From Lifted Endpoints to a Fiber Action
Let \(p:E\to X\) be a covering map and choose a base point \(x_0\in X\). A loop based at \(x_0\) lifts from any chosen point of the fiber \(p^{-1}(\{x_0\})\), and its endpoint lies in that same fiber. The key issue is whether this endpoint depends only on the loop’s homotopy class. Once that is settled, concatenation of loops should correspond to applying the endpoint assignment twice, in the same order as the paths are traversed.
A map \(p:E\to X\) is a covering map if it is surjective and every \(x\in X\) has an open neighborhood \(U\) such that \(p^{-1}(U)\) is a disjoint union of open sets, each mapped homeomorphically onto \(U\) by \(p\). Fix \(x_0\in X\), and write \(F=p^{-1}(\{x_0\})\) for the fiber over \(x_0\). A based loop is a continuous map \(\alpha:I\to X\) with \(\alpha(0)=\alpha(1)=x_0\). A homotopy relative to endpoints between based loops \(\alpha\) and \(\beta\) is a continuous map \(H:I\times I\to X\) with \(H(0,t)=\alpha(t)\), \(H(1,t)=\beta(t)\), and \(H(s,0)=H(s,1)=x_0\) for all \(s,t\in I\).
We use the path-lifting theorem and uniqueness of path lifts: a path in \(X\) has a unique lift once its initial point in the appropriate fiber is specified. We also use the homotopy lifting property for coverings: a lift of one edge of a homotopy \(I\times I\to X\) extends to a lift of the full homotopy. These are the earlier results on path lifting and homotopy lifting.
Why the Endpoint Depends Only on the Loop Class
The fiber \(F\) is discrete in its subspace topology. Indeed, for \(e\in F\), choose an evenly covered open neighborhood \(U\) of \(x_0\), and let \(V\) be the sheet over \(U\) containing \(e\). Since \(p|_V\) is one-to-one and \(p(e)=x_0\), the set \(V\cap F\) is exactly \(\{e\}\). Also, any continuous map from the connected interval \(I\) into a discrete space is constant. These facts make the two boundary edges of a lifted homotopy decisive.
Proof. Let \(H:I\times I\to X\) be a homotopy relative to endpoints from \(\alpha\) to \(\beta\), and let \(\widetilde{\alpha}_e\) be the lift of \(\alpha\) starting at \(e\). By the homotopy lifting property, there is a continuous lift \(\widetilde H:I\times I\to E\) with \(\widetilde H(0,t)=\widetilde{\alpha}_e(t)\). Since \(H(s,0)=x_0\), the map \(s\mapsto\widetilde H(s,0)\) takes values in \(F\). It is continuous, so it is constant; at \(s=0\) its value is \(e\), and therefore \(\widetilde H(s,0)=e\) for every \(s\). Thus the path \(t\mapsto\widetilde H(1,t)\) is the lift of \(\beta\) starting at \(e\), by uniqueness of path lifts. Similarly, \(H(s,1)=x_0\) implies that \(s\mapsto\widetilde H(s,1)\) is a continuous map into \(F\), hence constant. Its values at \(s=0\) and \(s=1\) are the endpoints of the lifts of \(\alpha\) and \(\beta\) starting at \(e\). Those endpoints are equal. \(\square\)
If both loops in a concatenation are replaced by homotopic representatives relative to endpoints, their concatenations are also homotopic relative to endpoints: concatenate the two homotopies, traversing the first on the first half of the parameter interval and the second on the second half. The shared value at the joining point is \(x_0\), so the resulting map is continuous. Thus the group product used below is compatible with the representative independence just proved.
Worked Example: Two Representatives with the Same Lifted Endpoint
Consider \(p:S^1\to S^1\), \(p(z)=z^2\), where \(S^1=\{z\in\mathbb C:|z|=1\}\) has the subspace topology from \(\mathbb C\). This is a covering: around any point of the target, choose an open arc short enough that its preimage consists of two disjoint open arcs; on each arc, squaring is a homeomorphism onto the chosen arc. Take \(x_0=1\), \(e=1\), and define \(\alpha(t)=e^{2\pi i t}\), \(\beta(t)=e^{2\pi i t^2}\).
The map \(H(s,t)=e^{2\pi i((1-s)t+st^2)}\) is a homotopy relative to endpoints: at \(s=0\) it is \(\alpha\), at \(s=1\) it is \(\beta\), and its values at \(t=0\) and \(t=1\) are both \(1\). For each fixed \(s\), a lift starting at \(1\) is \(\widetilde H(s,t)=e^{\pi i((1-s)t+st^2)}\), since squaring this expression gives \(H(s,t)\), and its value at \(t=0\) is \(1\). At \(t=1\), its value is \(e^{\pi i}=-1\), for every \(s\). In particular, the lift of \(\alpha\) and the lift of \(\beta\) both end at \(-1\).
Check: \(p(-1)=(-1)^2=1=x_0\), so the computed endpoint lies in the required fiber. The homotopy-lifting theorem guarantees endpoint invariance in general; the displayed lift verifies it directly in this example.
Concatenation Gives the Action Law
Proof. Let \(\widetilde{\alpha}_e\) be the lift of \(\alpha\) starting at \(e\), and put \(e_1=\widetilde{\alpha}_e(1)\). Let \(\widetilde{\beta}_{e_1}\) be the lift of \(\beta\) starting at \(e_1\). The concatenated loop is defined by \((\alpha*\beta)(t)=\alpha(2t)\) for \(0\le t\le 1/2\), and \((\alpha*\beta)(t)=\beta(2t-1)\) for \(1/2\le t\le1\). Define a path in \(E\) by \(\widetilde{\gamma}(t)=\widetilde{\alpha}_e(2t)\) on the first half and \(\widetilde{\gamma}(t)=\widetilde{\beta}_{e_1}(2t-1)\) on the second half. At \(t=1/2\), the first formula gives \(e_1\), and the second formula also gives \(e_1\), so \(\widetilde{\gamma}\) is continuous. Its projection under \(p\) is \(\alpha*\beta\), and it starts at \(e\). By uniqueness of path lifts, it is the lift of \(\alpha*\beta\) starting at \(e\). Its endpoint is \(\widetilde{\beta}_{e_1}(1)\), which proves the action law. The lift of the constant loop starting at \(e\) is the constant path at \(e\), so \(e\cdot[\,c\,]=e\). Together with representative independence, these equations give a right group action with the stated multiplication convention. \(\square\)
Worked Example: Lifting a Concatenation
Use \(p(z)=z^2\), \(x_0=e=1\), and the loops \(\alpha(t)=e^{2\pi it}\) and \(\beta(t)=e^{-2\pi it}\). The lift of \(\alpha\) starting at \(1\) is \(\widetilde{\alpha}_1(t)=e^{\pi it}\), because its square is \(e^{2\pi it}\); its endpoint is \(e^{\pi i}=-1\). The lift of \(\beta\) starting at \(-1\) is \(\widetilde{\beta}_{-1}(t)=-e^{-\pi it}\): its square is \(e^{-2\pi it}\), its initial value is \(-1\), and its endpoint is \(-e^{-\pi i}=1\).
Therefore \((1\cdot[\alpha])\cdot[\beta]=(-1)\cdot[\beta]=1\). The lift of \(\alpha*\beta\) starting at \(1\) is, on the two halves of the interval, \(e^{2\pi it}\) and \(-e^{-\pi i(2t-1)}\), respectively. At \(t=1/2\), the first expression is \(e^{\pi i}=-1\), and the second is \(-e^0=-1\); at \(t=1\), the second is \(-e^{-\pi i}=1\). Hence the concatenated lift also ends at \(1\), as the action law requires.
Check: Squaring the first-half lift gives \(e^{4\pi it}=\alpha(2t)\). Squaring the second-half lift gives \(e^{-2\pi i(2t-1)}=e^{-4\pi it+2\pi i}=e^{-4\pi it}=\beta(2t-1)\). Thus both pieces project to the specified concatenated loop.
Identity, Inverses, and a Common Pitfall
The action law also handles inverses. The reverse loop \(\alpha^{-1}(t)=\alpha(1-t)\) represents the inverse of \([\alpha]\) in the fundamental group. Since \([\alpha][\alpha^{-1}]\) is the identity class, the action law gives \((e\cdot[\alpha])\cdot[\alpha^{-1}]=e\). Indeed, lifting \(\alpha\) and then its reverse retraces the lifted path back to its starting point. No assumption that the fiber has finitely many points is needed.
Worked Example: A Trivial Covering
Let \(X=S^1\) with its usual topology, let \(E=S^1\times\{a,b\}\) with the product topology where \(\{a,b\}\) is discrete, and let \(p(z,j)=z\). This is a covering: for each open \(U\subseteq S^1\), its preimage is the disjoint union \(U\times\{a\}\) and \(U\times\{b\}\), and each projects homeomorphically onto \(U\). Choose \(x_0=1\), \(e=(1,b)\), and \(\gamma(t)=e^{2\pi it}\).
The lift of \(\gamma\) starting at \(e\) is \(\widetilde{\gamma}(t)=(e^{2\pi it},b)\). It projects to \(\gamma\), starts at \((1,b)\), and ends at \((1,b)\), so \(e\cdot[\gamma]=(1,b)\). More generally, for any loop \(\eta\) at \(1\), the path \((\eta(t),b)\) is its lift starting at \((1,b)\), and its endpoint is again \((1,b)\). The same calculation for a starting point \((1,a)\) keeps the second coordinate equal to \(a\). Thus every loop acts as the identity permutation on this fiber.
Check: For the displayed loop, \(p(\widetilde{\gamma}(t))=p(e^{2\pi it},b)=e^{2\pi it}=\gamma(t)\), and \(e^{2\pi i}=1\), so the endpoint is exactly the stated starting point.
Common pitfall: forgetting the fixed-endpoint condition. Homotopy lifting does not say that the lifted endpoints remain equal when the base homotopy is allowed to move its terminal point. For the same square covering, consider \(H(s,t)=e^{\pi ist}\). Starting at \(1\), its lift is \(\widetilde H(s,t)=e^{\pi ist/2}\), since \(\widetilde H(s,t)^2=H(s,t)\). The terminal point is \(\widetilde H(s,1)=e^{\pi is/2}\), which is \(1\) at \(s=0\) and \(i\) at \(s=1\). The endpoints differ because \(H(s,1)=e^{\pi is}\) moves with \(s\); the lifted terminal points do not lie in one fixed fiber. In the theorem, keeping both endpoints fixed is precisely what ensures that the lifted terminal points lie in a single discrete fiber.
The order convention matters as well. Here \([\alpha][\beta]\) means “traverse \(\alpha\), then \(\beta\),” so the action is written on the right and the law reads \((e\cdot[\alpha])\cdot[\beta]=e\cdot([\alpha][\beta])\). Reversing the convention for loop multiplication requires changing the written action law accordingly.
Check Your Understanding
In each question, keep track of the starting point of every lift and whether the homotopy fixes both endpoints.
- Why does the endpoint curve \(s\mapsto\widetilde H(s,1)\) lie in a single fiber when \(H(s,1)=x_0\)?
- For the square covering \(p(z)=z^2\), find the lift of \(t\mapsto e^{-2\pi it}\) starting at \(1\), and state its endpoint.
- Using the convention \([\alpha][\beta]=[\alpha*\beta]\), which loop is traversed first in the action law?
- What condition on a homotopy is missing if its lifted terminal points are not guaranteed to agree?
- For \(p(z,j)=z\), explain why the lift of a path starting at \((z_0,j)\) keeps its second coordinate equal to \(j\).