Degree of a Circle Map from Its Lift

From a Circle Map to an Integer

A continuous map from the circle to itself can travel around the circle while its input makes one trip around. The map’s values at the beginning and end of that trip agree, so those values alone do not record how many turns occurred. A lift to the real line keeps track of the turns: its net change over one trip is an integer, and that integer is the degree.

We use \(S^1=\{z\in\mathbb{C}:|z|=1\}\) with the subspace topology from \(\mathbb{C}\), \(I=[0,1]\) with its usual topology, and \(\mathbb{R}\) with its standard topology. Define \(p:\mathbb{R}\to S^1\) by \(p(t)=e^{2\pi i t}\). This is the standard exponential covering map. Its fibers satisfy \(p(a)=p(b)\) if and only if \(a-b\in\mathbb{Z}\).

Such a lift exists: the standard lifting criterion for a covering map says that a map \(g:X\to S^1\) from a path-connected, locally path-connected space lifts to \(\mathbb{R}\) once the image of its fundamental group lies in the image of the fundamental group of the covering space. For \(g=f\circ p\), the domain \(\mathbb{R}\) is contractible, so its fundamental group is trivial and the criterion applies. We use the same standard criterion below for a homotopy, whose parameter domain \(\mathbb{R}\times I\) is also contractible.

Why the Lift Defines a Unique Integer

Proof. Fix a lift \(F\). Since \(p(t+1)=p(t)\), the lift equation gives \(p(F(t+1))=f(p(t+1))=f(p(t))=p(F(t))\). The fiber property of \(p\) implies that \(\Delta(t)=F(t+1)-F(t)\) is an integer for every \(t\in\mathbb{R}\). The function \(\Delta\) is continuous because \(F\) is continuous. A continuous integer-valued function on \(\mathbb{R}\) is constant: if it took two different integer values, the intermediate value theorem would force it to take a noninteger value between them. Write its constant value as \(n\in\mathbb{Z}\).

Now let \(G\) be another lift of \(f\circ p\). For every \(t\), \(p(G(t))=p(F(t))\), so \(G(t)-F(t)\in\mathbb{Z}\). The continuous function \(G-F\) is therefore constant by the same intermediate-value argument; write \(G(t)=F(t)+k\) for a fixed \(k\in\mathbb{Z}\). It follows that \(G(t+1)-G(t)=(F(t+1)+k)-(F(t)+k)=F(t+1)-F(t)=n\). Thus the integer is constant in \(t\) and the same for every lift. \(\square\)

The choice of a starting height for a lift can change the lift by an integer, but it cannot change the lift’s displacement. This is the reason the definition gives an invariant of the map rather than an arbitrary feature of one chosen lift.

Homotopy Does Not Change Degree

Proof. Define \(G:\mathbb{R}\times I\to S^1\) by \(G(t,s)=H(p(t),s)\). The domain \(\mathbb{R}\times I\) is contractible, path-connected, and locally path-connected. The standard lifting criterion for the covering map \(p:\mathbb{R}\to S^1\) therefore gives a continuous lift \(\widetilde{H}:\mathbb{R}\times I\to\mathbb{R}\) satisfying \(p(\widetilde{H}(t,s))=G(t,s)\).

For every \((t,s)\), periodicity of \(p\) in the first variable gives \(G(t+1,s)=G(t,s)\). Consequently, \(p(\widetilde{H}(t+1,s))=p(\widetilde{H}(t,s))\), so the difference \(\Delta(t,s)=\widetilde{H}(t+1,s)-\widetilde{H}(t,s)\) is an integer. It is continuous on the connected space \(\mathbb{R}\times I\), and therefore it is constant by the intermediate-value argument used in the preceding theorem. For each fixed \(s\), the function \(t\mapsto\widetilde{H}(t,s)\) is a lift of \(H(\,\cdot\,,s)\circ p\). The constant difference \(\Delta(t,s)\) is thus the degree of that map. At \(s=0\) it equals \(\deg(f_0)\); at \(s=1\) it equals \(\deg(f_1)\). Since it is the same constant on all of \(\mathbb{R}\times I\), these degrees are equal. \(\square\)

This theorem says that degree is unchanged during a continuous deformation of circle maps. It does not, by itself, assert that degree is a complete test for whether two maps are homotopic; that stronger classification is not needed to define or use the invariant here.

A Common Pitfall: Equal Endpoints Do Not Mean Degree Zero

A circle map always has the same value at the start and end of a full trip in its input: \(p(0)=p(1)=1\), so \(f(p(0))=f(p(1))\). It is a mistake to infer degree zero from this equality. The degree measures the change in a real-valued lift, not the difference between the two endpoint values in \(S^1\).

For a concrete correction, use the map from the homotopy example at \(s=0\), given by \(f(p(t))=e^{2\pi i(2t)}\). Its endpoint values are both \(f(p(0))=e^0=1\) and \(f(p(1))=e^{4\pi i}=1\). A wrong calculation would subtract these equal circle values and report \(0\). But the lift \(F(t)=2t\) has displacement \(F(1)-F(0)=2-0=2\), so the degree is \(2\). The endpoint equality is compatible with this nonzero displacement because the exponential map identifies real numbers differing by an integer.

To compute a degree reliably, first give a lift and verify that it really lifts the map. Then calculate \(F(t+1)-F(t)\); the theorem ensures this is independent of \(t\) and of the lift. When a homotopy is available, the homotopy-invariance theorem allows the degree to be computed from any convenient member of that homotopy.