Continuity and Openness of Coordinate Projections

Coordinate Projections and the Product Topology

A point in a product space records one coordinate in each factor. A coordinate projection forgets all but one of those coordinates. The product topology is built so that restricting a single coordinate gives an open set in the product; this makes continuity of projections direct to verify. Openness also holds, but its proof uses the finite-support form of basic open sets. Surjectivity requires a separate check: a target coordinate can be prescribed only if there is a compatible choice of all the other coordinates.

Let \(I\) be a nonempty index set, and for each \(\alpha\in I\) let \((X_\alpha,\mathcal{T}_\alpha)\) be a topological space. Write \(X=\prod_{\alpha\in I}X_\alpha\), with the product topology. A point \(x\in X\) is a tuple with coordinate \(x_\alpha\in X_\alpha\) for each \(\alpha\in I\). For \(j\in I\), the \(j\)-th coordinate projection is the map \(\pi_j:X\to X_j\) defined by \(\pi_j(x)=x_j\).

Definition: A basic open set in the product topology is a set of the form \(B(F,(U_\alpha))=\{x\in X:x_\alpha\in U_\alpha\text{ for every }\alpha\in F\}\), where \(F\subseteq I\) is finite and each \(U_\alpha\in\mathcal{T}_\alpha\). Coordinates outside \(F\) are unrestricted. The product topology consists of unions of these basic open sets.

A set of this form may be empty, for instance if one of its restrictions is empty or if the product itself is empty. When it is nonempty, its image under a coordinate projection has a particularly simple description.

Proposition: If \(B(F,(U_\alpha))\) is nonempty, then \(\pi_j(B(F,(U_\alpha)))=U_j\) when \(j\in F\), and \(\pi_j(B(F,(U_\alpha)))=X_j\) when \(j\notin F\).

Proof. Fix a point \(x\in B(F,(U_\alpha))\). If \(j\in F\), every point of \(B(F,(U_\alpha))\) has \(j\)-th coordinate in \(U_j\), so the image is contained in \(U_j\). For any \(y\in U_j\), replace the \(j\)-th coordinate of \(x\) by \(y\), leaving every other coordinate unchanged. The resulting tuple is still in \(B(F,(U_\alpha))\) and projects to \(y\). Thus the image is exactly \(U_j\). If \(j\notin F\), the same replacement works for every \(y\in X_j\), and all projected values are in \(X_j\). Hence the image is \(X_j\). \(\square\)

The nonemptiness hypothesis matters in this proposition. If a basic open set is empty, its image is empty, even when \(U_j\) or \(X_j\) is nonempty.

Continuity and Openness

Theorem: For every \(j\in I\), the coordinate projection \(\pi_j:X\to X_j\) is continuous and open. These conclusions hold whether or not the factors are nonempty.

Proof. For continuity, let \(V\in\mathcal{T}_j\). By the definition of \(\pi_j\), \(\pi_j^{-1}(V)=\{x\in X:x_j\in V\}\). This is a basic open set, using the finite coordinate set \(F=\{j\}\) and restriction \(U_j=V\). Therefore the inverse image of every open set is open, which is the definition of continuity.

For openness, let \(W\) be open in \(X\). By the definition of the product topology, \(W\) is a union of basic open sets \(B_\lambda\). Images preserve unions, so \(\pi_j(W)=\bigcup_\lambda\pi_j(B_\lambda)\). Each \(B_\lambda\) is either empty, whose image is empty and open, or nonempty. In the latter case the proposition shows that its image is either one of the open restrictions \(U_j\) or the whole space \(X_j\), both open in \(X_j\). Thus \(\pi_j(W)\) is a union of open sets and is open. \(\square\)

This proof of openness does not require choosing a point in every factor of the product. For each nonempty basic set it uses just one point already in that set, then changes a single coordinate. The argument therefore applies even to arbitrary index sets.

Worked Example: Projections from the Plane

Give \(\mathbb{R}^2=\mathbb{R}\times\mathbb{R}\) the product topology, where each factor has the standard topology. Let \(W=(0,2)\times(-1,1)\), and compute its images under \(\pi_1\) and \(\pi_2\). Then compute the inverse image under \(\pi_1\) of \(V=(1/2,3/2)\).

The set \(W\) is a basic open set with restrictions \(U_1=(0,2)\) and \(U_2=(-1,1)\). It is nonempty, since \((1,0)\in W\). Because coordinate 1 is restricted, the proposition gives \(\pi_1(W)=(0,2)\). In detail, every \((x,y)\in W\) has \(x\in(0,2)\); conversely, for each \(x\in(0,2)\), the point \((x,0)\) belongs to \(W\) and projects to \(x\).

Coordinate 2 is also restricted, so \(\pi_2(W)=(-1,1)\). Indeed, every \((x,y)\in W\) has \(y\in(-1,1)\), and for each such \(y\), the point \((1,y)\in W\) projects to \(y\).

For the inverse image, the definition of \(\pi_1\) gives \(\pi_1^{-1}(V)=\{(x,y)\in\mathbb{R}^2:x\in(1/2,3/2)\}=(1/2,3/2)\times\mathbb{R}\). This is open in the product topology: it is a basic open set restricting only the first coordinate. As a check, \((1,100)\) belongs to this inverse image because its first coordinate lies in \(V\), regardless of its second coordinate; \((2,0)\) does not.

Inverse Images and Images in Products

The same calculations apply to any open subset of a factor, not just an interval. If \(V\) is open in \(X_j\), then its inverse image is the cylinder that restricts coordinate \(j\) to \(V\) and leaves every other coordinate unrestricted. In particular, inverse images of open sets under a projection are not usually just the open set \(V\): they are subsets of the whole product.

Formula: For \(V\in\mathcal{T}_j\), \(\pi_j^{-1}(V)=\{x\in X:x_j\in V\}\). For a nonempty basic open set \(B(F,(U_\alpha))\), its image is \(U_j\) if \(j\in F\), and \(X_j\) if \(j\notin F\). The image of an empty basic open set is empty.

Worked Example: A Cylinder in a Countable Product

Let \(X=\mathbb{R}^{\mathbb{N}}\) with the product topology, where every factor is \(\mathbb{R}\) with its standard topology. Consider \(W=\{x\in X:x_2\in(0,1),\ x_5\in(2,3)\}\). Compute \(\pi_2(W)\), \(\pi_3(W)\), and \(\pi_5^{-1}((2,3))\).

The restrictions in \(W\) are on the finite set \(F=\{2,5\}\), so \(W\) is basic open. It is nonempty: the sequence with \(x_2=1/2\), \(x_5=5/2\), and all other coordinates equal to \(0\) lies in \(W\). Since \(2\in F\), the image formula gives \(\pi_2(W)=(0,1)\). To check both inclusions, every point of \(W\) has second coordinate in \((0,1)\); given any \(t\in(0,1)\), set \(x_2=t\), \(x_5=5/2\), and all other coordinates to \(0\). This produces a point of \(W\) whose second coordinate is \(t\).

Coordinate 3 is not restricted, so \(\pi_3(W)=\mathbb{R}\). For any \(t\in\mathbb{R}\), take \(x_2=1/2\), \(x_5=5/2\), \(x_3=t\), and all other coordinates equal to \(0\). This sequence belongs to \(W\), and its third coordinate is \(t\). Thus every real number occurs in the image.

Finally, \(\pi_5^{-1}((2,3))=\{x\in X:x_5\in(2,3)\}\). This is a basic open cylinder restricting just coordinate 5, so it is open. The sequence with \(x_5=5/2\) and all other coordinates \(0\) belongs to it, while the sequence with \(x_5=3\) and all other coordinates \(0\) does not, because the interval has strict endpoints.

When Is a Projection Surjective?

Openness and continuity do not imply surjectivity. To project onto a prescribed value in \(X_j\), there must be at least one tuple of values available in all the other factors. This qualification is automatic for finite products of nonempty sets, but for arbitrary products the existence of such tuples is related to the axiom of choice.

Proposition: For \(j\in I\), \(\pi_j:X\to X_j\) is surjective if and only if either \(X_j=\varnothing\), or \(\prod_{\alpha\in I,\ \alpha\ne j}X_\alpha\ne\varnothing\). In particular, if \(X_j\ne\varnothing\), surjectivity requires the product of all the other factors to be nonempty.

Proof. If \(X_j=\varnothing\), then \(X=\varnothing\), and \(\pi_j\) is the map from the empty set to the empty set. Its image is the whole target \(\varnothing\), so it is surjective. Now suppose \(X_j\ne\varnothing\). If there is a tuple \(y=(y_\alpha)_{\alpha\ne j}\) in the product of the other factors, then for any \(t\in X_j\), combine \(t\) with the coordinates \(y_\alpha\) to obtain a point of \(X\) that projects to \(t\). Thus \(\pi_j\) is surjective. Conversely, if \(\pi_j\) is surjective, choose \(t\in X_j\), which is possible because \(X_j\ne\varnothing\). Surjectivity gives a point \(x\in X\) with \(x_j=t\); its other coordinates form a point of \(\prod_{\alpha\ne j}X_\alpha\). This product is therefore nonempty. \(\square\)

For a finite product in which every factor is nonempty, one can select one element from each of the finitely many other factors, so every projection is surjective. For an arbitrary family of nonempty factors, the assertion that their product is nonempty uses the axiom of choice in its full generality. The exact criterion above states what surjectivity means without silently assuming that such a tuple exists.

Worked Example: A Nonempty Target Does Not Ensure Surjectivity

Let \(X_1=\{a,b\}\) have the discrete topology, and let \(X_2=\varnothing\) have its unique topology \(\{\varnothing\}\). Consider the product \(X=X_1\times X_2\) and the first projection \(\pi_1:X\to X_1\).

There is no pair \((x_1,x_2)\) with \(x_1\in X_1\) and \(x_2\in X_2\), since \(X_2\) has no elements. Hence \(X=\varnothing\). The image of the projection is \(\pi_1(X)=\varnothing\), whereas its target is \(X_1=\{a,b\}\). In particular, neither \(a\) nor \(b\) is in the image, so \(\pi_1\) is not surjective even though its target factor is nonempty.

The map is still continuous: the inverse image of either open subset of \(X_1\) is \(\varnothing\), which is open in \(X\). It is also open: the only open subset of \(X\) is \(\varnothing\), whose image is \(\varnothing\), open in \(X_1\). This example separates the properties: continuity and openness hold, but surjectivity fails because the other factor has no point.

A Common Pitfall

A frequent error is to treat \(\pi_j^{-1}(V)\) as if it were just \(V\). These sets belong to different spaces: \(V\subseteq X_j\), whereas \(\pi_j^{-1}(V)\subseteq X\). For example, in \(\mathbb{R}\times\mathbb{R}\), the incorrect calculation \(\pi_1^{-1}((0,1))=(0,1)\) omits the second coordinate and is not even a subset of the domain \(\mathbb{R}^2\). The correct calculation is \(\pi_1^{-1}((0,1))=(0,1)\times\mathbb{R}\), because the projection tests only whether the first coordinate lies in \((0,1)\). This corrected set is open by the basic-set definition.

Another error is to assume that a projection is surjective merely because its target is nonempty. The empty-factor example shows the missing condition: other coordinates must be available as well. For finite products this is ensured when all factors are nonempty; for arbitrary products, nonemptiness of every factor alone does not provide a chosen tuple without an appropriate choice principle.

Question	Test or formula	Conclusion
Is \(\pi_j\) continuous?	\(\pi_j^{-1}(V)=\{x:x_j\in V\}\) for open \(V\subseteq X_j\)	Yes; this is a basic open set.
Is \(\pi_j\) open?	Check images of basic open sets and then take unions.	Yes; each image is open.
What is the image of a nonempty basic set?	\(U_j\) if \(j\) is restricted; \(X_j\) otherwise.	Use the finite-support image formula.
Is \(\pi_j\) surjective?	Check whether a tuple exists in the other factors, unless \(X_j=\varnothing\).	A nonempty target alone is insufficient.

Check Your Understanding

Use the basic-set description of the product topology to justify each answer.

In \(\mathbb{R}^3\) with the product topology, compute \(\pi_2^{-1}((1,4))\).
For a nonempty basic open set restricting coordinates 1 and 4, what is its image under \(\pi_3\)?
Prove directly from the definition that the image of any open set under a coordinate projection is open.
If \(X_1\) and \(X_2\) are both nonempty, explain why the projection \(X_1\times X_2\to X_1\) is surjective.
Give an example where a projection has a nonempty target but is not surjective, and identify the failed condition.