Changing the Order or Parentheses
A point in a product is a list of coordinates, and changing how that list is displayed should not change the information it contains. For example, \((x,y)\) and \((y,x)\) record the same two entries in a different order, while \(((x,y),z)\) and \((x,(y,z))\) differ only in their parentheses. The topologies also agree under the corresponding coordinate-preserving bijections: each map and its inverse are continuous in the product topologies.
We use the product topology throughout. For a family of spaces \((X_i,\mathcal{T}_i)\), \(i\in I\), a basic open set in \(\prod_{i\in I}X_i\) restricts open sets in only finitely many coordinates. More precisely, it has the form \(\bigcap_{i\in F}\pi_i^{-1}(U_i)\), where \(F\subseteq I\) is finite, \(U_i\in\mathcal{T}_i\), and \(\pi_i\) is the projection onto coordinate \(i\). The empty product, when \(I=\varnothing\), is a one-point space.
A Continuity Test for Product Targets
The key test is that a map into a product is continuous exactly when all of its coordinate maps are continuous. Here a coordinate map means the composition of the map with a projection.
Proof. Suppose first that \(f\) is continuous. Each projection \(\pi_i\) is continuous: the preimage under \(\pi_i\) of an open set \(U_i\subseteq X_i\) is the basic open set \(\pi_i^{-1}(U_i)\). Therefore \(\pi_i\circ f\) is continuous as a composition of continuous maps.
Conversely, suppose every \(\pi_i\circ f\) is continuous. A basic open set in the product has the form \(\bigcap_{i\in F}\pi_i^{-1}(U_i)\) for a finite set \(F\). Its preimage under \(f\) is \[ f^{-1}\left(\bigcap_{i\in F}\pi_i^{-1}(U_i)\right) =\bigcap_{i\in F}(\pi_i\circ f)^{-1}(U_i). \] Each set on the right is open in \(Z\) by the assumed continuity, and a finite intersection of open sets is open. Every open set in the product is a union of basic open sets, and preimages preserve unions. Thus the preimage of every open set is open, so \(f\) is continuous. If \(I=\varnothing\), the target is a one-point space, whose only subsets are open, and every map into it is continuous; the statement holds in that case as well. \(\square\)
Reindexing Product Factors
Proof. Define \(S:\prod_{j\in J}X_{\phi(j)}\to\prod_{i\in I}X_i\) by \(S((y_j)_{j\in J})=(y_{\phi^{-1}(i)})_{i\in I}\). For every \(x=(x_i)_{i\in I}\), the \(i\)-th coordinate of \(S(R(x))\) is \(R(x)_{\phi^{-1}(i)}=x_{\phi(\phi^{-1}(i))}=x_i\), so \(S(R(x))=x\). For every \(y=(y_j)_{j\in J}\), the \(j\)-th coordinate of \(R(S(y))\) is \(S(y)_{\phi(j)}=y_{\phi^{-1}(\phi(j))}=y_j\), so \(R(S(y))=y\). Thus \(R\) and \(S\) are inverse bijections.
For continuity of \(R\), fix \(j\in J\). The \(j\)-th coordinate map of \(R\) is \(x\mapsto R(x)_j=x_{\phi(j)}=\pi_{\phi(j)}(x)\), which is a continuous projection. The product continuity test shows that \(R\) is continuous. For continuity of \(S\), its \(i\)-th coordinate map is \(y\mapsto S(y)_i=y_{\phi^{-1}(i)}\), the projection onto coordinate \(\phi^{-1}(i)\) of its domain. This projection is continuous, so the same test shows that \(S\) is continuous. Hence \(R\) is a homeomorphism. If the index sets are empty, both products are one-point spaces and the unique map between them is a homeomorphism. \(\square\)
The proof uses no restriction that the index sets be finite. The crucial facts are that the assignment of coordinates is a bijection and that the product topology tests openness using finitely many coordinate restrictions.
Worked Example: Swapping Two Factors
Give \(\mathbb{R}\) and \(I=[0,1]\) their standard topologies, and give both \(\mathbb{R}\times I\) and \(I\times\mathbb{R}\) their product topologies. Define \(\tau:\mathbb{R}\times I\to I\times\mathbb{R}\) by \(\tau(x,t)=(t,x)\). The index assignment exchanges the two factors, so it is a bijection. Applying the map to \((3,\frac14)\) gives \(\tau(3,\frac14)=(\frac14,3)\).
A basic open set in the target is \(V\times U\), where \(V\) is open in \(I\) and \(U\) is open in \(\mathbb{R}\). Its preimage is \[ \tau^{-1}(V\times U)=U\times V, \] which is open in \(\mathbb{R}\times I\). Also, \(\tau^{-1}\) is the same swap, so the preimage under \(\tau^{-1}\) of a basic open set \(U\times V\) in the source is \(V\times U\), open in the target. Thus both directions are continuous. As a check on the inverse, \(\tau(\frac14,3)=(3,\frac14)\), recovering the original point.
Rebracketing Three Factors
Two-factor reordering handles changes in order; the same coordinate method handles changes in parentheses. Give each copy of \(\mathbb{R}\) its standard topology and give the iterated products their product topologies. Define \[ \alpha:(\mathbb{R}\times\mathbb{R})\times\mathbb{R}\longrightarrow \mathbb{R}\times(\mathbb{R}\times\mathbb{R}),\qquad \alpha((x,y),z)=(x,(y,z)). \] The inverse formula is \(\beta(x,(y,z))=((x,y),z)\). For example, \(\alpha((2,-1),4)=(2,(-1,4))\), and applying \(\beta\) gives \(((2,-1),4)\) again.
Proof. Applying \(\beta\) to \(\alpha((x,y),z)=(x,(y,z))\) gives \(((x,y),z)\), the original input. Applying \(\alpha\) to \(\beta(x,(y,z))=((x,y),z)\) gives \((x,(y,z))\), also the original input. Thus the displayed maps are inverse bijections.
To check continuity, take a basic open set \(U\times(V\times W)\) in the target, where \(U,V,W\) are open in \(\mathbb{R}\). Its preimage under \(\alpha\) is \((U\times V)\times W\), a basic open set in the source. Conversely, the preimage under \(\beta\) of a basic open set \((U\times V)\times W\) in the source is \(U\times(V\times W)\), open in the target. Since preimages of basic open sets are open, preimages of their unions are open as well. Both maps are continuous, and \(\alpha\) is a homeomorphism. \(\square\)
The parenthesized products have different formal descriptions as sets, but the coordinate formula identifies each entry with exactly one entry in the other product. Rebracketing is therefore not an assertion that the displayed objects are literally equal; it is a homeomorphism between them.
Regrouping Infinitely Many Pairs
Worked Example: Separating Two Coordinate Sequences
Give \(\mathbb{R}\times\mathbb{R}\) its product topology, give \((\mathbb{R}\times\mathbb{R})^{\mathbb{N}}\) the product topology, and give \(\mathbb{R}^{\mathbb{N}}\times\mathbb{R}^{\mathbb{N}}\) the product topology. A point of the first space is a sequence of pairs \(((a_n,b_n))_{n\in\mathbb{N}}\). Define \[ P(((a_n,b_n))_{n\in\mathbb{N}})=((a_n)_{n\in\mathbb{N}},(b_n)_{n\in\mathbb{N}}). \] The inverse sends a pair of sequences \(((a_n)_n,(b_n)_n)\) to \(((a_n,b_n))_n\). These formulas recover every coordinate: separating and then pairing returns \((a_n,b_n)\) at each \(n\); pairing and then separating returns \(a_n\) and \(b_n\) at each \(n\). Hence they are inverse bijections.
For continuity of \(P\), each coordinate of its first output sequence is the first-coordinate projection of the \(n\)-th pair in the input. This is a composition of continuous product projections. Each coordinate of its second output sequence is the second-coordinate projection of that same \(n\)-th pair, also continuous. The product continuity test, applied to the target product (including its two sequence factors), proves that \(P\) is continuous.
For continuity of the inverse, its \(n\)-th output pair is \((a_n,b_n)\). The maps selecting \(a_n\) and \(b_n\) from the two input sequences are continuous projections. The map into \(\mathbb{R}\times\mathbb{R}\) with these two coordinate maps is continuous by the product continuity test. This holds for every \(n\), so applying the test to the product of pairs proves that the inverse is continuous.
For a concrete point, set \(a_n=n\) and \(b_n=-n\) for every \(n\in\mathbb{N}\). The input begins \(((1,-1),(2,-2),(3,-3),\ldots)\), and its image is \(((1,2,3,\ldots),(-1,-2,-3,\ldots))\). Pairing these two output sequences gives \(((1,-1),(2,-2),(3,-3),\ldots)\), checking the inverse formula at each coordinate. Thus \(P\) is a homeomorphism.
A Common Pitfall: Continuity Is Not Enough
When checking a coordinate formula, continuity alone does not establish a homeomorphism. For instance, with the standard product topology on \(\mathbb{R}^2\), let \(F(x,y)=(x,x)\). Both coordinate maps are continuous projections, so the product continuity test proves that \(F\) is continuous. But the formula discards the second input coordinate: \(F(0,0)=(0,0)=F(0,1)\). Therefore \(F\) is not injective and cannot be a homeomorphism.
The reordering theorem avoids this failure by requiring a bijection of the index sets. That condition ensures every coordinate appears exactly once and supplies an explicit inverse. A reliable check is therefore to do both tasks: verify that the coordinate assignment is bijective, then use the product continuity test for the map and for its inverse.
Check Your Understanding
For each question, identify the coordinate formula and the topology being used before applying the product continuity test.
- Write the inverse of the map \((x,y,z)\mapsto(z,x,y)\) from \(\mathbb{R}^3\) to \(\mathbb{R}^3\), and explain why both maps are continuous.
- For spaces \(X,Y,Z\), write the coordinate formula for a homeomorphism from \(X\times(Y\times Z)\) to \((Z\times X)\times Y\).
- State which step of the reindexing proof would fail if the coordinate assignment were not injective.
- For the product of pairs in the worked example, describe the inverse image of a basic open set that restricts \(a_2\) to an open set \(U\subseteq\mathbb{R}\) and \(b_5\) to an open set \(V\subseteq\mathbb{R}\).
- Explain why a map from a one-point empty product to itself is a homeomorphism.