what makes the universal property work.
Check Your Understanding
Use the coordinate construction and identify where each hypothesis is needed.
- For \(g\in C(X,[0,1])\), verify directly that \(\pi_g|_{\beta X}\) extends \(g\).
- Explain why the equality \(\pi_u\circ j=u\) has the correct domains and codomains.
- In the proof, why does \(H^{-1}(j(Y))\) contain every point of \(\beta X\) once it contains \(e_X(X)\)?
- Which step in the uniqueness proof uses that \(Y\) is Hausdorff?
- Why does compactness of \(\beta\mathbb N\) rule out a continuous extension of the identity map from discrete \(\mathbb N\) to itself?