## MATH 422 Lecture Note #5 (2018 Spring)Fundamental theorems on covering spaces, part I

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

### Statements of fundamental theorems, part I

In what follows, suppose $p\colon E\to B$ is a covering map. For $b\in B$, $p^{-1}(b)$ is called the fiber over $b$. The subspace $p^{-1}(b)$ is always discrete. We will first state the theorems, postponing their proofs.

Defintion. Suppose $f\colon X\to B$ is a map. A map $\tilde f\colon X\to E$ is called a lift of $f$ if $p\circ \tilde f=f$. That is, the following diagram is commutative.

running…

Theorem (Uniqueness of a lift). Suppose $\tilde{f_1}$, $\tilde{f_2}\colon X\to E$ are lifts of $f$ such that $\tilde{f_1}(x_0)=\tilde{f_2}(x_0)$ for some $x_0\in X$. If $X$ is connected, then $\tilde{f_1}=\tilde{f_2}$ on $X$.

Theorem (Homotopy lifting property). Suppose $F\colon X\times I \to B$ is a homotopy from $f\colon X\to B$, that is, $F(x,0)=f(x)$ for $x\in X$. Then, for any lift $\tilde f\colon X\to E$ of $f$, there is a lift $\tilde F\colon X\times I \to E$ of $F$ which is from $\tilde f$, that is, $p\tilde F=F$ and $\tilde F(x,0)=\tilde f(x)$ for $x\in X$.

running…

The following two results are immediate consequences of the homotopy lifting property and uniqueness of a lift. (You should be able to figure out what $X$ is used when you apply them to prove the following!)

Theorem (Path lifting). Suppose $\alpha\colon I\to B$ is a path, and $e_0$ lies in the fiber $p^{-1}(\alpha(0))$ over $\alpha(0)$. Then there is a unique lift $\tilde\alpha\colon I\to E$ of $\alpha$ such that $\tilde\alpha(0)=e_0$.

Theorem (Path homotopy lifting). Suppose $F\colon I\times I\to B$ is a homotopy $F\colon \alpha\simeq \beta \rel\{0,1\}$, where $\alpha$ and $\beta\colon I\to B$ are paths in $B$. Then, for any lift $\tilde\alpha\colon I\to E$ of $\alpha$, there is a unique lift $\tilde F\colon I\times I\to E$ of $\alpha$ such that $\tilde F\colon \tilde\alpha \simeq \tilde\beta \rel \{0,1\}$ for some lift $\tilde\beta$ of $\beta$.

As a convention, we write $f\colon (X,x_0)\to (Y,y_0)$ when a map $f\colon X \to Y$ satisfies $f(x_0)=y_0$.

In what follows, we fix a basepoint $b_0\in B$ of the base space, and fix a basepoint $e_0$ of the total space which lies in the fiber $p^{-1}(b_0)$. That is, our covering is $p\colon (E,e_0) \to (B,b_0)$.

Theorem (Fundamental groups). The induced homomorphism $p_*\colon \pi_1(E,e_0) \to \pi_1(B,b_0)$ is injective.

Using the above theorem, we often identify $\pi_1(E,e_0)$ with the image $p_*\pi_1(E,e_0)$. This enables us to regard $\pi_1(E,e_0)$ as a subgroup of $\pi_1(B,b_0)$.

Theorem (Fiber and cosets). Suppose $E$ is path connected. Then there is a well-defined bijection

$p^{-1}(b_0) \xrightarrow{\approx} \{\text{right cosets of p_*\pi_1(E,e_0) in \pi_1(B,b_0)}\}$

given as follows: for $e\in p^{-1}(b_0)$, choose a path $\tilde\alpha\colon I \to E$ from $b_0$ to $e$, and let $\alpha=p\circ\tilde\alpha$. Then the coset $p_*\pi_1(E,e_0)\cdot [\alpha]$ corresponds to $e$.

### An application to the circle

Using the above result, we can compute $\pi_1(S^1,1)$.

Theorem. $\pi_1(S^1,1)$ is an infinite cyclic group. In fact,

$\begin{array}{cc@{}c@{}c} \Phi\colon &\pi_1(S^1,1) &\longrightarrow& \Z \\ &[\alpha] & \longmapsto & \tilde\alpha(1) \end{array}$

is an isomorphism, where $\tilde\alpha$ is a lift of $\alpha$, with respect to the covering $\exp\colon \R\to S^1$ given by $\exp(s)=e^{2\pi i s}$, such that $\tilde\alpha(0)=0$.

Proof. Let $b_0=1\in S^1$ and $e_0=0\in \exp^{-1}(b_0)=\Z$, and invoke the above theorem on the fiber and cosets. Since $\R$ is contractible, $\pi_1(\R,0)$ is trivial, and from this it follows that $\Phi$ is a well-defined bijection by the above theorem. So, it suffices to show that $\Phi$ is a homomorphism. For $[\alpha]$, $[\beta]\in \pi_1(S^1,1)$, choose lifts $\tilde\alpha$, $\tilde\beta\colon I\to \R$ such that $\tilde\alpha(0)=0$, $\tilde\beta(0)=0$. Define $\tilde\beta{}'\colon I\to \R$ by $\tilde\beta{}'(s)=\tilde\alpha(1)+\tilde\beta(s)$. Then the path product $\tilde\alpha \tilde\beta{}'$ is defined. Furthermore, since $\tilde\alpha(1)$ lies in the fiber $\exp^{-1}(1)=\Z$, $\tilde\beta{}'$ is a lift of $\beta$, and thus $\tilde\alpha \tilde\beta{}'$ is a lift of $\alpha\beta$ starting from $0=e_0\in \exp^{-1}(b_0)$. It follows that

$\Phi([\alpha][\beta]) = (\tilde\alpha \tilde\beta{}')(1) = \tilde\beta{}'(1) = \tilde\alpha(1)+\tilde\beta(1) = \Phi([\alpha])+\Phi([\beta]).$

This shows that $\Phi$ is a homomorphism. $\quad\square$

### Proofs of the theorems

In what follows, we continue assuming that $p\colon E\to B$ is a covering.

#### Uniqueness of a lift

Suppose $f\colon X\to B$ has two lifts $\tilde f$, $\tilde f{}' \colon X\to E$. Let $A=\{x\in X\mid \tilde f(x) = \tilde f{}'(x)\}$. We claim that both $A$ and $X\sm A$ are open subsets of $X$. From this claim, it follows that $X=A$ since $A$ is nonempty (recall that $x_0 \in A$ by the hypothesis) and $E$ is connected. The promised conclusion follows from this.

First, to prove that $A$ is open, fix an arbitrary $x\in A$. Choose an evenly covered neighborhood $U$ of $f(x)$ in $B$. Since $\tilde f(x)=\tilde f{}'(x)$ lies in $p^{-1}(U)$, there is a sheet $V\subset E$ over $U$ which contains $\tilde f(x)$. Let $W=\tilde f{}^{-1}(V) \cap \tilde f{}' ^{-1}(V)$. Observe $x\in W$. Since $\tilde f$ is a lift and $\tilde f(W) \subset V$, we have $(p|_ V) \circ (\tilde f|_ W) = f$. Since $p|_ V$ is a homeomorphism, it follows that $\tilde f|_ W = (p|_ V)^{-1} \circ f$. The same argument shows that $\tilde f{} '|_ W = (p|_ V)^{-1} \circ f$. So $\tilde f = \tilde f{}'$ on $W$. Since $W$ is open and $x\in W \subset A$, it follows that $A$ is open.

It remains to show that $X\sm A$ is open. Suppose $x\in X\sm A$. Choose an evenly covered neighborhood $U$ of $f(x)$ in $B$. Let $V$ be the sheet containing $\tilde f(x)$, and $V'$ be the sheet containing $\tilde f{}'(x)$. If $V=V'$, then since $p\tilde f(x)=f(x)=p\tilde f{}'(x)$ and $p$ is injective on each sheet, it follows that $\tilde f(x)=\tilde f{}'(x)$. Since this contracdicts the hypothesis that $x\in X\sm A$, $V$ and $V'$ are different sheets. Now, let $W=\tilde f{}^{-1}(V) \cap \tilde f{}' ^{-1}(V)$. $W$ is an open set containing $x$, and $W\subset X\sm A$. This shows that $X\sm A$ is open. $\quad\square$

#### Homotopy lifting property

Suppose $F\colon X\times I \to B$ and $\tilde f\colon X \to E$ are given and satisfy $p\tilde f(x)=F(x,0)$. The strategy is to construct a lift on $\{x\}\times I$ for each $x$, and then combine them to produce the desired lift defined on $X\times I$. In doing this, we will take care of the continuity.

Fix $x\in X$. For the purpose of constructing a lift on $\{x\}\times I$, we will once again divide the problem into smaller domain. The idea is that when the domain for a lifting problem is sufficient small, then it maps into an evenly covered open set, so that it can be lifted into a sheet. To make this rigorous, we proceed as follows. For each $t\in I$, choose an evenly covered open set $U_t\subset B$ which contains $F(x,t)$. Since $F$ is continuous at $(x,t)$, there is a neighborhood $W_t\subset X$ of $x$ and $J_t\subset I$ of $t$ such that $F(W_t\times J_t)\subset U_t$. Note that $\{U_t\}_ {t\in I}$ is an open cover of $I$. Since $I$ is compact, there is a subdivision $0=t_0 < t_1 < \cdots < t_n=1$ of $I$ such that for each $i$, $[t_i,t_{i+1}]\subset J_t(i)$ for some $t(i)$. Let $U_i=U_{t(i)}$ for brevity. Then we have $F(W_i\times[t_i,t_{i+1}])\subset U_i$.

We will choose an open neighborhood $W_x$ of $x$ and define a map $G_x\colon W_x\times I \to E$ which is a lift of $F|_ {W_x\times I}$. For now, let $W_x = \bigcap_{i=0}^{n-1} W_i$. (This $W_x$ will not be the promised neighborhood in general; indeed, in the following argument, we will replace $W_x$, at most $n-1$ times, with a smaller open neighborhood of $x$. This is completely fine for our goal!) First, define $G_x$ on $W_x\times \{t_0\} = W_x\times \{0\}$ by $G_x(z,0)=\tilde f(z)$, that is, $G_x|_ {W_x\times \{0\}} = \tilde f|_ {W_x}$. We will proceed inductively to extend the domain of $G_x$. Suppose $G_x$ has been defined on $W_x\times[0,t_i]$ and satisfies $p\circ (G_x|_ {W_x\times[0,t_i]})=F_{W_x\times[0,t_i]}$. Since $F(W_x\times \{t_i\})$ lies in $U_i$, there is a sheet $V$ over $U_i$ containing $G_x(x,t_i)$. By replacing $W_x$ with a smaller neighborhood of $x$ if necessary, we may ssume that $G_x(W_x,t_i) \subset V$, using the continuity of $G_x|_ {W_x\times \{t_i\}}$. Define $G_x$ on $W_x\times [t_i,t_{i+1}]$ by $G_x|_ {W_x\times [t_i,t_{i+1}]} = (p|_ V)^{-1} \circ F|_ {W_x\times [t_i,t_{i+1}]}$. This is a lift of $F|_ {W_x\times [t_i,t_{i+1}]}$ obviously, and moreover, it agrees with the earlier $G_x|_ {W_x\times[0,t_i]}$ on the common domain $W_x\times\{t_i\}$ since $p|_ V$ is one-to-one. Let $G_x|_ {W_x\times[0,t_{i+1}]}$ be the union of $G_x|_ {W_x\times [t_i,t_{i+1}]}$ and $G_x|_ {W_x\times[0,t_i]}$. It is is a well-defined lift of $F_ {W_x\times[0,t_{i+1}]}$. Repeating this, we obtain a lif t $G_x\colon W_x\times I \to E$ of $F|_ {W_x\times I}$, as promised.

Now, we define $G\colon X\times I \to E$ by $G|_ {W_x\times I} = G_x$. This is well-defined, by the following argument: if $z\in W_x\cap W_y$, then $G_x|_ {\{z\}\times I}$ and $G_y|_ {\{z\}\times I}$ are lifts of $F|_ {\{z\}\times I}$ and $G_x(z,0) = \tilde f(z) = G_y(z,0)$, and hence $G_x|_ {\{z\}\times I}=G_y|_ {\{z\}\times I}$ by the uniqueness of a lift. Since each $G_x$ is a lift and $G_x(z,0)=\tilde f(z)$, it follows that $G$ is a lift and satisfies $G(z,0)=\tilde f(z)$. $\quad\square$

#### Injectivity of $p_*\colon \pi_1(E,e_0)\to \pi_1(B,b_0)$

Let $\tilde\alpha$, $\tilde\beta\colon I\to E$ be loops in $E$ based at $e_0$. Suppose $p_*[\alpha]=p_*[\tilde\beta]$ in $\pi_1(B,b_0)$. That is, for the compositions $\alpha=p\tilde\alpha$ and $\beta=p\tilde\beta$, there is a homotopy $F\colon I\times I\to B$ from $\alpha$ to $\beta$ rel $\{0,1\}$. By the homotopy lifting property (for paths), there is a lift $\tilde F\colon I\times I \to E$ such that $F\colon \tilde\alpha\simeq \tilde\beta \rel\{0,1\}$. Therefore $[\tilde\alpha] = [\tilde\beta]$ in $\pi_1(E,e_0)$. $\quad\square$

#### Fiber and coset

Define

$\Phi\colon \pi_1(B,b_0)\to p^{-1}(b_0)$

by $\Phi([\alpha]) = \tilde\alpha(1)$, where $\tilde\alpha$ is the unique lift of $\alpha$ such that $\tilde\alpha(0)=e_0$. $\Phi$ is surjective, since for any $e\in p^{-1}(b_0)$, there is a path $\tilde\alpha$ in $E$ from $e_0$ to $e$ by the path connectedness of $E$, and for $\alpha=p\circ\tilde\alpha$, $\alpha$ is a loop based at $b_0$ in $B$ and our $\tilde\alpha$ is the lift of $\alpha$ satisfying $\tilde\alpha(0)=e_0$ and $\tilde\alpha(1)=e$. That is, $\Phi([\alpha])=e$.

Suppose $\Phi([\alpha])=\Phi([\beta])$. Then, for the lifts $\tilde\alpha$ and $\tilde\beta$ which start at $e_0$, we have $\tilde\alpha(1)=\tilde\beta(1)$. So the product $\tilde\alpha\cdot\tilde\beta{}^{-1}$ is defined. Moreover, $\tilde\alpha(\tilde\beta)^{-1}$ is a loop in $E$ based at $e_0$, and we have $[\alpha] = [\alpha\cdot\beta^{-1}]\cdot[\beta] = [\alpha] = p_*[\tilde\alpha\cdot \tilde\beta{}^{-1}]\cdot [\beta]$. This shows that the right cosets $p_*\pi_1(E,e_0)[\alpha]$ and $p_*\pi_1(E,e_0)[\beta]$ are equal. It follows that $\Phi$ gives rise to a bijection

$\{\text{right cosets of p_*\pi_1(E,e_0) in \pi_1(B,b_0)}\} \to p^{-1}(b_0). \quad\square$