Homeomorphisms Applied Coordinate by Coordinate
A homeomorphism changes the description of a space without changing its topological structure. If each factor of a product is replaced by a homeomorphic factor, the natural candidate for a homeomorphism of the whole products is to apply the given homeomorphism in each coordinate. The key point is that this works for arbitrary products when both products have the product topology. We will prove this using the basic open sets that restrict only finitely many coordinates.
Let \(I\) be an index set. For each \(i\in I\), let \((X_i,\mathcal{T}_i)\) and \((Y_i,\mathcal{U}_i)\) be topological spaces, and give \(\prod_{i\in I}X_i\) and \(\prod_{i\in I}Y_i\) their product topologies. A point \(x\) in the first product has coordinates \(x_i\in X_i\); the projection to coordinate \(i\) is \(\pi_i(x)=x_i\). A basic open set in a product restricts finitely many coordinates: it has the form \(\bigcap_{i\in F}\pi_i^{-1}(U_i)\), where \(F\subseteq I\) is finite and each \(U_i\) is open in its factor.
The formula defines a map even when some product has no points: in that case the corresponding function has empty domain. When \(I=\varnothing\), each empty product is, by convention, a singleton, and the unique map between these singletons is a homeomorphism. The main result below applies to every index set, including this case.
The Product Homeomorphism Theorem
Proof. For each \(i\), let \(h_i^{-1}:Y_i\to X_i\) be the inverse of \(h_i\), and define \(K:\prod_{i\in I}Y_i\to\prod_{i\in I}X_i\) by \(K(y)_i=h_i^{-1}(y_i)\). For each \(x\in\prod_{i\in I}X_i\) and each \(i\in I\), we have \(K(H(x))_i=h_i^{-1}(h_i(x_i))=x_i\), so \(K(H(x))=x\). For each \(y\in\prod_{i\in I}Y_i\), we also have \(H(K(y))_i=h_i(h_i^{-1}(y_i))=y_i\), so \(H(K(y))=y\). Thus \(H\) is a bijection and \(K=H^{-1}\).
To prove that \(H\) is continuous, consider a basic open set \(V=\bigcap_{i\in F}\pi_i^{-1}(U_i)\) in \(\prod_{i\in I}Y_i\), where \(F\) is finite and each \(U_i\) is open in \(Y_i\). The coordinate formula gives \(H^{-1}(V)=\bigcap_{i\in F}\pi_i^{-1}(h_i^{-1}(U_i))\), with the projections on the right taken from \(\prod_{i\in I}X_i\). Each \(h_i^{-1}(U_i)\) is open because \(h_i\) is continuous, and the intersection is finite. Hence \(H^{-1}(V)\) is open. Every open set in the product topology is a union of basic open sets, and preimages preserve unions, so the preimage of every open set is open.
For continuity of \(K\), take a basic open set \(W=\bigcap_{i\in F}\pi_i^{-1}(A_i)\) in \(\prod_{i\in I}X_i\), with \(A_i\) open in \(X_i\). Its preimage is \(K^{-1}(W)=\bigcap_{i\in F}\pi_i^{-1}((h_i^{-1})^{-1}(A_i))\), where these projections are from \(\prod_{i\in I}Y_i\). Since \(h_i^{-1}\) is continuous, each set \((h_i^{-1})^{-1}(A_i)\) is open in \(Y_i\); the finite intersection is therefore open. As before, preimages preserve unions, so \(K\) is continuous on all open sets. We have proved that \(H\) is bijective, continuous, and has continuous inverse \(K\), so it is a homeomorphism. If \(I=\varnothing\), the same conclusion holds directly because \(H\) is the unique map from a singleton to a singleton. \(\square\)
The proof uses the finite-coordinate form of basic open sets. No matter how many factors there are, a basic open condition involves only finitely many of them. On each restricted coordinate, continuity of the corresponding factor map gives the needed open preimage; the same reasoning applies to the inverse maps.
Worked Example: An Affine Map and a Cubic Map on \(\mathbb{R}^2\)
Give each copy of \(\mathbb{R}\) its standard topology and \(\mathbb{R}^2=\mathbb{R}\times\mathbb{R}\) the product topology. Define \(h_1(x)=2x+1\) and \(h_2(y)=y^3\). The first map has inverse \(h_1^{-1}(u)=(u-1)/2\). The second is a continuous, strictly increasing bijection \(\mathbb{R}\to\mathbb{R}\), with continuous inverse \(h_2^{-1}(v)=\sqrt[3]{v}\). Thus both factor maps are homeomorphisms.
The product map and its proposed inverse are \(H(x,y)=(2x+1,y^3)\) and \(K(u,v)=((u-1)/2,\sqrt[3]{v})\). For the input \((1,-2)\), the first coordinate of its image is \(2(1)+1=3\), and the second is \((-2)^3=-8\). Therefore \(H(1,-2)=(3,-8)\). Applying \(K\) gives \(((3-1)/2,\sqrt[3]{-8})=(2/2,-2)=(1,-2)\).
For a check in the other direction, start with \((3,-8)\). Applying \(K\) gives \((1,-2)\); applying \(H\) then gives \((2(1)+1,(-2)^3)=(3,-8)\). This verifies both inverse identities at the displayed point, while the theorem proves them for every point.
Infinite Products: The Same Coordinate Formula
The theorem is not limited to a finite number of factors. In an infinite product, the formula for the map and its inverse is still coordinatewise. Continuity remains a consequence of the product topology’s basic open sets, which restrict finitely many coordinates even when the index set is infinite.
Worked Example: From \(\mathbb{R}^{\mathbb{N}}\) to \((0,1)^{\mathbb{N}}\)
Give every factor its standard topology and both countable products their product topologies. Define \(\sigma:\mathbb{R}\to(0,1)\) by \(\sigma(t)=e^t/(1+e^t)\). Since \(e^t>0\), we have \(0<e^t/(1+e^t)<1\). Solving \(s=e^t/(1+e^t)\) for \(t\) gives \(s(1+e^t)=e^t\), then \(s=e^t(1-s)\), and hence \(t=\ln(s/(1-s))\). Thus the inverse is \(\sigma^{-1}(s)=\ln(s/(1-s))\), which is continuous for \(0<s<1\). The formula and inverse are continuous, so \(\sigma\) is a homeomorphism.
Apply \(\sigma\) in every coordinate to obtain \(H:\mathbb{R}^{\mathbb{N}}\to(0,1)^{\mathbb{N}}\). Its inverse applies \(s\mapsto\ln(s/(1-s))\) in every coordinate. To compute a point, let \(x_1=0\), \(x_2=\ln 3\), and \(x_3=-\ln 3\), with the remaining coordinates arbitrary. Then \(\sigma(x_1)=1/(1+1)=1/2\); \(\sigma(x_2)=3/(1+3)=3/4\); and \(\sigma(x_3)=(1/3)/(1+1/3)=(1/3)/(4/3)=1/4\). The first three coordinates of \(H(x)\) are therefore \((1/2,3/4,1/4)\).
Applying the inverse to these coordinates gives \(\ln((1/2)/(1-1/2))=\ln(1)=0\), \(\ln((3/4)/(1-3/4))=\ln(3)=\ln 3\), and \(\ln((1/4)/(1-1/4))=\ln(1/3)=-\ln 3\). So the original first three coordinates are recovered. The same inverse formula recovers every remaining coordinate, and the theorem supplies continuity for the maps on the entire products.
Worked Example: A Product of Tangent and Exponential Maps
Give \(X=(-\pi/2,\pi/2)\times\mathbb{R}\) and \(Y=\mathbb{R}\times(0,\infty)\) the product topologies, using the standard subspace topologies on the intervals. The map \(x\mapsto\tan x\) is a homeomorphism from \((-\pi/2,\pi/2)\) to \(\mathbb{R}\), with inverse \(u\mapsto\arctan u\). The map \(y\mapsto e^y\) is a homeomorphism from \(\mathbb{R}\) to \((0,\infty)\), with inverse \(v\mapsto\ln v\). Consequently their product map is \(H(x,y)=(\tan x,e^y)\), with inverse \(K(u,v)=(\arctan u,\ln v)\).
For \((x,y)=(\pi/4,\ln 2)\), the first output coordinate is \(\tan(\pi/4)=1\), and the second is \(e^{\ln 2}=2\). Hence \(H(\pi/4,\ln 2)=(1,2)\). Applying the inverse gives \(K(1,2)=(\arctan 1,\ln 2)=(\pi/4,\ln 2)\). Conversely, \(H(K(1,2))=H(\pi/4,\ln 2)=(1,2)\), so the calculation checks both compositions for this point.
Coordinatewise Composition
Proof. Let \(x\in\prod_{i\in I}X_i\). For each \(i\in I\), the \(i\)-th coordinate of \(G(H(x))\) is \(g_i(H(x)_i)\). By the definition of \(H\), \(H(x)_i=h_i(x_i)\), so this coordinate is \(g_i(h_i(x_i))=(g_i\circ h_i)(x_i)\). This is exactly the \(i\)-th coordinate of the product map defined by the family \((g_i\circ h_i)_{i\in I}\). Since product points are equal when all their coordinates are equal, the maps agree. \(\square\)
In particular, the inverse of a product of homeomorphisms is not usually found by repeating the original coordinate maps. It is obtained by taking the inverse in each factor. The next calculation illustrates a common mistake.
Common pitfall. For the affine factor \(h_1(x)=2x+1\), using \(2u+1\) as the inverse would give \(2(2x+1)+1=4x+3\), which is not \(x\); at \(x=1\), it gives \(7\), not \(1\). To solve \(u=2x+1\) for \(x\), subtract \(1\) to get \(u-1=2x\), then divide by \(2\), giving \(h_1^{-1}(u)=(u-1)/2\). The cubic coordinate likewise requires the inverse cube-root map, not another cubing.
A related distinction is that continuity and bijectivity alone do not define a homeomorphism: continuity of the inverse is also required. For example, let \(X=\{a,b\}\) have topology \(\{\varnothing,\{a\},X\}\), and let \(Y=\{a,b\}\) have the discrete topology, in which every subset is open. The identity map \(f:Y\to X\) is a continuous bijection because every subset of \(Y\) is open. Its inverse is not continuous: the open set \(\{b\}\) in \(Y\) has preimage \(\{b\}\) under \(f^{-1}:X\to Y\), and \(\{b\}\) is not open in \(X\). So \(f\) is not a homeomorphism. The product theorem avoids this failure by explicitly proving continuity of the coordinatewise inverse.
Coordinatewise homeomorphisms are useful when a product is easier to describe in different coordinates in each factor. The theorem ensures that these changes, even across infinitely many factors, preserve the topology of the whole product, provided the topology used is the product topology.
Check Your Understanding
For each question, specify the relevant factor maps and their inverses before considering the product map.
- If each \(h_i:X_i\to Y_i\) is a homeomorphism, what is the formula for the inverse of the coordinatewise product map?
- Why does the preimage of a basic open set in a product involve only finitely many factor maps in the continuity proof?
- For \(h(x)=3x-2\) on \(\mathbb{R}\), find \(h^{-1}\) and check both compositions.
- Use the coordinatewise theorem to describe a homeomorphism from \([0,1]^{\mathbb{N}}\) to \([2,5]^{\mathbb{N}}\), with product topologies.
- Give a reason a continuous bijection need not be a homeomorphism, and identify the missing condition.