The Torus as a Product of Quotient Circles

From Edge Pairings to a Product

In the square model of the torus, opposite edges are identified: the left edge is paired with the right edge, and the bottom edge with the top edge. These identifications happen independently in the two coordinates. The goal is to turn that observation into a homeomorphism, not just a correspondence between points. The key topological input is that products of open quotient maps are quotient maps.

Let \(I=[0,1]\) with its usual subspace topology from \(\mathbb{R}\). On \(I^2=I\times I\), use the product topology. Define an equivalence relation \(\sim\) by \[ (x,y)\sim(x',y')\quad\Longleftrightarrow\quad x-x'\in\mathbb{Z}\ \text{and}\ y-y'\in\mathbb{Z}. \] For points in \(I^2\), this says that each coordinate is either unchanged or is exchanged between 0 and 1. Write \(T=I^2/{\sim}\), with the quotient topology, and let \(q_T:I^2\to T\) be the quotient map. The edge-pairing torus is precisely this quotient.

The space \(C\) is a quotient circle: the map \([t]\mapsto(\cos(2\pi t),\sin(2\pi t))\) identifies it with the usual circle \(S^1\). In particular, \(C\) is Hausdorff. We will use this fact when proving that the natural bijection from \(T\) to \(C\times C\) is a homeomorphism.

The Product Quotient on the Real Plane

First we check the quotient-map input. Recall that a map is a quotient map when it is surjective and a subset of its target is open exactly when its preimage is open. An open surjection is a quotient map. The product theorem for open quotient maps says that if \(f:X\to Y\) and \(g:X'\to Y'\) are open quotient maps, then \(f\times g:X\times X'\to Y\times Y'\) is an open quotient map, with product topologies.

Proof. The map \(p\) is surjective by the definition of \(C\). If \(U\subseteq\mathbb{R}\) is open, then \[ p^{-1}(p(U))=\bigcup_{n\in\mathbb{Z}}(U+n). \] Each translate \(U+n\) is open in \(\mathbb{R}\), so their union is open. By the definition of the quotient topology, \(p(U)\) is open in \(C\). Thus \(p\) is an open surjection, hence a quotient map. The stated product result applies to \(p\) in both coordinates, so \(p\times p\) is an open quotient map. \(\square\)

The fibers of \(p\times p\) are exactly the equivalence classes for integer translation in each coordinate: \((x,y)\) and \((x',y')\) have the same image precisely when \(x-x'\in\mathbb{Z}\) and \(y-y'\in\mathbb{Z}\). Thus the product theorem gives the quotient topology associated with the two independent coordinate identifications on \(\mathbb{R}^2\).

The Homeomorphism

Proof. Define \(h:I^2\to C\times C\) by \(h(x,y)=(p(x),p(y))\). This is the restriction of the continuous map \(p\times p\), so it is continuous. If \((x,y)\sim(x',y')\), then \(x-x'\) and \(y-y'\) are integers, and therefore \(p(x)=p(x')\) and \(p(y)=p(y')\). Hence \(h\) is constant on each equivalence class. By the defining property of the quotient topology, there is a unique continuous map \(H:T\to C\times C\) satisfying \(H\circ q_T=h\).

We next prove that \(H\) is bijective. For surjectivity, take any \((c_1,c_2)\in C\times C\). Choose real representatives \(s,t\) with \(p(s)=c_1\) and \(p(t)=c_2\). Each real number differs by an integer from a number in \(I\), so there exist \(x,y\in I\) with \(p(x)=c_1\) and \(p(y)=c_2\). Thus \(H(q_T(x,y))=(c_1,c_2)\).

For injectivity, suppose \(H(q_T(x,y))=H(q_T(x',y'))\), where both points lie in \(I^2\). Equality of the first coordinates says \(p(x)=p(x')\), so \(x-x'\in\mathbb{Z}\); equality of the second says \(y-y'\in\mathbb{Z}\). The definition of \(\sim\) now gives \((x,y)\sim(x',y')\), and therefore \(q_T(x,y)=q_T(x',y')\). This proves injectivity.

It remains to check that the continuous bijection \(H\) is a homeomorphism. The square \(I^2\) is compact by the Heine–Borel theorem. Since \(q_T\) is continuous and surjective, \(T\) is compact: for any open cover of \(T\), its preimages cover \(I^2\); a finite subcover of \(I^2\) gives a finite subcover of \(T\). The target \(C\times C\) is Hausdorff. Indeed, \(C\) is Hausdorff, so distinct product points differ in at least one coordinate; disjoint open neighborhoods in that coordinate, multiplied by \(C\) in the other coordinate, separate them.

A continuous bijection from a compact space to a Hausdorff space is a homeomorphism. To recall why, a closed subset of a compact space is compact, its continuous image is compact, and compact subsets of a Hausdorff space are closed. Thus the continuous bijection \(H\) sends closed sets to closed sets, so its inverse is continuous. Applying this argument proves that \(H\) is a homeomorphism. \(\square\)

The product theorem has a specific role here: it establishes that the two integer-coordinate quotient maps together give the product quotient \(C\times C\), rather than some different topology on the set of coordinate classes. The compact-to-Hausdorff argument then verifies that restricting representatives to the square and making its edge identifications produces exactly that topology.

A Common Pitfall: The Interval Quotient Is Not Open

It is tempting to try to apply the product theorem directly to the map from \(I\) to \(I/{\sim}\) that identifies 0 with 1. That map is a quotient map, but it is not open. For example, \(U=[0,1/3)\) is open in the subspace \(I\): it is the intersection of \(I\) with the open real interval \((-1/4,1/3)\). Its image under the endpoint-identification map has preimage \[ [0,1/3)\cup\{1\}. \] This preimage is not open in \(I\), since it contains 1 but contains no interval of the form \((1-\delta,1]\) for \(\delta>0\). By the quotient topology, the image of \(U\) is not open.

The correction is to use the open quotient map \(p:\mathbb{R}\to\mathbb{R}/\mathbb{Z}\), for which integer translates of open sets remain open, and then apply the product theorem on \(\mathbb{R}^2\). The square is a convenient set of representatives, but its endpoint quotient should not be mistaken for an open map.