From Path Lifts to Homotopy Lifts
A homotopy is a continuously varying family of maps, so lifting each of its paths separately does not by itself prove that the lifts vary continuously. The key is to control a whole small rectangle of parameters at once. For a covering map, evenly covered neighborhoods provide local inverse branches; compactness of the interval allows us to divide each path into finitely many pieces where those branches apply.
Let \(p:E\to Y\) be a covering map. This means that for every \(y\in Y\), there is an open neighborhood \(V\) of \(y\) such that \(p^{-1}(V)\) is a union of pairwise disjoint open sets, called sheets, each mapped homeomorphically onto \(V\) by \(p\). For a sheet \(S\) over \(V\), write \(\phi_S:V\to S\) for the inverse of \(p|_S\). Let \(X\) be any topological space, let \(I=[0,1]\), and let \(H:X\times I\to Y\) be continuous.
The theorem places no compactness condition on \(X\). The compact space used in the proof is the parameter interval \(I\). The initial lift is essential: the theorem extends a specified lift at time \(0\); it does not assert that an arbitrary map \(H\) has a lift without that starting data.
Two Compactness Steps
We use a simple form of the tube lemma. If \(K\) is compact, \(G\) is open in \(X\times K\), and \(\{x_0\}\times K\subseteq G\), then some open neighborhood \(N\) of \(x_0\) satisfies \(N\times K\subseteq G\). Indeed, for each \(k\in K\), the product topology gives open neighborhoods \(N_k\) of \(x_0\) and \(W_k\) of \(k\) with \(N_k\times W_k\subseteq G\). Finitely many \(W_k\) cover \(K\); intersect the corresponding finitely many \(N_k\) to obtain \(N\).
We also need to subdivide an interval relative to an open cover. For completeness, if \(\mathcal U\) is an open cover of \(I\), there is a finite partition \(0=t_0<t_1<\cdots<t_m=1\) such that each \([t_{i-1},t_i]\) lies in some member of \(\mathcal U\). Here is a compactness argument. If no positive mesh size worked, for every \(n\in\mathbb N\) there would be a closed subinterval \(J_n\) of length at most \(1/n\) not contained in any member of \(\mathcal U\). Choose \(s_n\in J_n\). Compactness gives a subsequence \(s_{n_j}\) converging to some \(s\in I\). Choose \(U\in\mathcal U\) containing \(s\); since \(U\) is relatively open in \(I\), it contains all points of \(I\) sufficiently close to \(s\). For large \(j\), both \(s_{n_j}\) and every point of \(J_{n_j}\) are sufficiently close to \(s\), because \(s_{n_j}\to s\) and the length of \(J_{n_j}\) tends to zero. Then \(J_{n_j}\subseteq U\), a contradiction. A partition with mesh smaller than the resulting positive size has the stated property.
Constructing and Matching the Local Lifts
Fix \(x_0\in X\). We will construct a lift on a neighborhood of \(x_0\), for all times \(t\in I\). The sets \(H(\{x_0\}\times I)\) are covered by evenly covered neighborhoods in \(Y\). Their inverse images under the path \(t\mapsto H(x_0,t)\) form an open cover of \(I\). By the subdivision fact, choose a partition \(0=t_0<\cdots<t_m=1\) and evenly covered open sets \(V_i\subseteq Y\) such that $$ H(\{x_0\}\times[t_{i-1},t_i])\subseteq V_i \qquad (1\le i\le m). $$
For each \(i\), the set \(H^{-1}(V_i)\) is open in \(X\times I\) and contains \(\{x_0\}\times[t_{i-1},t_i]\). The tube lemma gives an open neighborhood \(N_i\) of \(x_0\) such that \(H(N_i\times[t_{i-1},t_i])\subseteq V_i\). Thus, after restricting \(x\) to \(N=\bigcap_{i=1}^m N_i\), every strip stays inside its chosen evenly covered neighborhood.
We now select the sheets in order. At \(t_0=0\), the point \(h_0(x_0)\) belongs to one sheet \(S_1\) over \(V_1\), since \(p(h_0(x_0))=H(x_0,0)\in V_1\). The sheet \(S_1\) is open, so continuity of \(h_0\) allows us to shrink \(N\) until \(h_0(x)\in S_1\) for every \(x\in N\). On the first strip define $$ \widetilde H(x,t)=\phi_{S_1}(H(x,t)), \qquad x\in N,\quad t\in[t_0,t_1]. $$ This is continuous, projects to \(H\), and agrees with \(h_0\) at \(t=0\).
Suppose the lift has been defined continuously through time \(t_{i-1}\) on a neighborhood of \(x_0\). Its endpoint \(x\mapsto\widetilde H(x,t_{i-1})\) is continuous. At \(x_0\), this endpoint lies in a sheet \(S_i\) over \(V_i\), because its image under \(p\) is \(H(x_0,t_{i-1})\in V_i\). Shrink the neighborhood so that every endpoint \(\widetilde H(x,t_{i-1})\) lies in \(S_i\), and so that the \(i\)-th strip still maps into \(V_i\). Define the lift on that strip by \(\phi_{S_i}\circ H\). At the shared edge \(t=t_{i-1}\), this formula agrees with the lift already constructed: both points lie in \(S_i\) and map under \(p\) to \(H(x,t_{i-1})\), and \(p|_{S_i}\) is one-to-one. Since there are finitely many strips, this process constructs a continuous lift on \(N\times I\).
To finish, we check that these local constructions give one well-defined lift on all of \(X\times I\). For each fixed \(x\), the construction is a lift of the path \(t\mapsto H(x,t)\) starting at \(h_0(x)\). Such a path lift is unique. To see uniqueness, suppose \(\alpha,\beta:I\to E\) are two lifts of the same path with \(\alpha(0)=\beta(0)\). The set of times where they agree is open: near an agreement time, choose an evenly covered neighborhood of the common projected point; by continuity both lifts lie in the same sheet for nearby times, where the inverse branch is unique. The set of times where they differ is also open: at a time of disagreement, the two points lie in different sheets over a common evenly covered neighborhood, and continuity keeps them in those disjoint sheets for nearby times. Thus the agreement set and its complement are open in the connected interval \(I\). Since the agreement set contains \(0\), it is all of \(I\). The local lifts therefore agree wherever their parameter neighborhoods overlap. They define a global lift, and around every \((x_0,t)\) it has the continuous local formula just constructed. Hence it is continuous on \(X\times I\). \(\square\)
Worked Circle-Covering Examples
Worked Example: A Rotating Homotopy
Take \(p:\mathbb R\to S^1\), \(p(u)=e^{2\pi i u}\), the standard covering, and \(X=I\). Set \(H(s,t)=e^{2\pi i(s+t/2)}\) and \(h_0(s)=s\). The initial condition holds because \(p(h_0(s))=e^{2\pi i s}=H(s,0)\). A lift is $$ \widetilde H(s,t)=s+\frac{t}{2}. $$ Indeed, \(p(\widetilde H(s,t))=e^{2\pi i(s+t/2)}=H(s,t)\), and \(\widetilde H(s,0)=s=h_0(s)\). At \(t=1\), the lifted value is \(s+1/2\); substituting gives \(p(s+1/2)=e^{2\pi i s}e^{\pi i}=-e^{2\pi i s}=H(s,1)\). The lift is continuous because it is a sum of continuous real-valued functions.
Worked Example: A Homotopy to a Constant Map
Again use \(p(u)=e^{2\pi i u}\). Define \(H(s,t)=e^{2\pi i((1-t)s+t/3)}\) and \(h_0(s)=s\). The lift is \(\widetilde H(s,t)=(1-t)s+t/3\). At \(t=0\) this equals \(s\), and applying \(p\) gives \(H(s,0)=e^{2\pi i s}\). At \(t=1\), it equals \(1/3\), independent of \(s\), and \(p(1/3)=e^{2\pi i/3}\). Thus the final map is constant at \(e^{2\pi i/3}\), since \(\cos(2\pi/3)=-1/2\) and \(\sin(2\pi/3)=\sqrt{3}/2\). For every \(s,t\), direct substitution gives \(p(\widetilde H(s,t))=H(s,t)\), so the formula verifies the lift on the whole rectangle.
Worked Example: Keeping Track of a Sheet
Let \(E=I\times\{0,1\}\) have the subspace topology from \(\mathbb R^2\), let \(Y=I\) have its standard topology, and define \(p(y,j)=y\). This is a covering: \(Y\) itself is evenly covered by the two open sheets \(I\times\{0\}\) and \(I\times\{1\}\), each mapped homeomorphically onto \(I\). Take \(X=I\), \(H(x,t)=x(1-t)\), and \(h_0(x)=(x,1)\). Then $$ \widetilde H(x,t)=(x(1-t),1). $$ Applying \(p\) gives \(x(1-t)=H(x,t)\), while \(\widetilde H(x,0)=(x,1)=h_0(x)\). The second coordinate remains \(1\), so this lift stays in the sheet chosen by the initial lift. The formula is continuous as a map into \(I\times\{0,1\}\), and substituting \(t=1\) gives \(\widetilde H(x,1)=(0,1)\) for every \(x\).
The Matching Condition and a Common Pitfall
The local inverse branch on a strip is not chosen independently of the preceding strip. Its sheet must contain the already constructed endpoint along their shared edge. This matching is what makes the piecewise lift continuous. The proof uses only finitely many strips for each fixed \(x_0\), but the partition and the sheets are allowed to depend on \(x_0\); the local lifts are then shown to agree by uniqueness of path lifts.
Common pitfall. For \(p:\mathbb R\to S^1\), \(p(u)=e^{2\pi i u}\), consider the constant path \(H(t)=1\) with starting lift \(0\). On the first half of the interval, choose the lift constantly equal to \(0\); on the second half, incorrectly choose it constantly equal to \(1\). Both choices project to \(1\), since \(p(0)=p(1)=1\). But at \(t=1/2\) the two formulas disagree, so the resulting function is discontinuous. The correction is to choose the sheet on the second strip containing the preceding endpoint \(0\); its inverse branch sends \(1\) to \(0\), and the lift is constantly \(0\) throughout. Merely checking \(p\circ\widetilde H=H\) on each strip does not check continuity at the seams.
Check Your Understanding
For each question, identify the initial lift and verify the projection condition as well as the endpoint condition.
- For \(p(u)=e^{2\pi i u}\), lift \(H(s,t)=e^{2\pi i(s-t/4)}\) from \(h_0(s)=s\). What is the lifted value at \(t=1\)?
- In the two-sheet covering \(p(y,j)=y\) of \(I\), what lift results from the initial map \(h_0(x)=(x,0)\) and homotopy \(H(x,t)=x(1-t)\)?
- Why does the tube lemma apply to each closed strip after the subdivision has been chosen?
- In the proof, where is uniqueness of path lifts used to establish continuity of the global lift?
- For the constant path \(H(t)=1\) under \(p(u)=e^{2\pi i u}\), which condition must the inverse branch on a new strip satisfy at the shared endpoint?