Definition : Let X and X˜ be topological spaces and let p: X˜ → X be a continuous map. An open subset U of X is said to be evenly covered by the map p if and only if p⁻¹ (U) is a disjoint union of open sets of X˜ each of which is mapped homeomorphically onto U by p. The map p: X˜ → X is said to be a covering map if p: X˜ → X is surjective and in addition every point of X is contained in some open set that is evenly covered by the map p.

If p: X˜ → X is a covering map, then we say that X˜ is a covering space of X.

Example Let S¹ be the unit circle in R² . Then the map p: R → S¹ defined by

                              p(t) = (cos 2πt,sin 2πt)

is a covering map. Indeed let n be a point of S¹. Consider the open set U in S¹ containing n defined by U = S¹ {−n}. Now n = (cos 2πt₀,sin 2πt₀)  for some t₀ ∈ R. Then p⁻¹ (U) is the union of the disjoint open sets J_n for all integers n, where

           Jn = {t ∈ R : t0 + n −1/2 < t < t₀ + n +1/2}.

Each of the open sets Jn is mapped homeomorphically onto U by the map p.

This shows that p: R → S¹ is a covering map.

Example The map p: C → C {0} defined by p(z) = exp(z) is a covering map. Indeed, given any θ ∈ [−π, π] let us define

              Uθ = {z ∈ C {0} : arg(−z) ≠θ}.

Then p⁻¹ (Uθ) is the disjoint union of the open sets

              {z ∈ C : |Im z − θ − 2πn| < π} ,

for all integers n, and p maps each of these open sets homeomorphically onto U_θ. Thus U_θ is evenly covered by the map p.

Example Consider the map α: (−2, 2) → S¹ , where α(t) = (cos 2πt,sin 2πt)  for all t ∈ (−2, 2). It can easily be shown that there is no open set U containing the point (1, 0) that is evenly covered by the map α. Indeed suppose that there were to exist such an open set U. Then there would exist some δ satisfying 0 < δ < ½ such that U_δ ⊂ U, where

                      U_δ = {(cos 2πt,sin 2πt) : −δ < t < δ}.

The open set Uδ would then be evenly covered by the map α. However the connected components of α⁻¹ (Uδ) are (−2, −2 +δ), (−1−δ, −1 +δ), (−δ, δ),  (1 − δ, 1 + δ) and (2 − δ, 2), and neither (−2, −2 + δ) nor (2 − δ, 2) is mapped homeomorphically onto U_δ by α.

Lemma 1.1 Let p: X˜ → X be a covering map. Then p(V ) is open in X for every open set V in X˜. In particular, a covering map p: X˜ → X is a homeomorphism if and only if it is a bijection.

Proof Let V be open in X˜, and let x ∈ p(V ). Then x = p(v) for some v ∈ V . Now there exists an open set U containing the point x 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 v; let U˜ be this open set, and let N_x = p(V ∩U˜). Now N_x is open in X, since V ∩U˜ is open in U˜ and p|U˜ is a homeomorphism from U˜ to U. Also x ∈ N_x and N_x ⊂ p(V ). It follows that p(V ) is the union of the open sets N_x as x ranges over all points of p(V ),  and thus p(V ) is itself an open set, as required. The result that a bijective covering map is a homeomorphism then follows directly from the fact that a continuous bijection is a homeomorphism if and only if it maps open sets to open sets.