Jae Choon Cha

POSTECH

From now on, we will often omit the symbol $*$ in a path product $\alpha*\beta$. Furthermore, as an abuse of notation, we will often denote the product of three paths by $\alpha\beta\gamma$, when we are interested in its homotopy class only. Note that while $\alpha(\beta\gamma)$ and $(\alpha\beta)\gamma$ are different as functions on $I$, they are homotopic rel $\{0,1\}$, by our earlier discussion on reparametrization. Furthermore, one can also use the partition $\{0,\frac13,\frac23,1\}$ consisting of subintervals of the same length, to define the product of three paths, which is again homotopic to them. Namely the path given by

is also homotopic rel $\{0,1\}$ to both of $\alpha(\beta\gamma)$ and $(\alpha\beta)\gamma$. We will often regard $\alpha\beta\gamma$ as this path. The same remark applies to the product of arbitrary number of paths in general.

### Change of the basepoint of the fundamental group

Recall that the definition of the fundamental group involves the choice of a point $x_0$, which we call a basepoint. A natural question is how $\pi_1(X,x_0)$ changes when $x_0$ varies. We begin with the following observation.

**Theorem.** Let $C$ be the path component of $X$ containing $x_0$. Then
$\pi_1(X,x_0)=\pi_1(C,x_0)$.

*Proof.* Since $I$ is path connected, any loop $\alpha\colon I\to X$ based at
$x_0$ is automatically in $C$. Similarly, any homotopy $F\colon I\times I \to X$ between two loops based at $x_0$ is in $C$. The conclusion follows from
this. $\quad\square$

The above theorem tells us that it suffices to consider a path component when we study the fundamental group. That is, it is enough to consider the case that $X$ is path connected. Suppose $x_0$ and $x_1$ are points in a path connected space $X$, and $\alpha$ is a path from $x_0$ to $x_1$, that is, $\alpha(0)=x_0$ and $\alpha(1)=x_1$. Define a function

by $\alpha_* ([\gamma]) = [\alpha \gamma \alpha^{-1}]$.

**Theorem.** $\alpha_*$ is a well-defined group isomorphism.

*Proof.* First, $\alpha$ is a well defined function, by appealing to the fact
that the composition of homotopic maps are homotopic. More precisely, if
$\gamma\simeq \gamma' \rel\{0,1\}$, then $\alpha \gamma \alpha^{-1} \simeq\alpha \gamma' \alpha^{-1} \rel\{0,1\}$.

Suppose $[\gamma]$ and $[\delta]$ are elements in $\pi_1(X,x_1)$. Then

This show that $\alpha_*$ is a group homomorphism.

It remains to show that $\alpha_*$ has an inverse. For this purpose, we will use the following lemma:

**Lemma.** (1) The homorophism $\alpha_*$ depends on the homotopy class of
$\alpha$,$\rel\{0,1\}$. That is, if $\alpha\simeq \beta \rel\{0,1\}$, then
$\alpha_* = \beta_*$.

(2) When $\alpha \beta$ is defined,
$(\alpha\beta)_ * = \alpha_ * \circ \beta_ *$.

*Proof.* For any $[\gamma]\in\pi_1(X,x_1)$, $\alpha\gamma\alpha^{-1} \simeq\beta\gamma\beta^{-1} \rel\{0,1\}$ since the path product is well-defined on
homotopy classes. This proves (1). Also, we have $[(\alpha \beta) \gamma (\alpha \beta)^{-1}] = [\alpha (\beta \gamma \beta^{-1}) \alpha^{-1}]$.
This shows (2). $\quad\square$

Now, returning to the proof of the theorem, consider the path $\alpha^{-1}$, which is from $x_1$ to $x_0$. It gives the homomorphism $(\alpha^{-1})_ *\colon \pi_1(X,x_0) \to \pi_1(X,x_1)$. We claim that $(\alpha^{-1})_ *$ is the inverse of $\alpha_ *$. Indeed, since $\alpha^{-1}*\alpha \simeq c_{x_1}\rel\{0,1\}$, we have $(\alpha^{-1})_ * \circ \alpha_ * = (\alpha^{-1}*\alpha)_* = (c_{x_0})_ * = \id_ {\pi_1(X,x_1)}$ by the above lemma. Exchanging $\alpha$ and $\alpha^{-1}$, we obtain $\alpha_ * \circ (\alpha^{-1})_ * = \id_{\pi_1(X,x_0)}$. This shows the claim, and completes the proof of the theorem. $\quad\square$

**Corollary.** If $X$ is path connected, then $\pi_1(X,x_0) \cong \pi_1(X,x_1)$
for any $x_0$, $x_1\in X$.

### Free homotopy of maps $S^1\to X$

In the above discussion, we showed that change of a basepoint does not alter the fundamental group up to isomorphism. That is, a choice of a basepoint has essentially no effect. This leads us to a natural question: would it be possible to remove basepoint in the theory of the fundamental group (at least for the case of path connected spaces)? In what follows, we will discuss a natural attempt toward this direction. Indeed, we will see that the answer is a bit sophisticated than one might guess.

Define $[Y,X]$ to be the set of homotopy classes of maps $Y\to X$. That is,

Denote the homotopy class of a map $f\colon Y\to X$ by $[f]\in [Y,X]$ as usual.

We are particularly interested in the case of $Y=S^1$. Observe that in the
definition of $[S^1,X]$, there is no restriction on the image of the basepoint
$1\in S^1$ under $f$. Also, the homotopy is not required to be relative to
anything. When two maps $f$, $g\colon S^1\to X$ are homotopic, not necessarily
relative to anything, we say that $f$ and $g$ are *freely homotopic*. So,
"freely homotopic" is identical with "homotopic", but we use the former term
when we want to emphasize that the homotopy is not relative to anything.

Our goal is to relate $[S^1,X]$ with the fundamental group. For this purpose, we will first construct a natural function between them, which will be denoted by $\phi$ in the below.

For a loop $\alpha\colon I\to X$, define $f_\alpha\colon S^1\to X$ by $f_\alpha(e^{2\pi si})=\alpha(s)$ for $0\le s\le 1$. It is straightforward to verify that $f_\alpha$ is a well-defined map using the property $\alpha(0)=\alpha(1)$. Define a function

by $\phi([\alpha]) = [f_\alpha]$. It is also straightforward to verify that this is well-defined, namely, $f_\alpha\simeq f_\beta$ if $\alpha\simeq \beta\rel\{0,1\}$.

**Lemma.** If $\alpha$ is a loop and $\beta$ is a path in $X$ such that
$\alpha(0)=\alpha(1)=\beta(0)$, then $f_\alpha$ and $f_{\beta^{-1}\alpha\beta}$
are freely homotopic.

Before proving the lemma, observe the following: the following lemma tells us
that $\phi$ is *not* injective unless $\pi_1(X,x_0)$ is not abelian. For,
whenever both $\alpha$ and $\beta$ are loops based at $x_0$, the lemma gives us
$\phi([\alpha])=\phi([\beta]^{-1}[\alpha][\beta])$, that is,
$[\alpha]^{-1}[\beta]^{-1}[\alpha][\beta]$ lies in the kernel of $\phi$; but
$[\alpha]^{-1}[\beta]^{-1}[\alpha][\beta]$ is nontrivial in general, if the
fundamental group is not abelian.

*Proof of Lemma.* Define a homotopy $F\colon I\times I \to X$ by

Indeed, it is straightforward to verify that $F$ is a homotopy from $\alpha$ to $\beta^{-1}\alpha\beta$, and that $F(0,t)=F(1,t)=\beta(t)$. (So, $F$ is not rel $\{0,1\}$ in general!)

Define $G\colon S^1\times I \to X$ by $G(e^{2\pi i s}) = F(s,t)$. Since $F(0,t)=F(1,t)$, $G$ is well-defined. Furthermore, $G\colon f_\alpha\simeq f_{\beta^{-1}\alpha \beta}$. $\quad\square$

While $\phi$ is not injective as observed above, it is still useful in comparing
$\pi_1(X,x_0)$ and $[S^1,X]$, as shown in the following result. Recall that two
elements $a$ and $b$ in a group $G$ are *conjugate* if $b=g^{-1}ag$ for some
$g\in G$. Conjugacy is an equivalence relation. A equivalence class is called
a *conjugacy class*. That is, $G/\text{conjugacy}$ is the set of conjugacy
classes.

**Theorem.** The function $\phi$ gives rise to a bijection

*Proof.* By the above lemma, $\phi([\alpha])=\phi([\alpha'])$ whenever
$[\alpha]$ and $[\alpha']$ are conjugate in $\pi_1(X,x_0)$. So $\phi$ induces a
well-defined function $\Phi\colon \pi_1(X,x_0)/\text{conjugacy} \to [S^1,X]$.
Namely, $\Phi$ is the map sending the conjugacy class of $[\alpha]\in\pi_1(X,x_0)$ to $\phi([\alpha]) = f_\gamma$. We will show that $\Phi$ is
bijective.

First we will prove the surjectivity. Suppose $f\colon S^1\to X$. Define $\gamma\colon I\to X$ by $\gamma(s)=f(e^{2\pi is})$. Then we have $f_\gamma=f$ by definition. Choose a path $\beta\colon I \to X$ from $f(1)$ to $x_0$. Then $f_\gamma \simeq f_{\beta^{-1}\gamma\beta}$ by the above lemma. Since $\beta^{-1}\gamma\beta$ is a loop based at $x_0$, we have $\phi([\beta^{-1}\gamma\beta]) = [f_{\beta^{-1}\gamma\beta}] = [f_\gamma]=[f]$. This shows the surjectivity.

It remains to prove the injectivity of $\Phi$ (not $\phi$!). Suppose $\phi([\alpha]) = \phi([\alpha'])$. We need to show that $[\alpha]$ and $[\alpha']$ are conjugate in $\pi_1(X,x_0)$. Since $f_\alpha \simeq f_{\alpha'}$, there is a homotopy $F\colon S^1\times I \to X$ such that $F(e^{2\pi i s},0) = \alpha(s)$, $F(e^{2\pi i s},1) = \alpha'(s)$. Define $G\colon I\times I \to X$ by $G(s,t)=F(e^{2\pi i s},t)$. Then $G(0,t)=G(1,t)$. Define $\beta\colon I\to X$ by $\beta(s)=G(0,s)$. (See the diagram below.) Since $\beta(0)=\alpha(0)=x_0=\alpha'(0)=\beta(1)$, $\beta$ is a loop based at $x_0$.

Let $b$, $a$, $l$, $r$ be the paths in $I\times I$ defined by $b(s)=(s,0)$, $a(s)=(s,1)$, $l(s)=(0,s)$, $r(s)=(1,s)$. Then, $G\circ b = \alpha$, $G\circ a= \alpha'$, $G\circ l = \beta = G\circ r$. (You must be able to visualize what these paths are in the above square diagram.) Since both paths $b$ and $lar^{-1}$ are from $(0,0)$ to $(1,0)$ in the convex set $I\times I$, they are homotopic rel $\{0,1\}$. It follows that the paths $G\circ b$ and $G\circ(lar^{-1})$ in $X$ are homotopic rel $\{0,1\}$. Since $G\circ b =\alpha$ and $G\circ(lar^{-1}) = (G\circ l)(G\circ a)(G\circ r^{-1}) =\beta\alpha'\beta^{-1}$, it follows that $[\alpha] =[\beta][\alpha'][\beta^{-1}]$. That is, $[\alpha]$ and $[\alpha']$ are conjugate in $\pi_1(X,x_0)$. This shows that $\Phi$ is injective. $\quad\square$