Why Winding Number Survives a Based Homotopy
A loop in the circle starts and ends at the same point, but its lift to the real line need not do so. The difference between the endpoints of that lift records how many times the loop winds around the circle, with sign. The key fact in this tutorial is that this integer cannot change during a homotopy that keeps the basepoint fixed. The proof has two ingredients: lifts of the homotopy vary continuously, and a continuous integer-valued function on an interval is constant.
Give \(I=[0,1]\) its usual topology, give \(S^1=\{z\in\mathbb{C}:|z|=1\}\) the subspace topology from \(\mathbb{C}\), and give \(\mathbb{R}\) its standard topology. Use the exponential covering map \(p:\mathbb{R}\to S^1\), \(p(u)=e^{2\pi i u}\). Its fibers are \(p^{-1}(z)=\{u+k:k\in\mathbb{Z}\}\) for any \(u\) with \(p(u)=z\).
The lift exists and is unique by the path-lifting and uniqueness results for covering spaces. The integer does not depend on which lift \(a\) of \(z_0\) is chosen: replacing \(a\) by \(a+k\), where \(k\in\mathbb{Z}\), replaces the entire lift by \(\widetilde{\gamma}+k\), so the endpoint difference remains \((\widetilde{\gamma}(1)+k)-(a+k)=n\). The winding number therefore measures endpoint displacement, not the absolute location of the lifted path.
Continuous Variation of Lifts
Proof. Every point of \(S^1\) has an open arc \(V\) that is evenly covered: \(p^{-1}(V)\) is a union of disjoint open sets, and on each such set \(p\) is a homeomorphism onto \(V\). The sets \(H^{-1}(V)\), as \(V\) ranges over these arcs, form an open cover of the compact square \(I\times I\). A finite subcover has a Lebesgue number: some \(\delta>0\) is small enough that every subset of the square of diameter less than \(\delta\) lies in one member of the subcover. This follows from compactness: otherwise choose sets of diameter tending to zero not contained in any member, choose a point from each, and pass to a convergent subsequence; an open member containing the limit eventually contains each entire small set, a contradiction.
Subdivide the square into a rectangular grid whose cells have diameter less than \(\delta\). The image under \(H\) of each cell lies in an evenly covered arc \(V\). Starting with the prescribed lift \(a\) along the bottom edge, extend the lift across the cells one at a time. On a cell, choose the sheet of \(p^{-1}(V)\) determined by the already chosen lift at its lower-left corner, and use the inverse of \(p\) on that sheet. This agrees with the lift already defined on the cell's bottom and left edges: on either edge the two maps are lifts of the same path and agree at the corner, so uniqueness of path lifts makes them agree along the whole edge. Thus the cell lifts fit together continuously. After all cells are treated, they give the required continuous \(\widetilde{H}\). \(\square\)
For a based homotopy of loops, the lemma supplies a continuous lift on the entire square, not merely separate lifts for individual loops. That shared lift is what allows endpoint displacement to be compared as the homotopy parameter varies.
Homotopy Invariance
Proof. Choose \(a\in\mathbb{R}\) with \(p(a)=z_0\), and use the lemma to lift \(H\) to a continuous \(\widetilde{H}\) satisfying \(\widetilde{H}(s,0)=a\). For each \(s\), the path \(t\mapsto\widetilde{H}(s,t)\) is the lift of \(\gamma_s\) starting at \(a\). Since \(H(s,1)=z_0\), its endpoint lies in \(p^{-1}(z_0)=a+\mathbb{Z}\). Consequently \[ n(s)=\widetilde{H}(s,1)-a \] is an integer for every \(s\). It is continuous in \(s\), because \(\widetilde{H}\) is continuous. If \(n\) took two different integer values, the intermediate value theorem would force it to take a noninteger value between them: for example, between distinct integers \(m\) and \(n\) lies a number of the form \(m+\tfrac12\) (or \(n+\tfrac12\), choosing the lower endpoint). This contradicts the fact that every value of \(n(s)\) is an integer. Thus \(n(s)\) is constant. By the definition of winding number, it equals \(\operatorname{wind}(\gamma_s)\) for each \(s\), proving the claim. \(\square\)
The fixed-basepoint condition matters twice: it makes every lifted path start at the same \(a\), and it ensures every lifted endpoint belongs to the same discrete set \(a+\mathbb{Z}\). The theorem does not say that arbitrary loops with different basepoints are directly related by this definition of based homotopy.
Worked Example: A Loop Deformed Without Changing Its Winding
For \(s,t\in I\), define \(H(s,t)=\exp(2\pi i(t+s\sin(2\pi t)))\). Each path is based at \(1\), since the phase is \(0\) at \(t=0\) and \(1\) at \(t=1\). A lift starting at \(0\) is \[ \widetilde{H}(s,t)=t+s\sin(2\pi t), \] because applying \(p\) gives \(H(s,t)\), and \(\widetilde{H}(s,0)=0\).
At the final endpoint, \(\sin(2\pi)=0\), so \(\widetilde{H}(s,1)=1+s\cdot0=1\). The endpoint displacement is \(1-0=1\) for every \(s\), hence each loop has winding number \(1\). As a numerical check at \(s=\tfrac12,t=\tfrac14\), the phase is \(\tfrac14+\tfrac12\sin(\tfrac{\pi}{2})=\tfrac14+\tfrac12=\tfrac34\), and \(H(\tfrac12,\tfrac14)=e^{2\pi i(3/4)}=-i\). The lift gives the same value, \(p(\tfrac34)=-i\).
Worked Example: A Loop Contracted to the Constant Loop
Set \(H(s,t)=\exp(2\pi i(1-s)\sin(2\pi t))\). At \(t=0\) and \(t=1\), the sine is zero, so every path is based at \(1\). The continuous lift starting at \(0\) is \(\widetilde{H}(s,t)=(1-s)\sin(2\pi t)\). At \(s=0\) this gives the loop \(\exp(2\pi i\sin(2\pi t))\); at \(s=1\) it gives the constant loop at \(1\).
For every \(s\), \(\widetilde{H}(s,0)=0\) and \(\widetilde{H}(s,1)=(1-s)\sin(2\pi)=0\), so the displacement is \(0-0=0\). Thus both the initial loop and the constant loop have winding number \(0\), as homotopy invariance requires. At \(s=\tfrac12,t=\tfrac14\), the lifted value is \((1-\tfrac12)\sin(\tfrac{\pi}{2})=\tfrac12\), and its image is \(e^{2\pi i(1/2)}=-1\). The formula for \(H\) gives the same value.
Worked Example: A Based Homotopy at a Different Point
Take the basepoint \(z_0=i=p(\tfrac14)\) and define \[ H(s,t)=\exp\!\left(2\pi i\left(\tfrac14+2t+s\,t(1-t)\right)\right). \] At \(t=0\), the phase is \(\tfrac14\); at \(t=1\), it is \(\tfrac14+2=\tfrac94\). Since \(\tfrac94-\tfrac14=2\) is an integer, both endpoints map to \(i\). A lift starting at \(a=\tfrac14\) is \(\widetilde{H}(s,t)=\tfrac14+2t+s\,t(1-t)\). Its final value is \(\tfrac14+2+s\cdot1\cdot0=\tfrac94\), so the displacement is \(\tfrac94-\tfrac14=2\) for every \(s\).
For a check at \(s=\tfrac12,t=\tfrac12\), the phase is \(\tfrac14+2(\tfrac12)+\tfrac12(\tfrac12)(\tfrac12)=\tfrac14+1+\tfrac18=\tfrac{11}{8}\). Therefore the image is \(e^{2\pi i(11/8)}=e^{3\pi i/4}=(-1+i)/\sqrt{2}\). This calculation uses the lift based at \(\tfrac14\); the endpoint difference, not the endpoint's absolute coordinate, gives the winding number.
A Common Pitfall: Comparing Endpoint Points Instead of Lift Endpoints
The standard loop \(\gamma(t)=e^{2\pi i t}\) begins and ends at \(1\). A mistaken calculation might subtract the endpoints on the circle and conclude that its winding number is \(1-1=0\). That subtraction is not the definition: winding number is calculated from a lift to \(\mathbb{R}\). The lift starting at \(0\) is \(\widetilde{\gamma}(t)=t\), since \(p(t)=e^{2\pi i t}=\gamma(t)\). Its lifted endpoints are \(0\) and \(1\), so its winding number is \(1-0=1\).
Using the principal argument does not fix the mistaken calculation. For this loop the principal argument returns to the same representative at the end, but it jumps when the path crosses the chosen argument cut; it is not a continuous lift on the whole interval. The theorem instead compares continuous lifts, whose endpoint displacement varies continuously and is constrained to lie in the integers. In particular, loops with different winding numbers cannot be based-homotopic: a based homotopy would force their winding numbers to be equal.
Check Your Understanding
Use the lifted endpoint displacement in each question, and state where the based-homotopy condition is needed.
- For \(H(s,t)=\exp(2\pi i(4t+s\,t(1-t)))\), find the winding number of each loop \(t\mapsto H(s,t)\).
- Why must the lift of a based homotopy start at one fixed real number rather than at an arbitrary lift depending on \(s\)?
- Show that the loops \(t\mapsto e^{2\pi i t}\) and \(t\mapsto e^{-2\pi i t}\) cannot be based-homotopic.
- For a loop based at \(z_0=p(a)\), explain why its lifted endpoint displacement is an integer even when \(a\) is not an integer.
- In the second worked example, compute the lifted value at \(s=\tfrac14,t=\tfrac14\), and check its image under \(p\).