Let p: X˜ → X be a covering map over a topological space X. Let Z be a topological space, and let f:Z → X be a continuous map from Z to X. A continuous map f˜:Z → X˜ is said to be a lift of the map f:Z → X if and only if p ◦f˜ = f. We shall prove various results concerning the existence and uniqueness of such lifts.

Proposition 1.2 Let p: X˜ → X be a covering map, let Z be a connected topological space, and let g:Z → X˜ and h:Z → X˜ be continuous maps.

Suppose that p ◦ g = p ◦ h and that g(z) = h(z) for some z ∈ Z. Then g = h.

 Proof Let Z₀ = {z ∈ Z : g(z) = h(z)}. Note that Z0 is non-empty, by hypothesis. We show that Z0 is both open and closed in Z.

Let z be a point of Z. There exists an open set U in X containing the point p(g(z)) which is evenly covered by the covering map p. Then p⁻¹ (U)  is a disjoint union of open sets, each of which is mapped homeomorphically onto U by the covering map p. One of these open sets contains g(z); let this set be denoted by U˜. Also one of these open sets contains h(z); let this open set be denoted by V˜ . Let N_z = g⁻¹ (U˜) ∩ h⁻¹ (V˜ ). Then N_z is an open set in Z containing z.

Consider the case when z ∈ Z₀. Then g(z) = h(z), and therefore V˜ = U˜.  It follows from this that both g and h map the open set N_z into U˜. But p ◦ g = p ◦ h, and p|U˜: U˜ → U is a homeomorphism. Therefore g|N_z = h|N_z, and thus N_z ⊂ Z₀. We have thus shown that, for each z ∈ Z₀, there exists an open set N_z such that z ∈ N_z and N_z ⊂ Z₀. We conclude that Z₀ is open.

Next consider the case when z ∈ Z Z₀. In this case U˜ ∩ V˜ = ∅, since g(z) ≠h(z). But g(Nz) ⊂ U˜ and h(Nz) ⊂ V˜ . Therefore g(z₀) ≠h(z₀) for all z₀ ∈ N_z, and thus N_z ⊂ Z₀. We have thus shown that, for each z ∈ Z₀,  there exists an open set N_z such that z ∈ N_z and N_z ⊂ Z Z₀. We conclude that Z Z₀ is open.

The subset Z₀ of Z is therefore both open and closed. Also Z₀ is nonempty by hypothesis. We deduce that Z0 = Z, since Z is connected. Thus g = h, as required.

Lemma 1.3 Let p: X˜ → X be a covering map, let Z be a topological space,  let A be a connected subset of Z, and let f:Z → X and g: A → X˜ be continuous maps with the property that p ◦ g = f|A. Suppose that f(Z) ⊂ U,  where U is an open subset of X that is evenly covered by the covering map p.

Then there exists a continuous map ˜f:Z → X˜ such that ˜f|A = g and p ◦f˜=f.

Proof The open set U is evenly covered by the covering map p, and therefore p^-1 (U) is a disjoint union of open sets, each of which is mapped homeomorphically onto U by the covering map p. One of these open sets contains g(a) for some a ∈ A; let this set be denoted by U˜. Let σ: U → U˜ be the inverse of the homeomorphism p|U˜: U˜ → U, and let f˜= σ ◦ f. Then p ◦f˜ = f.

Also p ◦f˜|A = p ◦ g and f˜(a) = g(a). It follows from Proposition 1.2 that f˜|A = g, since A is connected. Thus f˜:Z → X˜ is the required map.

Theorem 1.4 (Path Lifting Theorem) Let p: X˜ → X be a covering map, let γ: [0, 1] → X be a continuous path in X, and let w be a point of X˜ satisfying p(w) = γ(0). Then there exists a unique continuous path γ˜: [0, 1] → X˜ such that γ˜(0) = w and p ◦ γ˜ = γ.

Proof The map p: X˜ → X is a covering map; therefore there exists an open cover U of X such that each open set U belonging to X is evenly covered by the map p. Now the collection consisting of the preimages γ⁻¹ (U) of the open sets U belonging to U is an open cover of the interval [0, 1]. But [0, 1] is compact, by the Heine-Borel Theorem. It follows from the Lebesgue Lemma that there exists some δ > 0 such that every subinterval of length less than δ is mapped by γ into one of the open sets belonging to U. Partition the interval  [0, 1] into subintervals [t_i−1, t_i], where 0 = t₀ < t₁ < · · · < t_n−1 < t_n = 1, and where the length of each subinterval is less than δ. Then each subinterval  [t_i−1, t_i] is mapped by γ into some open set in X that is evenly covered by the map p. It follows from Lemma 1.3 that once γ˜ (ti−1) has been determined,  we can extend γ˜ continuously over the i^th subinterval [t_i−1, t_i]. Thus by extending γ˜ successively over [t₀, t₁], [t₁, t₂],. . ., [t_n−1, t_n], we can lift the path γ: [0, 1] → X to a path γ˜: [0, 1] → X˜ starting at w. The uniqueness of γ˜ follows from Proposition 1.2.

Theorem 1.5 (The Monodromy Theorem) Let p: X˜ → X be a covering map, let H: [0, 1] × [0, 1] → X be a continuous map, and let w be a point of X˜ satisfying p(w) = H(0, 0). Then there exists a unique continuous map H˜ : [0, 1] × [0, 1] → X˜ such that H˜ (0, 0) = w and p ◦ H˜ = H.

Proof The unit square [0, 1] × [0, 1] is compact. By applying the Lebesgue Lemma to an open cover of the square by preimages of evenly covered open sets in X (as in the proof of Theorem 1.4), we see that there exists some δ > 0 with the property that any square contained in [0, 1] × [0, 1] whose sides have length less than δ is mapped by H into some open set in X which is evenly covered by the covering map p. It follows from Lemma 1.3 that if the lift H˜ of H has already been determined over a corner, or along one side,  or along two adjacent sides of a square whose sides have length less than δ,  then H˜ can be extended over the whole of that square. Thus if we subdivide  [0, 1] × [0, 1] into squares S_j,k, where

and 1/n < δ, then we can construct a lift H˜ of H by defining H˜ (0, 0) = w,  and then successively extending H˜ in turn over each of these smaller squares.

(Indeed the map H˜ can be extended successively over the squares

S_1,1, S_1,2, . . . , S_1,n, S_2,1, S_2,2, . . . , S_2,n, S_3,1, . . . , S_n−1,n, . . . , S_n,1, S_n,2, . . . , S_n,n.)