The Homotopy Lifting Property

From Path Lifting to Homotopy Lifting

A homotopy is a continuous family of maps, with the parameter \(t\in I=[0,1]\) recording the family member. Path lifting handles one parameterized path at a time. The homotopy lifting property says that, for a covering map, an entire homotopy can be lifted continuously when its initial map has already been lifted. The initial lift matters: it specifies which sheet of the covering the lifted homotopy starts in.

Throughout, \(I=[0,1]\) has the subspace topology inherited from \(\mathbb{R}\). A covering map \(p:E\to B\) is a continuous surjection such that each \(b\in B\) has an open neighborhood \(U\) for which \(p^{-1}(U)\) is a union of pairwise disjoint open sets, each mapped homeomorphically onto \(U\) by \(p\). No connectedness assumption on \(E\), \(B\), or the parameter space is part of this definition.

Let \(X\) be a topological space, let \(H:X\times I\to B\) be a homotopy, and write \(H_t(x)=H(x,t)\) for its map at time \(t\). A lift of \(H_0\) is a continuous map \(\widetilde f_0:X\to E\) satisfying \(p\circ\widetilde f_0=H_0\). A lift of the whole homotopy must be a continuous map on \(X\times I\), not merely a collection of maps defined separately on \(X\) at each time.

For each \(t\in I\), define the parameter slice \(\widetilde H_t:X\to E\) by \(\widetilde H_t(x)=\widetilde H(x,t)\). Since \(p\circ\widetilde H=H\), each slice satisfies \(p\circ\widetilde H_t=H_t\). Thus it is a lift of the map \(H_t\). But the theorem asserts more than the existence of some lift of each \(H_t\): the slices must fit together as one continuous map of both variables, and the slice at time zero must equal the specified \(\widetilde f_0\).

Why the Initial Lift Determines Every Slice

The reason uniqueness holds is that, for each fixed \(x\), the map \(t\mapsto H(x,t)\) is a path in \(B\). Any homotopy lift restricts to a path lift of that path, starting at \(\widetilde f_0(x)\). We first record the path-lifting uniqueness fact used in this argument.

Proof. Let \(S=\{t\in I:\alpha(t)=\beta(t)\}\). It is nonempty because \(0\in S\). We show that both \(S\) and \(I\setminus S\) are open in \(I\). If \(t_0\in S\), choose an evenly covered open neighborhood \(U\) of \(\gamma(t_0)\). By continuity of \(\gamma\), there is a relatively open interval \(J\) around \(t_0\) with \(\gamma(J)\subseteq U\). The common point \(\alpha(t_0)=\beta(t_0)\) lies in one sheet \(V\) over \(U\). By continuity, after shrinking \(J\), both \(\alpha(J)\) and \(\beta(J)\) lie in \(V\). On \(V\), the map \(p\) is one-to-one, and \(p(\alpha(t))=\gamma(t)=p(\beta(t))\); hence \(\alpha(t)=\beta(t)\) throughout \(J\). So \(S\) is open.

If \(t_0\notin S\), then \(\alpha(t_0)\ne\beta(t_0)\), while both points map to \(\gamma(t_0)\). In an evenly covered neighborhood \(U\) of \(\gamma(t_0)\), they therefore lie in distinct sheets. Continuity lets us restrict to a relatively open interval \(J\) around \(t_0\) on which \(\gamma(J)\subseteq U\), \(\alpha(J)\) stays in the first sheet, and \(\beta(J)\) stays in the second. The sheets are disjoint, so \(\alpha(t)\ne\beta(t)\) for every \(t\in J\). Thus \(I\setminus S\) is open as well. The interval \(I\) is connected, so its nonempty subset \(S\), which is both open and closed in \(I\), must be all of \(I\). Therefore \(\alpha=\beta\). \(\square\)

Proof. Suppose \(\widetilde H\) and \(\widetilde G\) are two such lifts. Fix any \(x\in X\). The paths \(\alpha(t)=\widetilde H(x,t)\) and \(\beta(t)=\widetilde G(x,t)\) both lift the path \(\gamma(t)=H(x,t)\), since \(p(\widetilde H(x,t))=H(x,t)=p(\widetilde G(x,t))\). They have the same initial point, because \(\alpha(0)=\widetilde f_0(x)=\beta(0)\). The lemma gives \(\widetilde H(x,t)=\widetilde G(x,t)\) for all \(t\). This holds for every \(x\), so the two maps agree on all of \(X\times I\). If \(X\) is empty, the two maps also agree, since there is only one map from the empty set to \(E\). \(\square\)

In particular, once the initial lift is fixed, the lift at a later time cannot be chosen independently: for each \(x\), it is the value at that time of the unique path lift starting at \(\widetilde f_0(x)\). This pointwise description explains uniqueness. It does not, by itself, prove that the resulting values vary continuously with \(x\); continuity of the full lifted homotopy is part of the existence assertion.

Worked Lifts for the Circle Covering

Use the standard covering \(p:\mathbb{R}\to S^1\), where \(\mathbb{R}\) has its standard topology, \(S^1=\{(u,v)\in\mathbb{R}^2:u^2+v^2=1\}\) has the subspace topology, and $$ p(r)=(\cos(2\pi r),\sin(2\pi r)). $$ The identity \(p(r+n)=p(r)\) for \(n\in\mathbb{Z}\) is why a starting lift is needed: different real numbers can project to the same point of the circle.

A Common Pitfall: Lifting Slices Separately

It is not enough to choose a lift of every \(H_t\) separately. Those choices must combine to give a continuous map on \(X\times I\). For a concrete failure, let \(X=\{*\}\) be discrete, let \(H(*,t)=p(0)=(1,0)\) for every \(t\), and define a proposed choice of slice lifts by $$ \widetilde H(*,t)= \begin{cases} 0,&0\le t\le 1/2,\\ 1,&1/2<t\le1. \end{cases} $$ At every time, this is a lift of \(H_t\), because \(p(0)=(1,0)=p(1)\). It has the initial value \(0\) as well. But it is not continuous: for \(t_n=1/2+1/n\), \(n\ge3\), we have \(t_n\to1/2\) while \(\widetilde H(*,t_n)=1\) and \(\widetilde H(*,1/2)=0\). Thus these individually valid slice lifts do not form a homotopy lift. The homotopy lifting property concerns a continuous lift of the whole map \(H:X\times I\to B\).

For a covering map, the initial lift resolves the sheet ambiguity point by point, and path-lifting uniqueness then determines every parameter track. Existence supplies the continuous family of these tracks; uniqueness says there is no second continuous family with the same starting map. These are distinct parts of the property, and the examples show why both the initial condition and continuity in the homotopy parameter must be kept in view.