3 – Coverings of a Topological Space
289
We first prove the uniqueness of μ. In the neighbourhood of some t 0 ∈ I,
a lifting μ(t) of γ, being a continuous function of t, takes its values in a
neighbourhood of μ(t 0 ) mapped homeomorphically by p : X −→ B onto a
neighbourhood D of γ(t 0 ) satisfying axiom (R) . Hence μ = ϕ ◦ γ in D, where
ϕ is the unique section of X over D mapping γ(t 0 ) onto μ(t 0 ). It follows that
the set of point where two liftings coincide is open and closed in I, proving
uniqueness.
To prove the existence of μ, choose α ∈ X such that p(α) = γ(0) = a
and, as in § 4 du Chap. IV, consider all the couples (J, μ), where J ⊂ I is an
interval with initial point 0 and μ is a continuous map from J to X satisfying
p ◦ μ = γ on J, as well as μ(0) = α. The existence of such couples as well as
their “ coherence ” is clear – Take J sufficiently small so that X is trivial over
γ(J): if (J
, μ
) and (J”, μ”) are two such couples, then, by the uniqueness of
liftings, μ
= μ” in J
∩ J”. The union of all these J gives a couple (J 0 , μ 0 )
such that μ 0 cannot be extended beyond the terminal point b of J 0 . But as
X is trivial over the connected open neighbourhood D of γ(b) in B, if b
∈ J 0
is sufficiently near b for γ(b
) ∈ D to hold, then there is a section of X in
D equal to μ 0 (b
) at γ(b
), which makes it possible to extend μ 0 beyond b if
b < 1, and at b if b = 1. So b = 1.
Next, consider a homotopy σ : I × I = K −→ B between paths γ 0 and γ 1
of the statement, whose initial and terminal points and will be denoted by a
and b; hence
σ(s, 0) = a , σ(s, 1) = b for all s ,
(3.2)
σ(0, t) = γ 0 (t) , σ(1, t) = γ 1 (t) for all t .
Let α be the common initial point of μ 0 and μ 1 . The problem consists in
constructing a continuous map σ
: K −→ X satisfying p ◦ σ
= σ and
σ
(s, 0) = α for all s ,
(3.3)
σ
(0, t) = μ 0 (t) , σ
(1, t) = μ 1 (t) for all t .
In fact there is no need to require σ
(s, 0) to be independent of s, because
condition p ◦ σ
= σ shows that s → σ
(s, 0) is a continuous map from I to
the discrete space p
−1 (a), and so is constant. Conditions (3) could even be
replaced by the unique condition that σ
(0, 0) = α ; indeed, relation p◦σ
= σ
shows that (1) σ
(s, 0) takes its values in p
−1 (a), and so is constant, (2) σ
(0, t)
is a lifting of γ 0 having the same initial point as μ 0 , and so is equal to μ 0 ,
(3) σ
(1, t) is a lifting of γ 1 having the same initial point as μ 1 , and so is
equal to μ 1 .
Like in the previous case, the uniqueness of σ
is clear since K = I × I is
connected.
As σ(B) is compact and σ is uniformly continuous, (1) shows that there
exists r > 0 such that X is trivial over σ(D) for all discs D of radius r in
K. If we draw a grid on K consisting of the lines s = i/n and t = j/n, with
0 ≤ i, j ≤ n, where n is sufficiently large, then X is trivial over the images
289
We first prove the uniqueness of μ. In the neighbourhood of some t 0 ∈ I,
a lifting μ(t) of γ, being a continuous function of t, takes its values in a
neighbourhood of μ(t 0 ) mapped homeomorphically by p : X −→ B onto a
neighbourhood D of γ(t 0 ) satisfying axiom (R) . Hence μ = ϕ ◦ γ in D, where
ϕ is the unique section of X over D mapping γ(t 0 ) onto μ(t 0 ). It follows that
the set of point where two liftings coincide is open and closed in I, proving
uniqueness.
To prove the existence of μ, choose α ∈ X such that p(α) = γ(0) = a
and, as in § 4 du Chap. IV, consider all the couples (J, μ), where J ⊂ I is an
interval with initial point 0 and μ is a continuous map from J to X satisfying
p ◦ μ = γ on J, as well as μ(0) = α. The existence of such couples as well as
their “ coherence ” is clear – Take J sufficiently small so that X is trivial over
γ(J): if (J
, μ
) and (J”, μ”) are two such couples, then, by the uniqueness of
liftings, μ
= μ” in J
∩ J”. The union of all these J gives a couple (J 0 , μ 0 )
such that μ 0 cannot be extended beyond the terminal point b of J 0 . But as
X is trivial over the connected open neighbourhood D of γ(b) in B, if b
∈ J 0
is sufficiently near b for γ(b
) ∈ D to hold, then there is a section of X in
D equal to μ 0 (b
) at γ(b
), which makes it possible to extend μ 0 beyond b if
b < 1, and at b if b = 1. So b = 1.
Next, consider a homotopy σ : I × I = K −→ B between paths γ 0 and γ 1
of the statement, whose initial and terminal points and will be denoted by a
and b; hence
σ(s, 0) = a , σ(s, 1) = b for all s ,
(3.2)
σ(0, t) = γ 0 (t) , σ(1, t) = γ 1 (t) for all t .
Let α be the common initial point of μ 0 and μ 1 . The problem consists in
constructing a continuous map σ
: K −→ X satisfying p ◦ σ
= σ and
σ
(s, 0) = α for all s ,
(3.3)
σ
(0, t) = μ 0 (t) , σ
(1, t) = μ 1 (t) for all t .
In fact there is no need to require σ
(s, 0) to be independent of s, because
condition p ◦ σ
= σ shows that s → σ
(s, 0) is a continuous map from I to
the discrete space p
−1 (a), and so is constant. Conditions (3) could even be
replaced by the unique condition that σ
(0, 0) = α ; indeed, relation p◦σ
= σ
shows that (1) σ
(s, 0) takes its values in p
−1 (a), and so is constant, (2) σ
(0, t)
is a lifting of γ 0 having the same initial point as μ 0 , and so is equal to μ 0 ,
(3) σ
(1, t) is a lifting of γ 1 having the same initial point as μ 1 , and so is
equal to μ 1 .
Like in the previous case, the uniqueness of σ
is clear since K = I × I is
connected.
As σ(B) is compact and σ is uniformly continuous, (1) shows that there
exists r > 0 such that X is trivial over σ(D) for all discs D of radius r in
K. If we draw a grid on K consisting of the lines s = i/n and t = j/n, with
0 ≤ i, j ≤ n, where n is sufficiently large, then X is trivial over the images