288
X – The Riemann Surface of an Algebraic Function
taking a given value at a given point of D . If two sections ϕ and ψ defined
on the same set E ⊂ B are equal at a ∈ E, they are equal to the same ϕ i at
a : since D i is open in X, the values of ϕ and ψ in the neighbourhood of a
are in D i , and so ϕ(z) = ψ(z) = ϕ i (z) for all z ∈ E sufficiently near a.
The set of points of E where ϕ and ψ are equal is, therefore, open in E;
it is also closed since ϕ and ψ are continuous. As a result, two sections of X
over a connected subset of X are identical if they are equal at some point. For
example, if F is an algebraic function and if there are two uniform branches
f and g of F on a connected open subset U ⊂ C − S (using the notation of
the previous n
◦ ), then the existence of some a ∈ U where f (a) = g(a) implies
that f = g is all of U . So there are at most n uniform branches on U if
n = d
◦ (F) and, in fact, exactly n if U is simply connected (n
◦ 4, theorem 6).
Let us now show that if ϕ is a section of X over a connected subset E
of B, then ϕ(E) is a connected component of p
−1 (E). As (p
−1 (E), E, p) is
obviously a covering of E, it suffices to do so for E = B. Since z → (z, ϕ(z))
and (z, ζ) → z are continuous and mutually inverse, ϕ is a homeomorphism
from B onto Y = ϕ(B), which is, therefore, connected. By (R), Y is clearly
open in X. On the other hand, if a sequence ϕ(z n ) ∈ Y converges to a limit,
the z n = p[ϕ(z n )] converge to some a ∈ B, and so lim ϕ(z n ) = ϕ(a). As a
result, Y is open and closed in X, and connected as the image of B, qed.
On the other hand, if Y is a connected component of X, then p(Y ) is a
connected component of B, and so p(Y ) = B if B is connected. Firstly, Y
being open in X and p being a local homeomorphism, p(Y ) is open in B. Let
a ∈ B be a closure point of p(Y ) and D a connected open neighbourhood of
a in B satisfying (R). As D meets p(Y ), p
−1 (D) =
D i meets Y . If some
connected open D i , meets Y , then Y ∪ D i is a connected open set containing
Y . Hence D i ⊂ Y and a ∈ p(D i ) ⊂ p(Y ), so that p(Y ) is closed, qed.
Besides, Y is clearly a covering space of B: apply (R) by replacing X by
Y .
Finally suppose that, for any connected component Y of X, the map
p : Y −→ B is injective. It is then bijective if B is connected, and as p is a
local homeomorphism, it is a global homeomorphism from Y onto B. As a
result, Y is the image of B under a global section ϕ of X and the covering
(X, B, p) is trivial.
(iii) Path-lifting. Let γ : I −→ B be a continuous path in B, where
I = [0, 1]. A lifting of γ is a path μ : I −→ X such that p ◦ μ = γ (§ 4
of Chap. IV for the case of Log z). Such a lifting always exists, but there is
better still :
Theorem 2. Let (X, B, p) be a covering and γ : I −→ B a path in B. There
is a unique lifting μ from γ to X with a given initial point. If two paths γ 0
and γ 1 in B are fixed-endpoint homotopic and if μ 0 and μ 1 are liftings of γ 0
and γ 1 having the same initial point, then μ 0 and μ 1 have the same terminal
point and are homotopic.
X – The Riemann Surface of an Algebraic Function
taking a given value at a given point of D . If two sections ϕ and ψ defined
on the same set E ⊂ B are equal at a ∈ E, they are equal to the same ϕ i at
a : since D i is open in X, the values of ϕ and ψ in the neighbourhood of a
are in D i , and so ϕ(z) = ψ(z) = ϕ i (z) for all z ∈ E sufficiently near a.
The set of points of E where ϕ and ψ are equal is, therefore, open in E;
it is also closed since ϕ and ψ are continuous. As a result, two sections of X
over a connected subset of X are identical if they are equal at some point. For
example, if F is an algebraic function and if there are two uniform branches
f and g of F on a connected open subset U ⊂ C − S (using the notation of
the previous n
◦ ), then the existence of some a ∈ U where f (a) = g(a) implies
that f = g is all of U . So there are at most n uniform branches on U if
n = d
◦ (F) and, in fact, exactly n if U is simply connected (n
◦ 4, theorem 6).
Let us now show that if ϕ is a section of X over a connected subset E
of B, then ϕ(E) is a connected component of p
−1 (E). As (p
−1 (E), E, p) is
obviously a covering of E, it suffices to do so for E = B. Since z → (z, ϕ(z))
and (z, ζ) → z are continuous and mutually inverse, ϕ is a homeomorphism
from B onto Y = ϕ(B), which is, therefore, connected. By (R), Y is clearly
open in X. On the other hand, if a sequence ϕ(z n ) ∈ Y converges to a limit,
the z n = p[ϕ(z n )] converge to some a ∈ B, and so lim ϕ(z n ) = ϕ(a). As a
result, Y is open and closed in X, and connected as the image of B, qed.
On the other hand, if Y is a connected component of X, then p(Y ) is a
connected component of B, and so p(Y ) = B if B is connected. Firstly, Y
being open in X and p being a local homeomorphism, p(Y ) is open in B. Let
a ∈ B be a closure point of p(Y ) and D a connected open neighbourhood of
a in B satisfying (R). As D meets p(Y ), p
−1 (D) =
D i meets Y . If some
connected open D i , meets Y , then Y ∪ D i is a connected open set containing
Y . Hence D i ⊂ Y and a ∈ p(D i ) ⊂ p(Y ), so that p(Y ) is closed, qed.
Besides, Y is clearly a covering space of B: apply (R) by replacing X by
Y .
Finally suppose that, for any connected component Y of X, the map
p : Y −→ B is injective. It is then bijective if B is connected, and as p is a
local homeomorphism, it is a global homeomorphism from Y onto B. As a
result, Y is the image of B under a global section ϕ of X and the covering
(X, B, p) is trivial.
(iii) Path-lifting. Let γ : I −→ B be a continuous path in B, where
I = [0, 1]. A lifting of γ is a path μ : I −→ X such that p ◦ μ = γ (§ 4
of Chap. IV for the case of Log z). Such a lifting always exists, but there is
better still :
Theorem 2. Let (X, B, p) be a covering and γ : I −→ B a path in B. There
is a unique lifting μ from γ to X with a given initial point. If two paths γ 0
and γ 1 in B are fixed-endpoint homotopic and if μ 0 and μ 1 are liftings of γ 0
and γ 1 having the same initial point, then μ 0 and μ 1 have the same terminal
point and are homotopic.