MAS61015 Algebraic Topology

16. Applications of homology

Video (Theorems 16.1 and 16.2)

Theorem 16.1.

No sphere Sn is contractible. Moreover, if nm then Sn is not homotopy equivalent to Sm.


As homotopy equivalent spaces have isomorphic homology, it will suffice to prove that Hn(Sn)≄Hn(point) and that Hn(Sm)≄Hn(Sm) when nm. This is clear from the above calculations. ∎

Theorem 16.2.

Suppose that n,m0 with nm. Then n is not homeomorphic to m.

Note here that n and m are certainly homotopy equivalent, as they are both contractible. This is perfectly consistent: homeomorphism implies homotopy equivalence but not conversely.


Suppose that n,m0 and that we have a homeomorphism f:nm. We must show that n=m. If n=0 then n is a single point so m is a single point so m=0. Similarly, if m=0 then n=0. Thus, we can restrict attention to the case where n,m1.

Choose an and put b=f(a)m. It is easy to see that f restricts to give a homeomorphism f0:n{a}m{b}. We can now define maps


i(x) =x+a p(x) =(x-a)/x-a
j(y) =y+b q(y) =(y-b)/y-b.

A tiny adaptation of Proposition 9.12 shows that p and q are homotopy inverses for i and j respectively, so that all maps in the above diagrams are homotopy equivalences, so Sn and Sm are homotopy equivalent. It follows by Theorem 16.1 that n=m as required. ∎

Video (Theorem 16.3 to Lemma 16.5)

Theorem 16.3 (Brouwer Fixed Point Theorem).

Let f:BnBn be a continuous map (for some n>0). Then there is a point aBn such that f(a)=a.

The proof will rely on the following construction.

Definition 16.4.

Put Xn={(a,b)Bn×Bn|ab}. For (a,b)Xn we consider the map uab:n given by uab(t)=a+t(a-b), so uab traces out the straight line joining b to a, with uab(-1)=b and uab(0)=a.

It is geometrically clear that this line crosses the sphere Sn-1 at precisely two points, one with t-1 and the other with t0. We define m(a,b) to be the intersection point with t0.

Lemma 16.5.

The map m:XnSn-1 is continuous, and it satisfies m(a,b)=a if a=1.


It is possible to argue geometrically, but more efficient to just find the formula for the point c=m(a,b). Put v=a-b, so c=tv+a for some t0, and must satisfy c,c=1. Expanding this out, we get


The quadratic formula tells us that the positive root is


Note that the quantity under the square root is nonnegative, because squares are always nonnegative and aBn so 1-a20. Also, the vector v=a-b is nonzero by the definition of Xn so v2>0 so it is harmless to divide by v2. This shows that t+ is a well-defined continuous function of the pair (a,b). It follows that the function m(a,b)=a+t+b is also continuous.

It is clear from the geometry that if a=1 (so a lies on the unit sphere Sn-1) then m(a,b)=a. Alternatively, it is clear in this case that t=0 is a nonnegative root of our quadratic, so it must be the same as t+. ∎

Proof of Theorem 16.3.

Suppose, for a contradiction, that we have a continuous map f:BnBn with no fixed points. This means that for any aBn we have (a,f(a))Xn so we can define r(a)=m(a,f(a))Sn-1. If a=1, then this is just r(a)=a. This means that Sn-1 is a retract (and thus a homotopy retract) of Bn. As Bn is contractible, we can use Proposition 9.24 to see that Sn-1 is also contractible, but this contradicts Theorem 16.1. ∎

Video (Theorem 16.6)

Theorem 16.6 (Fundamental Theorem of Algebra).

Let p(x)[x] be a non-constant complex polynomial. Then p(x) has a complex root.

There are many different ways to prove this theorem. We will give a proof using homology.


Consider a non-constant polynomial p(x) of degree n>0, so


for some coefficients ai with an0. Suppose, for a contradiction that p(x) is never zero. Choose some very large radius R and define h:[0,1]×S1{0} by h(s,z)=p(Rsz)/p(Rs). As we are assuming that p is never zero, the division is valid and h(s,z) lies in {0} as required. Put f(z)=h(0,z)=1 and g(z)=h(1,z)=p(Rz)/p(R). As f is constant, the map f*:H1(S1)H1({0}) is zero. As h gives a homotopy between f and g, the map g*:H1(S1)H1({0}) must also be zero. However, as R is very large and an0, the term an(Rz)n will be much larger than all the other terms in p(Rz), so g(z)=p(Rz)/p(R)an(Rz)n/(anRn)=zn. Thus, if we put q(z)=zn then g(z) will be very close to q(z) for all zS1, so the straight line from g(z) to q(z) will never pass through 0, so g will be linearly homotopic to q, so q*=g*=0:H1(S1)H1({0}). However, it is clear from our earlier discussions in Section 11 that H1(S1)=H1({0})= and q* sends 1 to n0, so we have a contradiction. ∎