Lecture
49.3. Lemma. Let h, g : X → Y — be homotopic mappings, h(x0) = y0, g(x0) = y1, and
Φ : X × I → Y — a homotopy connecting them. Then for the path α : I → Y , α(t) = Φ(x0, t) with
beginning α(0) = h(x0) = y0 and end α(1) = g(x0) = y1 we have
g∗ = α# ◦ h∗.
Proof. We must show that
g∗([ϕ]) = α#(h∗([ϕ])) (49.1)
for any [ϕ] ∈ π(X, x0). Equality (49.1) is equivalent to the equality
[g ◦ ϕ] = [α]
−1
· [h ◦ ϕ] · [α]
or the equality
[α] · [g ◦ ϕ] = [h ◦ ϕ] · [α],
which we shall verify. We must show that α(g ◦ ϕ) ∼ (h ◦ ϕ)α.
Consider the mapping G = ϕ×id : I ×I → X ×I, and the paths in I ×I: δi
: I → I ×I, δi(s) = (s, i),
βi
: I → I × I, βi(t) = (i, t), i = 0, 1. Then
Φ ◦ G ◦ δ0 = h ◦ ϕ, Φ ◦ G ◦ δ1 = g ◦ ϕ,
Φ ◦ G ◦ βi = α, i = 0, 1.
Let us verify the first equality and the equality Φ ◦ G ◦ β0 = α. For any points s ∈ I, t ∈ I
(Φ ◦ G ◦ δ0)(s) = (Φ ◦ G)(s, 0) = Φ(ϕ(s), 0) = h(ϕ(s)) = (h ◦ ϕ)(s).
(Φ ◦ G ◦ β0)(t) = (Φ ◦ G)(0, t) = Φ(x0, t) = α(t).
The paths β0δ1 and δ0β1 in I × I have matching beginnings — (0, 0) and ends — (1, 1). Since
the square I ×I is simply connected (Example 48.13.2 of Lecture 15), the paths β0δ1 and δ0β1 are homotopic. Denote
the corresponding homotopy by H. Then Φ ◦ G ◦ H — is a based homotopy, connecting the paths
α(g ◦ ϕ) and (h ◦ ϕ)α.
Indeed, for any point s ∈ I
(Φ ◦ G ◦ H)(s, 0) = (Φ ◦ G ◦ (β0δ1))(s) = (Φ ◦ ((G ◦ β0)(G ◦ δ1)))(s) =
((Φ ◦ G ◦ β0)(Φ ◦ G ◦ δ1))(s) = (α(g ◦ ϕ))(s).
The equality Φ ◦ G ◦ H = (h ◦ ϕ)α is verified similarly. Let us check that the homotopy is based. For any
point t ∈ I
(Φ ◦ G ◦ H)(0, t) = (Φ ◦ G)(0, 0) = Φ(x0, 0) = h(x0) = y0,
and a similar check for the points (1, t) ⊂ I × I.
The continuity of the homotopy Φ ◦ G ◦ H is obvious.
49.4. Theorem. Let f : (X, x0) → (Y, y0) — be a homotopy equivalence. Then
f∗ : π(X, x0) → π(Y, y0)
is an isomorphism.
Proof. Let g : Y → X — be a homotopy inverse of f, g(y0) = x1, f(x1) = y1. We have
(X, x0)
f
−→ (Y, y0)
g
−→ (X, x1)
f
−→ (Y, y1).
The induced homomorphisms are defined
π(X, x0)
(fx0
)∗
−→ π(Y, y0)
g∗ −→ π(X, x1)
(fx1
)∗
−→ π(Y, y1).
Since the mapping g ◦ f : (X, x0) → (X, x1) is homotopic to the identity mapping idX, then
by Lemma 49.3 there exists a path α in X with beginning at the point x0 and end at the point x1, for which
g∗ ◦ (fx0
)∗ = (g ◦ f)∗ = α# ◦ (idX)∗ = α#. (49.2)
Thus g∗ ◦ (fx0
)∗ — is an isomorphism. Hence, g∗ — is surjective.
86
Since f ◦ g is homotopic to the identity mapping idY , then arguing similarly to the previous case, we have (f ◦ g)∗ = (fx1
)∗ ◦ g∗ — is an isomorphism. Hence, g∗ — is injective. Therefore, g∗
— is an isomorphism. From equality (49.2) we have (fx0
)∗ = g
−1
∗ ◦ α#. Hence (fx0
)∗ = f∗ — is an isomorphism.
49.5. Corollary. If h : (X, x0) → (Y, y0) — is a homeomorphism, then h∗ — is an isomorphism of π(X, x0)
onto π(Y, y0).
§ 50. Coverings. The fundamental group of the circle. The Brouwer fixed point theorem.
50.1. Definition. Let p : X → B — be a continuous surjective mapping. An open
subset O ⊂ B is called evenly covered, if the preimage p
−1O is the union of pairwise disjoint open subsets Vs, s ∈ S, for each of which the restriction p|Vs is a homeomorphism of Vs onto O. The elements of the family {Vs : s ∈ S} are called
the “sheets” of the covering p over O.
If every point b ∈ B has an evenly covered neighborhood, then p is called a covering map or a covering, X — is called the covering or total space,
B — the base.
50.2. Examples. 1. id : X → X.
2. prX : X × Y → X, where Y — is a discrete space. (X — is an evenly covered open
subset, X × {y}, y ∈ Y , — is a sheet of the covering.)
3. p : R → S
1
, p(t) = (cos(2πt),sin(2πt)). (For example, the preimage of the interval S
1 ∩ {(x, y) : x >
0} is the union of the intervals S
{(n −
1
4
, n +
1
4
) : n ∈ Z}, and the restriction of p to the interval
(n −
1
4
, n +
1
4
) — is a homeomorphism.)
4. The natural projection of S
n onto RP
n, n ∈ N.
50.3. Definition. Let p : X → B and f : A → B — be arbitrary mappings. A mapping g : A → X covers f or is a lift of it, if p ◦ g = f.
50.4. Path lifting theorem. Let p : X → B — be a covering, and p(x0) = b0. Then
any path ϕ : I → B with beginning at b0 admits a unique lift ϕ˜ : I → X — a path with
beginning at x0.
Proof. Let the cover u ∈ cov(B) be such that every element U ∈ u is evenly
covered. Divide the segment I into n segments Ik =
k
n
,
k+1
n
, k = 0, . . . , n − 1 so that every
set ϕ(Ik) is contained in some element of the cover u. For this it suffices that
n > 1
, where — is the Lebesgue number of the cover ϕ
−1u = {ϕ
−1
(U) : U ∈ u} (Lemma 37.1 (Lebesgue covering theorem) of Lecture 10). We shall successively define ϕ˜ — a lift of ϕ.
Set ϕ˜(0) = x0. ϕ(I0) lies in an evenly covered element U0 ∈ u. Take V0 — that
sheet of the covering over U0 to which the point x0 belongs. Define ϕ˜(t) for t ∈ I0 as
follows
ϕ˜(t) = (p|V0
)
−1
(ϕ(t)).
Since p|V0
: V0 → U0 is a homeomorphism, ϕ˜ is correctly defined and continuous on I0.
Let ϕ˜ be defined for t ∈ Ik, k < n − 1. ϕ(Ik+1) lies in an evenly covered element
Uk+1 ∈ u. Take Vk+1 — that sheet of the covering over Uk+1 to which the point ϕ˜(
k+1
n
)
belongs.
For t ∈ Ik+1 define ϕ˜(t) for t ∈ Ik+1 as follows
ϕ˜(t) = (p|Vk+1 )
−1
(ϕ(t)).
Since p|Vk+1 : Vk+1 → Uk+1 is a homeomorphism, ϕ˜ is correctly defined and continuous on Ik+1.
Thus the lift ϕ˜ is defined, whose continuity follows from item 13.4.6 of the properties of
mappings of Lecture 3.
Uniqueness of the lift. Let ϕ˜
0 — be a lift of ϕ. Then ϕ˜
0
(0) = ˜ϕ(0) = x0. Let ϕ˜
0
(t) = ˜ϕ(t)
for all 0 ≤ t ≤
k
n
. For t ∈ Ik we have ϕ˜(t) = (p|Vk
)
−1
(ϕ(t)). Since ϕ˜
0 — is a lift of ϕ, then
ϕ˜
0
(Ik) ⊂ W =
[
{Vs : Vs − a sheet of the covering over Uk}, where ϕ(Ik) ⊂ Uk.
The set ϕ˜
0
(Ik) is connected, and the set W is a union of open and pairwise disjoint subsets. Therefore ϕ˜
0
(Ik) is contained in a single sheet of the covering Vs, to which
the point ϕ˜
0
(
k
n
) belongs. But the same sheet contains the point ϕ˜(
k
n
). Hence ϕ˜
0
(t) = ˜ϕ(t)
for all 0 ≤ t ≤
k+1
n
since p|Vs — is a homeomorphism of Vs onto Uk.
87
50.5. Homotopy lifting theorem. Let p : X → B — be a covering, and p(x0) = b0.
Then for the mapping Φ : I × I → B, Φ(0, 0) = b0, there exists a unique lift
Φ : ˜ I × I → X such that Φ(0 ˜ , 0) = x0.
If Φ — is a homotopy of the paths ϕ0 = Φ|X×{0} and ϕ1 = Φ|X×{1}, then Φ˜ — is a homotopy of their lifts
ϕ˜0 and ϕ˜1.
Proof. Set Φ(0 ˜ , 0) = x0, and, using Theorem 50.4, define the mapping
Φ˜ on the set {0} × I ∪ I × {0} (we lift the paths Φ|{0}×I
: I → B and Φ|I×{0} : I → B).
Let the cover u ∈ cov(B) be such that every element U ∈ u is evenly covered. Using
the Lebesgue number of the cover ϕ
−1u = {ϕ
−1
(U) : U ∈ u}, divide the segment I into n segments
Ik =
k
n
,
k + 1
n
, k = 0, . . . , n − 1
and into m segments
Jl =
l
m
,
l + 1
m
, l = 0, . . . , m − 1
so that every set Φ(Ik ×Jl) is contained in some element of the cover u. We shall successively define Φ˜ — a lift of Φ. Starting from the rectangle I0 ×J0 we shall successively
for k = 1, . . . , n − 1 define the lift of Φ on the rectangles of the bottom row Ik × J0; then
the second row Ik × J1, k = 0, . . . , n − 1; and so on.
Set Φ(0 ˜ , 0) = x0. Φ(I0 × J0) lies in an evenly covered element U0,0 ∈ u. Take V0,0
— that sheet of the covering over U0,0 to which the point x0 belongs. Define Φ( ˜ t, t0
) for
(t, t0
) ∈ I0 × J0 as follows
Φ( ˜ t, t0
) = (p|V0,0
)
−1
(Φ(t, t0
)).
Since p|V0,0
: V0,0 → U0,0 is a homeomorphism, Φ˜ is correctly defined and continuous on I0 × J0.
Define Φ˜ on Ik0 × Jl0
, l0 ≤ m − 1, k0 ≤ n − 1, assuming that Φ˜ is defined for (t, t0
) ∈
Ik × Jl for l < l0 and for l = l0, k < k0, and on the set ({0} × I) ∪ (I × {0}). ϕ(Ik0 × Jl0
)
lies in an evenly covered element Uk0,l0 ∈ u. Take Vk0,l0 — that (unique) sheet
of the covering over Uk0,l0
, to which the connected set Φ(( ˜ {
k0
n
} × Il0
) ∪ (Ik0 × {l0+1
m })) belongs.
For (t, t0
) ∈ Ik0 × Jl0
define Φ( ˜ t, t0
) as follows
Φ( ˜ t, t0
) = (p|Ik0,l0
)
−1
(Φ(t, t0
)).
Since p|Uk0,l0
: Vk0,l0 → Uk0,l0
is a homeomorphism, Φ˜ is continuous on Ik0 × Jl0
. Thus
the lift Φ˜ is defined, whose continuity follows from item 13.4.6 of the properties of mappings
of Lecture 3.
The arguments for the proof of uniqueness of the lift Φ˜ are similar to those used
in Theorem 50.4.
Let Φ — be a homotopy of the paths ϕ0 and ϕ1. Then Φ({0}×I) = b0, and Φ( ˜ {0}×I) — is a connected subset
of the discrete space p
−1
(b0), containing the point x0. Hence, Φ( ˜ {0} × I) = x0. Similarly
one proves that Φ( ˜ {1} × I) is a single point. Thus, Φ˜ — is a homotopy of the paths ϕ˜0 and ϕ˜1.
50.6. Corollary. Let p : X → B — be a covering, and p(x0) = b0. Let ϕ and ψ — be paths in B with
beginning at b0, ϕ˜ and ψ˜ — their lifts with beginning at x0. If ϕ ∼ ψ (and, hence, ϕ(1) = ψ(1)), then
ϕ˜(1) = ψ˜(1).
50.7. Definition. Let p : X → B — be a covering, and p(x0) = b0. The mapping p? : π(B, b0) →
p
−1
(b0),
p?([ϕ]) = ˜ϕ(1),
where ϕ˜(0) = x0, we call the lifting correspondence.
By Corollary 50.6 it is correctly defined.
50.8. Theorem. Let p : X → B — be a covering, and p(x0) = b0.
If X is path-connected, then the lifting correspondence p? : π(B, b0) → p
−1
(b0) is surjective.
If X is simply connected, then p? is bijective.
Proof. If X is path-connected, and x1 ∈ p
−1
(b0), then there exists a path ϕ˜ : I → X with
beginning x0 and end x1. Then ϕ˜ is a lift of the loop ϕ = p ◦ ϕ˜. Hence, p?([ϕ]) = x1.
Let X be simply connected, and [ϕ], [ψ] ∈ π(B, b0) such that p?([ϕ]) = p?([ψ]). If ϕ˜ and ψ˜ — are lifts
of ϕ and ψ respectively, then, by the simple connectedness of X, there exists a homotopy Φ˜ between them. Then Φ = p ◦ Φ˜
— is a homotopy of the loops ϕ and ψ.
88
50.9. Theorem (fundamental group of the circle). The fundamental group of the circle is isomorphic to the (additive) group Z of integers.
Proof. Let p : R → S
1 — be the covering from Example 50.2.3, b0 = (1, 0). Then
p
−1
(1, 0) = Z. Set x0 = 0. Since R — is a simply connected space, the lifting correspondence
p? : π(S
1
, b0) → Z
is a bijection. Let us show that p? — is a homomorphism.
For [ϕ], [ψ] ∈ π(S
1
, b0) let ϕ˜ and ψ˜ — be lifts of ϕ and ψ respectively, and ϕ˜(1) = n and ψ˜(1) = m.
Then p?([ϕ]) = n, p?([ψ]) = m.
Let ψ˜0
(t) = n + ψ˜(t) — be a path in R. Since p(n + x) = (cos(2π(n + x)),sin(2π(n + x))) =
(cos(2πx),sin(2πx)) = p(x), then ψ˜0 — is a lift of the loop ψ with beginning at the point n. The product ϕ˜ψ˜0 of the paths ψ˜ and ψ˜0 in R is correctly defined. Moreover, the path ϕ˜ψ˜0
, with beginning at 0, — is a lift of the loop ϕψ.
Thus
p?([ϕ][ψ]) = p?([ϕψ]) = ˜ϕψ˜0
(1) = n + m = p?([ϕ]) + p?([ψ]).
50.10. Definition. Let A ⊂ X. A continuous mapping f : X → A, for which
f(a) = a, a ∈ A, is called a retraction of X onto A, and the subset A is called a retract of X.
Note that a continuous mapping f : X → X is a retraction onto its image f(X)
if and only if f ◦ f = f.
50.11. Proposition. There is no retraction of the closed disk B2 = {x = (x1, x2) :
x
2
1 + x
2
2 ≤ 1} onto the boundary circle S
1
.
Proof. Suppose that there exists a retraction r : B2 → S
1
, i : S
1 → B2 — the
natural embedding. Then r ◦ i = idS1 and r∗ ◦ i∗ — is an isomorphism by Theorem 49.2 of Lecture 15. Thus,
the homomorphism i∗ — is injective. But π1(S
1
) = Z, π1(B2
) = 0 (Example 48.13.2 of Lecture 15).
We obtain a contradiction.
50.12. Theorem. (Brouwer fixed point theorem.) Every mapping f : B2 → B2
has a fixed point (i.e. there exists a point x such that f(x) = x.)
Proof. Suppose that the mapping f : B2 → B2 has no fixed points. Then for every point x ∈ B2 the vector
−−−→
f(x)x is defined, and there exists k ≥ 0 such that
f(x) + k ·
−−−→
f(x)x ∈ S
1
. Denoting r(x) = f(x) + k ·
−−−→
f(x)x, we obtain a retraction r : B2 → S
1 of the disk B2 onto the boundary circle S
1
(its continuity should be checked independently). We obtain
a contradiction with Proposition 50.11.
Assignment N 16
1. Let the path α connect the points x0 and x1, the path γ connect the points x1 and x2. Prove that
(αγ)# = γ#α#.
2. Prove that if α ∼ γ, then α# = γ#.
3. Compute the fundamental group:
(1) S
n, n ∈ N, n ≥ 2;
(2) R
n \ {O}, n ∈ N, n ≥ 3.
4. Let X ⊂ R
n. Prove that if f : X → Y extends to a mapping ˜f : R
n → Y ,
then the homomorphism f∗ : π1(X, x0) → π1(Y, y0) is trivial.
5. Let the mapping f : X → Y be continuous, the points x0 and x1 connected by a path α, y0 = f(x0),
y1 = f(x1). Then
(f ◦ α)# ◦ (fx0
)∗ = (fx1
)∗ ◦ α#.
6. Let r : X → A — be a retraction, a ∈ A.
Prove the surjectivity of the homomorphism
r∗ : π1(X, a) → π1(A, a).
Prove for the embedding i : A → X the injectivity of the homomorphism
i∗ : π1(A, a) → π1(X, a).
7. Prove that p : R → S
1
, p(t) = (cos(2πt),sin(2πt)) — is a covering.
89
8. Prove that the natural projection of S
n onto RP
n — is a covering, n ∈ N.
9. Let p : X → B — be a covering.
Prove that if the base B is a Hausdorff (respectively regular, Tychonoff, locally compact Hausdorff) space, then X is a Hausdorff (respectively regular,
Tychonoff, locally compact Hausdorff) space.
If the base is compact, and p
−1
(b) is finite for every point b ∈ B, then X — is a compact space.
10 (Brouwer fixed point theorem.) Prove that every continuous mapping of the
closed disk B2 into itself has a fixed point.
11. Prove that under any homeomorphism of B2 points of the boundary S
1 map to points
of the boundary.
12. Let f : S
1 → X — be a continuous mapping. Prove that the following conditions are equivalent:
(a) f is null-homotopic;
(b) f extends to a continuous mapping ˜f : B2 → X;
(c) f∗ — is the trivial homomorphism.
13. Prove that R
2 is not homeomorphic to R
n for any n 6= 2, n ∈ N.
14. Compute the fundamental groups of the projective spaces RP
n, n ≥ 2.
Additional problems for Assignment N 16
15. Prove that the isomorphism α#, determined by the path α with beginning x0 and end x1, does not depend on the choice of path with beginning x0 and end x1 if and only if the fundamental
group π1(X, x0) is abelian.
16. Prove that a path-connected space that is the union of two open
simply connected sets, whose intersection is path-connected, is simply connected.
17. Let the sets U and V be open in X. Prove that if the sets U ∩ V and U ∪ V
are simply connected, then the sets U and V are simply connected.
18. Is the wedge S
2 ∨ S
2
simply connected? Is the wedge of simply connected spaces simply connected?
19. Give an example of a simply connected space that is not contractible.
20. Construct a covering:
(a) of the Möbius band by the cylinder;
(b) of the Klein bottle by the torus;
(c) of the Klein bottle by the plane;
(d) of the Klein bottle by the cylinder.
21. Prove that if p : X → B — is a covering, B0 ⊂ B, then for X0 = p
−1
(B0) p|X0
:
X0 → B0 is a covering.
22. Prove that if a path-connected space B has a non-one-sheeted path-connected
covering, then it is not simply connected.
23. Prove that any covering p : X → B with simply connected base and path-connected covering space is a homeomorphism.
24. Let p : X → B — be a covering, and p(x0) = b0. Then
(a) p∗ : π1(X, x0) → π1(B, b0) — is a monomorphism;
(b) for the group H = p∗(π1(X, x0)) the lifting correspondence
p? : π1(B, b0)/H → p
−1
(b0)
is injective. If X is path-connected, then p? — is a bijection.
(c) if ϕ is a loop in B, then [ϕ] ∈ H if and only if ϕ lifts to a
loop in X.
25. Let p : X → B and p
0
: X0 → B0 — be coverings. Prove that p × p
0
: X × X0 → B × B0
is a covering.
90
26. For the covering p × id : (R × R⊕) → S
1 × R⊕ (of the plane with a point removed), find a path
that is a lift of the path:
(a) ϕ(t) = (2 − t, 0);
(b) ψ(t) = ((1 + t) cos(2πt),(1 + t) sin(2πt));
(c) χ(t) = ϕψ(t).
27. For the covering p × p : R
2 → S
1 × S
1 of the torus, find a path that is a lift of the path:
ϕ(t) = (cos(2πt),sin(2πt)) × (cos(4πt),sin(4πt)).
28. Prove that a null-homotopic mapping f : S
1 → S
1 has a fixed point
(f(x) = x) and a point mapping to the antipodal point (i.e. f(x) = −x).
Let f : S
1 → S
1 be a continuous mapping such that f(−t) = −f(t) for every point
t ∈ S
1
. Prove that f is null-homotopic.
29. Prove that for any continuous function f : S
1 → R there exists a point t ∈ S
1
such that f(t) = f(−t).
30. Prove:
π1(X × Y,(x0, y0)) ∼= π1(X, x0) × π1(Y, y0).
Compute the fundamental group of: the torus T
2
; the solid torus S
1 ×B2
; S
1 ×S
2
; the cylinder S
1 ×R.
31. Compute the fundamental group of: the wedge of circles S
1 ∨ S
1
; the wedge of spheres S
2 ∨ S
2
;
the torus T
2
with a point removed.
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