Jae Choon Cha

POSTECH

In this course, spaces and maps designate topological spaces and continuous maps, unless stated otherwise. The unit circle in the complex plane $\C$ is denoted by $S^1$, and the unit interval $[0,1]$ is denoted by $I$.

### Homotopy

#### When are two mathematical objects the same?

In mathematics, we often regard two "different" objects as the same. From basic
abstract algebra, recall the following question: how many groups of order two
are there? The standard answer is that there is only one group of order two,
*up to isomorphisms*. That is, any two given groups of order two are
isomorphic. Here we see clearly that we regard two groups as the "same" if
there is an isomorphism between them. We do similarly for spaces as well: two
spaces are the same when there is a *homeomorphism* between them. For instance,
the coffee mug with a handle and the donut with a hole are identical in
topology, since there is a homeomorphism between them.

So, the notion of an isomorphism for groups, and the notion of a homeomorphism
for spaces, are essential in *defining* when two are the same. The two cases
are very similar. We consider a certain type of functions, namely isomorphisms
and homeomorphisms respectively, which preserve the concerned structure on our
objects: the group operation and topology in the above cases. (Indeed, this
generalizes to the notion of an *equivalence* in a *category*, not limited to
the case of groups and spaces, but we will not discuss it too much.)

In the definition of a homeomorphism, the notion of an inverse is essential. A map $f\colon X\to Y$ between spaces is a homeomorphism if and only if there is a map $g\colon Y\to X$ which is an inverse of $f$. Similarly, a homomorphism $f\colon G\to \Gamma$ between groups is an isomorphism if and only if there is a homomorphism $g\colon \Gamma\to G$ which is an inverse of $f$. (An alternative definition of a group isomorphism says that $f$ is an isomorphism if $f$ is a bijective homomorphism, but it is easily seen that this is equivalent to ours given above.)

Here, what is an inverse of $f\colon X\to Y$? It is $g\colon Y\to X$ satisfying $g\circ f=\mathrm{id}_ X$ and $f\circ g=\mathrm{id}_ Y$. Observe that here we have equalities, for maps this time. That is, the notion of the same objects is eventually reduced to the notion of the same maps (or homomorphisms in case of groups). Are $g\circ f$ and $f\circ g$ equal to the identity? If so, one is the inverse of another.

What we want to do with the notion of *homotopy* is to generalize the notion of
the same objects by generalizing the notion of the same maps.

#### Definition of homotopy

**Definition.** Two maps $f$, $g\colon X\to Y$ are said to be *homotopic* if there
is a map $F\colon X\times I \to Y$ satisfying that $F(x,0)=f(x)$ and
$F(x,1)=g(x)$ for all $x\in X$. We call $F$ is a *homotopy* from $f$ to $g$,
and write $F\colon f\simeq g$. Furthermore, for a subspace $A$ of $X$, if
$F(a,t)=f(a)$ for all $a\in A$ and $t\in I$, then we say that $f$ and $g$ are
*homotopic$\rel A$* and $F$ is a *homotopy$\rel A$*. We write
$F\colon f\simeq g \rel A$.

When we want to say only that there exists a homotopy from $f$ to $g$, without specifying what the homotopy is, we often write $f\simeq g$ or $f\simeq g \rel A$.

The idea is to use homotopy as a weaker notion of "being the same" for maps. For this purpose, we first verify that homotopy is an equivalence relation.

**Theorem.** For two spaces $X$ and $Y$, let $C(X,Y)$ be the set of maps
$f\colon X\to Y$. Suppose $A$ is a subset of $X$ (which is possibly empty).
Then $\simeq \rel A$ is an equivalence relation on $C(X,Y)$.

The proof is routine. We leave it as an exercise.

Now we are ready to define the promised equivalence relation of spaces.

**Definition.** Two spaces $X$ and $Y$ are *homotopy equivalent* if there exist
maps $f\colon X\to Y$ and $g\colon Y \to X$ satisfying $g\circ f\simeq\id_X$ and
$f\circ g \simeq \id_Y$. If it is the case, each of $f$ and $g$ is called a
*homotopy equivalence*, and $g$ is called the *homotopy inverse* of $f$. We
also say that $X$ and $Y$ have the same *homotopy type*.

**Definition.** A space $X$ is *contractible* if $X$ and a one-point space $\{*\}$ is homotopy equivalent.

The following lemmas give many examples of contractible spaces. When $x_0\in X$, define the *constant map* $c_{x_0}\colon X\to X$ by $c_{x_0}(x)=x_0$.

**Lemma.** $X$ is contractible if and only if the identity map $\id_X$ is homotopic to $c_{x_0}$ for some $x_0\in X$.

Proof. For the if direction, suppose $\id_X\simeq c_{x_0}$. Let $f\colon X\to \{*\}$ be the unique map, and define $g\colon \{*\} \to X$ by $g(*)=x_0$. Then $g\circ f$ is equal to $c_{x_0}$, and thus is homotopic to $\id_X$. Also, $f\circ g\colon \{*\} \to \{*\}$ is automatically equal to $\{*\}$. So $f$ and $g$ are homotopy equivalences. Conversely, suppose $X$ is contractible, that is, there is a map $f\colon X\to \{*\}$ which has a homotopy inverse $g\colon \{*\} \to X$. Let $x_0=g(*)$. Then $\id_X = g\circ f = c_{x_0}$. $\quad\square$

**Theorem.** Every convex subset $X$ in the Euclidean space $\R^N$ is
contractible.

*Proof.* Fix $x_0\in X$. Define $F\colon X\times I \to X$ by $F(x,t)=(1-t)x+tx_0$, using the vector space structure of $\R^N$. Since $X$ is convex, $F(x,t)\in X$ for each $x\in X$ and $t\in I$. Since $F(x,0)=x$ and $F(x,1)=x_0$, $F\colon \id_X \simeq c_{x_0}$. $\quad\square$

We remark that $F$ in the above proof is sometimes called a straight line homotopy.

As another fundamental property of homotopy, we state the behavior of homotopy under composition.

**Theorem.** Composition of homotopic maps are homotopic. More precisely, if
$f\simeq g \rel A$, $f'\simeq g' \rel B$ and $f(A)\subset B$, then
$f'\circ f \simeq g'\circ g \rel A$.

*Proof.* If $F\colon f\simeq g \rel A$ and $F'\colon f'\simeq g' \text{ rel }B$, then $G(x,t)=F'(F(x,t),t)$ is a desired homotopy. Detailed
verification, including the continuity, is left as an exercise. $\quad\square$

### Fundamental group

Let $X$ be a space. Fix $x_0 \in X$. Recall $S^1=\{x\in \mathbb{C} \mid |x|=1\}$ is the unit circle in the complex plane. In particular $1\in S^1$.

**Definition.** The fundamental group of $(X,x_0)$ is defined by

For now, $\pi_1(X,x_0)$ is merely a set, although it is called a group. A group structure will be given below. As a preliminary for the description of the group structure, note that $\pi_1(X,x_0)$ is in 1-1 correspondence with

under the association $f\leftrightarrow \alpha_f$ where $\alpha_f\colon I\to X$ is the map defined by $\alpha_f(s)=f(e^{2\pi s\sqrt{-1}})$. To give a proof, use the following two standard facts: (i) $S^1$ is homeomorphic to the quotient space $I/0\sim 1$, and (ii) for a quotient space $Y/\mathord{\sim}$ of a space $Y$, a map $f\colon Y \to Z$ induces a map $Y/\mathord{\sim} \to Z$ which fits into the following commutative diagram

if and only if $f(p)=f(q)$ whenever $p\sim q$; furthermore, every map $Y/\mathord{\sim} \to Z$ is induced by such a map $f$. Details are left as an exercise.

Recall that a map $\alpha\colon I\to X$ is called a *path in $X$*. When
$\alpha(0)=x_0=\alpha(1)$ as above, we call $\alpha$ a *loop in $X$ based at
$x_0$*. We will frequently use the second description given above, which is in
terms of loops, as the definition of $\pi_1(X,x_0)$. We denote the class of a
loop $\alpha$ based at $x_0$ by $[\alpha]\in \pi_1(X,x_0).$

#### Group operations on $\pi_1(X,x_0)$

First, we will establish a group structure on $\pi_1(X,x_0)$, and then discuss some examples.

**Definition.** For two paths $\alpha$ and $\beta$ in $X$ such that
$\alpha(1)=\beta(0)$, the *product path* $\alpha *\beta\colon I \to X$ is
defined by

The *inverse path* $\alpha^{-1}\colon I \to X$ is defined by

Intuitively, $\alpha*\beta$ can be viewed as the concatenation of $\alpha$ and $\beta$; the first half is equal to $\alpha$ with double speed (so that the whole $\alpha$ is done for $0\le s \le \frac12$), and the remaining half is equal to $\beta$ with double speed. The inverse path $\alpha^{-1}$ is the path $\alpha$ with reversed orientation. (Here, the word "orientation" is used as an informal word, not as a rigorously defined mathematical term, at least for now.)

The following lemma says that the product of homotopic ($\text{rel }\{0,1\}$) paths are homotopic ($\text{rel }\{0,1\}$), and the same holds for the inverse.

**Lemma.** Suppose $\alpha \simeq \alpha' \rel \{0,1\}$ and $\beta \simeq \beta' \rel \{0,1\}$. Whenever $\alpha* \beta$ is defined (that is,
$\alpha(0)=\beta(1)$), $\alpha'* \beta'$ is well defined too, and $\alpha* \beta \simeq \alpha'* \beta' \rel \{0,1\}$. In addition, $\alpha^{-1} \simeq (\alpha')^{-1} \rel \{0,1\}$.

**Proof.** For the first part, suppose $F\colon \alpha \simeq \alpha' \rel \{0,1\}$ and $G\colon \beta \simeq \beta' \rel \{0,1\}$ are homotopies. Define
$H\colon I\times I \to X$ by

Then it is straightforward to verify that $H$ is a well-defined (continuous) map by using the pasting lemma, and that $H$ is a homotopy, $\text{rel }\{0,1\}$, between $\alpha* \beta$ and $\alpha'* \beta'$. This proves the first assertion. For the assertion on the inverse, define $K\colon I\times I \to X$ by $K(s,t)=F(1-s,t)$. Then $K$ is a homotopy between $\alpha^{-1}$ and $(\alpha')^{-1} \rel \{0,1\}$. $\quad\square$

As a special case of our previous notation, let $c_{x_0}\colon I\to X$ be the constant path at $x_0$, that is, $c_{x_0}(s) = x_0$ for all $s$.

**Theorem.** The set $\pi_1(X,x_0)$ is a group under the group operation
$[\alpha]\cdot [\beta] = [\alpha*\beta]$. The class $[c_{x_0}]$ is the
identity. The inverse of $[\alpha]$ in the group $\pi_1(X,x_0)$ is equal to
$[\alpha^{-1}]$.

*Proof.* The two operations

and

are well-defined by the above lemma. So it remains to verify the three axioms of a group.

Associativity. Suppose $[\alpha]$, $[\beta]$, $[\gamma]\in \pi_1(X,x_0)$. We want to show that $(\alpha* \beta)* \gamma \simeq \alpha* (\beta* \gamma) \rel\{0,1\}$. The idea is to decompose the domain square $I\times I$ into three parts as in the following picture:

Motivated from the picture, define $H\colon I\times I \to I$ by

Then it is straightforward to check that $H$ is a desired homotopy.

Identity and inverse. Since arguments similar to the associativity case proves the desired properties, we will describe brief outlines only, and leave the details as an exercise to the readers. To verify $\alpha* c_{x_0} \simeq \alpha \simeq c_{x_0}* \alpha \rel\{0,1\}$, construct homotopies which correspond to the following pictures:

Also, to verify $\alpha* \alpha^{-1} \simeq c_{x_0} \rel\{0,1\}$, use a homotopy corresponding to the following:

Namely, define a homotopy by

By replacing $\alpha$ with $\alpha^{-1}$ and using $(\alpha^{-1})^{-1}=\alpha$ (verify this!), the remaining requirement $\alpha^{-1}* \alpha \simeq c_{x_0} \rel\{0,1\}$ follows from the above. $\quad\square$

#### Reparametrization

The following reparametrization trick is useful in proving that two paths are homotopic $\rel\{0,1\}$. We begin with a general lemma.

**Lemma.** Suppose $Y$ is a convex subset in $\R^N$, and $\phi$, $\psi\colon I\to Y$ are paths in $Y$ such that $\phi(t)=\psi(t)$ for $t=0,1$. Then
$\phi\simeq\psi \rel\{0,1\}$.

*Proof.* Once again, we use the straight line homotopy. Define
$F(s,t)=(1-t)\phi(s)+t\psi(s)$. Then $F(s,t)\in Y$, and thus $F\colon \phi\simeq\psi$. Since $F(s,t)=\phi(s)$ for $s=0,1$, $F$ is $\rel \{0,1\}$.
$\quad\square$

**Lemma.** Suppose $\phi$, $\psi\colon I\to I$ are maps satisfying
$\phi(t)=\psi(t)$ for $t=0,1$, and suppose $\alpha\colon I\to X$ is a path.
Then $\alpha\circ \phi \simeq \alpha\circ\psi \rel \{0,1\}$.

*Proof.* By the above lemma, $\phi\simeq\psi \rel\{0,1\}$ since $I$ is convex.
It follows that $\alpha\circ \phi \simeq \alpha\circ\psi \rel \{0,1\}$, using
the fact that composition of homotopic maps are homotopic. $\quad\square$

As an application, we will present an alternative proof of the group axioms of $\pi_1(X,x_0)$. To prove the associativity $(\alpha* \beta)* \gamma \simeq \alpha* (\beta* \gamma) \rel\{0,1\}$, define $\phi\colon I\to I$ by

Observe that $\phi$ sends the subdivision $\{0,\frac14,\frac12,1\}$ of the unit interval, which is used to define $(\alpha* \beta)* \gamma$, to the subdivision $\{0,\frac12,\frac34,1\}$ which is used to define $\alpha* (\beta* \gamma)$. Indeed, it is straightforward to verify that the composition $(\alpha* (\beta* \gamma))\circ \phi$ is equal to $(\alpha* \beta)* \gamma$. Since $\phi(0)=0$ and $\phi(1)=0$, we have

by the above lemma. This shows the associativity.

To show that $\alpha* c_{x_0}\simeq \alpha \rel\{0,1\}$, define $\phi\colon I\to I$ by

and use the property $\alpha\circ\phi = \alpha* c_{x_0}$. To show that $\alpha* \alpha^{-1} \simeq c_{x_0} \rel\{0,1\}$, define $\phi$, $\psi\colon I\to I$ by

and $\psi(s)=0$, and use the two facts $\alpha\circ \phi = \alpha* \alpha^{-1}$ and $\alpha\circ \phi = c_{x_0}$. We leave full details to the readers as an exercise.