3 – Coverings of a Topological Space
295
is open in X. Since, on the other hand, p maps every “ disc ” D(ζ) onto some
open set D, p maps every open subset of X, and in particular of D(ζ) onto
some open subset of B. As p : D(ζ) −→ D is bijective, it is a homeomorphism.
The fact that (X, B, p) is a covering of B is now clear.
(f) X is connected. Indeed, denoting by α the class of the constant path
t → a, if γ connects a to z ∈ B, the classes of the paths
γ s : t −→ γ(st) ,
s,t∈ I ,
define a path in X connecting α to the class ζ of γ.
To show that X is simply connected, observe that if there are two paths
μ and μ
in X with initial point α and the same terminal point ζ, then these
are the liftings of two paths γ and γ
with initial point a and terminal point
p(ζ) ; however, the endpoints of μ and μ
inX are, by definition, the homotopy
classes of γ and γ
; γ are γ
are, therefore, homotopic, hence so are μ and μ
as well (theorem 2). Taking into account the connectedness of X, we move
on from here to paths having an arbitrary initial point.
(g) Let (Y, B, q) be a connected covering of B and let α ∈ X, β ∈ Y be
such that q(β) = p(α). For all ζ ∈ X, there exists a path μ connecting α
to ζ. It is unique up to homotopy since X is simply connected. If γ is the
projection from μ onto B, then there is a unique lifting ν of γ to Y having
β as initial point. As γ is unique up to homotopy, the terminal point f (ζ) of
ν only depends on ζ. This gives the map f sought. Its uniqueness is due to
the fact that this is obviously the only way it could possibly be defined. That
(X, Y, f ) is a covering is more or less obvious.
If Y is simply connected, oppositely, there is a map g : Y −→ X such
that p ◦ g = q, g(β) = α. Clearly, f and g are mutually inverse. This gives
the isomorphism of the simply connected coverings considered, qed.
If B is a connected and locally arc-connected space, the connected and
simply connected covering (X, B, p) of theorem 4 is the universal covering
of B ; by part (g) of the proof, it “ dominates ” all the others. Example 1
above shows that due to the exponential map, C is a universal covering space
of C
∗ , while, due to the map t → e(t), R the universal covering space of T.
Exercise 3. Let G ⊂ C be a domain and ( ˜
G, G, p) its universal covering. Show that there is a complex analytic structure on ˜
G such that p is
a holomorphic submersion. Let f be a holomorphic function on G and ω f
the inverse image under p of the differential form f (z)dz. Show that there
is holomorphic function F on ˜
G such that dF = ω f . What about the case
G = C
∗ ?
(v) Coverings of a pointed disc. Let D
∗ be a pointed disc, i.e. either an
open disc in C with its centre removed, or else the exterior of a closed disc
in C (pointed disc “ centered at ∞ ”) . Using a conformal representation, we
need only consider the disc 0 < |z| < 1. Like all coverings of a Riemann
surface, a covering (Y, D
∗ , q) of D
∗ has a natural Riemann surface structure:
the analytic structure of the discs’ projections onto D
∗ is transferred to the
295
is open in X. Since, on the other hand, p maps every “ disc ” D(ζ) onto some
open set D, p maps every open subset of X, and in particular of D(ζ) onto
some open subset of B. As p : D(ζ) −→ D is bijective, it is a homeomorphism.
The fact that (X, B, p) is a covering of B is now clear.
(f) X is connected. Indeed, denoting by α the class of the constant path
t → a, if γ connects a to z ∈ B, the classes of the paths
γ s : t −→ γ(st) ,
s,t∈ I ,
define a path in X connecting α to the class ζ of γ.
To show that X is simply connected, observe that if there are two paths
μ and μ
in X with initial point α and the same terminal point ζ, then these
are the liftings of two paths γ and γ
with initial point a and terminal point
p(ζ) ; however, the endpoints of μ and μ
inX are, by definition, the homotopy
classes of γ and γ
; γ are γ
are, therefore, homotopic, hence so are μ and μ
as well (theorem 2). Taking into account the connectedness of X, we move
on from here to paths having an arbitrary initial point.
(g) Let (Y, B, q) be a connected covering of B and let α ∈ X, β ∈ Y be
such that q(β) = p(α). For all ζ ∈ X, there exists a path μ connecting α
to ζ. It is unique up to homotopy since X is simply connected. If γ is the
projection from μ onto B, then there is a unique lifting ν of γ to Y having
β as initial point. As γ is unique up to homotopy, the terminal point f (ζ) of
ν only depends on ζ. This gives the map f sought. Its uniqueness is due to
the fact that this is obviously the only way it could possibly be defined. That
(X, Y, f ) is a covering is more or less obvious.
If Y is simply connected, oppositely, there is a map g : Y −→ X such
that p ◦ g = q, g(β) = α. Clearly, f and g are mutually inverse. This gives
the isomorphism of the simply connected coverings considered, qed.
If B is a connected and locally arc-connected space, the connected and
simply connected covering (X, B, p) of theorem 4 is the universal covering
of B ; by part (g) of the proof, it “ dominates ” all the others. Example 1
above shows that due to the exponential map, C is a universal covering space
of C
∗ , while, due to the map t → e(t), R the universal covering space of T.
Exercise 3. Let G ⊂ C be a domain and ( ˜
G, G, p) its universal covering. Show that there is a complex analytic structure on ˜
G such that p is
a holomorphic submersion. Let f be a holomorphic function on G and ω f
the inverse image under p of the differential form f (z)dz. Show that there
is holomorphic function F on ˜
G such that dF = ω f . What about the case
G = C
∗ ?
(v) Coverings of a pointed disc. Let D
∗ be a pointed disc, i.e. either an
open disc in C with its centre removed, or else the exterior of a closed disc
in C (pointed disc “ centered at ∞ ”) . Using a conformal representation, we
need only consider the disc 0 < |z| < 1. Like all coverings of a Riemann
surface, a covering (Y, D
∗ , q) of D
∗ has a natural Riemann surface structure:
the analytic structure of the discs’ projections onto D
∗ is transferred to the