MAS61015 Algebraic Topology 15 Homology of spheres 17 The Snake Lemma

16. Applications of homology

Theorem 16.1.

No sphere $S^{n}$ is contractible. Moreover, if $n\neq m$ then $S^{n}$ is not homotopy equivalent to $S^{m}$ .

Proof.

As homotopy equivalent spaces have isomorphic homology, it will suffice to prove that $H_{n}(S^{n})\not\simeq H_{n}(\text{point})$ and that $H_{n}(S^{m})\not\simeq H_{n}(S^{m})$ when $n\neq m$ . This is clear from the above calculations. ∎

Theorem 16.2.

Suppose that $n,m\geq 0$ with $n\neq m$ . Then ${\mathbb{R}}^{n}$ is not homeomorphic to ${\mathbb{R}}^{m}$ .

Note here that ${\mathbb{R}}^{n}$ and ${\mathbb{R}}^{m}$ are certainly homotopy equivalent, as they are both contractible. This is perfectly consistent: homeomorphism implies homotopy equivalence but not conversely.

Proof.

Suppose that $n,m\geq 0$ and that we have a homeomorphism $f\colon{\mathbb{R}}^{n}\to{\mathbb{R}}^{m}$ . We must show that $n=m$ . If $n=0$ then ${\mathbb{R}}^{n}$ is a single point so ${\mathbb{R}}^{m}$ is a single point so $m=0$ . Similarly, if $m=0$ then $n=0$ . Thus, we can restrict attention to the case where $n,m\geq 1$ .

Choose $a\in{\mathbb{R}}^{n}$ and put $b=f(a)\in{\mathbb{R}}^{m}$ . It is easy to see that $f$ restricts to give a homeomorphism $f_{0}\colon{\mathbb{R}}^{n}\setminus\{a\}\to{\mathbb{R}}^{m}\setminus\{b\}$ . We can now define maps

	$\displaystyle i(x)$	$\displaystyle=x+a$	$\displaystyle p(x)$	$\displaystyle=(x-a)/\\|x-a\\|$
	$\displaystyle j(y)$	$\displaystyle=y+b$	$\displaystyle q(y)$	$\displaystyle=(y-b)/\\|y-b\\|.$

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 $S^{n}$ and $S^{m}$ 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 $f\colon B^{n}\to B^{n}$ be a continuous map (for some $n>0$ ). Then there is a point $a\in B^{n}$ such that $f(a)=a$ .

The proof will rely on the following construction.

Definition 16.4.

Put $X_{n}=\{(a,b)\in B^{n}\times B^{n}\;|\;a\neq b\}$ . For $(a,b)\in X_{n}$ we consider the map $u_{ab}\colon{\mathbb{R}}\to{\mathbb{R}}^{n}$ given by $u_{ab}(t)=a+t(a-b)$ , so $u_{ab}$ traces out the straight line joining $b$ to $a$ , with $u_{ab}(-1)=b$ and $u_{ab}(0)=a$ .

It is geometrically clear that this line crosses the sphere $S^{n-1}$ at precisely two points, one with $t\leq-1$ and the other with $t\geq 0$ . We define $m(a,b)$ to be the intersection point with $t\geq 0$ .

Lemma 16.5.

The map $m\colon X_{n}\to S^{n-1}$ is continuous, and it satisfies $m(a,b)=a$ if $\|a\|=1$ .

Proof.

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 $t\geq 0$ , and must satisfy $\langle c,c\rangle=1$ . Expanding this out, we get

\|v\|^{2}t^{2}+2\langle v,a\rangle t+\|a\|^{2}-1=0.

The quadratic formula tells us that the positive root is

t_{+}=(\sqrt{\langle v,a\rangle^{2}+(1-\|a\|^{2})\|v\|^{2}}-\langle v,a\rangle% )/\|v\|^{2}.

Note that the quantity under the square root is nonnegative, because squares are always nonnegative and $a\in B^{n}$ so $1-\|a\|^{2}\geq 0$ . Also, the vector $v=a-b$ is nonzero by the definition of $X_{n}$ so $\|v\|^{2}>0$ so it is harmless to divide by $\|v\|^{2}$ . 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 $S^{n-1}$ ) 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 $f\colon B^{n}\to B^{n}$ with no fixed points. This means that for any $a\in B^{n}$ we have $(a,f(a))\in X_{n}$ so we can define $r(a)=m(a,f(a))\in S^{n-1}$ . If $\|a\|=1$ , then this is just $r(a)=a$ . This means that $S^{n-1}$ is a retract (and thus a homotopy retract) of $B^{n}$ . As $B^{n}$ is contractible, we can use Proposition 9.24 to see that $S^{n-1}$ is also contractible, but this contradicts Theorem 16.1. ∎

Video (Theorem 16.6)

Theorem 16.6 (Fundamental Theorem of Algebra).

Let $p(x)\in{\mathbb{C}}[x]$ 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.

Proof.

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

p(x)=a_{0}+a_{1}x+\dotsb+a_{n}x^{n}

for some coefficients $a_{i}$ with $a_{n}\neq 0$ . Suppose, for a contradiction that $p(x)$ is never zero. Choose some very large radius $R$ and define $h\colon[0,1]\times S^{1}\to{\mathbb{C}}\setminus\{0\}$ 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 ${\mathbb{C}}\setminus\{0\}$ 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 $f_{*}\colon H_{1}(S^{1})\to H_{1}({\mathbb{C}}\setminus\{0\})$ is zero. As $h$ gives a homotopy between $f$ and $g$ , the map $g_{*}\colon H_{1}(S^{1})\to H_{1}({\mathbb{C}}\setminus\{0\})$ must also be zero. However, as $R$ is very large and $a_{n}\neq 0$ , the term $a_{n}(Rz)^{n}$ will be much larger than all the other terms in $p(Rz)$ , so $g(z)=p(Rz)/p(R)\approx a_{n}(Rz)^{n}/(a_{n}R^{n})=z^{n}$ . Thus, if we put $q(z)=z^{n}$ then $g(z)$ will be very close to $q(z)$ for all $z\in S^{1}$ , so the straight line from $g(z)$ to $q(z)$ will never pass through $0$ , so $g$ will be linearly homotopic to $q$ , so $q_{*}=g_{*}=0\colon H_{1}(S^{1})\to H_{1}({\mathbb{C}}\setminus\{0\})$ . However, it is clear from our earlier discussions in Section 11 that $H_{1}(S^{1})=H_{1}({\mathbb{C}}\setminus\{0\})={\mathbb{Z}}$ and $q_{*}$ sends $1$ to $n\neq 0$ , so we have a contradiction. ∎