Repeating One Point in Every Coordinate
A product space can store the same point in every coordinate. The resulting map is called the diagonal map. Its formula is simple, but its continuity illustrates a central feature of the product topology: a basic open set restricts only finitely many coordinates, even when the product has infinitely many factors.
Let \((X,\mathcal{T})\) be a topological space and let \(J\) be an index set. The repeated product \(X^J=\prod_{j\in J}X\) has the product topology. A point \(z\in X^J\) is a tuple \((z_j)_{j\in J}\), with \(z_j\in X\). A basic open set is specified by a finite subset \(F\subseteq J\) and open sets \(U_j\in\mathcal{T}\) for \(j\in F\): it consists of the tuples whose \(j\)-th coordinate belongs to \(U_j\) for every \(j\in F\). If \(F\) is empty, this basic open set is all of \(X^J\).
For a finite index set, this definition gives familiar maps such as \(x\mapsto(x,x)\) into \(X\times X\) and \(x\mapsto(x,x,x)\) into \(X^3\). For an infinite index set, the same rule applies: each input determines a tuple with that input in every coordinate. The topology on the product is still the product topology, so its basic open sets restrict only finitely many coordinates.
Continuity from Basic Open Sets
Proof. Consider a basic open set \(B\subseteq X^J\) specified by a finite subset \(F\subseteq J\) and open sets \(U_j\in\mathcal{T}\) for \(j\in F\). A point \(x\in X\) belongs to \(\Delta_J^{-1}(B)\) exactly when the tuple \(\Delta_J(x)\) belongs to \(B\). By the definition of \(B\), this happens exactly when \(x\in U_j\) for every \(j\in F\). Therefore \[ \Delta_J^{-1}(B)=\bigcap_{j\in F}U_j. \] If \(F\) is nonempty, this is a finite intersection of open sets and hence is open in \(X\); if \(F\) is empty, the intersection is \(X\), which is open. Every open set in the product is a union of basic open sets, and the preimage of a union is the union of the preimages. Thus the preimage of every open set in \(X^J\) is open, proving continuity. For each \(j\in J\) and \(x\in X\), the \(j\)-th coordinate of \(\Delta_J(x)\) is \(x\), so \((\pi_j\circ\Delta_J)(x)=x\). This proves \(\pi_j\circ\Delta_J=\operatorname{id}_X\). \(\square\)
The displayed preimage formula is the essential computation. It does not depend on how many coordinates the product has: only the finite set \(F\) used to specify the basic open set enters the intersection. The formula also gives a direct continuity proof, without needing a separate continuity test for maps into products.
Worked Example: The Diagonal in the Plane
Give \(\mathbb{R}\) its standard topology and \(\mathbb{R}^2=\mathbb{R}\times\mathbb{R}\) the product topology. Consider \(\Delta_{\{1,2\}}:\mathbb{R}\to\mathbb{R}^2\), so \(\Delta_{\{1,2\}}(x)=(x,x)\). For the open rectangle \(B=(-2,4)\times(1,5)\), compute its preimage.
A real number \(x\) belongs to the preimage exactly when its first coordinate \(x\) lies in \((-2,4)\) and its second coordinate \(x\) lies in \((1,5)\). Hence \[ \Delta_{\{1,2\}}^{-1}(B)=(-2,4)\cap(1,5)=(1,4). \] The answer is open in \(\mathbb{R}\), as continuity requires. The coordinate maps are \(\pi_1\circ\Delta_{\{1,2\}}:x\mapsto x\) and \(\pi_2\circ\Delta_{\{1,2\}}:x\mapsto x\); both are \(\operatorname{id}_{\mathbb{R}}\).
Check: If \(x=2\), then \(2\in(-2,4)\) and \(2\in(1,5)\), so \((2,2)\in B\) and \(2\in\Delta_{\{1,2\}}^{-1}(B)\). If \(x=0\), then \(0\notin(1,5)\), so \((0,0)\notin B\) and \(0\notin(1,4)\). These membership tests agree with the computed preimage.
Worked Example: A Diagonal in a Countable Product
Give every factor \(\mathbb{R}\) its standard topology and give \(\mathbb{R}^{\mathbb{N}}\) the product topology. Let \(\Delta_{\mathbb{N}}(x)=(x,x,\ldots)\). Consider the basic open set \(B\) that restricts coordinates \(2,5,9\) to \((-1,3)\), \((0,4)\), and \((1,6)\), respectively, and imposes no restrictions on other coordinates.
The point \(x\) is in \(\Delta_{\mathbb{N}}^{-1}(B)\) precisely when the same value \(x\) lies in all three specified intervals. The computation is \[ \Delta_{\mathbb{N}}^{-1}(B)=(-1,3)\cap(0,4)\cap(1,6)=(1,3). \] The endpoint \(1\) is excluded because each interval is open, and the endpoint \(3\) is excluded because \((-1,3)\) does not contain \(3\). The result is open. For every \(n\in\mathbb{N}\), the \(n\)-th coordinate map is \(\pi_n\circ\Delta_{\mathbb{N}}=\operatorname{id}_{\mathbb{R}}\), including coordinates other than \(2,5,9\).
Check: Substituting \(x=2\), we find \(2\in(-1,3)\), \(2\in(0,4)\), and \(2\in(1,6)\); thus \(\Delta_{\mathbb{N}}(2)\in B\). Substituting \(x=1\), we find \(1\notin(1,6)\); thus \(\Delta_{\mathbb{N}}(1)\notin B\). These tests confirm both the interval intersection and its endpoint strictness.
The Diagonal Map and Its Image
When \(J\) is nonempty, the diagonal map also gives a precise way to recognize the original space inside the product. Its image is the subset \[ D=\Delta_J(X)=\{(x)_{j\in J}:x\in X\}\subseteq X^J, \] consisting of the constant tuples. We give \(D\) the subspace topology inherited from \(X^J\).
Proof. The map \(\Delta_J:X\to D\) is surjective by the definition of \(D\). To prove injectivity, choose \(j\in J\). If \(\Delta_J(x)=\Delta_J(y)\), then their \(j\)-th coordinates are equal, so \(x=y\). Thus \(\Delta_J\) is a bijection onto \(D\). It is continuous as a map into \(D\): an open subset of \(D\) has the form \(D\cap V\) for an open set \(V\subseteq X^J\), and \(\Delta_J^{-1}(D\cap V)=\Delta_J^{-1}(V)\), which is open by the theorem. Fix \(j\in J\). The restriction \(\pi_j|_D:D\to X\) is continuous, because for any open \(U\subseteq X\), its preimage is \(D\cap\pi_j^{-1}(U)\), open in the subspace \(D\); here \(\pi_j^{-1}(U)\) is a basic open set in the product. On every point of \(D\), the \(j\)-th coordinate recovers the value repeated in that tuple. Therefore \((\pi_j|_D)\circ\Delta_J=\operatorname{id}_X\) and \(\Delta_J\circ(\pi_j|_D)=\operatorname{id}_D\). Thus \(\pi_j|_D\) is the continuous inverse of \(\Delta_J\), proving that \(\Delta_J\) is a homeomorphism. \(\square\)
The condition \(J\ne\varnothing\) matters for this statement: the proof uses a coordinate to recover the input. If \(J=\varnothing\), the empty product \(X^\varnothing\) is a one-point space, and there are no coordinate projections. The unique map \(X\to X^\varnothing\) is continuous, but it need not be injective or a homeomorphism onto its image. For example, if \(X\) has two distinct points, both map to the one point of \(X^\varnothing\).
Worked Example: A Diagonal from a Two-Point Space
Let \(X=\{a,b\}\) have the topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\), and give \(X^3\) the product topology. The diagonal map is \(\Delta_{\{1,2,3\}}(a)=(a,a,a)\) and \(\Delta_{\{1,2,3\}}(b)=(b,b,b)\). Consider the basic open set \(B=\{a\}\times X\times\{a\}\). Each factor set is open in \(X\), so \(B\) is open in the product.
A diagonal tuple lies in \(B\) exactly when its first and third coordinates equal \(a\), which happens for \(\Delta_{\{1,2,3\}}(a)\) and not for \(\Delta_{\{1,2,3\}}(b)\). Consequently, \[ \Delta_{\{1,2,3\}}^{-1}(B)=\{a\}. \] This is open in \(X\). The three coordinate maps are each \(\operatorname{id}_X\): for example, the first coordinate sends \((a,a,a)\) to \(a\) and \((b,b,b)\) to \(b\), and the second and third coordinates do the same.
Check: The diagonal image is \(D=\{(a,a,a),(b,b,b)\}\). The restriction of the first projection sends \((a,a,a)\) to \(a\) and \((b,b,b)\) to \(b\), so it is the inverse of \(\Delta_{\{1,2,3\}}:X\to D\), as the proposition states.
A Common Pitfall: Restricting Every Coordinate
A frequent mistake is to treat an infinite collection of coordinate restrictions as if it were a basic open set in the product topology. For example, in \(\mathbb{R}^{\mathbb{N}}\), consider the set \[ C=\{(x_n)_{n\in\mathbb{N}}:x_n\in(-1/n,1/n)\text{ for every }n\in\mathbb{N}\}. \] It may be tempting to call \(C\) a basic open set and calculate its diagonal preimage as \(\bigcap_{n\in\mathbb{N}}(-1/n,1/n)=\{0\}\), then use this as a continuity test. The intersection calculation is correct, but the continuity argument is not: \(C\) is not open in the product topology.
To verify that \(C\) is not open, consider its point \((0,0,\ldots)\). Any basic open neighborhood of this point restricts only a finite set \(F\subseteq\mathbb{N}\) of coordinates. Choose \(k\in\mathbb{N}\setminus F\), which is possible because \(\mathbb{N}\) is infinite. The tuple with \(k\)-th coordinate \(1\) and all other coordinates \(0\) belongs to that basic neighborhood: it agrees with the zero tuple in every restricted coordinate. But it does not belong to \(C\), since \(1\notin(-1/k,1/k)\). Thus no basic open neighborhood of the zero tuple lies inside \(C\). Continuity requires open sets in the target to have open preimages; a preimage calculation for a set that is not open does not establish or refute continuity.
For an empty index set \(J\), \(X^J\) is the one-point space and the unique map \(X\to X^J\) is continuous because the preimages of its open sets are either \(X\) or \(\varnothing\), both open. For every nonempty finite or infinite \(J\), the finite-coordinate calculation proves continuity of \(\Delta_J\), and every coordinate map recovers the original input.
Check Your Understanding
Use the basic-open-set computation and the coordinate formula to answer the following.
- For the diagonal map \(\mathbb{R}\to\mathbb{R}^3\), compute the preimage of \((-2,1)\times(0,4)\times(-1,3)\).
- For a nonempty index set \(J\), prove that \(\Delta_J\) is injective by using one coordinate.
- In \(\mathbb{R}^{\mathbb{N}}\) with the product topology, what is the preimage under \(\Delta_{\mathbb{N}}\) of the basic open set restricting coordinate \(4\) to \((2,5)\) and coordinate \(8\) to \((1,3)\)?
- Why does the proposition identifying \(X\) homeomorphically with \(\Delta_J(X)\) require \(J\ne\varnothing\)?
- Explain why a set defined by restrictions in every coordinate cannot automatically be treated as a basic open set in an infinite product.