3 – Coverings of a Topological Space
291
over H 3 equal to ϕ 2 over the image of the common side of the two squares
K 2 and K 1 , and so on, and define σ
i = ϕ i ◦ σ in K i .
Suppose that the existence of a map σ
in K 1 ∪ . . . ∪ K i coinciding with
σ
j in each K j , j ≤ i has been proved. The definition of σ
will extend to
K 1 ∪ . . . ∪ K i+1 if and only if σ
i+1 = σ
j in K i+1 ∩ K j , where j ≤ i, when
this intersection is non-empty. Let us do it for i = 10 and j = 5 (figure !).
By construction, the partial liftings σ
10 and σ
9 are equal in K 10 ∩ K 9 , but
the induction hypothesis shows that σ
9 and σ
5 are equal in the set K 9 ∩ K 5
reduced to a point and hence non-empty. As this point also belongs to K 10 ,
σ
10 and σ
5 are equal at this point, and so in K 10 ∩ K 5 as expected.
This argument shows that there is continuous map σ
: I × I −→ X such
that p ◦ σ
= σ. As it satisfies σ
(0, 0) = α, this proves the theorem.
An immediate corollary is that all covering maps over I are trivial, as
they have global sections : take B = I and for σ take the identity map from
I to B. The same result holds for I
n , where n ∈ N.
More importantly :
Corollary 1. Let (X, B, p) be a covering and suppose that X is connected
and simply connected.
11 Let γ 0 and γ 1 be two paths in B with the same
endpoints and let μ 0 and μ 1 be liftings to X of γ 0 and γ 1 having the same
initial point. γ 0 and γ 1 are homotopic if and only if μ 0 and μ 1 have the same
terminal point.
If the condition holds, the two liftings are homotopic, since X is simply
connected, and thus that is also the case of the given paths in B. The converse,
which makes no assumptions on X, is the second statement of theorem 2.
In particular, a closed path γ in B is homotopic to a point if and only if
some (and hence all) lifting of γ to X is closed, of course provided that X is
simply connected.
To state the next corollary which solves an essential problem in Cauchy
theory, take B = C
∗ and consider the path
u : t −→ r. exp(2πit)
in B, i.e. the circle centered at 0 and of radius r > 0 traveled once counterclockwise ; let nu be the path t → r. exp(2πint), i.e. the circle centered at 0
traveled n times counterclockwise if n ≥ 0, or −n times clockwise if n < 0 ;
obviously he choice of r has no impact on the homotopy class of nu in C
∗ ,
the latter being all that matters to us in what follows. Hence r = 1 may be
assumed.
Corollary 2. For any closed path γ in C
∗ , there is a unique integer n such
that γ is homotopic to nu : t → e(nt).
11 i.e. having the following property: two arbitrary paths with the same endpoints
are always fixed end-point homotopic.
291
over H 3 equal to ϕ 2 over the image of the common side of the two squares
K 2 and K 1 , and so on, and define σ
i = ϕ i ◦ σ in K i .
Suppose that the existence of a map σ
in K 1 ∪ . . . ∪ K i coinciding with
σ
j in each K j , j ≤ i has been proved. The definition of σ
will extend to
K 1 ∪ . . . ∪ K i+1 if and only if σ
i+1 = σ
j in K i+1 ∩ K j , where j ≤ i, when
this intersection is non-empty. Let us do it for i = 10 and j = 5 (figure !).
By construction, the partial liftings σ
10 and σ
9 are equal in K 10 ∩ K 9 , but
the induction hypothesis shows that σ
9 and σ
5 are equal in the set K 9 ∩ K 5
reduced to a point and hence non-empty. As this point also belongs to K 10 ,
σ
10 and σ
5 are equal at this point, and so in K 10 ∩ K 5 as expected.
This argument shows that there is continuous map σ
: I × I −→ X such
that p ◦ σ
= σ. As it satisfies σ
(0, 0) = α, this proves the theorem.
An immediate corollary is that all covering maps over I are trivial, as
they have global sections : take B = I and for σ take the identity map from
I to B. The same result holds for I
n , where n ∈ N.
More importantly :
Corollary 1. Let (X, B, p) be a covering and suppose that X is connected
and simply connected.
11 Let γ 0 and γ 1 be two paths in B with the same
endpoints and let μ 0 and μ 1 be liftings to X of γ 0 and γ 1 having the same
initial point. γ 0 and γ 1 are homotopic if and only if μ 0 and μ 1 have the same
terminal point.
If the condition holds, the two liftings are homotopic, since X is simply
connected, and thus that is also the case of the given paths in B. The converse,
which makes no assumptions on X, is the second statement of theorem 2.
In particular, a closed path γ in B is homotopic to a point if and only if
some (and hence all) lifting of γ to X is closed, of course provided that X is
simply connected.
To state the next corollary which solves an essential problem in Cauchy
theory, take B = C
∗ and consider the path
u : t −→ r. exp(2πit)
in B, i.e. the circle centered at 0 and of radius r > 0 traveled once counterclockwise ; let nu be the path t → r. exp(2πint), i.e. the circle centered at 0
traveled n times counterclockwise if n ≥ 0, or −n times clockwise if n < 0 ;
obviously he choice of r has no impact on the homotopy class of nu in C
∗ ,
the latter being all that matters to us in what follows. Hence r = 1 may be
assumed.
Corollary 2. For any closed path γ in C
∗ , there is a unique integer n such
that γ is homotopic to nu : t → e(nt).
11 i.e. having the following property: two arbitrary paths with the same endpoints
are always fixed end-point homotopic.