Now that we have constructed homology groups, it is natural to ask what sorts of information these “algebraic pictures” of spaces can yield. This unit describes some of the most basic things that can be done with the subject. The importance of homology groups in analyzing homotopy classes of maps from one space to another are illustrated by two fundamental results whose proofs appear in most comprehensive (as opposed to introductory) texts on algebraic topology, and they can be found in Hatcher.
SPECIAL CASE OF HOPF’S THEOREM. LetP be a finiten-dimensional polyhedron such that Hn−1(P) has no elements of finite order. Then there is a 1−1 correspondence between the
set of homotopy classes[P, Sn] and the algebraic homomorphisms from H
n(P) to Hn(Sn)∼=Z.
There is also a version of Hopf’s Theorem forn-dimensional polyhedra for whichHn−1(P) has
elements of finite order, but we do not have the background needed to state it here. Since the result obviously also holds if P is merely homeomorphic to a polyhedron, it follows that two continuous maps from Sn to itself are homotopic if and only if they induce the same homomorphism from Hn(Sn) ∼= Z to itself; such a homomorphism is determined by its value on a generator and thus
determines a number called thedegree. We shall look at this concept further in Section V.1. SIMPLY CONNECTED CASE OF J. H. C. WHITEHEAD’S THEOREM. Suppose that P and Qare finite simply connectedpolyhedra andf :P →Qis a continuous map such that for each i≥0 the induced map of homology f∗ :Hi(P)→Hi(Q) is an isomorphism. Thenf is a
homotopy equivalence.
The converse is an immediate consequence of the functoriality and homotopy invariance of homology groups. There are versions of Whitehead’s Theorem for connected finite polyhedra that are not simply connected, but once again we do not have the background needed to formulate such a result here. However, it is important to note that the non-simply connected case requires stronger hypotheses than the condition that f defines isomorphisms of ordinary homology groups (specifi- cally, one needs to know that f induces an isomorphism of fundamental groups and isomorphisms on the homology groups of the universal covering spaces forP andQ).
V.1 : Degree theory (Hatcher,§ 2.2)
Definition. Ifn >0 andf :Sn→Sn is a continuous mapping, then thedegreeoff is the unique
integer dsuch that the mapf∗:Hn(Sn)→Hn(Sn) is multiplication byd (recall thatHn(Sn)∼=Z
and every homomorphism of the latter to itself is multiplication by some integer). Several properties of the degree are immediate:
(1) If f is the identity, then the degree of f is 1. (2) If f is a constant map, then the degree of f is 0. (3) If f and g are homotopic, then their degrees are equal.
(4) If f and g are continuous maps from Sn to itself, then the degree offog is equal to the
degree off times the degree ofg.
(5) If h is a homeomorphism of Sn to itself, then the degree of h and h−1 is ±1, and the degree ofhofoh−1 is equal to the degree off.
(6) If n= 1 and f(z) =zm (complex arithmetic), then the degree of f is equal tom.
The last property is the only one which is nontrivial. It follows because (a) the map f∗ from π1(S1,1)∼=Zis multiplication bym, (b) the Hurewicz map fromπ1(S1,1) toH1(S1) is an isomor- phism, (c) the Hurewicz map defines a natural transformation of functors from the fundamental group to 1-dimensional singular homology.
For all n≥2, there is a standard recursive process for constructing continuous maps from Sn to itself with arbitrary degree.
PROPOSITION 1. Let f : Sn−1 → Sn−1 be a continuous mapping of degree d, and let Σ(f) :Sn→Sn be defined on(x, t)∈Sn ⊂Rn×R by
Σ(f) x, t = p1−t2f(x), t . Then the degree ofΣ(f) is also equal tod.
COROLLARY 2. Ifn≥1andd is an arbitrary integer, then there exists a continuous mapping g:Sn→Sn whose degree is equal tod.
The case n= 1 of the corollary is just (6), above, and the proposition supplies the inductive step to show that if the corollary is true for (n−1) then it is also true forn.
Proof of Proposition 1. We should check first that the map Σ(f) is continuous. This is immediate from the formula for all points except the north and south poles, and at the latter one can check directly that if ε >0 then we can take δ =ε.
DefineDn
+ andDn− to be the subsets ofSn on which the last coordinates are nonnegative and nonpositive respectively. It follows immediately thatSn is formed fromSn−1 by attaching twon- cells corresponding to Dn
±. This and the vanishing of the homology of disks in positive dimensions imply that all the arrows in the diagram below are isomorphisms:
H∗−1(Sn−1)←H∗(D+n, Sn−1)→H∗(Sn, Dn−)←H∗(S
n
)
Furthermore, the mappings f and Σ(f) determine homomorphisms from each of these homology groups to themselves such that the following diagram commutes:
Hn−1(Sn−1) ∼ = ←−−−−− Hn(Dn+, Sn−1) ∼ = −−−−−→ Hn(Sn, D−n) ∼ = ←−−−−− H∗(Sn) yf∗ yΣ(f)∗ yΣ(f)∗ yΣ(f)∗ Hn−1(Sn−1) ∼ = ←−−−−− Hn(Dn+, Sn−1) ∼ = −−−−−→ Hn(Sn, D−n) ∼ = ←−−−−− H∗(Sn)
It follows immediately that the degrees of f and Σ(f) must be equal. Here is another basic property:
Proof. If the image off does not include some point p, thenf∗ has a factorization of the form Hn(Sn) → Hn(Sn− {p}) → Hn(Sn)
and this homomorphism is trivial because the middle group is zero.
Linear algebra and degree theory We shall start with orthogonal transformations.
PROPOSITION 4. Suppose thatT is an orthogonal linear transformation ofRn, where n≥2,
and let fT : Sn−1 → Sn−1 be the corresponding homeomorphism of Sn−1. Then the degree of fT
is equal to the determinant ofT.
Sketch of proof. We shall use a basic fact about orthogonal matrices; namely, if A is an orthogonal matrix then there is another orthogonal matrix B such that B ·A·B−1 is equal to a block sum of 2×2 rotation matrices plus a block sum of 1×1 matrices such that at most one of the latter has an entry of −1 (and the rest must have entries of 1).
Every 2×2 rotation matrix can be joined to the identity by a path consisting entirely of 2×2 rotation matrices. Therefore it follows that fT is homotopic to fS, whereS is a diagonal matrix
with at most one entry equal to −1 and all others equal to 1. Clearly the degrees offS and fT are
equal, and likewise the determinants of S and T must be equal (by continuity of the determinant and the fact that its value for an orthogonal matrix is always ±1). Thus the proof reduces to showing that the degree of fS is equal to−1 if there is a negative diagonal entry and is equal to 1
if there are no negative diagonal entries. — In fact, the second statement is obvious since T and fT are identity mappings in this case.
Therefore everything reduces to showing that the degree of fS is equal to −1. We can use
Proposition 2 to show that the result is true for all n if it is true for n = 2, and the truth of the result when n = 2 follows immediately from Property (6) of degrees that was stated at the beginning of this document.
We shall now consider an arbitrary invertible linear transformationT from Rn to itself. Such
a map is a homeomorphism and thus extends to a mapT• of one point compactifications fromSn
to itself.
THEOREM 5. In the setting above, the degree ofT•is equal to the sign of the determinant of T.
The proof of this result requires some additional input.
LEMMA 6. Suppose that we are given a continuous curve Tt defined fort ∈ [0,1] and taking
values in the set of all invertible linear transformations on Rn (equivalently, invertible n ×n
matrices). ThenT•
0 is homotopic toT1•.
Proof of Lemma 6. We would like to define a homotopy by the formulaHt =Tt•, and we can
do so if and only if the latter is continuous at every point of{∞} ×[0,1]. The latter in turn reduces to showing the following: For each t ∈ [0,1] and M > 0 there are numbers δ >0 and P > 0 such that|s−t|< δ and |v| ≥P imply |Ts(v)| ≥M.
LetkTkbe the usual norm of a linear transformation given by the maximum value of|T|on the unit sphere. It follows immediately that the norm is a continuous function in (the matrix entries associated to) T. It follows that
and since the inverse operation is also continuous it follows thatkT−1
s kis a continuous function ofs.
In particular, if kTt−1k=B >0 then we can findδ >0 such that|s−t|< δ implies kTs−1k> B/2,
and hence if|v|>2M/B and |t−s|< δ thenTs(v)| ≥M, as required.
Proof of Theorem 5. Both the degree ofT•and the sign of the determinant are homomorphisms from invertible matrices to { ±1}, and therefore it will suffice to prove the theorem for a set of linear transformations which generate all the invertible linear transformations. Not surprisingly, we shall take this set to be the linear transformations given by the elementary matrices.
LetEi,j denote then×n matrix which has a 1 in the (i, j) entry and zeros elsewhere. Then
the function sendingt∈ [0,1] to I+tEi,j defines a curve from the elementary matrix I+Ei,j to
the identity. Therefore the associated linear transformation determines a map which is homotopic to the identity, and consequently the degree and determinant sign agree for elementary linear transformations given by adding a multiple of one row to another.
Similarly, ifD(k, r) is a diagonal matrix which has ones except in thekth position and a positive real numberrin the latter position, then there is a continuous straight line curve joining the matrix in question to the identity, and this matrix takes values in the group of invertible diagonal matrices. It follows that the degree and determinant sign agree for elementary linear transformations given by multiplying one row by a positive constant.
We are now left with elementary matrices given by either multiplying one row by −1 or by interchanging two rows. These two types of matrices are similar, so both the degrees and determinant signs are equal in each case. Therefore it will suffice to check that the degree and determinant sign agree when one considers an elementary matrix given by multiplying a single row by −1.
By Proposition 2 and the invariance of our numerical invariants under similarity, it will suffice to consider the case wheren= 2 and we are multiplying the second row by−1. LetW ⊂R2 be the open disk of radius 2 about the origin, so that there is a canonical homeomorphism fromW − {0}
to S1×(0,2). Now the map T• sends S2− {0} to itself and likewise forW and S1. Excision and homotopy invariance now yield the following chain of isomorphic homology groups:
H1(S1) ← H1(W − {0}) → H2(W, W − {0}) ← H2(S2, S2− {0}) −→ H2(S2)
As in Proposition 3, one has associated maps of homology groups to form a corresponding com- mutative diagram, and from this diagram one sees that the degree of T• is equal to the degree of the map determined byT• onS1. Since the map onS1 is merely the mapping sendingz toz−1, it follows that the degree is equal to−1, and of course this is the same as the sign of the determinant.
The Fundamental Theorem of Algebra
One can use degree theory to prove the Fundamental Theorem of Algebra. All proofs of the latter involve some analysis and plane topology, and one advantage of the degree-theoretic proof is that the role of topology is particularly easy to recognize. This proof can also be generalized to obtain a generalization of the Fundamental Theorem of Algebra to polynomials with quaternionic coefficients (this was done by Eilenberg and Niven in the nineteen forties).
We start with an argument that is similar to the proof in the last part of Theorem 5.
PROPOSITION 7. The mapψm of the complex plane sendingz to zm (where m is a positive integer) extends continuously to a map of one point compactifications sending the point at infinity to itself, and the degree of the compactified map is equal to m.
Proof. The existence of a continuous extension follows because ifM >0 then|z|> M1/m implies |zm|> M.
It follows that ψm sends C− {0} to itself. Of course, the map also sends S1 to itself and this map has degree m, so a diagram chase plus the naturality of the Hurewicz homomorphism imply that ψm
∗ is multiplication by m on H1(C− {0}) ∼=Z. Diagram chases now show that ψ∗ is multiplication by mon
H2(C,C− {0}) ∼= H2(S2, S2− {0}) ∼= H2(S2) and thus the degree of the compactified map is equal tom.
The following result is standard.
PROPOSITION 8. If p is a nonconstant monic polynomial, then p extends continuously to a map of one point compactifications sending the point at infinity to itself.
Sketch of proof. We need to show that if M > 0 then there is some ρ > 0 such that |z| > ρ implies |p(z)|> M. One easy way of doing this is to begin by writing pas follows:
p(z) = zm·1 + an−1
z + · · · + a0 zn
If we write the expression inside the parentheses as 1 +b(z), then it is clear that if|z|is sufficiently large (say |z| > N) then |b(z)| < 12. It follows immediately that if M >0 and |z| >2M1/m+N then|p(z)|> M.
The Fundamental Theorem of Algebra will now be a consequence of Proposition 3 and the following generalization of Proposition 8:
PROPOSITION 9. If p is a nonconstant monic polynomial of degree m≥1, then the degree of the compactified mapp• is equal tom.
Proof. It will suffice to show thatp• is homotopic to (ψm)•.
Define a homotopy fromψm to pon the set where |z| ≥N+ 1 by ht(z) =zm(1 +t b(z) ). By
the Tietze Extension Theorem, one can extend this to a homotopy over all ofC. As in the previous argument, ifM >0 and |z| >2M1/m+N + 1 then |ht(z)| > M for all t. One can then argue as
in the first paragraph of the proof of Lemma 6 to show that p• is homotopic to (ψm)•.
V.2 : Classical theorems of Jordan and Brouwer (Hatcher, § 2.B; Munkres, 61–64)
Most of this has become standard in algebraic topology texts, and we shall quote Hatcher as appropriate. The following result corresponds to the first half of Proposition 2B.1 on page 169 of that reference.
PROPOSITION 1. If A ⊂ Sn is homeomorphic to Dk for some k < n, then the H i(A) is
infinite cyclic ifi= 0 and trivial otherwise.
Since Hatcher’s statement involves reduced homology and this concept has not yet been dis- cussed in these notes, we shall do so now. There are (at least) two ways of looking at the reduced
homology of a space X. If P is a space with one point andc : X →P is the constant map, then the reduced homology H∗(Xe ) may be viewed as the kernel of the homomorphism c∗ in homology. If X is nonempty and b:P → X maps the point inP to an arbitrary point in X, thencob is the
identity onP, and it follows that there is a direct sum decomposition H∗(X) ∼= He∗(X) ⊕ H∗(P) . This has the following consequences:
(1) If i6= 0, thenHi(X)∼=He∗(X).
(2) If i−0, then Hi(X) =∼ H∗(X)e ⊕Z. — In particular, X is arcwise connected if and
only ifH0(X) is trivial.e
It follows immediately that ifX is a nonempty space and b∈X, then the reduced homology of X is isomorphic to the homology of the pair (X,{b}) (verify this!). Using this description, one can prove the following result which is needed in Hatcher’s (standard) proof of Proposition 1:
PROPOSITION 2. (Reduced Mayer-Vietoris Sequence in singular homology) Let X be a topological space, and letU and V be open subsets such that X =U∪V and U∩V is nonempty. Denote the inclusions ofU andV inX byiU andiv respectively, and denote the inclusions ofU∩V
inU and V by gU and gV respectively. Then there is a long exact sequence as in Theorem IV.4.5
in which ordinary homology groups are replaced by reduced homology groups.
Sketch of proof. Letb∈U ∩V. Then there is a short exact sequence of chain complexes 0−→S∗(U ∩V,{b})−→S∗(U,{b})⊕S∗(V,{b})−→S∗U(X,{b})−→0
analogous to the one which appears in the proof of Theorem IV.4.5, and the long exact homol- ogy sequence of this short exact sequence of chain complexes will be the reduced Mayer-Vietoris sequence.
Note on the proof of Proposition 1. In order to use the relative Mayer-Vietoris sequence it is necessary to know from the start thatAis a proper subset ofSn; however,A cannot be equal to
Sn because the homology groups of A and Sn are not isomorphic.
We shall state the Jordan-Brouwer Separation Theorem in a slightly more detailed version than the one in Hatcher:
THEOREM 3. (Jordan-Brouwer Separation Theorem. ) Let n≥2, and suppose that A⊂Sn
is homeomorphic to Sn−1. Then Sn−A contains two components, and A is the frontier of each
component.
Note on the proof. The existence of two components is shown in the second half of Hatcher’s previously cited Proposition 2B.1 (q.v.).
It remains to prove that points ofAare limit points of each components. Suppose that Sn−A
is the union of the two open, connected, disjoint subsetsU and V.
Assume that not every point ofAis a limit point of bothU andV. Without loss of generality, it is enough to consider the case wherex∈Ais not a limit point ofV. Sincex6∈V, it follows that