From a Subgroup Condition to a Candidate Lift
Suppose a map \(f:X\to Y\) is to be lifted through a covering map \(p:E\to Y\). A lift is a map \(\widetilde f:X\to E\) satisfying \(p\circ\widetilde f=f\). If we choose base points \(x_0\in X\) and \(e_0\in E\) with \(p(e_0)=f(x_0)\), paths offer a natural way to try to define \(\widetilde f(x)\): travel from \(x_0\) to \(x\), apply \(f\), lift the resulting path starting at \(e_0\), and take its endpoint.
The difficulty is that there may be many paths from \(x_0\) to \(x\). The key hypothesis is an inclusion between the subgroups induced on fundamental groups. This tutorial proves that the inclusion makes the endpoint independent of the chosen path. It constructs a function, but does not establish its continuity; that requires a separate argument.
We use two standard results proved in the earlier material on covering spaces. First, a path in \(Y\) has a unique lift to \(E\) once its starting point over the initial endpoint is specified. Second, if two paths in \(Y\) with the same endpoints are homotopic while their endpoints remain fixed, their lifts starting at the same point have the same endpoint. The second statement is endpoint invariance under path homotopy.
For paths \(\alpha,\beta\) with \(\alpha(1)=\beta(0)\), write \(\alpha*\beta\) for their concatenation, traversing \(\alpha\) on the first half of \(I\) and \(\beta\) on the second half. Write \(\overline{\alpha}(t)=\alpha(1-t)\) for the reversed path. These conventions let us compare two choices of path to the same point.
Why the Endpoint Does Not Depend on the Path
Proof. The subgroup condition for this particular loop means there is a loop \(\gamma\) at \(e_0\) such that \([f\circ\omega]=[p\circ\gamma]\) in \(\pi_1(Y,f(x_0))\). Thus \(f\circ\omega\) and \(p\circ\gamma\) are homotopic with endpoints fixed. The path \(\gamma\) is a lift of \(p\circ\gamma\) starting at \(e_0\), and ends at \(e_0\). By endpoint invariance under path homotopy, the lift of \(f\circ\omega\) starting at \(e_0\) has the same endpoint. \(\square\)
Proof. Since \(X\) is path-connected, there is at least one path from \(x_0\) to each \(x\). Let \(\alpha\) and \(\beta\) be any two such paths. Their concatenation \(\omega=\alpha*\overline{\beta}\) is a loop at \(x_0\). By the subgroup hypothesis, \([f\circ\omega]\in p_*(\pi_1(E,e_0))\), so the lemma says that the lift of \(f\circ\omega\) starting at \(e_0\) ends at \(e_0\).
Let \(A\) be the lift of \(f\circ\alpha\) starting at \(e_0\), and let \(B\) be the lift of \(f\circ\beta\) starting at \(e_0\). Write \(a=A(1)\) and \(b=B(1)\). The lift of \(f\circ\omega=(f\circ\alpha)*(f\circ\overline{\beta})\) first follows \(A\), ending at \(a\), and then follows the lift of \(f\circ\overline{\beta}\) starting at \(a\). Its final endpoint is \(e_0\). Reverse that second lifted segment. It is a lift of \(f\circ\beta\) starting at \(e_0\) and ending at \(a\). By uniqueness of path lifts, it must be \(B\), whose endpoint is \(b\). Therefore \(a=b\). The two paths chosen to define \(\widetilde f(x)\) give the same endpoint. Since \(x\), \(\alpha\), and \(\beta\) were arbitrary, the rule defines a function on all of \(X\). \(\square\)
Proof. For \(x\in X\), choose a path \(\alpha\) from \(x_0\) to \(x\), and let \(A\) be the lift of \(f\circ\alpha\) starting at \(e_0\). By the definition of a lift, \(p(A(1))=(f\circ\alpha)(1)=f(x)\). Since \(\widetilde f(x)=A(1)\), this proves \(p(\widetilde f(x))=f(x)\). For \(x=x_0\), choose the constant path at \(x_0\). Its image has constant lift at \(e_0\), so \(\widetilde f(x_0)=e_0\). \(\square\)
Worked Constructions
Worked Example: A Degree-Six Map Through a Three-Fold Cover
Use the standard topology on \(S^1=\{z\in\mathbb C:|z|=1\}\). Let \(p_3:S^1\to S^1\) be \(p_3(w)=w^3\), and let \(f_6:S^1\to S^1\) be \(f_6(z)=z^6\). Take \(x_0=e_0=1\). The standard calculation \(\pi_1(S^1,1)\cong\mathbb Z\) identifies \((p_3)_*\) with multiplication by \(3\) and \((f_6)_*\) with multiplication by \(6\). Thus \(6\mathbb Z\subseteq3\mathbb Z\), and the subgroup condition holds.
For \(z=e^{2\pi i t}\), choose the path \(\alpha_t(s)=e^{2\pi i st}\), where \(0\leq s\leq1\). Its image under \(f_6\) is \(e^{2\pi i6st}\). A lift through \(p_3\) starting at \(1\) is \(A(s)=e^{2\pi i2st}\), since $$ p_3(A(s))=\bigl(e^{2\pi i2st}\bigr)^3=e^{2\pi i6st}. $$ The endpoint rule therefore gives \(\widetilde f(e^{2\pi i t})=e^{2\pi i2t}=z^2\).
Check: if \(e^{2\pi i t}=e^{2\pi i(t+k)}\) for \(k\in\mathbb Z\), the proposed value changes by the factor \(e^{2\pi i2k}=1\), so the formula does not depend on the angle representative. Also \(p_3(\widetilde f(z))=(z^2)^3=z^6=f_6(z)\). The theorem guarantees that using any other path from \(1\) to \(z\) gives the same endpoint.
Worked Example: Lifting a Map from an Interval
Give \(I=[0,1]\) and \(S^1\) their standard topologies. Let \(p:\mathbb R\to S^1\) be \(p(u)=e^{2\pi i u}\), and let \(f:I\to S^1\) be \(f(t)=e^{2\pi i3t}\). Use \(x_0=0\) and \(e_0=0\). The interval is path-connected. Its fundamental group is trivial: any loop \(\omega\) at \(0\) contracts through the based homotopy \(H(s,r)=(1-r)\omega(s)\), which stays in \(I\) and keeps both endpoints at \(0\). Hence the subgroup condition holds.
For a point \(x\in I\), choose \(\alpha_x(s)=sx\). Then \(f(\alpha_x(s))=e^{2\pi i3sx}\), whose lift starting at \(0\) is \(A_x(s)=3sx\), because \(p(A_x(s))=e^{2\pi i3sx}\). The endpoint rule gives \(\widetilde f(x)=3x\). For example, at \(x=2/3\), the lifted path ends at \(3(2/3)=2\), and \(p(2)=e^{2\pi i2}=1=f(2/3)\).
Check: for every \(x\in I\), \(p(\widetilde f(x))=e^{2\pi i(3x)}=f(x)\), and \(\widetilde f(0)=0\). The theorem ensures that even if a different path in \(I\) from \(0\) to \(x\) is chosen, the resulting lifted endpoint is still \(3x\).
What Fails Without the Inclusion
Worked Example: Two Paths Give Different Endpoints
Use \(S^1\) with its standard topology, let \(p_2:S^1\to S^1\) be \(p_2(w)=w^2\), and let \(f:S^1\to S^1\) be the identity map. Set \(x_0=e_0=1\). On fundamental groups, \(f_*\) is multiplication by \(1\), while \((p_2)_*\) is multiplication by \(2\). Thus \(f_*(\mathbb Z)=\mathbb Z\) is not contained in \((p_2)_*(\mathbb Z)=2\mathbb Z\).
To reach \(x=1\), one possible path is the constant path \(\alpha(s)=1\). Its image lifts from \(1\) to the constant path at \(1\), so the endpoint is \(1\). Another path to the same point is \(\beta(s)=e^{2\pi i s}\). Its image under \(f\) is itself, and a lift through \(p_2\) starting at \(1\) is \(B(s)=e^{\pi i s}\), because $$ p_2(B(s))=\bigl(e^{\pi i s}\bigr)^2=e^{2\pi i s}. $$ This lift ends at \(B(1)=e^{\pi i}=-1\). The two paths have the same initial and final points, but their lifted image paths end at \(1\) and \(-1\), respectively.
Check: both proposed endpoints lie over \(f(1)=1\), since \(p_2(1)=1\) and \(p_2(-1)=(-1)^2=1\). They are nevertheless different points of the fiber. The path rule therefore cannot define a single value at \(x=1\).
Common pitfall. A common error is to assume that any loop in the source, after applying \(f\), must lift to a loop at the chosen point \(e_0\). The calculation above disproves that: the source loop \(\beta\) returns to \(1\), but its image lift from \(1\) ends at \(-1\). The subgroup condition is precisely what rules out this obstruction. A related error is to conclude that the path-based function is continuous just because it is well defined. The argument here proves independence of choices and the equation \(p\circ\widetilde f=f\); continuity needs its own proof.
The construction separates the lifting problem into two tasks. First, the subgroup inclusion controls loops and makes the endpoint assignment unambiguous. Then one must examine the local behavior of the resulting function to establish continuity. The first task is global: it compares all based loops. It is what allows a local path-lifting operation to define a function on the whole path-connected space.
Check Your Understanding
For each question, identify the role of the base point, the chosen path, and the subgroup hypothesis.
- In the path construction, where is path-connectedness used, and which part of the theorem would fail to address points in another path component?
- Suppose two paths \(\alpha,\beta\) run from \(x_0\) to \(x\). Write a based loop using \(\alpha\), \(\beta\), and path reversal that compares them.
- For the maps \(p_4(w)=w^4\) and \(f_8(z)=z^8\) on \(S^1\), check the subgroup inclusion and find the formula for the candidate lift.
- In the degree-one map through \(p_2\), which loop witnesses failure of the subgroup condition, and what is the endpoint of its lift from \(1\)?
- Which property of the constructed function is proved by the endpoint argument, and which important property is not established here?