4.4.16 Remark The “Mayer-Vietoris principle” indeed yields a long exact sequence in cohomology. One nice application is to show that the de Rham cohomology spaces of a compact manifold are always finite-dimensional.
4.5. DEGREE THEORY 109 4.5.2 Lemma Letωbe ann-form onRnwith support contained in the open cube Csuch thatR
Rnω = 0. Then there exists an(n−1)-formηonRnwith support contained inCsuch thatdη=ω.
Proof. Poincaré’s lemma (4.4.6) yieldsηwith dη =ω but does not give information about the support ofη. Instead, writeω=f dx1∧ · · · ∧dxnfor f ∈C∞(Rn). Thensuppf ⊂CandR
Rnf dx1· · ·dxn = 0, sof =P
i ∂fi
∂xi as in Lemma 4.5.1, and thusω =dηwhereη =P
i(−1)i−1fidx1∧ · · · ∧dxˆi∧
· · · ∧dxn.
4.5.3 Proposition If M is a compact connected orientable smooth manifold of dimensionn, thenHn(M) =R.
Proof. By compactness, there is a finite cover{U1, . . . , Um}by coordinate neighborhoods diffeomorphic to the open cubeC. Letω0be a bumpn-form as in Example 4.3.3(c) with support contained inU1 and total integral 1.
Thenω0 defines a non-zero cohomology class in Hn(M). We shall prove that anyn-formωonM is cohomologous to a multiple ofω0, namely,ω = c ω0 +dη for somec ∈ R and some (n−1)-formη. Using a partition of unity{ρi}subordinate to{Ui}, we can writeω=Pm
i=1ρiωwhereρiωis an n-form with support inUi. By linearity, it suffices to prove the result for ρiω, so we may assume from the outset that the support ofω is contained inUk, for somek= 1, . . . , m.
Owing to the connectedness ofM, we can find a chainUi1, . . . , Uir such thatUi1 =U1,Uir =UkandUij ∩Uij+1 6=∅for allj= 1, . . . , r−1. For all j = 1, . . . , r−1, choose ann-formωj with support inUij ∩Uij+1 and total integral1. Nowω0−ω1has support inUi1 =U1 and total integral zero, so by Lemma 4.5.2, there existsη1with support inU1such that
ω0−ω1 =dη1.
Next,ω1−ω2has support inUi2 and total integral zero, so the lemma yields η2with support inUi2 such that
ω1−ω2 =dη2.
Continuing, we findηj with support inUij such that ωj−1−ωj =dηj
for allj= 1, . . . , r−1. Adding up, we get ω0−ωr−1=dη whereη=Pr−1
j=1ηj. NowUir =Ukcontains the support ofωandωr−1, and ω−cωr−1 has total integral zero, wherec=R
Mω. By applying the lemma again,
ω−cωr−1 =dζ
and hence
ω =cω0+d(ζ−cη)
as required.
4.5.4 Corollary LetM be a compact connected oriented smooth manifold of di-mensionn. Then “integration overM”
Z
M
:Hn(M)→R
is a well defined linear isomorphism which is positive precisely on the cohomology classes defined by nowhere vanishingn-forms belonging to the orientation ofM.
Proof. By Stokes’ formula, the integral of an exact form is zero, so the integral of ann-form depends only on its cohomology class and thus the map is well defined. By the theorem,Hn(M)is one dimensional and there exist bumpn-forms with non-zero integral, so the map is an isomorphism.
Letωbe a nowhere vanishingn-form belonging to the orientation ofM, choose an oriented atlas{(Uα, ϕα = (xα1, . . . , xαn))}and a partition of unity {ρα}subordinate to{Uα}. Thenω=P
αραω, whereραωhas support inUα and on which its local representation is of the formfαdxα1 ∧ · · · ∧dxαnfor a non-negative smooth functionfαonUα. It follows that
Z
M
ω =X
α
Z
ϕα(Uα)
(fα◦ϕ−1α )dx1· · ·dxn>0
sincefα ≥ 0 and it is positive somewhere. Conversely, ifω′ is ann-form withR
Mω′ >0, thenω′is cohomologous tocω, wherec=R
Mω′/R
Mω >0, andcωandωare nowhere vanishingn-forms defining the same orientation
onM.
Letf :M →N be a smooth map between compact connected oriented manifolds of the same dimension n. Let ωM, ωN be n-forms on M, N, respectively, with total integral one. Thenf∗ : Hn(N) → Hn(M) carries [ωN] to a multiple of [ωM]; this number is called the degreeof f, denoted degf. It follows from Proposition 4.4.1 that homotopic maps have the same degree.
4.5.5 Remark In caseN =Sn, Hopf’s degree theorem [GP10, Chap. 3, §6]
asserts that non-homotopic maps have different degrees. For the casen= 1, see Problem 22.
4.5.6 Proposition Letf :M →N be a smooth.
a. The degree off is an integer.
b. For allω∈Ωn(N), Z
M
f∗ω= (degf) Z
N
ω
4.5. DEGREE THEORY 111 c. Ifq∈N is a regular value off, then
degf = X
p∈f−1(q)
sgn(detdfp) (finite sum)
Proof. (b) follows from the commutativity of the diagram
Hn(N) f∗
> Hn(M)
R R
N∨
degf >R R
∨ M
and (a) follows from (c). Let us prove (c).
Consider first the case in whichqis a point outside the image off. Since f(M)is compact, we can find a bumpn-formαonN with total integral one and support disjoint fromf(M). It follows from (b) thatdegf =R
Mf∗α= 0. Sincef−1(q) =∅, (c) is proved in this case.
Suppose now q lies in the image of f. Since q is a regular value and dimM = dimN,f is a local diffeomorphism at eachp ∈f−1(q). In partic-ular,f−1(q)is discrete and thus finite, due to the compactness ofM. Write f−1(q) = {p1, . . . , pm}and choose open neighborhoods U˜i of pi andVi of q such that f : ˜Ui → Vi is a diffeomorphism for all i = 1, . . . , m. Setting V =∩mi=1ViandUi = ˜Ui∩f−1(V), nowf :Ui →V is a diffeomorphism for alli. Moreover,f(M \ ∪mi=1U˜i)is a compact subset ofN disjoint fromq, so by further shrinkingV we can ensure thatf−1(V) =∪mi=1Ui.
Letαbe ann-form onN with total integral one and support contained in V. Then f∗α is an n-form on M with support in ∪mi=1Ui. In view of Exercise 4.1.8
Z
Ui
f∗α= sgn(detdfpi) Z
V
α= sgn(detdfpi)
where we consider the determinant of the Jacobian matrix off atpirelative to orientation-preserving local charts around pi and q, so its sign is +1 if dfpi :TpiM → TqN preserves orientation and−1if it reverses orientation.
It follows that degf =
Z
M
f∗α= Xp
i=1
Z
Ui
f∗α= Xp
i=1
sgn(detdfpi),
as desired.
4.5.7 Corollary The degree of a non-surjective map is zero.
4.5.8 Remark Sard’s theorem is the statement that the set of critical values of a smooth map has measure zero, see [GP10, App. 1] for a proof. In par-ticular the set of regular values is open and dense. It follows from Sard’s theorem that in Proposition 4.5.6 there always exists a regular value off.
More generally, ifMhas finitely many connected componentsM1, . . . , Mr, the degree off :M → N can still be defined as the sum of the degrees of the restrictionsf :Mi →N, and Proposition 4.5.6 remains true.
4.5.9 Example ConsiderS1 as the set of unit complex numbers. Thenf : S1 → S1 given byf(z) = zn is smooth and has degree n, which we can show as follows. Recall the angular form dθ = −y dx +x dy generates H1(S1). Removal of one point does not change the integral below on the left hand side, and h : (0,2π) → S1 \ {1}, h(x) = eix is an orientation-preserving diffeomorphism, so
Z
S1
f∗dθ = Z 2π
0
h∗f∗dθ= Z 2π
0
(f ◦h)∗dθ
where(f ◦h)∗dθis exact on(0,2π)and in fact equal ton dxby direct com-putation. Therefore
Z
S1
f∗dθ = Z 2π
0
n dx= 2πn=n· Z
S1
dθ, as we wished.
4.5.10 Example Let f : S1 → R2 be a smooth map. Its image is a closed curve in the plane. Fix a point q not in the curve. The winding number W(f, q)off with respect toqis the degree of the mapu:S1→S1given by
u(x) = f(x)−q
||f(x)−q||.
Note thatW(f, q1) = W(f, q2)ifq1 andq2 lie in the same connected com-ponent of the complement of the image off.
Introducing the complex variablez=x+iywe have
−y dx+x dy x2+y2 =ℑ
1 zdz
(compare Examples 4.3.3(d)). Using this formula, it is easy to arrive at the complex integral for the winding number,
(4.5.11) W(f, q) = 1
2πi Z
C
dz z−q whereCis the image off (Cauchy 1825).
4.6. THE BORSUK-ULAM THEOREM 113