Building a Map Between Product Spaces
A map between product spaces can be specified one coordinate at a time. If the \(i\)-th output coordinate depends only on the \(i\)-th input coordinate, then continuity of all the coordinate maps guarantees continuity of the resulting map. Surjectivity and openness require closer attention: empty products, infinitely many factors, and coordinate maps that are not onto can all affect the answer.
Let \(I\) be an index set, let \((X_i,\mathcal{T}_i)\) and \((Y_i,\mathcal{S}_i)\) be topological spaces for each \(i\in I\), and give \(\prod_{i\in I}X_i\) and \(\prod_{i\in I}Y_i\) their product topologies. A basic open set in a product has the form \(\prod_{i\in I}U_i\), where each \(U_i\) is open in \(X_i\) and \(U_i=X_i\) except at finitely many indices. The same description applies to the \(Y_i\) product.
Continuity of the Product Map
Proof. Let \(V=\prod_{i\in I}V_i\) be a basic open set in \(\prod_{i\in I}Y_i\), so each \(V_i\) is open and \(V_i=Y_i\) except at finitely many indices. By the coordinate definition of \(F\), \(F^{-1}(V)=\prod_{i\in I}f_i^{-1}(V_i)\). Each \(f_i^{-1}(V_i)\) is open in \(X_i\), by continuity of \(f_i\), and for every index with \(V_i=Y_i\) its preimage is \(X_i\). Thus \(F^{-1}(V)\) is basic open in the product topology. Every open set in the target is a union of basic open sets, and preimages preserve unions. Therefore the preimage of every open set is open, which proves that \(F\) is continuous. \(\square\)
The converse needs a condition. Write \(\pi_i^X:\prod_jX_j\to X_i\) and \(\pi_i^Y:\prod_jY_j\to Y_i\) for the coordinate projections. The identity \(\pi_i^Y\circ F=f_i\circ\pi_i^X\) alone does not let us recover continuity of \(f_i\) from continuity of \(F\): the projection \(\pi_i^X\) need not have a continuous section unless the other input coordinates can be filled.
Proof. One direction is the theorem. For the converse, assume \(F\) is continuous and fix \(i\). Choose a tuple \(c=(c_j)_{j\ne i}\in\prod_{j\ne i}X_j\). Define \(s_i:X_i\to\prod_jX_j\) by setting the \(i\)-th coordinate equal to the input and all other coordinates equal to \(c_j\). This section is continuous: the preimage of a basic open product is either empty or an open subset of \(X_i\), obtained by intersecting the finitely many restrictions that involve the \(i\)-th coordinate with the restrictions tested at the fixed \(c_j\). The projection \(\pi_i^Y\) is continuous because the preimage of an open subset \(V_i\subseteq Y_i\) is the open cylinder with \(i\)-th coordinate \(V_i\). Hence \(f_i=\pi_i^Y\circ F\circ s_i\) is continuous. This applies to every \(i\). \(\square\)
If the source product is empty, \(F\) is continuous automatically, even if some coordinate map is not continuous. For example, take \(I=\{1,2\}\), \(X_1=Y_1=\mathbb{R}\), \(f_1(x)=x^2\), \(X_2=\varnothing\), and \(Y_2=\mathbb{R}\), with the usual topology on each copy of \(\mathbb{R}\). The source product is empty, so its map to \(\mathbb{R}^2\) is continuous, but \(f_1\) is not continuous? In fact \(x^2\) is continuous, so this does not demonstrate failure. Instead take \(f_1:\mathbb{R}\to\mathbb{R}\) to be \(f_1(0)=1\) and \(f_1(x)=0\) for \(x\ne0\), which is not continuous at \(0\). The product map from the empty source is still continuous. The failure occurs because the complementary product for \(i=1\) is empty.
Worked Example: A Continuous, Open, Surjective Map
Give \(\mathbb{R}\times\mathbb{R}\) the product topology, where both factors have the standard topology, and define \(F(x,y)=(2x,y^3)\). Its coordinate maps are \(f_1(x)=2x\) and \(f_2(y)=y^3\).
Both coordinate maps are continuous real functions, so the theorem gives continuity of \(F\). To check surjectivity explicitly, let \((u,v)\in\mathbb{R}^2\). The choices \(x=u/2\) and \(y=\sqrt[3]{v}\) give \(F(x,y)=(u,v)\). Thus \(F\) is onto.
Each coordinate map is open. For \(f_1\), the image of an open interval \((a,b)\) is \((2a,2b)\), which is open; every open subset of \(\mathbb{R}\) is a union of open intervals, so its image is open. For \(f_2\), the cube function is strictly increasing and has continuous inverse \(v\mapsto\sqrt[3]{v}\); equivalently, the image of an interval \((a,b)\) is \((a^3,b^3)\). Thus it too maps open sets to open sets.
A basic open rectangle \(U\times V\) has image \(f_1(U)\times f_2(V)\), an open rectangle. Every open subset of \(\mathbb{R}^2\) is a union of basic open rectangles, and images preserve unions. Therefore \(F\) is open. As a direct check, the rectangle \((0,1)\times(1,2)\) maps to \((0,2)\times(1,8)\), which is open.
Surjectivity: Finite and Arbitrary Products
For a finite family of factors, the product map is onto exactly when each coordinate map is onto. For arbitrary families, the direction from coordinatewise surjectivity to surjectivity of the product map may require the axiom of choice: a preimage must be selected in every coordinate.
Proof. Suppose \(F\) is onto and choose a tuple \(y^0\in\prod_iY_i\), which exists by the nonemptiness hypothesis. Fix \(i\in I\) and \(y_i\in Y_i\). Replace the \(i\)-th coordinate of \(y^0\) by \(y_i\), leaving its other coordinates unchanged. This gives a tuple in the target product. A preimage under \(F\) has \(i\)-th coordinate mapped by \(f_i\) to \(y_i\), so \(f_i\) is onto. Conversely, for a given target tuple \((y_i)\), surjectivity of each \(f_i\) gives a nonempty set \(f_i^{-1}(\{y_i\})\) of possible preimages at coordinate \(i\). Choosing one such preimage for every \(i\) produces a source tuple mapped to \((y_i)\). This simultaneous choice is automatic for finite \(I\), and for arbitrary \(I\) it uses the axiom of choice. If the target product is empty, the first implication has no coordinate conclusion; surjectivity onto the empty set is vacuous whenever the map exists. \(\square\)
Worked Example: Surjectivity Can Fail in One Coordinate
Give both copies of \(\mathbb{R}\) the standard topology and define \(G:\mathbb{R}^2\to\mathbb{R}^2\) by \(G(x,y)=(x^2,y)\). The maps \(x\mapsto x^2\) and \(y\mapsto y\) are continuous, so \(G\) is continuous.
The image is exactly \([0,\infty)\times\mathbb{R}\). Indeed, every output has first coordinate \(x^2\ge0\); conversely, for \(u\ge0\) and any \(v\in\mathbb{R}\), \(G(\sqrt{u},v)=(u,v)\). In particular, \((-1,0)\) is not in the image, so \(G\) is not surjective. The coordinate map \(x\mapsto x^2\) is not onto \(\mathbb{R}\), in agreement with the proposition.
When Is the Product Map Open?
A map is open if it sends every open set to an open set. For finitely many factors, open coordinate maps induce an open product map. For infinitely many factors, coordinate openness alone is not enough: an unrestricted coordinate in a basic open set still contributes the full range \(f_i(X_i)\), which may be smaller than \(Y_i\).
Proof. Under the axiom of choice, the image of a product \(\prod_i A_i\) under \(F\) is \(\prod_i f_i(A_i)\): membership from left to right follows coordinatewise, and for the reverse inclusion choose a preimage in each coordinate. A basic open set in the source is \(\prod_iU_i\), with \(U_i=X_i\) except at finitely many indices. If every \(f_i\) is open and all but finitely many \(f_i(X_i)\) equal \(Y_i\), then \(\prod_i f_i(U_i)\) is a basic open set in the target: every coordinate set is open, and only finitely many differ from \(Y_i\). Thus images of basic open sets are open, and images of their unions are open. This proves sufficiency.
For necessity, put \(R_i=f_i(X_i)\). The range product \(\prod_iR_i=F(\prod_iX_i)\) is nonempty and open because \(F\) is open. Projecting this open product onto coordinate \(i\) shows \(R_i\) is open: coordinate projections are open, since the projection of a basic product open set is either empty or open. Also, a nonempty open product \(\prod_iR_i\) must have \(R_i=Y_i\) except at finitely many indices. To see this, choose \(r\in\prod_iR_i\) and a basic open neighborhood \(\prod_iV_i\) of \(r\) contained in \(\prod_iR_i\). At every index outside the finite set where \(V_i\ne Y_i\), the neighborhood places no restriction, so \(Y_i=V_i\subseteq R_i\), hence \(R_i=Y_i\).
Finally fix \(i\) and an open set \(U_i\subseteq X_i\). The cylinder with \(i\)-th coordinate \(U_i\) and all other coordinates unrestricted is open. Its image is \(f_i(U_i)\times\prod_{j\ne i}R_j\), an open set. The complementary product \(\prod_{j\ne i}R_j\) is nonempty because \(\prod_jR_j\) is nonempty. Projecting this image onto \(Y_i\) therefore gives \(f_i(U_i)\), which must be open. This proves that \(f_i\) is open for every \(i\). \(\square\)
Worked Example: Open Coordinate Maps Need Not Give an Open Infinite Product Map
For each \(n\in\mathbb{N}\), let \(X_n=(0,\infty)\) with its subspace topology from the standard real line, let \(Y_n=\mathbb{R}\) with its standard topology, and let \(f_n:X_n\to Y_n\) be inclusion. Each \(f_n\) is continuous, open, and not surjective: an open subset of \((0,\infty)\) is open in \(\mathbb{R}\), while \(f_n(X_n)=(0,\infty)\ne\mathbb{R}\).
The induced map \(F:\prod_n(0,\infty)\to\prod_n\mathbb{R}\) is the inclusion of the positive sequences. Its image is not open. The sequence \(r=(1,1,\ldots)\) belongs to the image. Any basic neighborhood of \(r\) restricts only finitely many coordinates. Choose an index \(k\) outside those restrictions. The sequence equal to \(r\) except with \(k\)-th coordinate \(-1\) is in that neighborhood but not in the image. Thus no basic neighborhood of \(r\) lies inside the image, so the image is not open.
A Common Pitfall
Wrong claim: “A product of open maps is always open.” The preceding example disproves this for infinitely many factors: every coordinate inclusion is open, but the image of the whole source product is not open. The missing condition is that the coordinate ranges must equal the target factors at all but finitely many indices (under the nonempty-range hypothesis in the theorem).
There is also a frequent error in finite products. For \(G(x,y)=(x^2,y)\), the open rectangle \((-1,1)\times(0,2)\) has image \([0,1)\times(0,2)\), not \((0,1)\times(0,2)\): the value \(0\) is attained by \(x=0\). The image is not open in \(\mathbb{R}^2\), since every neighborhood of a point \((0,y)\) in it contains points with negative first coordinate. This computation also shows why continuity and surjectivity do not imply openness.
Check Your Understanding
Use the product topology throughout, and state any conditions you need when the family of factors is infinite.
- Prove that the product of finitely many continuous maps is continuous directly from preimages of basic open rectangles.
- For \(H:\mathbb{R}^2\to\mathbb{R}^2\) given by \(H(x,y)=(x,y^3)\), determine whether \(H\) is continuous, open, and surjective.
- Explain why continuity of a product map need not imply continuity of every coordinate map when a complementary source product is empty.
- For arbitrary \(I\), which direction of the surjectivity criterion may require the axiom of choice, and why?
- Give an example of a continuous surjective product map that is not open, and verify the image of a specific open set.