Continuous Functions as Embedding Coordinates

From Functions to Coordinates

A continuous real-valued function can record one feature of a point in a space. A family of such functions records many features at once: each point is sent to the list of its function values. The product topology makes this construction useful even when the family is infinite. The key distinction is that distinguishing points guarantees injectivity, but preserving the full topology requires enough coordinates to detect neighborhoods as well.

Let \((X,\mathcal{T})\) be a topological space, let \(I\) be an index set, and give \([0,1]\) its usual subspace topology from \(\mathbb{R}\). Give \([0,1]^I\) the product topology. Its basic open sets restrict only finitely many coordinates: they have the form \(\bigcap_{i\in F}\pi_i^{-1}(V_i)\), where \(F\subseteq I\) is finite, each \(V_i\) is open in \([0,1]\), and \(\pi_i\) is the \(i\)th projection. When \(F=\varnothing\), this intersection is the whole product.

The coordinate map is continuous. Indeed, the inverse image of a basic open set is \(\Phi^{-1}\bigl(\bigcap_{i\in F}\pi_i^{-1}(V_i)\bigr)=\bigcap_{i\in F}f_i^{-1}(V_i)\). Each set \(f_i^{-1}(V_i)\) is open in \(X\) by continuity, and a finite intersection of open sets is open. Since every open set in the product is a union of basic open sets, its inverse image is open too. This argument also covers \(F=\varnothing\), whose inverse image is \(X\).

Point Separation and Injectivity

Proof. Suppose first that \(\Phi\) is injective. Take any distinct \(x,y\in X\). Then \(\Phi(x)\ne\Phi(y)\), so the two indexed lists of coordinate values differ in at least one coordinate: for some \(i\in I\), \(f_i(x)\ne f_i(y)\). Thus the family separates points. Conversely, suppose the family separates points and \(\Phi(x)=\Phi(y)\). Equality of these lists means \(f_i(x)=f_i(y)\) for every \(i\in I\). If \(x\ne y\), point separation would give an index where these values differ, a contradiction. Hence \(x=y\), and \(\Phi\) is injective. If \(X\) has fewer than two points, both conditions hold without requiring any pair of distinct points to be tested. \(\square\)

Injectivity says that the coordinates distinguish points, but not that they distinguish nearby points from points far away. Continuity of \(\Phi\) gives one direction of topological control: open sets in the product pull back to open sets in \(X\). For an embedding, the topology on \(X\) must also be recoverable from those coordinate restrictions.

Proof. Suppose \(\Phi\) is an embedding. It is injective by the definition of embedding. For an open set \(U\subseteq X\), the set \(\Phi(U)\) is open in the subspace \(\Phi(X)\), because the inverse of \(\Phi:X\to\Phi(X)\) is continuous. Thus \(\Phi(U)=G\cap\Phi(X)\) for some open set \(G\subseteq[0,1]^I\). If \(x\in U\), then \(\Phi(x)\in G\). By the definition of the product topology, there is a basic open set \(C=\bigcap_{i\in F}\pi_i^{-1}(V_i)\) with \(\Phi(x)\in C\subseteq G\), for some finite \(F\subseteq I\). The membership \(\Phi(x)\in C\) gives \(f_i(x)\in V_i\) for each \(i\in F\). If \(y\in\bigcap_{i\in F}f_i^{-1}(V_i)\), then \(\Phi(y)\in C\subseteq G\). Also \(\Phi(y)\in\Phi(X)\), so \(\Phi(y)\in G\cap\Phi(X)=\Phi(U)\). Therefore \(y\in U\), as required.

Conversely, suppose \(\Phi\) is injective and the stated neighborhood condition holds. We already know that \(\Phi\) is continuous. It remains to show that its inverse from \(\Phi(X)\) to \(X\) is continuous. Let \(U\) be open in \(X\). For every \(x\in U\), the condition supplies a basic coordinate neighborhood whose inverse image contains \(x\) and is contained in \(U\). That inverse image is open, because it is a finite intersection of inverse images under continuous functions. Taking the union of these open sets over \(x\in U\) shows that \(U\) is open in the topology generated by the coordinates. More directly, each such set is the inverse image under \(\Phi\) of a basic open set in the product, and its image under the injective map \(\Phi\) is open relative to \(\Phi(X)\). Their union is \(\Phi(U)\). Hence \(\Phi(U)\) is relatively open in \(\Phi(X)\), so the inverse map is continuous. Thus \(\Phi\) is an embedding. If \(U=\varnothing\), its image is empty and is relatively open; the argument also covers \(X=\varnothing\). \(\square\)

The finite-coordinate requirement is essential to the criterion: basic neighborhoods in the product topology restrict finitely many coordinates, not all coordinates at once. It says that every open neighborhood in \(X\) can be refined locally by conditions on finitely many function values.

Three Worked Examples

What the Coordinates Must Preserve

The examples separate three claims that are easy to conflate. Continuity of all coordinates makes the coordinate map continuous. Point separation makes it injective. To obtain an embedding, the coordinate neighborhoods must additionally recover every open neighborhood in \(X\). An injective continuous map can fail that last test, as the discrete-interval example shows.

In practice, the embedding criterion gives a direct way to test a proposed family. Start with an arbitrary open set \(U\) and a point \(x\in U\). Find finitely many coordinates and open value ranges containing their values at \(x\), then verify that the simultaneous coordinate restrictions keep every point inside \(U\). A single coordinate may suffice, as it does on the real line above; in other examples, several coordinates may be needed.

Common pitfall. Do not infer that an injective continuous coordinate map is automatically an embedding. Injectivity only says that distinct points have different coordinate lists. It does not ensure that an open set in the domain corresponds to a relatively open set in the image. To establish an embedding, prove continuity of the inverse by checking the local finite-coordinate condition, or equivalently show that every open subset of the domain has relatively open image.