Chapter 5. Proof of the Arc Translation Lemma
1
Equivalence of embeddings
Given two maps h and g in E(R2 ), we say that they are strongly equivalent
if there is a disk D with
h(D) ∩ D = ∅ and h = g on R2 \ D.
Sometimes D will be called a disk of modification.
First of all we observe that the roles of h and g can be interchanged.
Actually h and g must coincide on ∂D and so h(∂D) = g(∂D). If we apply
Lemma ?? 5 in Chapter 3 with Γ = ∂D we are lead to h(D) = g(D) and
so the conditions h(D) ∩ D = ∅ and g(D) ∩ D = ∅ are equivalent. Another
consequence of the definition of strong equivalence is the coincidence between
the sets of fixed points. The common set of fixed points will be denoted by
F := Fix(h) = Fix(g).
The sets F and D are closed and disjoint and so it is possible to a find a
neighborhood U of F which is also disjoint with D. Once we know that h
and g coincide in a neighborhood of F we can apply the excision property
and conclude that
deg(id − h, Ω) = deg(id − g, Ω)
for any Ω bounded and open subset of R2 with ∂Ω ∩ F = ∅.
The maps h, g ∈ E(R2 ) are freely equivalent if there exists a chain h =
h1 , h2 , . . . , hk = g in E(R2 ) such that hi and hi+1 are strongly equivalent for
each i = 1, . . . , k − 1.
The set of fixed points and the degree are also preserved under free
equivalence.
To get some familiarity with these notions we discuss an example. Consider a disk D = [a, b] × [0, 1] and let φ be a homeomorphism of D satisfying
φ(p) = p if p ∈ ∂D, φ(p) = p if p ∈ int(D).
1
For instance φ could be the time map of a flow on D whose equilibria are
precisely the points in ∂D.
Figure 1: A flow inducing φ
The translation T (x, y) = (x + 1, y) will be modified to the map TD in E(R2 )
defined by
TD = T ◦ φ on D, TD = T outside D.
If b − a < 1 the maps T and TD are strongly equivalent with D as a disk
of modification. It is interesting to observe that T and TD are not strongly
equivalent if b − a ≥ 1. This is a consequence of the following
Exercise 1 Given h, g ∈ E(R2 ), the set N is the closure of
{p ∈ R2 : h(p) = g(p)}.
Prove that if h(N ) ∩ N = ∅ then h and g are not strongly equivalent.
Next we consider the disks
1
1
D1 = [0, ] × [0, 1], D2 = [ , 1] × [0, 1]
2
2
and corresponding homeomorphisms φ1 and φ2 as before. A new map T̂ is
defined by
T̂ = T ◦ φ1 on D1 , T̂ = T ◦ φ2 on D2 , T̂ = T outside D1 ∪ D2 .
The chain h1 = T , h2 = TD1 , h3 = T̂ shows that T and T̂ are freely
equivalent. However they are not strongly equivalent.
2
2
Compression of translation arcs
In this Section we show how to contract a translation arc via free modifications. Throughout the Section α = pq
is an arc and p is a given point
and
in α̇ = α \ {p, q}. This point splits the arc α in two sub-arcs α = pp
α = p q.
Proposition 2 Assume that α is a translation arc for some h ∈ E(R2 ) and
U is a neighborhood of α . Then there exists g ∈ E(R2 ) which is strongly
equivalent to h with disk of modification contained in U and such that
g(α ) = h(α), g(p ) = q, g(q) = h(q).
Figure 2: From α to α
Notice that α is a translation arc for g.
Before proving this Proposition we state a simpler result which seems
obvious.
Lemma 3 Let D be a disk with α ⊂ int(D). Then there exists φ ∈ H(R2 )
with
φ(p) = p , φ(q) = q, φ(α) = α , φ = id outside D.
Proof. We use that arcs are tamely imbedded in the plane. This means
that we can find ψ ∈ H(R2 ) with
1
ψ(α) = [0, 1] × {0}, ψ(p) = (0, 0), ψ(p ) = ( , 0), ψ(q) = (1, 0).
2
For the special arc [0, 1] × {0} it is easy to find an explicit homeomorphism
φ0 in the conditions required by the Lemma and such that φ0 = id outside
the disk ψ(D). Now we can go back to α with φ = ψ −1 ◦ φ0 ◦ ψ.
3
Proof of Proposition 2. Since α is a translation arc for h we know that
h(α ) ∩ α = ∅. This allows us to find a disk D with α ⊂ int(D), D ⊂ U
and h(D) ∩ D = ∅. This disk can be obtained by inflating α . The previous
Lemma produces a homeomorphism φ contracting α to α . The embedding
g = h ◦ φ−1 is strongly equivalent to h with D as a disk of modification.
In the next result we show how to compress the image of a translation
arc. At first sight it seems that such a result could be obtained by applying
Proposition 2 to h−1 . We must have some care since we are dealing with
embeddings and not with homeomorphims.
Proposition 4 Assume that α is a translation arc for some h ∈ E(R2 )
and h(q) ∈ α. Given V neighborhood of α there exists g ∈ E(R2 ) strongly
equivalent to h, with a disk of modification contained in V and such that
g(α) = h(α ), g(p) = q, g(q) = h(p ).
Figure 3: From h(α) to h(α )
Proof. The basic properties of embeddings which were discussed at the
beginning of Chapter 3??? imply that if ∆ is a disk with α ⊂ int(∆),
∆ ⊂ V then h(∆) = D is also a disk with h(α ) ⊂ int(D). The arcs α and
h(α ) are disjoint and so we can select D small enough so that D ∩ ∆ = ∅.
We apply Lemma 3 and find φ compressing h(α) onto h(α ) and such that
φ = id outside D. The searched map is g = φ ◦ h.
3
Reduction to periodic orbits
Assume that α = pq
is a translation arc for h ∈ E(R2 ). We are going to
associate an index ν = ν(α) to this arc. It takes the values ν = 2, 3, . . . , ∞
4
according to
• ν = 2 whenever h(q) ∈ α
• 3 ≤ ν < ∞ whenever h(q) ∈ α and
hk (α) ∩ α = ∅ if 2 ≤ k < ν − 1,
hν−1 (α) ∩ α = ∅
• ν = ∞ whenever hk (α) ∩ α = ∅, k ≥ 2.
Notice that this definition is consistent because the condition h(q) ∈ α
implies that h2 (α) ∩ α = ∅.
Exercise 5 Compute ν in the following cases
h(x, y) = (x + 1, y),
h(z) = eiΘ z, 0 < Θ ≤ π,
α = [0, 1] × {0}
α = {eiθ : 0 ≤ θ ≤ Θ}.
In the next figure we illustrate two possible situations having the same index.
We will show that, after a free modification, we can reduce to the periodic
Figure 4: Two cases of index ν = 5
case.
Proposition 6 Assume that α is a translation arc for h ∈ E(R2 ) and ν <
∞. Then there exist g ∈ E(R2 ), which is freely equivalent to h, a periodic
orbit with minimum period ν,
P0 , P1 = g(P0 ), . . . , Pν = g ν (P0 ) = P0 ,
5
and a translation arc for g denoted by β = P
0 P1 such that
Γ = β ∪ g(β) ∪ . . . ∪ g ν−1 (β)
is a Jordan curve contained in α ∪ h(α) ∪ . . . ∪ hν−1 (α).
We first state a preliminary result whose proof is immediate.
Lemma 7 Assume that a0 , a1 , . . . , ar−1 are r ≥ 3 different points which are
connected by arcs
γ0 = a
0 a1 , γ1 = a
1 a2 , . . . , γr−1 = a
r−1 a0
satisfying
γ0 ∩ γ1 = {a1 }, . . . , γr−2 ∩ γr−1 = {ar−1 }, γr−1 ∩ γ0 = {a0 }
and
γj ∩ γk = ∅ if 2 ≤ |j − k| < r − 1.
Then Γ = γ0 ∪ γ1 ∪ . . . ∪ γr−1 is a Jordan curve.
Proof of Proposition 6. Assume first that ν = 2 so that the point p =
h(q) lies in α \ {q}. If p = p there is no need to modify h since {p, q} is
a periodic orbit and α ∪ h(α) is a Jordan curve. Assume now that p ∈ α̇,
q. We apply Proposition
then Γ = α ∪ h(α) is a Jordan curve with α = p
2 and compress α to α . In this way we obtain g, strongly equivalent to h
and having the periodic orbit {p , q} and the translation arc β = α .
Assume from now on that 3 ≤ ν < ∞. We define the arcs
α0 = α, α1 = h(α), . . . , αν−1 = hν−1 (α)
and observe that they satisfy
αj ∩ αk = ∅ if 2 ≤ |j − k| < ν − 1
and
α0 ∩ α1 = {h(p)}, α1 ∩ α2 = {h2 (p)}, . . . , αν−2 ∩ αν−1 = {hν−1 (p)}.
This is a consequence of the definition of ν if we use that h is one-to-one
and that α is a translation arc.
The arcs α0 , α1 , . . . , αν−1 are almost in the conditions of Lemma 7,
excepting that α0 and αν−1 can intersect many times. We select p as the
6
Figure 5: p is the first point in αν−1 lying also in α
first point in the arc αν−1 which encounters α0 . In principle the point p
can be anywhere on α \ {q}. In particular it could coincide with p. The
same can be said about the location of p with respect to αν−1 \ {hν−1 (p)}.
From now on we assume that
p = p, hν (p).
This assumption contains the most difficult cases.
Consider the sub-arc of α from p to q,
q
α0 = p
and the sub-arc of αν−1 from hν−1 (p) to p ,
αν−1
= hν−1
(p)p .
are in the conditions of
By construction the arcs α0 , α1 , . . . , αν−2 , αν−1
Lemma 7 and
Γ = α0 ∪ α1 ∪ . . . ∪ αν−2 ∪ αν−1
is a Jordan curve. The successive images of α are also translation arcs for h.
In particular αν−2 is a translation arc with end points hν−2 (p) and hν−2 (q).
Since ν > 2 we know that h(q) ∈ α and this implies that h(hν−2 (q)) ∈ αν−2 .
We are now in the position to apply Proposition 4 with α = αν−2 . We
ν−1 (p), where r is such
= hν−2
(p)r, αν−2
= rh
employ the notations αν−2
7
that h(r) = p . In this way we obtain g1 ∈ E(R2 ) strongly equivalent to h
and such that
.
g1 (αν−2 ) = αν−1
The disk of modification has to contain αν−2
in its interior but we can select
it small enough so that it does not intersect α0 ∪ α1 ∪ . . . ∪ αν−3 . The maps
h and g1 coincide in these arcs and so
g1 (α0 ) = α1 , . . . , g1 (αν−3 ) = αν−2 .
The arc α is also a translation arc for g1 and we can apply Proposition 2
and obtain g ∈ E(R2 ) strongly equivalent to g1 and such that
g(α0 ) = α1 .
The disk of modification is chosen in a neighborhood of α0 which does not
intersect α1 ∪ . . . ∪ αν−2 . This leads to
.
g(α1 ) = α2 , . . . , g(αν−2 ) = αν−1
This completes the proof of the Proposition under the assumption p =
p, hν (p).
Exercise 8 Adapt the previous proof to the cases p = p = hν (p) and p =
hν (p) = p. What can be said about the case p = p = hν (p)?
A consequence of Proposition 6 is that the set of fixed points F = Fix(g) =
Fix(h) is disjoint with Γ. This follows easily from
Exercise 9 Assume that X is a set and f : X → X is a one-to-one mapping. If A is a subset of X without fixed points then f n (A) ∩ Fix(f ) = ∅ for
each n = 0, 1, 2, . . .
The maps h and g are freely equivalent and this implies that
deg(id − h, Ri (Γ)) = deg(id − g, Ri (Γ)).
The curve Γ is not always invariant under g. In fact
g(Γ) = g(β) ∪ . . . ∪ g ν−1 (β) ∪ g ν (β)
and we cannot guarantee that g ν (β) lies in Γ. At this moment it is convenient
to present a short summary of the next steps towards the proof of the Arc
Translation Lemma. The idea will be to construct a continuous deformation
8
Figure 6: A deformation towards the invariant curve
of g, {gt }t∈[0,1] , in such a way that g0 = g and Γ is invariant under g1 . This
will be done in such a way that id − g and id − g1 are homotopic on Ri (Γ)
and so
deg(id − g, Ri (Γ)) = deg(id − g1 , Ri (Γ)).
Finally we will show that the degree inside an invariant curve is 1.
In the next Section we develop the tools to construct the deformation
{gt }.
4
Lemmas on isotopies
The exterior of the unit disk is denoted by
E = {p ∈ R2 : ||p|| ≥ 1}.
The boundary of E is the unit circle ∂E = S1 . The class of mappings
h : E → E which are continuous and one-to-one will be denoted by E(E).
Two maps h0 , h1 ∈ E(E) are isotopic relative to ∂E if h0 = h1 on ∂E and
there exists a continuous map
H : [0, 1] × E → E, (t, p) → Ht (p)
such that
• H0 = h0 , H1 = h1
• Ht ∈ E(E) for each t ∈ [0, 1]
• Ht (p) = h0 (p) = h1 (p) if p ∈ ∂E and t ∈ [0, 1].
9
Lemma 10 (Alexander’s isotopy). Assume that h ∈ E(E) and h = id
on ∂E. Then h is isotopic to id relative to ∂E.
Proof. Define
Ht (p) =
1
t h(tp)
p
if t||p|| ≥ 1
if t||p|| ≤ 1.
To check that the map H satisfies the required properties is more or less
automatic. It is perhaps more interesting to understand this isotopy geometrically. For each t ∈]0, 1] we distinguish the sets
1
1
At = {p ∈ E : 1 ≤ ||p|| ≤ }, Bt = {p ∈ E : ||p|| ≥ }
t
t
and observe that Ht = id on At and Ht = Dt−1 ◦ h ◦ Dt on Bt . Here Dt is the
linear contraction Dt (p) = tp. As the time t grows from 0 to 1 the annulus
At shrinks until collapsing on ∂E.
Figure 7: The isotopy H
We go back to the whole plane R2 and employ the notion of isotopy in
E(R2 ) relative to an arc α. The reader can state the precise definition.
Lemma 11 Let α be an arc in R2 and h ∈ E (R2 ) satisfies h = id on α.
Then h is isotopic to id relative to α.
It is important to observe that we are assuming that h is orientation preserving. Without this assumption the Lemma is false, as it is shown by the
example
h(x, y) = (x, −y), α = [0, 1] × {0}.
10
We justify this assertion by a contradiction argument. If Ht were an isotopy
from h to id relative to α, then Ht (0, 0) = (0, 0) for each t and the homotopy
property of the degree would imply that deg(h, U) = deg(id, U) on any open
and bounded neighborhood of the origin. This is absurd since deg(h, U) =
−1.
Lemma 11 is, according to Brown, a ”Folk Theorem”. Brown sketched
the proof as follows: ”cut” open the arc α to form a disk and apply the
Alexander isotopy to the exterior of the disk. Next we provide the details
of this procedure.
Proof of Lemma 11. First of all we take φ ∈ H(R2 ) mapping α onto the
segment β = [0, 1] × {0}. It is enough to prove the result for g = φ ◦ h ◦ φ−1
and the arc β. We cut the segment to form a disk. This means that we are
going to distinguish the two sides of β in R2 \ β. To this end we consider
the topological space
X = (R2 \ β̇) ∪ {(r, +) : r ∈ β̇} ∪ {(r, −) : r ∈ β̇}.
The points p = (0, 0), q = (1, 0) are the end points of β. The definition
Figure 8: An open cut of the arc
of the topology in X is natural if it must produce the continuity of the
projection π : X → R2 and to make X homeomorphic to E. We just notice
that a sequence {pn } in R2 \ β converges to (r, +) in X if and only if pn → r
and yn > 0 for n large enough, were pn = (xn , yn ).1
Given r ∈ β̇ we consider the disk
D = {p ∈ R2 : ||p − r|| ≤ }.
We can choose small enough so that g(D) is contained in the vertical strip
β × R. This is just a consequence of the continuity of g at r. Define
H + = {(x, y) : y > 0},
1
H − = {(x, y) : y < 0}
The reader who is familiar with Carathéodory’s theory of prime ends will recognize
this construction (see [?]).
11
D+ = D ∩ H + ,
D− = D ∩ H − .
We claim that one of the possibilities below holds:
(i)
g(D+ ) ⊂ H + ,
g(D− ) ⊂ H −
(ii)
g(D+ ) ⊂ H − ,
g(D− ) ⊂ H + .
To prove this claim we first notice that g(D+ ) and g(D− ) cannot intersect
the axis y = 0. This is a consequence of g(D) ⊂ β ×R and g = id on β ×{0}.
The sets g(D+ ) and g(D− ) are connected and contained in H + ∪H − . Each of
them must be either in H + or in H − . It remains to prove that both of them
cannot lie in the same half-plane. To this end we proceed by contradiction.
If g(D+ ) and g(D− ) were in the same half-plane, then r should belong to the
boundary of the disk g(D). But g is an open map and so r ∈ int(D) should
imply r = g(r) ∈ g(int(D)) = int(g(D)). Then r would be simultaneously on
the boundary and in the interior of the disk g(D) and this is the searched
contradiction. Once the claim has been proved it is possible to define a
Figure 9: The orientation reversing case
continuous and one-to-one map
ĝ : X → X
satisfying π ◦ ĝ = g ◦ π. Namely g = ĝ on R2 \ β̇ and ĝ(r, ±) = (r, ±) in the
case (i), ĝ(r, ±) = (r, ∓) in the case (ii).
In the space X we consider the arcs
C± = {(r, ±) : r ∈ β̇} ∪ {p , q }.
The Jordan curve Γ = C+ ∪ C− is invariant under ĝ. In the case (i) all the
points in Γ are fixed. Since the pairs (E, S1 ) and (X, Γ) are homeomorphic
we can apply Lemma 10 and obtain an isotopy Ĥt : X → X between id
and ĝ. This isotopy is relative to Γ and so it induces an isotopy Ht in R2
between id and g. Indeed,
Ht = Ĥt on R2 \ β̇,
12
Ht = id on β.
This argument completes the proof in the case (i). It remains to prove
that the case (ii) cannot hold and it is here where we shall use that h
and g = φ ◦ h ◦ φ−1 are orientation preserving (see Lemma ?? in Chapter
3????). We first notice that the symmetry S(x, y) = (x, −y) induces a
homeomorphism of X which will be denoted by Ŝ. It satisfies S ◦ π = π ◦ Ŝ.
If g is in case (ii) then Ŝ ◦ ĝ = id on Γ. We apply again Lemma 10 but
this time to Ŝ ◦ ĝ and find an isotopy Ĥt : X → X between id and Ŝ ◦ ĝ
and relative to Γ. It induces an isotopy Ht : R2 → R2 between id and S ◦ g
which is relative to β. Given any open and bounded neighborhood U of
the origin we observe that deg(S ◦ Ht , U) is independent of t. This is so
because (S ◦ Ht )−1 (0) = 0 and one can apply the invariance by homotopies.
In particular we observe that S ◦ H1 = g, S ◦ H0 = S and so
deg(g, U) = deg(S, U) = −1.
This is excluded since g belongs to E (R2 ).
5
Invariant curves and computation of the degree
The goal of this Section is to prove the following result.
Proposition 12 Assume that h ∈ E (R2 ) and γ is an arc in R2 such that
Fix(h) ∩ γ = ∅ and Γ = γ ∪ h(γ) is a Jordan curve. Then Fix(h) ∩ Γ = ∅
and
deg(id − h, Ri (Γ)) = 1.
Up to now arcs have been denoted by α and now we have changed to γ.
The intention is to indicate that γ is not necessarily a translation arc. For
instance, if h is a rotation of 90 degrees then any arc γ = {eiθ : θ ∈ [0, Θ]}
with 2π > Θ ≥ 3π
2 is in the conditions of the Proposition but it is not a
translation arc.
Exercise 13 Construct an example showing that the above result cannot be
extended to orientation reversing embeddings. Hint: the examples presented
in Chapter 3???????? can be useful.
Given h ∈ E(R2 ) we say that Γ ⊂ R2 is an invariant curve if it is a Jordan
curve satisfying h(Γ) = Γ. We say that the invariant curve has no fixed
points if Fix(h) ∩ Γ = ∅. Before proving the Proposition 12 we need to
present some properties of invariant curves.
13
Lemma 14 Assume that h ∈ E(R2 ) and Γ is an invariant curve without
fixed points. Then
deg(id − h, Ri (Γ)) = 1.
Proof. The disk Γ ∪ Ri (Γ) is transformed onto the unit disk D via some
φ ∈ H(R2 ). The embedding ĥ = φ◦h◦φ−1 has the unit circle as an invariant
curve and Fix(ĥ) ∩ ∂D = ∅. We define the homotopy
H(p, λ) = p − λĥ(p)
and observe that it cannot vanish if p ∈ ∂D and λ ∈ [0, 1]. The key fact is
that ||ĥ(p)|| = 1 for every p ∈ ∂D. The topological character of the fixed
point index together with the homotopy property imply that
deg(id − h, Ri (Γ)) = deg(id − ĥ, int(D)) = deg(id, int(D)) = 1.
In the previous Lemma we did not impose the preservation of orientation
for h, however this is a consequence of the assumptions.
Lemma 15 Assume that h ∈ E(R2 ) has an invariant curve without fixed
points. Then h ∈ E (R2 ).
To prepare the proof of this Lemma we recall some facts about homeomorphisms of the circle.
Given k ∈ H(S1 ) there exists a lift k̃ : R → R such that
k(eiθ ) = eik̃(θ) for each θ ∈ R.
The lift is a homeomorphism of the real line and it satisfies
k̃(θ + 2π) = k̃(θ) ± 2π
where the sign + appears for increasing k̃ and the sign − for decreasing k̃.
If we assume that k has no fixed points then
k̃(θ) − θ − 2kπ = 0
for each θ ∈ R and k ∈ Z. This is impossible if k̃ is decreasing and so we
conclude that k̃ is increasing if Fix(k) = ∅.
Proof of Lemma 15. Assume first that Γ = S1 . The restriction of h to S1
produces a map k ∈ H(S1 ). By assumption it has no fixed points and so
h(eiθ ) = eik̃(θ)
14
with k̃ : R → R an increasing homeomorphism. Consider the map in the
plane
ĥ(reiθ ) = reik̃(θ) .
It belongs to H(R2 ) and satisfies
h = ĥ
on S1 .
In particular,
deg(h, int(D)) = deg(ĥ, int(D)).
We construct an isotopy between ĥ and id which is easily expressed in polar
coordinates,
θ1 = tθ + (1 − t)k̃(θ)
Ht :
r1 = r.
The map Ht is a homeomorphism because k̃ is increasing. We also observe
that H0 = ĥ, H1 = id and Ht−1 (0) = 0 for each t. The homotopy property
implies that
deg(ĥ, int(D)) = deg(id, int(D)) = 1.
This completes the proof for Γ = S1 .
Assume now that Γ is an arbitrary Jordan curve. Then we can find
φ ∈ H(R2 ) with φ(Γ) = S1 . The unit circle becomes an invariant curve
without fixed points for the map φ ◦ h ◦ φ−1 . The previous discussions imply
that φ ◦ h ◦ φ−1 is in E (R2 ) and this property is transferred to h via Lemma
?? in Chapter 3?????????.
Proof of Proposition 12. This result would be a consequence of Lemma
14 if Γ were invariant. In general Γ will not be invariant since h(γ) can
be mapped outside Γ. From the Exercise 9 we know that h has no fixed
points on Γ. Therefore we can construct a homeomorphism of Γ, say k :
Γ → Γ, without fixed points and such that k = h on γ. To do this we
have some freedom. In fact it is sufficient to observe that we can map
the arc σ = Γ − γ̇ homeomorphically onto σ1 = Γ \ h(γ̇). Of course the
end points must be transformed appropriately so that the resulting k is
continuous. By construction we know that k has no fixed points on γ and so
we can apply again Exercise 9 and conclude that there are no fixed points
on γ ∪ k(γ) = Γ. Next we apply the Jordan-Schoenflies Theorem and extend
k to a homeomorphism φ ∈ H(R2 ). The curve Γ is invariant under φ and
fixed point free. We can apply Lemma 15 and deduce that φ ∈ H (R2 ).
From Lemma ?? we deduce that h ◦ φ−1 ∈ E (R2 ) and it is at this moment
that we have used the assumption h ∈ E (R2 ). The map h ◦ φ−1 coincides
15
with the identity on γ and so it is possible to apply Lemma 11 and find an
isotopy {Gt } with G0 = φ−1 ◦ h, G1 = id, Gt = id on γ. The family of maps
Ht = φ ◦ Gt define an isotopy between h and φ. Since Ht = φ = h on γ we
know that Ht has no fixed points in γ. Once again we apply Exercise 9 to
deduce that Ht has no fixed points on γ ∪ Ht (γ) = γ ∪ φ(γ) = γ ∪ k(γ) = Γ.
In particular
deg(id − h, Ri (Γ)) = deg(id − φ, Ri (Γ)).
The conclusion follows after applying Lemma 14 to φ.
6
Brouwer’s Lemma and some consequences
Let us prove Brouwer’s Lemma as stated in Chapter 3???????. We are
given h ∈ E (R2 ) and a translation arc α with hn (α) ∩ α = ∅ for some
n ≥ 2. This means that the index of this arc is finite, say ν ≥ 2. We
apply Proposition 6 and find g ∈ E(R2 ), freely equivalent to h, with a
periodic orbit P0 , P1 , . . . , Pν−1 and a translation arc β = P
0 P1 such that Γ =
ν−1
β ∪g(β)∪. . .∪g
(β) is a Jordan curve contained in α∪h(α)∪. . .∪hν−1 (α).
We notice that the equivalence of h and g implies that they coincide outside a
compact set and so also g belongs to E (R2 ). Next we shall apply Proposition
12 to the map g and the arc γ = β ∪ g(β) ∪ . . . ∪ g ν−2 (β). We observe that
Fix(g) ∩ γ = ∅ because γ is a translation arc. Also, γ ∪ g(γ) = Γ is a Jordan
curve and so we conclude that deg(id − g, Ri (Γ)) = 1. The properties of
equivalent embeddings imply that deg(id − g, Ri (Γ)) = deg(id − h, Ri (Γ))
and this completes the proof.
The previous proof provides extra information about the curve Γ. It
satisfies
Γ ⊂ α ∪ h(α) ∪ . . . ∪ hν−1 (α).
This fact can be useful for the location of fixed points. We finish with the
following consequence.
Theorem 16 Assume that h ∈ E (R2 ) and α is a translation arc with finite
index ν ≥ 2. Then h has a fixed point in some bounded component of
R2 \ (α ∪ h(α) ∪ . . . ∪ hν−1 (α)).
Proof. We know that h has a fixed point p in Ri (Γ) with Γ ⊂ α ∪ h(α) ∪
. . . ∪ hν−1 (α). We will prove that p belongs to some connected component
of R2 \ [α ∪ h(α) ∪ . . . ∪ hν−1 (α)]. Otherwise p and ∞ should lie in the same
bounded component of S2 \ [α ∪ h(α) ∪ . . . ∪ hν−1 (α)], say Ω. Here we are
using the Riemann sphere S2 = R2 ∪ {∞}. The connected set Ω is contained
16
in S2 \ Γ and so, by Jordan’s Theorem, p should lie in Re (Γ). This is absurd
since we started with a fixed point p lying in Ri (Γ).
References
[1]
17