A Reliable Order for Lifting Problems
A lifting problem asks whether a map \(f:X\to Y\) can be written as \(f=p\circ\widetilde f\), where \(p:E\to Y\) is a covering map and \(\widetilde f:X\to E\) is the proposed lift. The most efficient approach is not to begin with formulas. First check that the specified points are compatible with \(p\); then decide whether the problem is local, as for a path, or global, as for a map from a space with nontrivial loops. Finally, verify the lift by composing with \(p\), and check endpoints when the domain is an interval.
For a based problem, the data include \(x_0\in X\), \(y_0\in Y\), and \(e_0\in E\), with \(f(x_0)=y_0\). A lift based at \(e_0\) must also satisfy \(\widetilde f(x_0)=e_0\). Necessarily \(p(e_0)=y_0\), since \(p(\widetilde f(x_0))=f(x_0)\). This elementary check can rule out a lift before any further calculation.
Proof. Let \(D=\{x\in X:\widetilde f_1(x)=\widetilde f_2(x)\}\). We show that \(D\) and \(X\setminus D\) are open. Fix \(x\in X\), and choose an evenly covered neighborhood \(U\) of \(f(x)\). By continuity, there is an open neighborhood \(V\) of \(x\) with \(f(V)\subseteq U\). Each lift on \(V\) lies in one sheet over \(U\) after restricting to a sufficiently small neighborhood of \(x\): the sheet containing its value is open, and its inverse under the lift is an open neighborhood of \(x\). On that neighborhood, the lift is the inverse of \(p\) on that sheet composed with \(f\). If the two lift values at \(x\) agree, the sheets are the same, so these inverse formulas agree nearby. If the values differ, they lie in different sheets, so the formulas differ nearby. Thus \(D\) and its complement are open. Since \(X\) is connected, \(D\) is either empty or all of \(X\). It is nonempty because the lifts agree at one point, so \(D=X\). \(\square\)
Path Problems: Follow the Sheets
For a path \(\alpha:I\to Y\), where \(I=[0,1]\) has its usual topology, the path-lifting theorem says that for each \(e_0\in p^{-1}(\alpha(0))\), there is a unique lift \(\widetilde\alpha:I\to E\) with \(\widetilde\alpha(0)=e_0\). The theorem applies because \(p\) is a covering map. Evenly covered neighborhoods explain the local calculation: while \(\alpha\) stays in one such neighborhood \(U\), the lift is obtained by applying the inverse of \(p\) on the sheet selected by the current lift value. Subdivision of the interval lets us carry out this procedure along the whole path.
For computations below, let \(C_m=\mathbb{R}/m\mathbb{Z}\) have the quotient topology, and write \([t]_m\) for the class of \(t\). Consider \(p:C_4\to C_2\), \(p([t]_4)=[t]_2\). This map is well defined because changing \(t\) by a multiple of \(4\) does not change its class modulo \(2\). It is a covering map: an arc in \(C_2\) of length less than \(1\) has preimage consisting of two disjoint arcs in \(C_4\), and \(p\) maps each of those arcs homeomorphically onto the original arc.
Worked Example: A Path Lift Computed on a Sheet
Take \(\alpha:I\to C_2\) defined by \(\alpha(t)=[\tfrac32+t]_2\), and prescribe the starting point \(\widetilde\alpha(0)=[\tfrac32]_4\). The formula \(\widetilde\alpha(t)=[\tfrac32+t]_4\) gives a lift, since for every \(t\in I\),
Its endpoint is \(\widetilde\alpha(1)=[\tfrac52]_4\), and projecting gives \(p([\tfrac52]_4)=[\tfrac52]_2=[\tfrac12]_2=\alpha(1)\). The path is not a loop: \(\alpha(0)=[\tfrac32]_2\), whereas \(\alpha(1)=[\tfrac52]_2=[\tfrac12]_2\), and these classes differ because their representatives differ by \(1\), not a multiple of \(2\). Thus there is no reason to require the lifted endpoint to equal the lifted starting point.
Map Problems: Apply the Subgroup Test
For a map from a connected, locally path-connected space, checking a few local sheets is not enough: loops in the domain can impose a global obstruction. The fundamental-group criterion handles that obstruction. For a based space \((X,x_0)\), \(\pi_1(X,x_0)\) denotes the group of based loops up to based homotopy, with multiplication by concatenation. A based map \(f:(X,x_0)\to(Y,y_0)\) induces a homomorphism \(f_*:\pi_1(X,x_0)\to\pi_1(Y,y_0)\).
This criterion is a standard result established in the earlier material on the fundamental-group criterion for lifting maps. Its hypotheses matter: the base points must match, and the stated connectedness and local path-connectedness assumptions on \(X\) are part of the theorem. The image \(p_*(\pi_1(E,e_0))\) consists of the based loop classes in \(Y\) that arise by projecting based loops in \(E\).
Worked Example: Rejecting a Map by Subgroup Inclusion
Suppose a based lifting problem has \(\pi_1(X,x_0)\cong\mathbb{Z}\), \(\pi_1(Y,y_0)\cong\mathbb{Z}\), and, in these identifications, \(f_*\) is multiplication by \(6\), while \(p_*(\pi_1(E,e_0))=4\mathbb{Z}\). These are stipulated group data for the problem. The image of \(f_*\) is \(6\mathbb{Z}\). The subgroup test asks whether \(6\mathbb{Z}\subseteq4\mathbb{Z}\). But \(6\in6\mathbb{Z}\), and \(6\notin4\mathbb{Z}\), since there is no integer \(k\) with \(6=4k\). Therefore the inclusion fails and the based lift does not exist. This conclusion is valid only after the base-point condition \(f(x_0)=p(e_0)\) has also been checked; assume it holds here.
Check: the failure is not based on comparing the sizes of the subgroups. It is witnessed by the specific element \(6\), which belongs to the proposed image but not to the covering subgroup.
Loop Endpoints as a Direct Test
For a based loop \(\gamma:I\to Y\) with \(\gamma(0)=\gamma(1)=y_0\), lift it from \(e_0\in p^{-1}(y_0)\). The lift need not be a loop. The following criterion relates its endpoint to the subgroup test.
Proof. If the lift \(\widetilde\gamma\) ends at \(e_0\), it is a based loop in \(E\), and \(p\circ\widetilde\gamma=\gamma\). Therefore \([\gamma]=p_*([\widetilde\gamma])\), so \([\gamma]\) belongs to the indicated image. Conversely, suppose \([\gamma]=p_*([\lambda])\) for a based loop \(\lambda\) in \(E\) at \(e_0\). Then \(\gamma\) is based-homotopic to \(p\circ\lambda\). The lift of \(p\circ\lambda\) starting at \(e_0\) is \(\lambda\), by uniqueness of path lifts, and ends at \(e_0\). Endpoint invariance under based path homotopy, established in the earlier material on endpoint invariance, implies that the lift of \(\gamma\) from \(e_0\) has the same endpoint. Hence it ends at \(e_0\). \(\square\)
Worked Example: Check the Loop Before Its Lift
Define \(f:I\to C_2\) by \(f(t)=[3t]_2\). Its endpoints are \(f(0)=[0]_2\) and \(f(1)=[3]_2=[1]_2\), which are different. Thus \(f\) is a path, not a based loop at \([0]_2\), and the closed lift criterion for based loops cannot be applied to it.
For a genuine loop, use \(\gamma(t)=[2t]_2\). Now \(\gamma(0)=[0]_2\) and \(\gamma(1)=[2]_2=[0]_2\). Lift it through \(p:C_4\to C_2\) from \([0]_4\). The formula \(\widetilde\gamma(t)=[2t]_4\) satisfies \(p(\widetilde\gamma(t))=[2t]_2=\gamma(t)\), and its endpoint is \([2]_4\), not \([0]_4\). The lift is not closed. The endpoint projects correctly: \(p([2]_4)=[2]_2=[0]_2=\gamma(1)\). This is consistent with the closed lift criterion: \([\gamma]\) is not in the image subgroup for this chosen covering and starting point. The endpoint computation itself is enough to decide this particular path-lifting question; no subgroup calculation is needed.
Common Pitfall: A Formula Does Not Make a Loop
A common error is to write down a path formula and call it a loop without checking its endpoint. For \(f(t)=[3t]_2\) on \(I\), the incorrect calculation is \([3]_2=[0]_2\), which would require \(3\) to be a multiple of \(2\). In fact \(3=2+1\), so \([3]_2=[1]_2\ne[0]_2\). The correction is to compare the classes modulo \(2\) explicitly. If a loop is required, choose a formula whose endpoint difference is a multiple of \(2\), such as \(\gamma(t)=[2t]_2\), and then recompute the lift endpoint from its own formula.
A second useful discipline is to separate the methods. For a path, use path lifting and follow the sheets, then check both projected endpoints. For a map from a connected, locally path-connected domain, first check the base point and then use the subgroup criterion. A continuous bijection or a plausible formula is not, by itself, a lifting argument: the defining identity \(p\circ\widetilde f=f\) and all specified base-point conditions must be verified.
Check Your Understanding
For each problem, identify the applicable test before carrying out any calculation.
- For \(p:C_4\to C_2\), can a lift based at \([0]_4\) project to a map whose base-point value is \([1]_2\)? Explain.
- For \(\alpha(t)=[t]_2\) on \(I\), determine whether it is a loop based at \([0]_2\). If lifted from \([0]_4\) by the formula \([t]_4\), compute its endpoint and project that endpoint.
- In a based map problem, suppose \(f_*(\pi_1(X,x_0))=5\mathbb{Z}\) and \(p_*(\pi_1(E,e_0))=10\mathbb{Z}\), with compatible base points and the hypotheses of the lifting criterion satisfied. Does a lift exist? Give a specific element witnessing your conclusion.
- State the endpoint condition equivalent to \([\gamma]\in p_*(\pi_1(E,e_0))\) for a based loop \(\gamma\).
- Why is the subgroup criterion not applicable to \(f(t)=[3t]_2\) on \(I\) as a based loop at \([0]_2\)?