## MATH 422 Lecture Note #2 (2018 Spring)Change of basepoint and free homotopy

Jae Choon Cha
POSTECH $\def\id{\operatorname{id}}\def\rel{\text{ rel }}\def\R{\mathbb{R}}\def\C{\mathbb{C}}$

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

$s\mapsto \begin{cases} \alpha(3s) &\text{for }0\le s \le \frac13\\ \beta(3s-1) &\text{for }\frac13 \le s \le \frac23\\ \gamma(3s-2) &\text{for }\frac23 \le s \le 1 \end{cases}$

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

$\alpha_* \colon \pi_1(X,x_1)\to \pi_1(X,x_0)$

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

\begin{aligned} \alpha_* ([\gamma]\cdot[\delta]) &= \alpha_* ([\gamma \delta]) \\ & = [\alpha \gamma \delta \alpha^{-1}] \\ & = [\alpha \gamma \alpha^{-1} \alpha \delta \alpha^{-1}] \\ & = [\alpha \gamma \alpha^{-1}] \cdot [\alpha \delta \alpha^{-1}] \\ & = \alpha_ ([\gamma]) \cdot \alpha_ ([\delta]). \end{aligned}

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,

$[Y,X] = \{\text{maps } f\colon Y\to X\}/\simeq.$

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

$\phi\colon \pi_1(X,x_0) \to [S^1,X]$

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

$F(s,t)=\begin{cases} \beta(3(\frac t3-s)) & \text{for }0\le s \le \frac t3, \\ \alpha\big(\frac{s-t/3}{1-2t/3}\big) & \text{for }\frac t3 \le s \le 1-\frac t3, \\ \beta(3(s-\frac{3-t}{3})) & \text{for }1-\frac t3 \le s \le 1. \end{cases}$
It may be helpful to compare this definition with the following picture:

running…

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

$\Phi\colon \pi_1(X,x_0)/\text{conjugacy} \stackrel{\approx}{\to} [S^1,X].$

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 $x_0$.

running…

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. $\quad\square$