9. Homotopy

Video (Definition 9.1 to Example 9.6)

We would like to make the following definition:

Definition 9.1(Bogus).

Let $X$ and $Y$ be topological spaces, and let $f,g\colon X\to Y$ be two continuous maps from $X$ to $Y$, or in other words, two elements of $\operatorname{Top}(X,Y)$. We say that $f$ and $g$ are homotopic if there is a path $u\colon[0,1]\to\operatorname{Top}(X,Y)$ from $f$ to $g$, so that $f$ and $g$ lie in the same path component of $\operatorname{Top}(X,Y)$. This is an equivalence relation, by Proposition 5.9; the equivalence classes are called homotopy classes of maps. We define $[X,Y]=\pi_{0}\operatorname{Top}(X,Y)$, 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 $\operatorname{Top}(X,Y)$, 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 $\operatorname{Top}(X,Y)$, but this involves many subtle technicalities. We therefore reformulate the definition in a different way.

Definition 9.2.

Let $X$ and $Y$ be topological spaces, and let $f,g\colon X\to Y$ be two continuous maps from $X$ to $Y$. A homotopy from $f$ to $g$ is a continuous map $h\colon[0,1]\times X\to Y$ 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.

Proposition 9.3.

The relation of being homotopic is an equivalence relation.

Proof.

This is closely analogous to Proposition 5.9.

• (a)

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.

• (b)

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.

• (c)

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 $k*h\colon[0,1]\times X\to Y$ by

 $(k*h)(t,x)=\begin{cases}k(2t,x)&\text{ if }0\leq t\leq\tfrac{1}{2}\\ h(2t-1,x)&\text{ if }\tfrac{1}{2}\leq t\leq 1.\end{cases}$

The two clauses are consistent, because when $t=\tfrac{1}{2}$ 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 $[0,\tfrac{1}{2}]\times X$ and $[\tfrac{1}{2},1]\times X$. 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.

Definition 9.4.

The equivalence class $[f]$ of a continuous map $f\colon X\to Y$ is called the homotopy class of $f$. We define $[X,Y]=\operatorname{Top}(X,Y)/\equiv$, so this is the set of all homotopy classes.

Example 9.5.

Consider a continuous map $f\colon S^{1}\to{\mathbb{C}}\setminus\{0\}$. 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 ${\mathbb{C}}\setminus\{0\}$ are homotopic iff they have the same winding number. It follows that the map $n\colon\operatorname{Top}(S^{1},{\mathbb{C}}\setminus\{0\})\to{\mathbb{Z}}$ induces a bijection $[S^{1},{\mathbb{C}}\setminus\{0\}]\to{\mathbb{Z}}$.

Example 9.6.

Let $X$ be an arbitrary topological space, and let $Y$ be a subset of ${\mathbb{R}}^{n}$, with the subspace topology. Let $f,g\colon X\to Y$ be continuous maps. We can then define a map $h\colon[0,1]\times X\to{\mathbb{R}}^{n}$ by

 $h(t,x)=(1-t)\,f(x)+t\,g(x).$

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 $h(t,x)=(1-t)\,f(x)+t\,g(x)$ 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.

Video (Proposition 9.7 to Definition 9.9)

Proposition 9.7.

Suppose we have continuous maps

Suppose that $f_{0}$ is homotopic to $f_{1}$, and that $g_{0}$ is homotopic to $g_{1}$. Then $g_{1}\circ f_{1}$ is homotopic to $g_{0}\circ f_{0}$.

Proof.

We are assuming that $f_{0}$ is homotopic to $f_{1}$, which means that there is a homotopy $h\colon[0,1]\times X\to Y$ 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 $k\colon[0,1]\times Y\to Z$ with $k(0,y)=g_{0}(y)$ and $k(1,y)=g_{1}(y)$ for all $y\in Y$. We can therefore define $m\colon[0,1]\times X\to Z$ 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 $p\colon[0,1]\times X\to[0,1]$ be the projection, and note that $m$ can be written as the composite

 $[0,1]\times X\xrightarrow{\langle p,h\rangle}[0,1]\times Y\xrightarrow{k}Z.$

Here $p$ is continuous by Lemma 7.10, so $\langle p,h\rangle$ is continuous by Proposition 7.11, so $m$ is continuous by Proposition 3.24. ∎

Corollary 9.8.

For $u\in[X,Y]$ and $v\in[Y,Z]$, there is a well-defined composite $v\circ u\in[X,Z]$, given by $v\circ u=[g\circ f]$ for any choice of maps $f,g$ with $u=[f]$ and $v=[g]$.

Proof.

Immediate from the proposition. ∎

Definition 9.9.

We can now define a new category $\operatorname{hTop}$, called the homotopy category. The objects are topological spaces (just as for the category $\operatorname{Top}$), but the morphisms are now homotopy classes of maps, so

 $\operatorname{hTop}(X,Y)=\operatorname{Top}(X,Y)/\equiv=[X,Y].$

The composition rule is given by Corollary 9.8. The identity morphism in $\operatorname{hTop}$ for an object $X$ is just the homotopy class $[\operatorname{id}_{X}]$ corresponding to the identity function.

Video (Definition 9.10 to Proposition 9.20)

Definition 9.10.

A continuous map $f\colon X\to Y$ 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 $g\colon Y\to X$ such that $g\circ f\equiv\operatorname{id}_{X}$ and $f\circ g\equiv\operatorname{id}_{Y}$. 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.

Lemma 9.11.

Every homeomorphism is a homotopy equivalence.

Proof.

Let $f\colon X\to Y$ be a homeomorphism, with inverse $g\colon Y\to X$. This means that $g\circ f=\operatorname{id}_{X}$ and $f\circ g=\operatorname{id}_{Y}$. As $g\circ f$ is equal to $\operatorname{id}_{X}$, it is certainly homotopic to $\operatorname{id}_{X}$. As $f\circ g$ is equal to $\operatorname{id}_{Y}$, it is certainly homotopic to $\operatorname{id}_{Y}$. Thus, $g$ is also a homotopy inverse for $f$, so $f$ is a homotopy equivalence. ∎

A basic example is as follows:

Proposition 9.12.

For $n>0$, the sphere $X=S^{n-1}$ is homotopy equivalent to the space $Y={\mathbb{R}}^{n}\setminus\{0\}$.

Proof.

Recall that $X=S^{n-1}=\{x\in{\mathbb{R}}^{n}\;|\;\|x\|=1\}$. We define $f\colon X\to Y$ to be the inclusion function, so we just have $f(x)=x$. We define $g\colon Y\to X$ by $g(y)=y/\|y\|$. (It is important here that $Y$ is the space of nonzero vectors in ${\mathbb{R}}^{n}$, so $\|y\|>0$ and it is valid to divide by $\|y\|$.) Note that for $x\in X$ we have $\|x\|=1$ so $g(f(x))=x/\|x\|=x$, so $g\circ f=\operatorname{id}_{X}$. As $g\circ f$ is equal to $\operatorname{id}_{X}$, it is certainly homotopic to $\operatorname{id}_{X}$. In the opposite direction, however, the composite $g\circ f$ is not equal to the identity. Nonetheless, we can define $h\colon[0,1]\times Y\to{\mathbb{R}}^{n}$ by

 $h(t,y)=(1-t)\frac{y}{\|y\|}+ty=\left((1-t)\|y\|^{-1}+t\right)y.$

Here $(1-t)\|y\|^{-1}+t$ is always strictly positive (for $0\leq t\leq 1$), 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 $\operatorname{id}_{Y}$. This proves that $g$ is a homotopy inverse for $f$, so that $f$ is a homotopy equivalence.

Definition 9.13.

Let $X$ be a topological space, and let $a$ be a point in $X$. A contraction of $X$ to $a$ is a continuous map $h\colon[0,1]\times X\to X$ 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 $a\in X$).

Example 9.14.

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 $B^{n}$, $\Delta_{n}$ and $[0,1]^{n}$ are all contractible.

Example 9.15.

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.

Definition 9.16.

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).

Proposition 9.17.

A space $X$ is contractible iff it is homotopy equivalent to $1$.

Proof.

Suppose that $X$ is homotopy equivalent to $1$. This means that we have maps $f\colon X\to 1$ and $g\colon 1\to X$, and a homotopy $h$ from $g\circ f$ to $\operatorname{id}_{X}$, and a homotopy $k$ from $f\circ g$ to $\operatorname{id}_{1}$. 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 $\operatorname{id}_{X}$, 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 $1$. ∎

Remark 9.18.

The map $[0,1]\to 1$ is the most basic example of a homotopy equivalence that is not a homeomorphism.

Proposition 9.19.

The relation of being homotopy equivalent is an equivalence relation.

Proof.

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 $f\colon X\xrightarrow{}Y$ and a homotopy inverse $g\colon Y\xrightarrow{}X$, so $fg\colon Y\xrightarrow{}Y$ and $gf\colon X\xrightarrow{}X$ 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

 $degf=d(eg)f\simeq d1_{Y}f=df\simeq 1_{X}$
 $gfde=g(fd)e\simeq g1_{Y}e=ge\simeq 1_{Z}.$

This proves that $gf\colon X\xrightarrow{}Z$ is a homotopy equivalence, with homotopy inverse $de\colon Z\xrightarrow{}X$. This in turn shows that our relation is transitive, as required. ∎

Proposition 9.20.

Suppose that $X_{0}$ is homotopy equivalent to $Y_{0}$ and $X_{1}$ is homotopy equivalent to $Y_{1}$. Then $X_{0}\times X_{1}$ is homotopy equivalent to $Y_{0}\times Y_{1}$.

Proof.

For $i=0,1$ we let $f_{i}\colon X_{i}\to Y_{i}$ be a homotopy equivalence, with homotopy inverse $g_{i}\colon Y_{i}\to X_{i}$. This means that we have homotopies $h_{i}$ from $g_{i}\circ f_{i}$ to $\operatorname{id}_{X_{i}}$, and homotopies $k_{i}$ from $f_{i}\circ g_{i}$ to $\operatorname{id}_{Y_{i}}$. Now make the following definitions:

 $\displaystyle f$ $\displaystyle\colon X_{0}\times X_{1}\to Y_{0}\times Y_{1}$ $\displaystyle f(x_{0},x_{1})$ $\displaystyle=(f_{0}(x_{0}),f_{1}(x_{1}))$ $\displaystyle g$ $\displaystyle\colon Y_{0}\times Y_{1}\to X_{0}\times X_{1}$ $\displaystyle g(y_{0},y_{1})$ $\displaystyle=(g_{0}(y_{0}),g_{1}(y_{1}))$ $\displaystyle h$ $\displaystyle\colon[0,1]\times X_{0}\times X_{1}\to X_{0}\times X_{1}$ $\displaystyle h(t,x_{0},x_{1})$ $\displaystyle=(h_{0}(t,x_{0}),\;h_{1}(t,x_{1}))$ $\displaystyle k$ $\displaystyle\colon[0,1]\times Y_{0}\times Y_{1}\to Y_{0}\times Y_{1}$ $\displaystyle k(t,y_{0},y_{1})$ $\displaystyle=(k_{0}(t,y_{0}),\;k_{1}(t,y_{1})).$

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. ∎

Example 9.21.

The solid torus, the Möbius band, and ${\mathbb{C}}\setminus\{0\}$ 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 $(2\cos(\theta),2\sin(\theta),0)$. 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 $f\colon S^{1}\xrightarrow{}M^{\prime}$ and $g\colon M^{\prime}\xrightarrow{}S^{1}$ and $h\colon I\times M^{\prime}\xrightarrow{}M^{\prime}$ by

 $\displaystyle f(z)$ $\displaystyle=(z,0)$ $\displaystyle g(z,w)$ $\displaystyle=z$ $\displaystyle h(t,(z,w))$ $\displaystyle=(z,tw).$

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 ${\mathbb{C}}$ can be identified with ${\mathbb{R}}^{2}$, so ${\mathbb{C}}\setminus\{0\}$ is also homotopy equivalent to $S^{1}$.

Definition 9.22.

Let $X$ and $Y$ be topological spaces. We say that $X$ is a homotopy retract of $Y$ if there exist continuous maps $X\xrightarrow{f}Y\xrightarrow{g}X$ such that $g\circ f$ is homotopic to $\operatorname{id}_{X}$. (We make no assumption about $f\circ g$.) Any pair $(f,g)$ with this property will be called a homotopy retraction pair for $(X,Y)$.

Example 9.23.

Put

 $\displaystyle X$ $\displaystyle={\mathbb{R}}^{2}\setminus\{0\}$ $\displaystyle Y$ $\displaystyle=\text{ figure eight }=\{(x,y)\in{\mathbb{C}}\;|\;(x-1)^{2}+y^{2}% =1\text{ or }(x+1)^{2}+y^{2}=1\}.$

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 $X\xrightarrow{f}Y\xrightarrow{g}X$ by

 $\displaystyle f(x,y)$ $\displaystyle=(1,0)+(x,y)/\|(x,y)\|$ $\displaystyle g(x,y)$ $\displaystyle=(x-1,y).$

The composite $g\circ f\colon X\to X$ is just $(g\circ f)(x,y)=(x,y)/\|(x,y)\|$. The straight line joining $(x,y)$ to $(x,y)/\|(x,y)\|$ does not pass through the origin, so $g\circ f$ is homotopic to the identity as required. (The map $f\circ g\colon Y\to Y$ is not homotopic to the identity, so we do not have a homotopy equivalence, but that is not important.)

Proposition 9.24.

Suppose that $X$ is a homotopy retract of $Y$ and that $Y$ is contractible. Then $X$ is also contractible.

Proof.

By hypothesis, we have continuous maps

such that $gf$, $qp$ and $pq$ are homotopic to the respective identity maps. Now put $m=pf\colon X\to 1$ and $n=gq\colon 1\to X$. We then have $nm=gqpf\simeq gf\simeq\operatorname{id}\colon X\to X$. Also, $mn$ is a map from $1$ to $1$, and the only map from $1$ to $1$ is the identity, so $mn=\operatorname{id}$. Thus, $m$ and $n$ give a homotopy equivalence from $X$ to $1$, proving that $X$ is contractible. ∎

Proposition 9.25.

Suppose we have two continuous maps $f,g\colon X\to Y$, giving maps $f_{*},g_{*}\colon\pi_{0}(X)\to\pi_{0}(Y)$ as in Proposition 5.20. If $f$ is homotopic to $g$, then $f_{*}=g_{*}$.

Proof.

Let $h\colon[0,1]\times X\to Y$ 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 $u(t)=h(t,a)$. ∎