8 VIN DE SILVA, JOEL W. ROBBIN AND DIETMAR A. SALAMON
Definition 2.1 (Traces). Fix two (not necessarily distinct) intersection points
x, y α β.
(i) Let w : Σ \ β) Z be a two-chain. The triple Λ = (x, y, w) is called an
(α, β)-trace if there exists an element u D(x, y) such that w is given by (1).
In this case Λ =: Λu is also called the (α, β)-trace of u and we sometimes write
wu := w.
(ii) Let Λ = (x, y, w) be an (α, β)-trace. The triple ∂Λ := (x, y, ∂w) is called the
boundary of Λ.
(iii) A one-chain ν : \ β) \ α) Z is called an (x, y)-trace if there exist
smooth curves γα : [0, 1] α and γβ : [0, 1] β such that γα(0) = γβ(0) = x,
γα(1) = γβ(1) = y, γα and γβ are homotopic in Σ with fixed endpoints, and
(3) ν(z) =
deg(γα,z), for z α \ β,
deg(γβ,z), for z β \ α.
Remark 2.2. Assume Σ is simply connected. Then the condition on γα and
γβ to be homotopic with fixed endpoints is redundant. Moreover, if x = y then a
one-chain ν is an (x, y)-trace if and only if the restrictions
να := ν|α\β, νβ := −ν|β\α
are constant. If x = y and α, β are embedded circles and A, B denote the positively
oriented arcs from x to y in α, β, then a one-chain ν is an (x, y)-trace if and only if
να|α\(A∪β) = να|A\β 1
and
νβ|β\(B∪α) = νβ|B\α 1.
In particular, when walking along α or β, the function ν only changes its value at
x and y.
Lemma 2.3. Let x, y α∩β and u D(x, y). Then the boundary of the (α, β)-
trace Λu of u is the triple ∂Λu = (x, y, ν), where ν is given by (2). In other words,
if w is given by (1) and ν is given by (2) then ν = ∂w.
Proof. Choose an embedding γ : [−1, 1] Σ such that u is transverse to
γ, γ(t) Σ \ β) for t = 0, γ(−1), γ(1) are regular values of u, γ(0) α \ β
is a regular value of u|D∩R, and γ intersects α transversally at t = 0 such that
orientations match in
Tγ(0)Σ = Tγ(0)α R ˙ γ (0).
Denote Γ := γ([−1, 1]). Then
u−1(Γ)
D is a 1-dimensional submanifold with
boundary
∂u−1(Γ)
=
u−1(γ(−1))

u−1(γ(1))

(
u−1(γ(0))
R)
)
.
If z u−1(Γ) then
im du(z) + Tu(z)Γ = Tu(z)Σ,
Tzu−1(Γ)
=
du(z)−1Tu(z)Γ.
We orient
u−1(Γ)
such that the orientations match in
Tu(z)Σ = Tu(z)Γ
du(z)iTzu−1(Γ).
In other words, if z
u−1(Γ)
and u(z) = γ(t), then a nonzero tangent vector
ζ
Tzu−1(Γ)
is positive if and only if the pair γ (t),du(z)iζ) is a positive basis
of Tγ(t)Σ. Then the boundary orientation of
u−1(Γ)
at the elements of
u−1(γ(1))
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