Standard Coverings of the Circle and Torus

Three Standard Maps

The circle and torus have simple descriptions as quotients of Euclidean spaces: points that differ by integers in the appropriate coordinates map to the same point. The standard maps make this identification explicit. To verify that they are covering maps, it is not enough to count the points in a fiber: we must also find neighborhoods whose preimages split into disjoint open sets, each mapping homeomorphically onto the neighborhood.

Write \(\mathbb{C}\) with its usual topology, let \(S^1=\{z\in\mathbb{C}:|z|=1\}\) have the subspace topology, and give \(T^2=S^1\times S^1\) the product topology. Define \(p:\mathbb{R}\to S^1\) and \(P:\mathbb{R}^2\to T^2\) by

For an integer \(n\), define \(q_n:S^1\to S^1\) by \(q_n(z)=z^n\). For negative \(n\), this uses the usual reciprocal \(z^n=1/z^{-n}\); this is defined on \(S^1\), since its elements are nonzero.

The Real Line Covering the Circle

The map \(p\) is continuous because \(t\mapsto e^{2\pi i t}\) is continuous, and it is surjective because every point of \(S^1\) has the form \(e^{i\theta}=p(\theta/(2\pi))\). Its fibers are infinite: periodicity gives the same circle point whenever the input changes by an integer.

Proof. The equality \(p(u)=p(t)\) holds exactly when \(e^{2\pi i(u-t)}=1\), which holds exactly when \(u-t\in\mathbb{Z}\). Thus the fiber formula follows. If \(u,v\in J\) and \(p(u)=p(v)\), then \(u-v\in\mathbb{Z}\); since \(|u-v|<1\), it follows that \(u=v\). Hence \(p|_J\) is injective. To see that \(p(J)\) is open, take any \(t\in J\), and choose \(\eta>0\) so that \((t-\eta,t+\eta)\subseteq J\) and \(\eta<1/2\). Its image is the open arc \(\{z\in S^1:\operatorname{Re}(e^{-2\pi it}z)>\cos(2\pi\eta)\}\). These arcs around points of \(J\) show that \(p(J)\) is open. The restriction \(p|_J\) is a homeomorphism onto \(p(J)\): it is a continuous bijection, and it is open because each open subset of \(J\) is a union of small intervals whose images are open arcs. Periodicity gives \(p^{-1}(p(J))=\bigcup_{k\in\mathbb{Z}}(J+k)\). These intervals are open and pairwise disjoint: if \(J+k\) and \(J+l\) intersect for distinct integers \(k,l\), then two points of \(J\) differ by the nonzero integer \(k-l\), contradicting \(b-a<1\). Translation by \(k\) does not change \(p\), so every restriction to \(J+k\) is a homeomorphism onto \(p(J)\). \(\square\)

Integer Powers on the Circle

The identity \(q_n(p(t))=p(nt)\) connects the power maps to the real-line map. For \(n\ne0\), each target has exactly \(|n|\) preimages. The sign of \(n\) changes the orientation of the local parametrization when \(n<0\), but not the number of sheets.

Proof. For \(n\ne0\), continuity follows from continuity of multiplication and, for negative powers, continuity of \(z\mapsto 1/z\) on \(S^1\). The map is surjective: if \(y=p(s)\), then \(x=p(s/n)\) satisfies \(q_n(x)=p(s)=y\). To count the fiber over \(p(s)\), write any candidate point as \(p(t)\). It belongs to the fiber exactly when \(p(nt)=p(s)\), or \(nt=s+k\) for some \(k\in\mathbb{Z}\). Thus the fiber consists of \(p((s+k)/n)\) for integer \(k\). Two such points agree exactly when their indices differ by a multiple of \(n\), so there are \(|n|\) distinct points, represented by \(k=0,\ldots,|n|-1\).

To find an evenly covered neighborhood, fix \(p(s)\) and choose \(0<\varepsilon<1/2\). Let \(J=(s-\varepsilon,s+\varepsilon)\) and \(V=p(J)\), an open arc. For \(k=0,\ldots,|n|-1\), set \(I_k=((s+k)/n-\varepsilon/|n|,(s+k)/n+\varepsilon/|n|)\) and \(W_k=p(I_k)\). Each \(I_k\) has length \(2\varepsilon/|n|<1\), so \(p|_{I_k}\) is a homeomorphism onto \(W_k\). The equation \(q_n(p(t))=p(nt)\) shows that \(q_n(W_k)=V\): as \(t\) runs through \(I_k\), \(nt\) runs through \(J+k\) if \(n>0\), or through the same interval in reverse order if \(n<0\). Since \(p(J+k)=p(J)\), the image is \(V\). The restriction \(q_n|_{W_k}\) is a homeomorphism: in these interval coordinates it is the affine map \(t\mapsto nt-k\), followed by \(p|_J\), with the sign of \(n\) allowing reversal. Finally, if \(q_n(p(t))\in V\), then \(nt\in J+m\) for some integer \(m\), so \(t\) belongs to the interval centered at \((s+m)/n\) with radius \(\varepsilon/|n|\). Changing \(m\) by a multiple of \(n\) translates this interval by an integer and therefore gives the same arc in \(S^1\); the remaining possibilities are precisely the \(|n|\) arcs \(W_k\). They are pairwise disjoint because their centers differ, modulo integers, by at least \(1/|n|\), while the sum of their radii is \(2\varepsilon/|n|<1/|n|\). Thus \(q_n^{-1}(V)\) is exactly the disjoint union of these sheets. For \(n=0\), \(q_0(z)=1\) for all \(z\), so it is not surjective onto \(S^1\) and hence is not a covering map. \(\square\)

The Plane Covering the Torus

The product map \(P\) applies the circle covering in each coordinate. The topology on \(T^2\) is the product topology, so products of open arcs form open neighborhoods. Their preimages are products of the corresponding intervals, which are open rectangles in \(\mathbb{R}^2\).

Proof. The map is continuous by continuity of \(p\) in each coordinate and the definition of the product topology. It is surjective because each coordinate of a point in \(T^2\) is \(p\) of some real number. The fiber formula follows by applying the fiber formula for \(p\) separately to each coordinate. For an evenly covered neighborhood, choose intervals \(J=(a-\varepsilon,a+\varepsilon)\) and \(K=(b-\delta,b+\delta)\), with \(0<\varepsilon,\delta<1/2\), and put \(V=p(J)\times p(K)\). This is open in \(T^2\). Its preimage is \(P^{-1}(V)=\bigcup_{k,l\in\mathbb{Z}}(J+k)\times(K+l)\). The rectangles are open and pairwise disjoint: if two intersect, their first-coordinate intervals must agree in index and their second-coordinate intervals must agree in index, since each interval has length less than \(1\). On each rectangle, \(P\) is the product of the homeomorphisms \(p|_{J+k}\) and \(p|_{K+l}\), so it is a homeomorphism onto \(V\). This verifies the covering-map definition. \(\square\)

Common Pitfall: Counting Fibers Is Not Enough

A finite fiber does not by itself prove that a map is a covering. A covering also requires neighborhoods whose full preimages split into open sheets, each mapped homeomorphically onto the neighborhood. For the standard maps, the interval and rectangle calculations above establish this local condition, rather than relying only on fiber counts.

There is also a specific error to avoid for integer powers: treating \(n=0\) as if it had \(|n|\) sheets. The expression \(|0|=0\) is not a count of a fiber. In fact \(q_0(z)=1\) for every \(z\in S^1\). Its fiber over \(1\) is the entire circle, while the fiber over any other point is empty; moreover the map is not surjective. The covering result therefore applies to \(n\ne0\), exactly as stated.

These examples display three related but distinct fiber patterns: integer translates for the real-line covering, \(|n|\) points for a nonzero power map, and integer lattices in the plane for the torus covering. In each case, the explicit evenly covered neighborhoods supply the local structure required by the definition.