Why Continuity Needs a Local Argument
The path construction of a lift assigns a point of the covering space to each point of the domain by lifting a path from a chosen basepoint. That construction gives the desired pointwise equation, but a pointwise definition does not by itself show that the resulting map is continuous. The key is to examine nearby points using paths that remain in one suitable neighborhood. On that neighborhood, the lift is given by a single inverse branch of the covering map.
Let \(p:E\to Y\) be a covering map: every point \(y\in Y\) has an open neighborhood \(V\) for which \(p^{-1}(V)\) is a disjoint union of open sets \(W_\lambda\), and each restriction \(p|_{W_\lambda}:W_\lambda\to V\) is a homeomorphism. Such a \(V\) is evenly covered, and the sets \(W_\lambda\) are its sheets. Write \(s_\lambda:V\to W_\lambda\) for the inverse of \(p|_{W_\lambda}\).
We use the path-lifting and homotopy-lifting results established earlier. In particular, a path in \(Y\) has a unique lift once its starting point in the appropriate fiber is specified, and homotopic paths with fixed endpoints have lifts with the same endpoint when their lifts start at the same point.
The Path-Defined Map and Its Continuity
Assume \(X\) is path-connected and locally path-connected, \(f:X\to Y\) is continuous, \(x_0\in X\), and \(e_0\in E\) satisfies \(p(e_0)=f(x_0)\). The subgroup condition in the lifting criterion is \(f_*(\pi_1(X,x_0))\subseteq p_*(\pi_1(E,e_0))\). Here \(\pi_1(X,x_0)\) denotes the group of based loops up to homotopy with endpoints fixed, and the induced maps send a loop class to the class of its composition with the given map.
Under this condition, the path construction defines \(\widetilde f(x)\) as follows: choose a path \(\alpha\) from \(x_0\) to \(x\), lift \(f\circ\alpha\) starting at \(e_0\), and take the endpoint of that lift. The subgroup condition, together with path and homotopy lifting, ensures that the endpoint is independent of the choice of \(\alpha\). Thus this construction gives a well-defined map with \(p\circ\widetilde f=f\). We now prove the additional fact needed here: this map is continuous.
Proof. Fix \(x\in X\). Choose an evenly covered open neighborhood \(V\) of \(f(x)\). Since \(f\) is continuous, \(f^{-1}(V)\) is an open neighborhood of \(x\). Local path-connectedness gives a path-connected open set \(U\) such that \(x\in U\subseteq f^{-1}(V)\). Let \(W\) be the sheet over \(V\) containing \(\widetilde f(x)\), and let \(s:V\to W\) be the inverse of \(p|_W\).
For any \(z\in U\), choose a path \(\beta\) in \(U\) from \(x\) to \(z\), which exists because \(U\) is path-connected. Choose a path \(\alpha\) from \(x_0\) to \(x\). The concatenated path \(\alpha*\beta\) runs from \(x_0\) to \(z\). By the definition of the path-defined lift, lifting \(f\circ(\alpha*\beta)\) from \(e_0\) ends at \(\widetilde f(z)\). After the segment \(f\circ\alpha\), the lift is at \(\widetilde f(x)\). The remaining segment \(f\circ\beta\) stays in \(V\), because \(\beta(I)\subseteq U\subseteq f^{-1}(V)\).
The path \(s\circ f\circ\beta\) lies in \(W\), starts at \(\widetilde f(x)\), and projects to \(f\circ\beta\), since \(p\circ s\) is the identity on \(V\). By uniqueness of path lifts, it is the lift of \(f\circ\beta\) starting at \(\widetilde f(x)\). Therefore its endpoint is \(\widetilde f(z)\), and \(\widetilde f(z)=s(f(z))\) for every \(z\in U\). Thus \(\widetilde f|_U=s\circ f|_U\), a composition of continuous maps. Each \(x\in X\) has such an open neighborhood \(U\). For any open set \(G\subseteq E\), the set \(\widetilde f^{-1}(G)\cap U\) is open in \(U\), hence in \(X\), for each such \(U\); these neighborhoods cover \(X\), so \(\widetilde f^{-1}(G)\) is open. Hence \(\widetilde f\) is continuous. \(\square\)
The proof uses path-connectedness of \(U\) to compare the values at \(x\) and \(z\) by a path within \(U\). It uses openness of \(U\) to obtain a neighborhood on which continuity can be checked. Local path-connectedness supplies both features at once. The sheet \(W\) matters because the inverse \(s\) is defined on that one sheet; the argument shows that the lift cannot switch sheets while the base path remains in \(V\).
Worked Applications of the Local Formula
Worked Example: A Lift Along an Interval
Let \(p:\mathbb R\to S^1\) be \(p(u)=e^{2\pi i u}\), where \(S^1=\{z\in\mathbb C:|z|=1\}\), and let \(I=[0,1]\) have its usual subspace topology from \(\mathbb R\). Define \(f(t)=e^{2\pi i(3t+1/6)}\), and choose the starting lift \(u_0=1/6\) over \(f(0)\).
The interval \(I\) is path-connected and locally path-connected, and its fundamental group is trivial, so the subgroup condition holds. The path-defined lift is \(F(t)=3t+1/6\): it starts at \(F(0)=1/6\), and \(p(F(t))=e^{2\pi i(3t+1/6)}=f(t)\).
To see the local continuity mechanism explicitly, fix \(t_0\in I\). Take the arc \(V\) on \(S^1\) consisting of points whose angular distance from \(f(t_0)\) is less than \(\pi/4\). Set \(U=(t_0-1/48,t_0+1/48)\cap I\), an open neighborhood of \(t_0\) in \(I\). For \(t\in U\), \(|t-t_0|<1/48\), so the change in the angle \(2\pi(3t+1/6)\) is less than \(2\pi(3/48)=\pi/8\). Hence \(f(U)\subseteq V\). The inverse branch over \(V\) whose sheet contains \(F(t_0)\) has value \(3t+1/6\) on \(f(t)\) for \(t\in U\). Thus \(F\) agrees on \(U\) with that inverse branch composed with \(f\).
Check: \(p(F(t))=e^{2\pi i(3t+1/6)}=f(t)\) for every \(t\in I\), and \(F(0)=1/6\), the chosen starting lift.
Worked Example: A Lift Between Circle Coverings
Give both circles their usual topology. Let \(p:S^1\to S^1\) be \(p(z)=z^3\), and let \(f:S^1\to S^1\) be \(f(z)=z^6\), with basepoint \(1\) and chosen lift \(1\). The map \(p\) is a covering: sufficiently short open arcs in the target have three disjoint open arcs over them, and \(p\) maps each such arc homeomorphically onto the target arc.
Using the standard identification \(\pi_1(S^1,1)=\mathbb Z\), the map \(f\) sends an integer \(n\) to \(6n\), while \(p\) sends \(n\) to \(3n\). Thus \(f_*(\mathbb Z)=6\mathbb Z\subseteq3\mathbb Z=p_*(\mathbb Z)\), as required. For a path from \(1\) to \(z\), its image under \(f\) lifts under \(p\) starting at \(1\); the resulting endpoint is \(z^2\). Indeed, if \(z=e^{2\pi i t}\) along a path parameterized by \(t\), then \(f(z)=e^{2\pi i6t}\), and \(e^{2\pi i2t}\) projects to it because \((e^{2\pi i2t})^3=e^{2\pi i6t}\). The constructed lift is therefore \(F(z)=z^2\).
Near each \(z_0\in S^1\), choose an open arc \(U\) around \(z_0\) small enough that \(f(U)\) lies in an evenly covered arc \(V\). The local formula proved above says \(F|_U=s\circ f|_U\), where \(s\) is the inverse branch whose sheet contains \(F(z_0)\). This proves continuity locally, while the formula for the lift also gives the direct check \(p(F(z))=(z^2)^3=z^6=f(z)\).
Check: At the basepoint, \(F(1)=1^2=1\). For every \(z\in S^1\), the displayed power calculation verifies both the lift equation and the chosen basepoint condition.
Worked Example: A Product Covering with Two Sheets
Let \(E=\mathbb R\times\{0,1\}\), with the product topology where \(\{0,1\}\) is discrete, and let \(p:E\to\mathbb R\) be \(p(y,j)=y\). This is a covering: for any open interval \(V\subseteq\mathbb R\), its preimage is the disjoint union \(V\times\{0\}\) and \(V\times\{1\}\), and each projects homeomorphically onto \(V\).
Take \(X=\mathbb R\), \(f(x)=x^2\), basepoint \(x_0=0\), and starting lift \((0,1)\). Since \(\mathbb R\) is simply connected, the subgroup condition holds. Lifting any path \(f\circ\alpha\) while starting in the sheet with second coordinate \(1\) keeps that coordinate equal to \(1\). Consequently the path-defined lift is \(F(x)=(x^2,1)\).
For an explicit neighborhood check at any \(x_0\in\mathbb R\), let \(\delta=1/(2|x_0|+2)\), \(U=(x_0-\delta,x_0+\delta)\), and \(V=(x_0^2-1,x_0^2+1)\). If \(y\in U\), then \(|y-x_0|<\delta\leq1\), so \(|y+x_0|\leq2|x_0|+1\). Therefore \(|y^2-x_0^2|=|y-x_0||y+x_0|<\delta(2|x_0|+1)=(2|x_0|+1)/(2|x_0|+2)<1\). Thus \(f(U)\subseteq V\), and on \(U\) the lift is the inverse branch \(s(y)=(y,1)\) composed with \(f\).
Check: \(p(F(x))=p(x^2,1)=x^2=f(x)\), and \(F(0)=(0,1)\), as prescribed.
A Common Pitfall
The equation \(p\circ g=f\) says that \(g\) is a lift as a set map; it does not imply that \(g\) is continuous. For a concrete example, take \(p:\mathbb R\to S^1\), \(p(u)=e^{2\pi i u}\), and the constant map \(f:I\to S^1\) with \(f(t)=1\). Define \(g(t)=0\) for \(t\leq1/2\), and \(g(t)=1\) for \(t>1/2\). Then \(p(g(t))=1\) in both cases, so \(p\circ g=f\). But for every integer \(n>2\), \(g(1/2+1/n)=1\), whereas \(g(1/2)=0\). Since \(1/2+1/n\to1/2\), \(g\) is not continuous.
The error would be to conclude “\(p\circ g=f\), therefore \(g\) is continuous.” The path-defined lift starting at \(0\), by contrast, is the constant lift \(F(t)=0\): the lift of the constant path at \(1\) starting at \(0\) stays at \(0\). The local inverse-branch argument proves continuity of this constructed lift; the projection equation alone does not.
The practical point is that the subgroup condition and path construction first produce a well-defined candidate lift. Local path-connectedness then provides open sets where paths can be compared, and the covering property turns that comparison into one continuous inverse branch. This is the step that upgrades the pointwise construction to a continuous map.
Check Your Understanding
For each question, identify the neighborhood or sheet used in the argument.
- In the continuity proof, where is continuity of \(f\) used, and what open set does it produce?
- Why must the path from \(x\) to \(z\) stay inside \(U\subseteq f^{-1}(V)\)?
- Which sheet contains the lifted path starting at \(\widetilde f(x)\), and why does uniqueness of path lifts apply?
- For the covering \(p(z)=z^4\) and map \(f(z)=z^8\) on \(S^1\), check the subgroup inclusion and give the lift based at \(1\).
- Explain why the discontinuous map \(g\) in the pitfall still satisfies the lift equation.