Constructing Path Lifts by Subdivision

From Local Sheets to a Lift on the Whole Interval

A covering map provides local inverse branches, but a path may pass through many evenly covered neighborhoods. To construct a lift on the entire interval, we need to divide the path into pieces, each contained in one such neighborhood. On each piece we use the inverse branch determined by the lift already chosen at its starting point. This tutorial gives the subdivision argument and verifies why the pieces fit continuously.

Let \(p:E\to B\) be a covering map. This means that \(p\) is a continuous surjection and every \(b\in B\) has an open neighborhood \(U\) such that \(p^{-1}(U)\) is a disjoint union of open sets, called sheets, each mapped homeomorphically onto \(U\) by \(p\). Such a \(U\) is an evenly covered neighborhood. A path in \(B\) is a continuous map \(\gamma:I\to B\), where \(I=[0,1]\) has its usual topology. A lift of \(\gamma\) is a continuous map \(\widetilde{\gamma}:I\to E\) satisfying \(p\circ\widetilde{\gamma}=\gamma\).

Finding a Subdivision Suitable for the Path

Choose evenly covered neighborhoods of points in the image of \(\gamma\). Their inverse images under \(\gamma\) form an open cover of \(I\). We need a finite subdivision for which each entire closed subinterval lies in one member of this cover. The following interval argument supplies it.

Proof. For each \(t\in I\), choose \(V_t\in\mathcal{V}\) containing \(t\). Since \(V_t\) is open in the subspace \(I\), there is \(r_t>0\) such that \((t-2r_t,t+2r_t)\cap I\subseteq V_t\). The sets \((t-r_t,t+r_t)\cap I\), as \(t\) varies, cover \(I\). The interval \(I\) is compact by the Heine–Borel theorem, so finitely many of these smaller neighborhoods cover \(I\); write their centers and radii as \(c_1,\ldots,c_m\) and \(r_1,\ldots,r_m\). Set \(\delta=\min\{r_1,\ldots,r_m\}>0\). Divide \(I\) into finitely many equal closed intervals, each of length strictly less than \(\delta\). For any one such interval \(J\), choose \(s\in J\). Some smaller neighborhood \((c_k-r_k,c_k+r_k)\cap I\) contains \(s\). For every \(u\in J\), \(|u-c_k|\leq |u-s|+|s-c_k|<\delta+r_k\leq2r_k\). Thus \(J\subseteq(c_k-2r_k,c_k+2r_k)\cap I\subseteq V_{c_k}\). This works for every interval in the division, proving the lemma. \(\square\)

For a path \(\gamma\), take an evenly covered neighborhood \(U\) around each point of \(\gamma(I)\). The sets \(\gamma^{-1}(U)\) are open in \(I\), because \(\gamma\) is continuous, and they cover \(I\). Apply the subdivision lemma to this cover. Each resulting subinterval is then mapped by \(\gamma\) into one evenly covered neighborhood. The subdivision is chosen for the whole path before the sheet choices begin.

Constructing the Lift Successively

Proof. Use the subdivision just obtained, so for each \(i=1,\ldots,n\) there is an evenly covered neighborhood \(U_i\) with \(\gamma([t_{i-1},t_i])\subseteq U_i\). We define the lift in order, starting with \(\widetilde{\gamma}(0)=e_0\). Suppose it has been defined through \(t_{i-1}\), with \(p(\widetilde{\gamma}(t_{i-1}))=\gamma(t_{i-1})\). Since \(\gamma(t_{i-1})\in U_i\), the point \(\widetilde{\gamma}(t_{i-1})\) lies in \(p^{-1}(U_i)\), hence in exactly one sheet \(S_i\) over \(U_i\). The restriction \(p|_{S_i}:S_i\to U_i\) is a homeomorphism, so its inverse is defined on \(U_i\). On the new subinterval set \(\widetilde{\gamma}(t)=(p|_{S_i})^{-1}(\gamma(t))\) for \(t\in[t_{i-1},t_i]\). This is continuous there, and at the left endpoint it equals \(\widetilde{\gamma}(t_{i-1})\), because that point belongs to \(S_i\) and maps to \(\gamma(t_{i-1})\). Also \(p(\widetilde{\gamma}(t))=\gamma(t)\) throughout this interval. Repeating the construction defines the lift on all of \(I\), with the required initial value and projection identity. Finally, the definitions agree at each shared endpoint. Each subinterval is closed in \(I\), and there are finitely many of them; the finite pasting lemma therefore gives continuity on \(I\). In this application, the pasting lemma follows by taking any closed \(F\subseteq E\): the preimage of \(F\) under each continuous piece is closed in its closed subinterval, hence closed in \(I\), and the preimage under the pasted map is the finite union of those closed sets. Thus the constructed lift is continuous. \(\square\)

The sheet choice is not arbitrary once the lift has reached a subdivision point. It must be the sheet containing the point already constructed. This condition makes the next inverse branch agree with the existing lift at the join.

Why the Sheet Choice Matters

Common pitfall: choosing each inverse branch independently. In the one-turn example, the first segment ends at lift value \(1/4\), which projects to \(i\). On the next arc, the sheet \((1/8,5/8)\) contains \(1/4\), so its inverse branch continues the lift. The other sheet over that same arc, \((-7/8,-3/8)\), gives the same projected point \(i\) the lift value \(-3/4\). Choosing this other branch for the second segment would make the value jump from \(1/4\) to \(-3/4\) at \(t=1/4\). Both values project to \(i\), but the piecewise map would not be continuous. The condition “choose the sheet containing the current lift point” prevents this error.

The method works because it separates two tasks. Compactness of the parameter interval supplies finitely many pieces on which local inverse branches are available. The covering property supplies those branches, and the prescribed initial point determines the first sheet; matching the current point determines each later sheet. Continuity on the whole interval then follows from the agreement at the finitely many subdivision points.