Why Lifted Endpoints Are Invariant
A covering map can send different points of its domain to the same point of its base. Thus two paths with the same initial and terminal points need not have lifts with the same terminal point, even if the lifts begin together. The missing condition is a homotopy between the paths that keeps both endpoints fixed. Under that condition, the homotopy lifting property gives a continuously varying family of lifted paths. Its terminal points vary continuously inside one fiber, and a covering fiber is discrete. That combination forces the terminal point to remain fixed.
Let \(I=[0,1]\) with its standard topology, and let \(p:E\to B\) be a covering map between topological spaces. This means that every \(b\in B\) has an open neighborhood \(V\) such that \(p^{-1}(V)\) is a union of pairwise disjoint open sets, each mapped homeomorphically onto \(V\) by \(p\). A lift of a map \(f:Z\to B\) is a map \(\widetilde f:Z\to E\) satisfying \(p\circ\widetilde f=f\).
The homotopy lifting property for a covering says that if a homotopy \(H:Z\times I\to B\) and a lift \(\widetilde f:Z\to E\) of \(H(\,\cdot\,,0)\) are given, there is a lift \(\widetilde H:Z\times I\to E\) with \(\widetilde H(\,\cdot\,,0)=\widetilde f\). We use this property with \(Z=I\). We also use uniqueness of path lifts: two lifts of the same path that agree at one parameter value agree everywhere. Both properties are established in the earlier material on covering maps.
The Endpoint Invariance Theorem
Proof. Let \(H\) be a homotopy relative to endpoints from \(\alpha\) to \(\beta\). By the homotopy lifting property, there is a lift \(\widetilde H:I\times I\to E\) with \(\widetilde H(s,0)=\widetilde\alpha(s)\). Consider the left edge \(t\mapsto\widetilde H(0,t)\). It lifts the constant path \(t\mapsto x_0\), since \(p(\widetilde H(0,t))=H(0,t)=x_0\). At \(t=0\) this edge starts at \(\widetilde\alpha(0)\). The constant path in \(E\) at \(\widetilde\alpha(0)\) is also a lift of that base path, so uniqueness of path lifts gives \(\widetilde H(0,t)=\widetilde\alpha(0)\) for every \(t\).
Now consider the endpoint track \(\gamma(t)=\widetilde H(1,t)\). It is continuous because \(\widetilde H\) is continuous. For every \(t\), \(p(\gamma(t))=H(1,t)=x_1\), so \(\gamma(t)\) lies in the fiber \(p^{-1}(\{x_1\})\).
This fiber is discrete in its subspace topology. Indeed, for any \(e\in p^{-1}(\{x_1\})\), choose an evenly covered open neighborhood \(V\) of \(x_1\). The sheet \(W\) over \(V\) containing \(e\) is open in \(E\), and \(p|_W\) is one-to-one. Hence \(W\cap p^{-1}(\{x_1\})=\{e\}\), showing that \(\{e\}\) is open in the fiber.
The interval \(I\) is connected, so its continuous image \(\gamma(I)\) is connected. A connected subset of a discrete space has at most one point: if it contained a point \(e\) and another point, its singleton \(\{e\}\) and its complement in that subset would be disjoint, nonempty, relatively open sets separating it. Therefore \(\gamma\) is constant. At \(t=0\), \(\gamma(0)=\widetilde\alpha(1)\); at \(t=1\), \(\widetilde H(s,1)\) is a lift of \(\beta\) starting at \(\widetilde\alpha(0)=\widetilde\beta(0)\). Uniqueness of path lifts gives \(\widetilde H(s,1)=\widetilde\beta(s)\), so \(\gamma(1)=\widetilde\beta(1)\). Since \(\gamma\) is constant, \(\widetilde\alpha(1)=\widetilde\beta(1)\). \(\square\)
The two fixed-edge conditions play different roles. Fixing the initial endpoint makes the left edge of the lifted homotopy constant, so the final lifted path starts at the prescribed initial point. Fixing the terminal endpoint puts the entire endpoint track in one fiber. Without the latter condition, the endpoint track need not lie in a single fiber at all.
Worked Calculations
Worked Example: A Homotopy in the Circle
Give two distinct paths from \(1\) to \(i\) in \(S^1\) that are homotopic relative to endpoints, and compute the endpoints of their lifts under the covering \(p:\mathbb{R}\to S^1\), \(p(u)=e^{2\pi i u}\). Here \(\mathbb{R}\) has its standard topology and \(S^1\) has the subspace topology from \(\mathbb{C}\).
Set $$ H(s,t)=e^{2\pi i\left(s/4+(t/5)\sin(\pi s)\right)}. $$ At \(s=0\), the exponent is \(0/4+(t/5)\sin(0)=0\), so \(H(0,t)=e^0=1\). At \(s=1\), it is \(1/4+(t/5)\sin(\pi)=1/4\), so \(H(1,t)=e^{2\pi i/4}=i\). Thus the paths \(\alpha(s)=H(s,0)\) and \(\beta(s)=H(s,1)\) have fixed endpoints \(1\) and \(i\).
Their lifts starting at \(0\) are $$ \widetilde\alpha(s)=s/4,\qquad \widetilde\beta(s)=s/4+(1/5)\sin(\pi s). $$ Applying \(p\) to each expression gives \(\alpha(s)\) and \(\beta(s)\), respectively, and both lifts start at \(0\). At \(s=1\), \(\widetilde\alpha(1)=1/4\), while \(\widetilde\beta(1)=1/4+(1/5)\sin(\pi)=1/4\). The common endpoint lies over \(i\), since \(p(1/4)=e^{2\pi i/4}=i\), as required.
Worked Example: A Homotopy in the Torus
Use the standard covering \(p:\mathbb{R}^2\to T^2=S^1\times S^1\), where \(\mathbb{R}^2\) has its Euclidean topology and \(T^2\) has the product topology. The map is \(p(u,v)=(e^{2\pi i u},e^{2\pi i v})\). Consider the homotopy $$ H(s,t)=\left(e^{2\pi i(s/3+(t/6)\sin(\pi s))},\ e^{2\pi i(s/4-(t/8)\sin(\pi s))}\right). $$
At \(s=0\), both exponents are \(0\), so \(H(0,t)=(1,1)\). At \(s=1\), \(\sin(\pi)=0\), so \(H(1,t)=(e^{2\pi i/3},e^{2\pi i/4})\) for every \(t\). The paths \(\alpha(s)=H(s,0)\) and \(\beta(s)=H(s,1)\) are therefore homotopic relative to endpoints.
The lifts that start at \((0,0)\) are $$ \widetilde\alpha(s)=(s/3,s/4),\qquad \widetilde\beta(s)=(s/3+(1/6)\sin(\pi s),\ s/4-(1/8)\sin(\pi s)). $$ Applying \(p\) to these coordinate formulas gives the corresponding paths, and substitution of \(s=0\) gives \((0,0)\) for both lifts. At \(s=1\), each sine term is zero, so both endpoints are \((1/3,1/4)\). Checking the projection gives \(p(1/3,1/4)=(e^{2\pi i/3},e^{2\pi i/4})\), exactly the common terminal point of the base paths.
Worked Example: Different Lift Endpoints Obstruct a Homotopy
Consider the upper and lower semicircle paths from \(1\) to \(-1\): $$ \alpha(s)=e^{\pi i s},\qquad \beta(s)=e^{-\pi i s}. $$ Under the same covering \(p(u)=e^{2\pi i u}\), their lifts starting at \(0\) are \(\widetilde\alpha(s)=s/2\) and \(\widetilde\beta(s)=-s/2\). Indeed, \(p(s/2)=e^{\pi i s}=\alpha(s)\) and \(p(-s/2)=e^{-\pi i s}=\beta(s)\); both lifts start at \(0\).
The first lift ends at \(1/2\), and \(p(1/2)=e^{\pi i}=-1\). The second ends at \(-1/2\), and \(p(-1/2)=e^{-\pi i}=-1\). Thus the base paths have the same endpoints, but their lifts from \(0\) have different terminal points. By the theorem, the paths cannot be homotopic relative to endpoints. The unequal lifted endpoints are a checkable obstruction, not merely a difference in how the paths are drawn.
A Common Mistake and Its Correction
Common pitfall. One might argue, “Both semicircle paths start at \(1\) and end at \(-1\), so their lifts from \(0\) must end at the same point.” The calculation above disproves that claim: the endpoints are \(1/2\) and \(-1/2\). The covering map identifies these two real numbers because \(e^{2\pi i(1/2)}=e^{\pi i}=-1\) and \(e^{2\pi i(-1/2)}=e^{-\pi i}=-1\). What forces equality is not equality of the base endpoints alone, but a homotopy that keeps both endpoints fixed. In this example the different lift endpoints show that no such homotopy exists.
A related error is to say that the fiber is discrete and therefore a continuous endpoint track must be constant, without checking that the track stays in one fiber. The theorem supplies that check through \(H(1,t)=x_1\) for every \(t\). If the terminal point of the base path were allowed to move during the homotopy, the lifted endpoints could lie over different base points, and discreteness of a single fiber would not apply.
What the Result Provides
Endpoint invariance is useful when a path is replaced by a homotopic path but a specific starting point in the covering space is fixed. It guarantees that the resulting lifted path has the same terminal point, so computations can use a simpler representative without changing the lift endpoint. It also provides a practical obstruction: if two paths have lifts from the same initial point with different terminal points, they cannot be homotopic relative to endpoints.
The theorem is one-way as stated. It does not claim that equal terminal points of lifts always produce a homotopy relative to endpoints; that converse requires additional structure or hypotheses and is not used here. The argument depends precisely on homotopy lifting, uniqueness of path lifts, and the fact that covering fibers are discrete.
Check Your Understanding
For each question, track the hypotheses as well as the lift endpoints.
- In the circle example, what are the starting and terminal points of each lifted path, and which equalities use \(\sin(0)=\sin(\pi)=0\)?
- Explain why the endpoint track \(t\mapsto\widetilde H(1,t)\) lies in one fiber when the terminal endpoint of the base homotopy is fixed.
- For \(p(u)=e^{2\pi i u}\), find the lift from \(0\) of the path \(s\mapsto e^{2\pi i(2s/3)}\), and state its terminal point.
- Why do the upper and lower semicircle paths fail to be homotopic relative to endpoints, even though they have the same endpoints in \(S^1\)?
- Which part of the endpoint invariance proof fails if \(H(1,t)\) is allowed to vary with \(t\)?