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

As homotopy equivalent spaces have isomorphic homology, it will suffice to prove that ${H}_{n}({S}^{n})\simeq \u0338{H}_{n}(\text{point})$ and that ${H}_{n}({S}^{m})\simeq \u0338{H}_{n}({S}^{m})$ when $n\ne m$. This is clear from the above calculations. ∎

Suppose that $n\mathrm{,}m\mathrm{\ge}\mathrm{0}$ with $n\mathrm{\ne}m$. 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.

Suppose that $n,m\ge 0$ and that we have a homeomorphism $f:{\mathbb{R}}^{n}\to {\mathbb{R}}^{m}$. 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 $n,m\ge 1$.

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 ${f}_{0}:{\mathbb{R}}^{n}\setminus \{a\}\to {\mathbb{R}}^{m}\setminus \{b\}$. We can now define maps

by

$i(x)$ | $=x+a$ | $p(x)$ | $=(x-a)/\parallel x-a\parallel $ | ||

$j(y)$ | $=y+b$ | $q(y)$ | $=(y-b)/\parallel y-b\parallel .$ |

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. ∎

Let $f\mathrm{:}{B}^{n}\mathrm{\to}{B}^{n}$ be a continuous map (for some $n\mathrm{>}\mathrm{0}$). Then there is a point $a\mathrm{\in}{B}^{n}$ such that $f\mathit{}\mathrm{(}a\mathrm{)}\mathrm{=}a$.

The proof will rely on the following construction.

Put ${X}_{n}=\{(a,b)\in {B}^{n}\times {B}^{n}|a\ne b\}$. For $(a,b)\in {X}_{n}$ we consider the map ${u}_{ab}:\mathbb{R}\to {\mathbb{R}}^{n}$ 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 $t\le -1$ and the other with $t\ge 0$. We define $m(a,b)$ to be the intersection point with $t\ge 0$.

The map $m\mathrm{:}{X}_{n}\mathrm{\to}{S}^{n\mathrm{-}\mathrm{1}}$ is continuous, and it satisfies $m\mathit{}\mathrm{(}a\mathrm{,}b\mathrm{)}\mathrm{=}a$ if $\mathrm{\parallel}a\mathrm{\parallel}\mathrm{=}\mathrm{1}$.

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 $t\ge 0$, and must satisfy $\u27e8c,c\u27e9=1$. Expanding this out, we get

$${\parallel v\parallel}^{2}{t}^{2}+2\u27e8v,a\u27e9t+{\parallel a\parallel}^{2}-1=0.$$ |

The quadratic formula tells us that the positive root is

$${t}_{+}=(\sqrt{{\u27e8v,a\u27e9}^{2}+(1-{\parallel a\parallel}^{2}){\parallel v\parallel}^{2}}-\u27e8v,a\u27e9)/{\parallel v\parallel}^{2}.$$ |

Note that the quantity under the square root is nonnegative, because squares are always nonnegative and $a\in {B}^{n}$ so $1-{\parallel a\parallel}^{2}\ge 0$. Also, the vector $v=a-b$ is nonzero by the definition of ${X}_{n}$ so ${\parallel v\parallel}^{2}>0$ so it is harmless to divide by ${\parallel v\parallel}^{2}$. 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 $\parallel a\parallel =1$ (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 ${t}_{+}$. ∎

Suppose, for a contradiction, that we have a continuous map $f:{B}^{n}\to {B}^{n}$ 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 $\parallel a\parallel =1$, 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. ∎

Let $p\mathit{}\mathrm{(}x\mathrm{)}\mathrm{\in}\u2102\mathit{}\mathrm{[}x\mathrm{]}$ be a non-constant complex polynomial. Then $p\mathit{}\mathrm{(}x\mathrm{)}$ 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

$$p(x)={a}_{0}+{a}_{1}x+\mathrm{\cdots}+{a}_{n}{x}^{n}$$ |

for some coefficients ${a}_{i}$ with ${a}_{n}\ne 0$. Suppose, for a contradiction that $p(x)$ is never zero. Choose some very large radius $R$ and define $h:[0,1]\times {S}^{1}\to \u2102\setminus \{0\}$ 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 $\u2102\setminus \{0\}$ 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 ${f}_{*}:{H}_{1}({S}^{1})\to {H}_{1}(\u2102\setminus \{0\})$ is zero. As $h$ gives a homotopy between $f$ and $g$, the map ${g}_{*}:{H}_{1}({S}^{1})\to {H}_{1}(\u2102\setminus \{0\})$ must also be zero. However, as $R$ is very large and ${a}_{n}\ne 0$, 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 ${q}_{*}={g}_{*}=0:{H}_{1}({S}^{1})\to {H}_{1}(\u2102\setminus \{0\})$. However, it is clear from our earlier discussions in Section 11 that ${H}_{1}({S}^{1})={H}_{1}(\u2102\setminus \{0\})=\mathbb{Z}$ and ${q}_{*}$ sends $1$ to $n\ne 0$, so we have a contradiction. ∎