The Circle as the Real Line Modulo Integers

From Integer Translations to a Circle

A point on a circle can be described by an angle, but angles that differ by a full turn describe the same point. Using turns rather than radians, a full turn has length \(1\): the parameters \(t\) and \(t+1\) give the same point on the unit circle. The quotient of the real line by these integer shifts records exactly this identification. We will prove that the resulting quotient space has the same topology as the circle, not merely the same collection of points.

Let \(\mathbb{R}\) have its standard topology, and let \(S^1=\{z\in\mathbb{C}:|z|=1\}\) have the subspace topology inherited from the usual topology on \(\mathbb{C}\). Define a relation on \(\mathbb{R}\) by \(x\sim y\) if and only if \(x-y\in\mathbb{Z}\). Write \([x]=\{x+n:n\in\mathbb{Z}\}\) for the equivalence class of \(x\), and let \(\mathbb{R}/\mathbb{Z}\) be the set of these classes.

Thus an open set in the quotient is tested by looking at all of its representatives in \(\mathbb{R}\). In particular, \(q^{-1}(W)\) is a union of entire equivalence classes: if it contains \(x\), it contains \(x+n\) for every \(n\in\mathbb{Z}\). Such a union is called saturated for this equivalence relation.

The Exponential Map and Its Fibers

Consider the map \(e:\mathbb{R}\to S^1\) given by \(e(t)=e^{2\pi i t}=\cos(2\pi t)+i\sin(2\pi t)\). It is continuous because its real and imaginary parts are continuous, and its values have modulus \(1\). The standard parametrization of the unit circle shows it is surjective: if \(z=\cos\theta+i\sin\theta\), then \(z=e(\theta/(2\pi))\).

Proof. Using the exponential identity, \(e(s)=e(t)\) if and only if \(e(s-t)=1\). For a real number \(u\), \(e(u)=1\) exactly when \(\cos(2\pi u)=1\) and \(\sin(2\pi u)=0\). These two equalities hold exactly when \(2\pi u=2\pi n\) for some \(n\in\mathbb{Z}\), or equivalently \(u=n\in\mathbb{Z}\). Taking \(u=s-t\) proves both directions. \(\square\)

This proposition says that the fibers of \(e\) are precisely the equivalence classes used to form \(\mathbb{R}/\mathbb{Z}\). Therefore the formula \(\overline{e}([t])=e(t)\) defines a map \(\overline{e}:\mathbb{R}/\mathbb{Z}\to S^1\): choosing another representative \(t+n\) does not change the value, since \(e(t+n)=e(t)e(n)=e(t)\). It is bijective. It is onto because \(e\) is onto; if \(\overline{e}([s])=\overline{e}([t])\), the proposition gives \(s-t\in\mathbb{Z}\), hence \([s]=[t]\).

Why the Induced Map Preserves the Topology

A continuous bijection need not be a homeomorphism, so the fiber calculation alone is not enough. We will show that \(e\) is a quotient map. The key step is proving that it is open: it sends open subsets of \(\mathbb{R}\) to open subsets of \(S^1\).

Proof. Surjectivity was established using the parametrization of \(S^1\). To prove openness, let \(U\subseteq\mathbb{R}\) be open. For each \(t\in U\), choose \(\varepsilon\) with \(0<\varepsilon<1/2\) and \((t-\varepsilon,t+\varepsilon)\subseteq U\). The image of this interval is \[ \{z\in S^1:\operatorname{Re}(z\overline{e(t)})>\cos(2\pi\varepsilon)\}. \] Indeed, writing \(z=e(s)\), the inequality becomes \(\cos(2\pi(s-t))>\cos(2\pi\varepsilon)\). Reduce \(s-t\) modulo integers to a number \(u\in[-1/2,1/2]\). Since cosine is strictly decreasing on \([0,\pi]\), the inequality holds exactly when \(|u|<\varepsilon\), which says \(z=e(s)\) for some \(s\in(t-\varepsilon,t+\varepsilon)\). The displayed set is open in \(S^1\), because it is the intersection of \(S^1\) with the inverse image of an open interval under the continuous real-valued function \(z\mapsto\operatorname{Re}(z\overline{e(t)})\). These open arcs, as \(t\) ranges over \(U\), cover \(e(U)\) and lie in \(e(U)\), so \(e(U)\) is open.

Finally, a quotient map is a continuous surjection \(f:X\to Y\) such that \(V\subseteq Y\) is open if and only if \(f^{-1}(V)\) is open in \(X\). For an open surjection this condition follows directly: continuity gives the forward implication, and if \(f^{-1}(V)\) is open, then \(V=f(f^{-1}(V))\) is open because \(f\) is open. Applying this to \(e\) proves the proposition. \(\square\)

Proof. We have already proved that \(\overline{e}\) is a bijection. It remains to prove that it and its inverse are continuous. The quotient topology gives a set \(W\subseteq\mathbb{R}/\mathbb{Z}\) as open exactly when \(q^{-1}(W)\) is open in \(\mathbb{R}\). For any \(V\subseteq S^1\), \[ q^{-1}(\overline{e}^{-1}(V))=e^{-1}(V), \] because \(e=\overline{e}\circ q\). Since \(q\) is a quotient map by the definition of the quotient topology, and \(e\) is a quotient map by the proposition, we obtain \[ \overline{e}^{-1}(V)\text{ is open}\quad\Longleftrightarrow\quad e^{-1}(V)\text{ is open}\quad\Longleftrightarrow\quad V\text{ is open}. \] In particular, \(\overline{e}\) is continuous. The same equivalence shows that \(\overline{e}\) is itself a quotient map. A bijective quotient map is a homeomorphism: if \(W\) is open in its domain, bijectivity gives \(\overline{e}^{-1}(\overline{e}(W))=W\), so the quotient condition implies that \(\overline{e}(W)\) is open. Thus \(\overline{e}\) is open, and its inverse is continuous. \(\square\)

A Common Mistake and Why the Quotient Matters

Common pitfall. The exponential map \(e:\mathbb{R}\to S^1\) is not one-to-one, so it is incorrect to call \(e\) a homeomorphism. For example, \(e(0)=1\) and \(e(1)=1\), although \(0\ne1\). The correction is to identify precisely the points with the same image: the fiber proposition says they are exactly the points differing by an integer. After making that identification, the induced map \(\overline{e}\) is bijective and, by the quotient-map argument, a homeomorphism.

It is also important to match the scale of the exponential to the equivalence relation. The map used here is \(e^{2\pi i t}\), whose period is \(1\). For the different map \(f(t)=e^{it}\), \(f(0)=1\) but \(f(1)=\cos(1)+i\sin(1)\ne1\); that map has period \(2\pi\), not \(1\). The factor \(2\pi\) is exactly what makes integer translations correspond to complete turns of the circle.

The construction gives a useful model: points of the circle are real parameters considered modulo whole turns, and the quotient topology ensures that this identification agrees with the usual topology on \(S^1\). The proof used two essential facts, not just a visual picture: the fibers of the exponential map are the equivalence classes, and the exponential map is open as well as continuous.