We would like to make the following definition:
Let $X$ and $Y$ be topological spaces, and let $f,g:X\to Y$ be two continuous maps from $X$ to $Y$, or in other words, two elements of $\mathrm{Top}(X,Y)$. We say that $f$ and $g$ are homotopic if there is a path $u:[0,1]\to \mathrm{Top}(X,Y)$ from $f$ to $g$, so that $f$ and $g$ lie in the same path component of $\mathrm{Top}(X,Y)$. This is an equivalence relation, by Proposition 5.9; the equivalence classes are called homotopy classes of maps. We define $[X,Y]={\pi}_{0}\mathrm{Top}(X,Y)$, so $[X,Y]$ is the set of all homotopy classes of maps from $X$ to $Y$.
This is bogus, because we have not introduced a topology on the set $\mathrm{Top}(X,Y)$, so it is not meaningful to talk about continuous paths. As we mentioned in Remark 7.28, it is possible to introduce an appropriate topology on $\mathrm{Top}(X,Y)$, but this involves many subtle technicalities. We therefore reformulate the definition in a different way.
Let $X$ and $Y$ be topological spaces, and let $f,g:X\to Y$ be two continuous maps from $X$ to $Y$. A homotopy from $f$ to $g$ is a continuous map $h:[0,1]\times X\to Y$ such that $h(0,x)=f(x)$ and $h(1,x)=g(x)$ for all $x\in X$. We say that $f$ and $g$ are homotopic if there is a homotopy between them, and we write $f\equiv g$ in this case.
The relation of being homotopic is an equivalence relation.
This is closely analogous to Proposition 5.9.
We can define a homotopy $h$ from $f$ to $f$ by $h(t,x)=f(x)$ for all $t$. This proves that $f\equiv f$, so the relation is reflexive.
Suppose that $f\equiv g$, so we can choose a homotopy $h$ from $f$ to $g$. The function $\overline{h}(t,x)=h(1-t,x)$ then gives a homotopy from $g$ to $f$, proving that $g\equiv f$. Thus, the relation is symmetric.
Now suppose that $e\equiv f$ and $f\equiv g$, so we can choose a homotopy $k$ from $e$ to $f$, and another homotopy $h$ from $f$ to $g$. We then define $k*h:[0,1]\times X\to Y$ by
$$(k*h)(t,x)=\{\begin{array}{cc}k(2t,x)\hfill & \text{if}0\le t\le \frac{1}{2}\hfill \\ h(2t-1,x)\hfill & \text{if}\frac{1}{2}\le t\le 1.\hfill \end{array}$$ |
The two clauses are consistent, because when $t=\frac{1}{2}$ we have
$$k(2t,x)=k(1,x)=f(x)=h(0,x)=h(2t-1,x).$$ |
The combined map $k*h$ is easily seen to be continuous on when restricted to the sets $[0,\frac{1}{2}]\times X$ and $[\frac{1}{2},1]\times X$. These sets are closed (in the product topology), and their union is all of $[0,1]\times X$, so the full map $k*h$ is continuous by Proposition 3.35. It gives a homotopy from $e$ to $g$, proving that the homotopy relation is transitive.
∎
The equivalence class $[f]$ of a continuous map $f:X\to Y$ is called the homotopy class of $f$. We define $[X,Y]=\mathrm{Top}(X,Y)/\equiv $, so this is the set of all homotopy classes.
Consider a continuous map $f:{S}^{1}\to \u2102\setminus \{0\}$. Then $f$ represents a closed loop in the complex plane, which cannot pass through the origin, but which can wind around the origin. Let $n(f)$ be the total number of times the curve winds around the origin, with anticlockwise turns counting positively, and clockwise turns counting negatively. We call this the winding number of $f$. (We will formulate the definition more carefully at a later stage.)
It can be shown that two maps from ${S}^{1}$ to $\u2102\setminus \{0\}$ are homotopic iff they have the same winding number. It follows that the map $n:\mathrm{Top}({S}^{1},\u2102\setminus \{0\})\to \mathbb{Z}$ induces a bijection $[{S}^{1},\u2102\setminus \{0\}]\to \mathbb{Z}$.
Let $X$ be an arbitrary topological space, and let $Y$ be a subset of ${\mathbb{R}}^{n}$, with the subspace topology. Let $f,g:X\to Y$ be continuous maps. We can then define a map $h:[0,1]\times X\to {\mathbb{R}}^{n}$ by
$$h(t,x)=(1-t)f(x)+tg(x).$$ |
In general, we have no right to expect that this will land in the subspace $Y\subseteq {\mathbb{R}}^{n}$, so it will probably not provide a homotopy between $f$ and $g$. However, in some special cases we may be able to prove that $h([0,1]\times X)\subseteq Y$, and in that case we do get a homotopy from $f$ to $g$, which we will call a straight line homotopy.
For the most extreme example of this, suppose that $Y$ is convex (as in Example 5.11), so that for all $a,b\in Y$, the line segment from $a$ to $b$ is contained wholly in $Y$. Let $f$ and $g$ be continuous maps from $X$ to $Y$, and put $h(t,x)=(1-t)f(x)+tg(x)$ as before. Then $h(t,x)$ lies on the line segment from $f(x)\in Y$ to $g(x)\in Y$, so $h(t,x)\in Y$ for all $t$ and $x$. Thus, $h$ gives a homotopy from $f$ to $g$, proving that $[f]=[g]$ in $[X,Y]$. As $f$ and $g$ were arbitrary, it follows that $|[X,Y]|=1$.
We next check that the notion of homotopy is compatible with composition.
Suppose we have continuous maps
Suppose that ${f}_{\mathrm{0}}$ is homotopic to ${f}_{\mathrm{1}}$, and that ${g}_{\mathrm{0}}$ is homotopic to ${g}_{\mathrm{1}}$. Then ${g}_{\mathrm{1}}\mathrm{\circ}{f}_{\mathrm{1}}$ is homotopic to ${g}_{\mathrm{0}}\mathrm{\circ}{f}_{\mathrm{0}}$.
We are assuming that ${f}_{0}$ is homotopic to ${f}_{1}$, which means that there is a homotopy $h:[0,1]\times X\to Y$ with $h(0,x)={f}_{0}(x)$ and $h(1,x)={f}_{1}(x)$ for all $x\in X$. We are also assuming that ${g}_{0}$ is homotopic to ${g}_{1}$, which means that there is a homotopy $k:[0,1]\times Y\to Z$ with $k(0,y)={g}_{0}(y)$ and $k(1,y)={g}_{1}(y)$ for all $y\in Y$. We can therefore define $m:[0,1]\times X\to Z$ by
$$m(t,x)=k(t,h(t,x)).$$ |
This satisfies $m(0,x)=k(0,h(0,x))=k(0,{f}_{0}(x))={g}_{0}({f}_{0}(x))$ and $m(1,x)=k(1,h(1,x))=k(1,{f}_{1}(x))={g}_{1}({f}_{1}(x))$. Thus, $m$ gives the required homotopy, provided that we can check that it is continuous. The tidiest proof of continuity is as follows: we let $p:[0,1]\times X\to [0,1]$ be the projection, and note that $m$ can be written as the composite
$$[0,1]\times X\stackrel{\u27e8p,h\u27e9}{\to}[0,1]\times Y\stackrel{\mathit{k}}{\to}Z.$$ |
Here $p$ is continuous by Lemma 7.10, so $\u27e8p,h\u27e9$ is continuous by Proposition 7.11, so $m$ is continuous by Proposition 3.24. ∎
For $u\mathrm{\in}\mathrm{[}X\mathrm{,}Y\mathrm{]}$ and $v\mathrm{\in}\mathrm{[}Y\mathrm{,}Z\mathrm{]}$, there is a well-defined composite $v\mathrm{\circ}u\mathrm{\in}\mathrm{[}X\mathrm{,}Z\mathrm{]}$, given by $v\mathrm{\circ}u\mathrm{=}\mathrm{[}g\mathrm{\circ}f\mathrm{]}$ for any choice of maps $f\mathrm{,}g$ with $u\mathrm{=}\mathrm{[}f\mathrm{]}$ and $v\mathrm{=}\mathrm{[}g\mathrm{]}$.
Immediate from the proposition. ∎
We can now define a new category $\mathrm{hTop}$, called the homotopy category. The objects are topological spaces (just as for the category $\mathrm{Top}$), but the morphisms are now homotopy classes of maps, so
$$\mathrm{hTop}(X,Y)=\mathrm{Top}(X,Y)/\equiv =[X,Y].$$ |
The composition rule is given by Corollary 9.8. The identity morphism in $\mathrm{hTop}$ for an object $X$ is just the homotopy class $[{\mathrm{id}}_{X}]$ corresponding to the identity function.
A continuous map $f:X\to Y$ is a homotopy equivalence if the corresponding homotopy class $[f]$ is an isomorphism in the homotopy category. Explicitly, this means that there must exist a continuous map $g:Y\to X$ such that $g\circ f\equiv {\mathrm{id}}_{X}$ and $f\circ g\equiv {\mathrm{id}}_{Y}$. Any such map $g$ is called a homotopy inverse for $f$. If there exists a homotopy equivalence from $X$ to $Y$, we will say that $X$ and $Y$ are homotopy equivalent. We will sometimes use the notation $X\cong Y$ to indicate this.
Every homeomorphism is a homotopy equivalence.
Let $f:X\to Y$ be a homeomorphism, with inverse $g:Y\to X$. This means that $g\circ f={\mathrm{id}}_{X}$ and $f\circ g={\mathrm{id}}_{Y}$. As $g\circ f$ is equal to ${\mathrm{id}}_{X}$, it is certainly homotopic to ${\mathrm{id}}_{X}$. As $f\circ g$ is equal to ${\mathrm{id}}_{Y}$, it is certainly homotopic to ${\mathrm{id}}_{Y}$. Thus, $g$ is also a homotopy inverse for $f$, so $f$ is a homotopy equivalence. ∎
A basic example is as follows:
For $n\mathrm{>}\mathrm{0}$, the sphere $X\mathrm{=}{S}^{n\mathrm{-}\mathrm{1}}$ is homotopy equivalent to the space $Y\mathrm{=}{\mathbb{R}}^{n}\mathrm{\setminus}\mathrm{\{}\mathrm{0}\mathrm{\}}$.
Recall that $X={S}^{n-1}=\{x\in {\mathbb{R}}^{n}|\parallel x\parallel =1\}$. We define $f:X\to Y$ to be the inclusion function, so we just have $f(x)=x$. We define $g:Y\to X$ by $g(y)=y/\parallel y\parallel $. (It is important here that $Y$ is the space of nonzero vectors in ${\mathbb{R}}^{n}$, so $\parallel y\parallel >0$ and it is valid to divide by $\parallel y\parallel $.) Note that for $x\in X$ we have $\parallel x\parallel =1$ so $g(f(x))=x/\parallel x\parallel =x$, so $g\circ f={\mathrm{id}}_{X}$. As $g\circ f$ is equal to ${\mathrm{id}}_{X}$, it is certainly homotopic to ${\mathrm{id}}_{X}$. In the opposite direction, however, the composite $g\circ f$ is not equal to the identity. Nonetheless, we can define $h:[0,1]\times Y\to {\mathbb{R}}^{n}$ by
$$h(t,y)=(1-t)\frac{y}{\parallel y\parallel}+ty=\left((1-t){\parallel y\parallel}^{-1}+t\right)y.$$ |
Here $(1-t){\parallel y\parallel}^{-1}+t$ is always strictly positive (for $0\le t\le 1$), so $h(t,y)$ can never be zero, so we can regard $h$ as a map $[0,1]\times Y\to Y$. Now $h(0,y)=f(g(y))$ and $h(1,y)=y$, so $h$ is a homotopy from $g\circ f$ to ${\mathrm{id}}_{Y}$. This proves that $g$ is a homotopy inverse for $f$, so that $f$ is a homotopy equivalence.
∎
Let $X$ be a topological space, and let $a$ be a point in $X$. A contraction of $X$ to $a$ is a continuous map $h:[0,1]\times X\to X$ such that $h(0,x)=a$ for all $x$, and $h(1,x)=x$ for all $x$. We say that $X$ is contractible if it has a contraction (to some point $a\in X$).
Let $X\subseteq {\mathbb{R}}^{n}$ be a convex space, and let $a$ be a point in $X$. We can then define a linear contraction of $X$ to $a$ by $h(t,x)=(1-t)a+tx$, so $X$ is contractible. In particular, the spaces ${B}^{n}$, ${\mathrm{\Delta}}_{n}$ and ${[0,1]}^{n}$ are all contractible.
The sphere ${S}^{2}$ is not contractible, as you will probably agree if you try to imagine a contraction. However, at the moment we do not have any way of proving this.
We write $1$ for the set $\{0\}$ (or any other set with precisely one element). We regard this as a topological space using the discrete topology (which is in fact the only possible topology in this case).
A space $X$ is contractible iff it is homotopy equivalent to $\mathrm{1}$.
Suppose that $X$ is homotopy equivalent to $1$. This means that we have maps $f:X\to 1$ and $g:1\to X$, and a homotopy $h$ from $g\circ f$ to ${\mathrm{id}}_{X}$, and a homotopy $k$ from $f\circ g$ to ${\mathrm{id}}_{1}$. As $1=\{0\}$ there is no choice about $f$: we must have $f(x)=0$ for all $x$. Similarly, there is no choice about $k$: we must have $k(t,0)=0$ for all $t$. However, there is some choice about $g$ and $h$. Put $a=g(0)\in X$. We then have $g(f(x))=g(0)=a$ for all $x$. Moreover, $h$ is a homotopy from $g\circ f$ to ${\mathrm{id}}_{X}$, so we have $h(0,x)=g(f(x))=a$ and $h(1,x)=x$ for all $x$. Thus, $h$ is a contraction t $a$, showing that $X$ is contractible.
Conversely, suppose that we start from the assumption that $X$ is contractible. All the above steps can be reversed in a straightforward way to prove that $X$ is homotopy equivalent to $1$. ∎
The map $[0,1]\to 1$ is the most basic example of a homotopy equivalence that is not a homeomorphism.
The relation of being homotopy equivalent is an equivalence relation.
The identity function from $X$ to itself is clearly a homotopy equivalence, so the relation is reflexive. Suppose that $X$ is homotopy equivalent to $Y$, so we can choose a homotopy equivalence $f:X\stackrel{}{\to}Y$ and a homotopy inverse $g:Y\stackrel{}{\to}X$, so $fg:Y\stackrel{}{\to}Y$ and $gf:X\stackrel{}{\to}X$ are homotopic to the respective identity maps. This means that $g$ is a homotopy equivalence with homotopy inverse $f$, so $Y$ is homotopy equivalent to $X$. This proves that our relation is symmetric. Finally, suppose that $X$ is homotopy equivalent to $Y$ and $Y$ is homotopy equivalent to $Z$. Let $d$ and $e$ be homotopy inverses for $f$ and $g$, so $df\simeq {1}_{X}$ and $fd\simeq {1}_{Y}\simeq eg$ and $ge\simeq {1}_{Z}$. Using Proposition 9.7, we deduce that
$$degf=d(eg)f\simeq d{1}_{Y}f=df\simeq {1}_{X}$$ |
$$gfde=g(fd)e\simeq g{1}_{Y}e=ge\simeq {1}_{Z}.$$ |
This proves that $gf:X\stackrel{}{\to}Z$ is a homotopy equivalence, with homotopy inverse $de:Z\stackrel{}{\to}X$. This in turn shows that our relation is transitive, as required. ∎
Suppose that ${X}_{\mathrm{0}}$ is homotopy equivalent to ${Y}_{\mathrm{0}}$ and ${X}_{\mathrm{1}}$ is homotopy equivalent to ${Y}_{\mathrm{1}}$. Then ${X}_{\mathrm{0}}\mathrm{\times}{X}_{\mathrm{1}}$ is homotopy equivalent to ${Y}_{\mathrm{0}}\mathrm{\times}{Y}_{\mathrm{1}}$.
For $i=0,1$ we let ${f}_{i}:{X}_{i}\to {Y}_{i}$ be a homotopy equivalence, with homotopy inverse ${g}_{i}:{Y}_{i}\to {X}_{i}$. This means that we have homotopies ${h}_{i}$ from ${g}_{i}\circ {f}_{i}$ to ${\mathrm{id}}_{{X}_{i}}$, and homotopies ${k}_{i}$ from ${f}_{i}\circ {g}_{i}$ to ${\mathrm{id}}_{{Y}_{i}}$. Now make the following definitions:
$f$ | $:{X}_{0}\times {X}_{1}\to {Y}_{0}\times {Y}_{1}$ | $f({x}_{0},{x}_{1})$ | $=({f}_{0}({x}_{0}),{f}_{1}({x}_{1}))$ | ||
$g$ | $:{Y}_{0}\times {Y}_{1}\to {X}_{0}\times {X}_{1}$ | $g({y}_{0},{y}_{1})$ | $=({g}_{0}({y}_{0}),{g}_{1}({y}_{1}))$ | ||
$h$ | $:[0,1]\times {X}_{0}\times {X}_{1}\to {X}_{0}\times {X}_{1}$ | $h(t,{x}_{0},{x}_{1})$ | $=({h}_{0}(t,{x}_{0}),{h}_{1}(t,{x}_{1}))$ | ||
$k$ | $:[0,1]\times {Y}_{0}\times {Y}_{1}\to {Y}_{0}\times {Y}_{1}$ | $k(t,{y}_{0},{y}_{1})$ | $=({k}_{0}(t,{y}_{0}),{k}_{1}(t,{y}_{1})).$ |
We find that $h$ gives a homotopy from $g\circ f$ to the identity, and $k$ gives a homotopy from $f\circ g$ to the identity, so we have a homotopy equivalence between ${X}_{0}\times {X}_{1}$ and ${Y}_{0}\times {Y}_{1}$, as required. ∎
The solid torus, the Möbius band, and $\u2102\setminus \{0\}$ are all homotopy equivalent to ${S}^{1}$. To explain this in more detail, let $D$ be the vertical disc in the $xz$ plane of radius $1$ centred at $(2,0,0)$. The “solid torus” is the space obtained by revolving $D$ around the $z$-axis; this is easily seen to be homeomorphic to ${S}^{1}\times {D}^{2}$. Now ${D}^{2}$ is convex, so it is homotopy equivalent to $1$. It follows that ${S}^{1}\times {D}^{2}$ is homotopy equivalent to ${S}^{1}\times 1={S}^{1}$.
Next, for $\theta \in [0,2\pi ]$, let ${P}_{\theta}$ be the vertical plane through the $z$-axis in ${\mathbb{R}}^{3}$ that has angle $\theta $ with the $xz$-plane. Let ${D}_{\theta}$ be the intersection of ${P}_{\theta}$ with the solid torus, which is a vertical disc of radius $1$ centred at $(2\mathrm{cos}(\theta ),2\mathrm{sin}(\theta ),0)$. Let ${I}_{\theta}$ be the diameter of ${D}_{\theta}$ that makes an angle of $\theta /4$ to the vertical, and let $M$ be the union of all the sets ${D}_{\theta}$. This is a version of the Möbius band. It is homeomorphic to the space
$${M}^{\prime}=\{(z,w)\in {S}^{1}\times {B}^{2}|{w}^{2}/z\text{is real and nonnegative}\}.$$ |
Define $f:{S}^{1}\stackrel{}{\to}{M}^{\prime}$ and $g:{M}^{\prime}\stackrel{}{\to}{S}^{1}$ and $h:I\times {M}^{\prime}\stackrel{}{\to}{M}^{\prime}$ by
$f(z)$ | $=(z,0)$ | ||
$g(z,w)$ | $=z$ | ||
$h(t,(z,w))$ | $=(z,tw).$ |
Then $gf=1$ and $h$ is a homotopy from $fg$ to $1$, so $g$ is a homotopy inverse for $f$, so ${M}^{\prime}$ is homotopy equivalent to ${S}^{1}$ as claimed.
Finally, Proposition 9.12 tells us that ${\mathbb{R}}^{2}\setminus \{0\}$ is homotopy equivalent to ${S}^{1}$, and $\u2102$ can be identified with ${\mathbb{R}}^{2}$, so $\u2102\setminus \{0\}$ is also homotopy equivalent to ${S}^{1}$.
Let $X$ and $Y$ be topological spaces. We say that $X$ is a homotopy retract of $Y$ if there exist continuous maps $X\stackrel{\mathit{f}}{\to}Y\stackrel{\mathit{g}}{\to}X$ such that $g\circ f$ is homotopic to ${\mathrm{id}}_{X}$. (We make no assumption about $f\circ g$.) Any pair $(f,g)$ with this property will be called a homotopy retraction pair for $(X,Y)$.
Put
$X$ | $={\mathbb{R}}^{2}\setminus \{0\}$ | ||
$Y$ | $=\text{figure eight}=\{(x,y)\in \u2102|{(x-1)}^{2}+{y}^{2}=1\text{or}{(x+1)}^{2}+{y}^{2}=1\}.$ |
Note that $X$ is two-dimensional whereas $Y$ is one-dimensional so there is no injective continuous map from $X$ to $Y$, so $X$ is not an actual retract of $Y$. However, it is a homotopy retract of $Y$. To see this, define maps $X\stackrel{\mathit{f}}{\to}Y\stackrel{\mathit{g}}{\to}X$ by
$f(x,y)$ | $=(1,0)+(x,y)/\parallel (x,y)\parallel $ | ||
$g(x,y)$ | $=(x-1,y).$ |
The composite $g\circ f:X\to X$ is just $(g\circ f)(x,y)=(x,y)/\parallel (x,y)\parallel $. The straight line joining $(x,y)$ to $(x,y)/\parallel (x,y)\parallel $ does not pass through the origin, so $g\circ f$ is homotopic to the identity as required. (The map $f\circ g:Y\to Y$ is not homotopic to the identity, so we do not have a homotopy equivalence, but that is not important.)
Suppose that $X$ is a homotopy retract of $Y$ and that $Y$ is contractible. Then $X$ is also contractible.
By hypothesis, we have continuous maps
such that $gf$, $qp$ and $pq$ are homotopic to the respective identity maps. Now put $m=pf:X\to 1$ and $n=gq:1\to X$. We then have $nm=gqpf\simeq gf\simeq \mathrm{id}:X\to X$. Also, $mn$ is a map from $1$ to $1$, and the only map from $1$ to $1$ is the identity, so $mn=\mathrm{id}$. Thus, $m$ and $n$ give a homotopy equivalence from $X$ to $1$, proving that $X$ is contractible. ∎
Suppose we have two continuous maps $f\mathrm{,}g\mathrm{:}X\mathrm{\to}Y$, giving maps ${f}_{\mathrm{*}}\mathrm{,}{g}_{\mathrm{*}}\mathrm{:}{\pi}_{\mathrm{0}}\mathit{}\mathrm{(}X\mathrm{)}\mathrm{\to}{\pi}_{\mathrm{0}}\mathit{}\mathrm{(}Y\mathrm{)}$ as in Proposition 5.20. If $f$ is homotopic to $g$, then ${f}_{\mathrm{*}}\mathrm{=}{g}_{\mathrm{*}}$.
Let $h:[0,1]\times X\to Y$ be a homotopy between $f$ and $g$, so $h(0,x)=f(x)$ and $h(1,x)=g(x)$ for all $x$. For any $a\in X$ we have ${f}_{*}[a]=[f(a)]$ and ${g}_{*}[a]=[g(a)]$. We need to prove that these are the same. Equivalently, we need to find a path $u$ from $f(a)$ to $g(a)$ in $Y$. We can just take $u(t)=h(t,a)$. ∎