Diagnosing Lift, Orientation, and Extension Errors

Keep the Conventions Visible

Degree arguments for maps of the circle are short, but several different choices enter them: a lift has a starting value, a parametrization has an orientation, and concatenation changes how the parameter interval is traversed. Errors often come from treating one of these choices as if it were part of the degree. We will use the exponential covering map and keep each choice explicit.

Give \(S^1=\{z\in\mathbb{C}:|z|=1\}\) its subspace topology from \(\mathbb{C}\), and give \(\mathbb{R}\) its standard topology. Write \(p:\mathbb{R}\to S^1\) for \(p(t)=e^{2\pi i t}\). A lift of a path \(\gamma:[0,1]\to S^1\) is a path \(\theta:[0,1]\to\mathbb{R}\) such that \(p(\theta(t))=\gamma(t)\) for every \(t\). We use the standard path-lifting property of this covering map: after choosing \(\theta(0)\) with \(p(\theta(0))=\gamma(0)\), there is a unique lift with that starting value. This is the lifting property used in the earlier material on the fundamental group of the circle.

The definition uses the difference of two lift values, not either value separately. If \(\theta\) and \(\widetilde{\theta}\) are lifts of the same path, then \(p(\theta(t))=p(\widetilde{\theta}(t))\), so \(\theta(t)-\widetilde{\theta}(t)\) is integer-valued. It is continuous, and a continuous integer-valued function on the interval is constant. Thus changing the lift adds the same integer to both endpoints and leaves their difference unchanged.

Orientation and Concatenation

For a loop \(\gamma\), define its reversal by \(\overline{\gamma}(t)=\gamma(1-t)\). Reversing the direction of traversal changes the sign of degree. For loops \(\gamma\) and \(\eta\) based at the same point, their concatenation \(\gamma*\eta\) traverses \(\gamma\) first and \(\eta\) second, each at twice its original speed. Its degree is the sum of their degrees.

Proof. Let \(\theta\) lift \(\gamma\), and put \(n=\theta(1)-\theta(0)\). The path \(t\mapsto\theta(1-t)\) lifts \(\overline{\gamma}\), and its endpoint difference is \(\theta(0)-\theta(1)=-n\). For concatenation, choose lifts \(\theta\) of \(\gamma\) and \(\phi\) of \(\eta\). Since the loops have the same basepoint, \(\phi(0)-\theta(0)\) is an integer. Add the integer \(\theta(1)-\phi(0)\) to \(\phi\), so that the adjusted lift \(\psi\) satisfies \(\psi(0)=\theta(1)\). A lift of \(\gamma*\eta\) is \(\theta(2t)\) for \(0\le t\le1/2\), followed by \(\psi(2t-1)\) for \(1/2\le t\le1\). It is continuous at \(t=1/2\), and its endpoint difference is \(\psi(1)-\theta(0)=(\psi(1)-\psi(0))+(\theta(1)-\theta(0))=\deg(\eta)+\deg(\gamma)\). For domain reflection, the parametrized loop of \(f\circ r\) is \(t\mapsto f(p(-t))\), which is the reversal, up to a shift of the loop parameter, of the loop for \(f\). A shift changes the starting value of a lift but not its endpoint difference. For target reflection, \(p(-\theta(t))=\overline{p(\theta(t))}\), so negating a lift negates its endpoint difference. \(\square\)

Map Extensions and Homotopies

An extension of \(f:S^1\to S^1\) over the disk is a continuous map \(F:D^2\to S^1\) whose restriction to the boundary \(S^1\) is \(f\). The disk has its usual subspace topology from \(\mathbb{R}^2\), with boundary the unit circle. The criterion below requires no orientation convention beyond the parametrization already fixed.

Proof. First suppose \(F:D^2\to S^1\) extends \(f\). The maps \(\gamma_s(t)=F((1-s)p(t))\), for \(0\le s\le1\), give a homotopy of loops from \(t\mapsto f(p(t))\) to the constant loop at \(F(0)\). Degree is invariant under homotopy of loops; this is the standard homotopy invariance of the fundamental-group class, established in the earlier material on the fundamental group. A constant loop has a constant lift and degree \(0\), so \(\deg(f)=0\).

Conversely, suppose \(\deg(f)=0\). Let \(\theta:[0,1]\to\mathbb{R}\) lift \(t\mapsto f(p(t))\), and put \(a=\theta(0)\). The degree condition says \(\theta(1)-\theta(0)=0\), so \(\theta(1)=\theta(0)=a\). For \(0\le r\le1\) and \(0\le t\le1\), define \(F(rp(t))=p((1-r)a+r\theta(t))\). At \(r=1\) this equals \(p(\theta(t))=f(p(t))\). At \(r=0\) it equals \(p(a)\), independent of \(t\). At the parameter seam, \(t=0\) and \(t=1\) give the same value because \(\theta(0)=\theta(1)\). Thus the formula is well-defined on the disk in polar coordinates, including its center. It is continuous away from the center by continuity of the formula. At the center, \(\theta\) is bounded on \([0,1]\); hence \((1-r)a+r\theta(t)\to a\) uniformly in \(t\) as \(r\to0\), so the formula is continuous there as well. This gives an extension of \(f\). \(\square\)

A homotopy is a map defined on \(S^1\times[0,1]\); it is not, by itself, an extension over \(D^2\). A homotopy from \(f\) to a constant map does produce an extension: identify the constant-end circle of the homotopy with the disk’s center, using the homotopy parameter as the radial coordinate. The constant endpoint is essential for this construction to be well-defined at the center.

For contrast, \(H(p(t),s)=p(t)\) is a homotopy that is constant in \(s\), but its endpoint map is still the identity \(f(p(t))=p(t)\). The lift \(t\mapsto t\) gives degree \(1-0=1\), so the identity has no extension over the disk. Merely having a homotopy on the boundary does not provide a disk extension; the homotopy must end at a constant map. Also, extending a map and extending a homotopy are different tasks: the first asks for a map on \(D^2\) with prescribed boundary values, while the second asks for a map on \(D^2\times[0,1]\) matching prescribed data throughout a time interval. A proof of one statement should not silently substitute for the other.

A Reliable Diagnostic Checklist

Before accepting a degree argument, check the object being lifted and the direction in which its parameter runs. Use endpoint differences, not endpoint values; make the concatenation lift continuous at the joining time; and state whether the domain or target orientation has been reversed. For an extension claim, identify the map on the boundary and verify that its degree is zero. If a homotopy is being used to build an extension, check explicitly that its terminal loop is constant.

Common pitfall. For \(f(z)=z^2\), the lift \(2t+4\) begins at \(4\) and ends at \(6\). The incorrect value \(6\) is not the degree: the correct difference is \(6-4=2\). Reversing the domain then gives degree \(-2\), not \(+2\), and the nonzero degree means neither map extends over the disk. Each correction follows from the relevant definition rather than from a visual intuition about the circle.