https://www.math.uchicago.edu/~may/VIGRE/VIGRE2007/REUPapers/FINALFULL/Nadathur.pdf
https://pi.math.cornell.edu/~hatcher/AT/AT.pdf
# Covering Transformations
Given a covering map $p : E \to B$, it is of some interest to consider the set of all equivalences of this covering space with $itself$. Such an equivalence is called a $\textit{covering transformation}$. Composites and inverses of covering transformations are covering transformations, so this set forms a group; it is called the $\textit{group of covering transformations}$ and denoted $\mathcal{C}(E, p, B)$.

Throughout this section, we shall assume that $p : E \to B$ is a covering map with $p(e_0) = b_0$; and we shall let $H_0 = p_*(\pi_1(E, e_0))$. We shall show that the group $\mathcal{C}(E, p, B)$ is completely determined by the group $\pi_1(B, b_0)$ and the subgroup $H_0$. Specifically, we shall show that if $N(H_0)$ is the largest subgroup of $\pi_1(B, b_0)$ of which $H_0$ is a normal subgroup, then $\mathcal{C}(E, p, B)$ is isomorphic to $N(H_0)/H_0$.

We define $N(H_0)$ formally as follows:

$\textbf{Definition.}$
If $H$ is a subgroup of the group $G$, then the normalizer of $H$ in $G$ is the subset of $G$ defined by the equation
\begin{equation}
N(H) = \{ g \mid gHg^{-1} = H \}.
\end{equation}

It is easy to see that $N(H)$ is a subgroup of $G$. It follows from the definition that it contains $H$ as a normal subgroup and is the largest such subgroup of $G$.

$\textbf{Definition.}$
Given $p: E \to B$ with $p(e_0) = b_0$, let $F$ be the set $F = p^{-1}(b_0)$. Let
\begin{equation}
\Phi: \pi_1(B, b_0)/H_0 \longrightarrow F
\end{equation}be the lifting correspondence of Theorem 54.6; it is a bijection. Define also a correspondence
\begin{equation}
\psi: \mathcal{C}(E,p,B) \longrightarrow F
\end{equation}by setting $\psi(h) = h(e_0)$ for each covering transformation $h: E \to E$. Since $h$ is uniquely determined once its value at $e_0$ is known, the correspondence $\psi$ is injective.

$\textbf{Lemma}$
The image of the map $\psi$ equals the image under $\Phi$ of the subgroup $N(H_0)/H_0$ of $\pi_1(B, b_0)/H_0$.

$\textbf{Proof}$
Recall that the lifting correspondence $\phi : \pi_1(B, b_0) \to F$ is defined as follows: Given a loop $\alpha$ in $B$ at $b_0$, let $\gamma$ be its lift to $E$ beginning at $e_0$; let $e_1 = \gamma(1)$, and define $\phi$ by setting $\phi([\alpha]) = e_1$. To prove the lemma, we need to show that there is a covering transformation $h : E \to E$ with $h(e_0) = e_1$ if and only if $[\alpha] \in N(H_0)$.

This is easy. Lemma 79.1 tells us that $h$ exists if and only if $H_0 = H_1$, where $H_1 = p_*(\pi_1(E, e_1))$. And Lemma 79.3 tells us that $[\alpha] * H_1 * [\alpha]^{-1} = H_0$. Hence $h$ exists if and only if $[\alpha] \in N(H_0)$.

$\textbf{Theorem}$
The bijection
\begin{equation}
\Phi^{-1} \circ \psi : \mathcal{C}(E,p,B) \to N(H_0)/H_0
\end{equation}is an isomorphism of groups.

$\textbf{Proof:}$
We need only show that $\Phi^{-1} \circ \psi$ is a homomorphism. Let $h, k : E \rightarrow E$ be covering transformations. Let $h(e_0) = e_1$ and $k(e_0) = e_2$; then
\begin{equation}
\psi(h) = e_1 \quad and \quad \psi(k) = e_2,
\end{equation}by definition. Choose paths $\gamma$ and $\delta$ in $E$ from $e_0$ to $e_1$ and $e_2$, respectively. If $\alpha = p \circ \gamma$ and $\beta = p \circ \delta$, then
\begin{equation}
\Phi([\alpha]H_0) = e_1 \quad and \quad \Phi([\beta]H_0) = e_2,
\end{equation}by definition. Let $e_3 = h(k(e_0))$; then $\psi(h \circ k) = e_3$. We show that
\begin{equation}
\Phi([\alpha * \beta]H_0) = e_3,
\end{equation}and the proof is complete.

Since $\delta$ is a path from $e_0$ to $e_2$, the path $h \circ \delta$ is a path from $h(e_0) = e_1$ to $h(e_2) = h(k(e_0)) = e_3$. See Figure 81.1.Then the product $\gamma * (h \circ \delta)$ is defined and is a path from $e_0$ to $e_3$. It is a lifting of $\alpha * \beta$, since $p \circ \gamma = a$ and $p \circ h \circ \delta = p \circ \delta = \beta$. Therefore, $\Phi([\alpha * \beta]H_0) = e_3$, as desired.


$\textbf{Corollary}$
The group $H_0$ is a normal subgroup of $\pi_1(B, b_0)$ if and only if for every pair of points $e_1$ and $e_2$ of $E$, there is a covering transformation $h : E \rightarrow E$ with $h(e_1) = e_2$. In this case, there is an isomorphism
\begin{equation}
\Phi^{-1} \circ \psi : \mathcal{C}(E, p, B) \rightarrow \pi_1(B, b_0)/H_0.
\end{equation}

$\textbf{Corollary}$
Let $p : E \rightarrow B$ be a covering map. If $E$ is simply connected, then
\begin{equation}
\mathcal{C}(E, p, B) \cong \pi_1(B, b_0).
\end{equation}

If $H_0$ is a normal subgroup of $\pi_1(B, b_0)$, then $p: E \to B$ is called a $\textit{regular covering map}$.

$\textbf{EXAMPLE 1.}$
Because the fundamental group of the circle is abelian, every covering of $S^1$ is regular. If $p: \mathbb{R} \to S^1$ is the standard covering map, for instance, the covering transformations are the homeomorphisms $x \to x + n$. The group of such transformations is isomorphic to $\mathbb{Z}$.

$\textbf{EXAMPLE 2}$
For an example at the other extreme, consider the covering space of the figure eight indicated in Figure 81.2. In this case, we show that the group $\mathcal{C}(E, p, B)$ is trivial.

In general, if $h: E \to E$ is a covering transformation, then any loop in the base space that lifts to a loop in $E$ at $e_0$ also lifts to a loop when the lift begins at $h(e_0)$. In the present case, a loop that generates the fundamental group of $A$ lifts to a non-loop when the lift is based at $e_0$ and lifts to a loop when it is based at any other point of $p^{-1}(b_0)$ lying on the $y$-axis. Similarly, a loop that generates the fundamental group of $B$ lifts to a non-loop beginning at $e_0$ and to a loop beginning at any other point of $p^{-1}(b_0)$ lying on the $x$-axis. It follows that $h(e_0) = e_0$, so that $h$ is the identity map.

$\textbf{Definition.}$
Let $X$ be a space, and let $G$ be a subgroup of the group of homeomorphisms of $X$ with itself. The orbit space $X/G$ is defined to be the quotient space obtained from $X$ by means of the equivalence relation $x \sim g(x)$ for all $x \in X$ and all $g \in G$. The equivalence class of $x$ is called the orbit of $x$.

$\textbf{Definition.}$
If $G$ is a group of homeomorphisms of $X$, the action of $G$ on $X$ is said to be $\textit{properly discontinuous}$ if for every $x \in X$ there is a neighborhood $U$ of $x$ such that $g(U)$ is disjoint from $U$ whenever $g \neq e$. (Here $e$ is the identity element of $G$.) ==It follows that $g_0(U)$ and $g_1(U)$ are disjoint whenever $g_1 \neq g_0$, for otherwise $U$ and $g_0^{-1}g_1(U)$ would not be disjoint.==

$\textbf{Theorem}$
Let $X$ be path connected and locally path connected; let $G$ be a group of homeomorphisms of $X$. The quotient map $\pi: X \to X/G$ is a covering map if and only if the action of $G$ is properly discontinuous. In this case, the covering map $\pi$ is regular and $G$ is its group of covering transformations.

$\textbf{Proof:}$
We show $\pi$ is an open map. If $U$ is open in $X$, then $\pi^{-1}\pi(U)$ ==is the union of the open sets $g(U)$ of $X$, for $g \in G$.== Hence $\pi^{-1}\pi(U)$ is open in $X$, so that $\pi(U)$ is open in $X/G$ by definition. Thus $\pi$ is open.

$\textit{Step 1.}$ We suppose that the action of $G$ is properly discontinuous and show that $\pi$ is a covering map. Given $x \in X$, let $U$ be a neighborhood of $x$ such that $g_0(U)$ and $g_1(U)$ are disjoint whenever $g_0 \neq g_1$. Then $\pi(U)$ is evenly covered by $\pi$. Indeed, $\pi^{-1}\pi(U)$ equals the union of the disjoint open sets $g(U)$, for $g \in G$, ==each of which contains at most one point of each orbit.== Therefore, ==the map $g(U) \to \pi(U)$ obtained by restricting $\pi$ is bijective;== being continuous and open, it is a homeomorphism. The sets $g(U)$, for $g \in G$, thus form a partition of $\pi^{-1}\pi(U)$ into slices.

$\textit{Step 2.}$ Suppose now that $\pi$ is a covering map and show that the action of $G$ is properly discontinuous. Given $x \in X$, let $V$ be a neighborhood of $\pi(x)$ that is evenly covered by $\pi$. Partition $\pi^{-1}(V)$ into slices; let $U_a$ be the slice containing $x$. Given $g \in G$ with $g \neq e$, the set $g(U_a)$ must be disjoint from $U_a$, ==for otherwise, two points of $U_a$ would belong to the same orbit and the restriction of $\pi$ to $U_a$ would not be injective.== It follows that the action of $G$ is properly discontinuous.

$\textit{Step 3.}$ Show that if $\pi$ is a covering map, then $G$ is its group of covering transformations and $\pi$ is regular. Certainly, any $g \in G$ is a covering transformation, ==for $\pi \circ g = \pi$ because the orbit of $g(x)$ equals the orbit of $x$==. On the other hand, let $h$ be a covering transformation with $h(x_1) = x_2$, say. Because $\pi \circ h = \pi$, ==the points $x_1$ and $x_2$ map to the same point under $\pi$==; therefore, there is an element $g \in G$ such that $g(x_1) = x_2$. The uniqueness part of Theorem 79.2 then implies that $h = g$.

It follows that $\pi$ is regular. Indeed, for any two points $x_1$ and $x_2$ lying in the same orbit, there is an element $g \in G$ such that $g(x_1) = x_2$. Then Corollary 81.3 applies.

$\textbf{Theorem}$
If $p : X \to B$ is a regular covering map and $G$ is its group of covering transformations, then there is a homeomorphism $k : X/G \to B$ such that $p = k \circ \pi$, where $\pi : X \to X/G$ is the projection.

$\textbf{Proof}$
If $g$ is a covering transformation, then $p(g(x)) = p(x)$ by definition. ==Hence $p$ is constant on each orbit==, so it induces a continuous map $k$ of the quotient space $X/G$ into $B$. On the other hand, $p$ is a quotient map because it is continuous, surjective, and open. ==Because $p$ is regular, any two points of $X$ belong to the same orbit under the action of $G$. Therefore, $\pi$ induces a continuous map $B \to X/G$ that is an inverse for $k$.==

$\textbf{EXAMPLE 3.}$
Let $X$ be the cylinder $S^1 \times I$; let $h : X \to X$ be the homeomorphism $h(x, t) = (-x, t)$; and let $k : X \to X$ be the homeomorphism $k(x, t) = (-x, 1 - t)$. The groups $G_1 = \{e, h\}$ and $G_2 = \{e, k\}$ are isomorphic to the integers modulo 2; both act properly discontinuously on $X$. But $X/G_1$ is homeomorphic to $X$, while $X/G_2$ is homeomorphic to the Möbius band, as you can check. See Figure 81.3.


# Existence of Covering Spaces

$\textbf{Definition.}$
A space $B$ is said to be $\textit{semilocally simply connected}$ if for each $b \in B$, there is a neighborhood $U$ of $b$ such that the homomorphism
\begin{equation}
i_* : \pi_1(U, b) \to \pi_1(B, b)
\end{equation}induced by inclusion is trivial.

$\textbf{Theorem}$
Let $B$ be path connected, locally path connected, and semilocally simply connected. Let $b_0 \in B$. Given a subgroup $H$ of $\pi_1(B, b_0)$, there exists a covering map $p : E \to B$ and a point $e_0 \in p^{-1}(b_0)$ such that
\begin{equation}
p_*(\pi_1(E, e_0)) = H.
\end{equation}

$\textbf{Proof}$

$\textbf{Step 1.}$ $Construction$ of $E$. The procedure for constructing $E$ is reminiscent of the procedure used in complex analysis for constructing Riemann surfaces. Let $\mathcal{P}$ denote the set of all paths in $B$ beginning at $b_0$. Define an equivalence relation on $\mathcal{P}$ by setting $\alpha \sim \beta$ if $\alpha$ and $\beta$ end at the same point of $B$ and
\begin{equation}
\alpha * \overline{\beta} \in H.
\end{equation}This relation is easily seen to be an equivalence relation. We will denote the equivalence class of the path $\alpha$ by $\alpha^{\#}$.

Let $E$ denote the collection of equivalence classes, and define $p : E \to B$ by the equation
\begin{equation}
p(\alpha^{\#}) = \alpha(1).
\end{equation}Since $B$ is path connected, $p$ is surjective. We shall topologize $E$ so that $p$ is a covering map.

We first note two facts:
(1) $[\alpha] = [\beta]$, then $\alpha^{\#} = \beta^{\#}$,
(2) If $\alpha^{\#} = \beta^{\#}$, then $(\alpha * \delta)^{\#} = (\beta * \delta)^{\#}$ for any paths $\delta$ in $B$ beginning at $\alpha(1)$.
The first follows by noting that if $[\alpha] = [\beta]$, then $[\alpha * \bar{\beta}]$ is the identity element, which belongs to $H$. The second follows by noting that $\alpha * \delta$ and $\beta * \delta$ end at the same point of $B$, and
\begin{equation}
[(\alpha * \delta)* \overline{(\beta * \delta)}
] = [(\alpha * \delta)*(\bar{\delta} * \bar{\beta})] = [\alpha * \bar{\beta}],
\end{equation}which belongs to $H$ by hypothesis.

$\textbf{Step 2.}$ Topologizing E. One way to topologize $E$ is to give $\mathcal{P}$ the compact-open topology (see Chapter 7) and E the corresponding quotient topology. But we can topologize E directly as follows:

Let $\alpha$ be any element of $\mathcal{P}$ and let $U$ be any path-connected neighborhood of
$\alpha(1)$. Define
\begin{equation}
B(U, \alpha) = \{ (\alpha * \delta)^{\#} \mid \delta \text{ is a path in } U \text{ beginning at } \alpha(1) \}.
\end{equation}

Note that $\alpha^{\#}$ is an element of $B(U, \alpha)$, since if $b = \alpha(1)$, then $\alpha^{\#} = (\alpha * e_b)^{\#}$,this element belongs to $B(U, \alpha)$ by definition. We assert that the sets $B(U, \alpha)$
form a basis for a topology on $E$.

First, we show that if $\beta^{\#} \in B(U, \alpha)$, then $\alpha^{\#} \in B(U, \beta)$, and $B(U, \alpha) = B(U, \beta)$.

If $\beta^{\#} \in B(U, \alpha)$, then $\beta^{\#} = (\alpha * \delta)^{\#}$ for some path $\delta$ in $U$.Then

\begin{align*}
(\beta * \bar{\delta})^{\#} &= ((\alpha * \delta) * \bar{\delta})^{\#} && \text{by (2)}, \\
&= \alpha^{\#} && \text{by (1)}.
\end{align*}
So that $\alpha^{\#} \subseteq B(U, \beta)$ by definition.We show first that $B(U, \beta) \subseteq B(U, \alpha)$. Note that the general element of $B(U, \beta)$ is of the form $(\beta * \gamma)^{\#}$, where $\gamma$ is a path in $U$. Then note that
\begin{align*}
(\beta * \gamma)^{\#} &= ((\alpha * \delta) * \gamma)^{\#} \\
&= (\alpha * (\gamma * \delta))^{\#}
\end{align*}

which belongs to $B(U, \alpha)$ by definition. Symmetry gives the inclusion $B(U, \alpha) \subseteq B(U, \beta)$ as well.

Now we show the sets $B(U, \alpha)$ form a basis. If $\beta^{\#}$ belongs to the intersection $B(U_1, \alpha_1) \cap B(U_2, \alpha_2)$, we need merely choose a path-connected neighborhood $V$ of $\beta(1)$ contained in $U_1 \cap U_2$. The inclusion
\begin{equation}
B(V, \beta) \subset B(U_1, \beta) \cap B(U_2, \beta)
\end{equation}follows from the definition of these sets, and the right side of the equation equals $B(U_1, \alpha_1) \cap B(U_2, \alpha_2)$ by the result just proved.

$\textbf{Step 3.}$ The map $p$ is continuous and open. It is easy to see that $p$ is open, for the image of the basis element $B(U, \alpha)$ is the open subset $U$ of $B$: Given $x \in U$, we choose a path $\beta$ in $U$ from $\alpha(1)$ to $x$; then $(\alpha * \delta)^{\#}$ is in $B(U, \alpha)$ and $p((\alpha * \delta)^{\#}) = x$.

To show that $p$ is continuous, let us take an element $\alpha^{\#}$ of $E$ and a neighborhood $W$ of $p(\alpha^{\#})$. Choose a path-connected neighborhood $U$ of the point $p(\alpha^{\#}) = \alpha(1)$ lying in $W$. Then $B(U, \alpha)$ is a neighborhood of $\alpha^{\#}$ that $p$ maps into $W$. Thus $p$ is continuous at $\alpha^{\#}$.

$\textbf{Step 4.}$ Every point of $B$ has a neighborhood that is evenly covered by $p$. Given $b_1 \in B$, choose $U$ to be a path-connected neighborhood of $b_1$ that satisfies the further condition that the homomorphism $\pi_1(U, b_1) \to \pi_1(B, b_1)$ induced by inclusion is trivial. We assert that $U$ is evenly covered by $p$.

First, we show that $p^{-1}(U)$ equals the union of the sets $B(U, \alpha)$, as $\alpha$ ranges over all paths in $B$ from $b_0$ to $b_1$. Since $p$ maps each set $B(U, \alpha)$ onto $U$, it is clear that $p^{-1}(U)$ contains this union. On the other hand, if $\beta^{\#}$ belongs to $p^{-1}(U)$, then $\beta(1) \in U$. Choose a path $\delta$ in $U$ from $b_1$ to $\beta(1)$ and let $\alpha$ be the path $\beta * \bar{\delta}$ from $b_0$ to $b_1$. Then $[\beta] = [\alpha * \delta]$, so that $\beta^{\#} = (\alpha * \delta)^{\#}$,which belongs to $B(U, \alpha)$. Thus $p^{-1}(U)$ is contained in the union of the sets $B(U, \alpha)$.

Second, note that distinct sets $B(U, \alpha)$ are disjoint. For if $\beta^{\#}$ belongs to $B(U, \alpha_1) \cap
B(U, \alpha_2)$, then $B(U, \alpha_1) = B(U, \beta) = B(U, \alpha_2)$, by Step 2.

Third, we show that $p$ defines a bijective map of $B(U, \alpha)$ with $U$. It follows that $p|_{B(U, \alpha)}$ is a homeomorphism, being bijective and continuous and open. We already know that $p$ maps $B(U, \alpha)$ onto $U$. To prove injectivity, suppose that
\begin{equation}
p((\alpha * \delta_1)^{\#}) = p((\alpha * \delta_2)^{\#}),
\end{equation}where $\delta_1$ and $\delta_2$ are paths in $U$. Then $\delta_1(1) = \delta_2(1)$. Because the homomorphism $\pi_1(U, b_1) \to \pi_1(B, b_1)$ induced by inclusion is trivial, $\delta_1 * \bar{\delta_2}$ is path homotopic in $B$ to the constant loop. Then $[\alpha * \delta_1 ] = [\alpha * \delta_2 ]$, so that $(\alpha * \delta_1)^{\#} = (\alpha * \delta_2)^{\#}$, as desired.

It follows that $p: E \to B$ is a covering map in the sense used in earlier chapters. To show it is a covering map in the sense used in this chapter, we must show $E$ is path connected. This we shall do shortly.

$\textbf{Step 5.}$ Lifting a path in $B$. Let $e_0$ denote the equivalence class of the constant path at $b_0$; then $p(e_0) = b_0$ by definition. Given a path $\alpha$ in $B$ beginning at $b_0$, we calculate its lift to a path in $E$ beginning at $e_0$ and show that this lift ends at $\alpha^{\#}$.

To begin, given $c \in [0, 1]$, let $\alpha_c: I \to B$ denote the path defined by the equation
\begin{equation}
\alpha_c(t) = \alpha(tc) \quad for \quad 0 \leq t \leq 1.
\end{equation}Then $\alpha_c$ is the "portion" of $\alpha$ runs from $\alpha(0)$ to $\alpha(c)$. In particular, $\alpha_0$ is the constant path at $b_0$, and $\alpha_1$ is the path $\alpha$ itself. We define $\tilde{\alpha}: I \to E$ by the equation
\begin{equation}
\tilde{\alpha}(c) = (\alpha_c)^{\#}
\end{equation}and show that $\tilde{\alpha}$ is continuous. Then $\tilde{\alpha}$ is a lift of $\alpha$, since $p(\tilde{\alpha}(c)) = \alpha_c(1) = \alpha(c)$; furthermore, $\tilde{\alpha}$ begins at $(\alpha_0)^{\#} = e_0$ and ends at $(\alpha_1)^{\#} = \alpha^{\#}$.

To verify continuity, we introduce the following notation. Given $0 \leq c < d \leq 1$, let $\delta_{c,d}$ denote the path that equals the positive linear map of $I$ onto $[c, d]$ followed by $\alpha$. Note that the paths $\alpha_{d}$ and $\alpha_{c} * \delta_{c,d}$ are path homotopic because one is just a reparametrization of the other. See Figure 82.2.


We now verify continuity of $\tilde{\alpha}$ at the point $c$ of $[0, 1]$. Let $W$ be a basis element in $E$ about the point $\tilde{\alpha}(c)$. Then $W$ equals $B(U, \alpha_c)$ for some path-connected neighborhood $U$ of $\alpha(c)$. Choose $\epsilon > 0$ so that for $|c - t| < \epsilon$, the point $\alpha(t)$ lies in $U$.We show that if $d$ is a point of $[0, 1]$ with $|c - d| < \epsilon$, then $\tilde{\alpha}(d) \in W$; this proves continuity of $\tilde{\alpha}$ at $c$.

So suppose $|c - d| < \epsilon$. Take first the case where $d > c$. Set $\delta = delta_{c,d}$; then since$[\alpha_d] = [\alpha_c * \delta],$we have
\begin{equation}
\tilde{\alpha}(d) = (\alpha_d)^{\#} = (\alpha_c * \delta)^{\#}.
\end{equation}Since $\delta$ lies in $U$, we have $\tilde{\alpha}(d) \in B(U, \alpha_c)$, as desired. If $d < c$, set $\delta = \delta_{d,c}$ and proceed similarly.

$\textbf{Step 6.}$ The map $p: E \to B$ is a covering map. We need only verify that $E$ is path connected, and this is easy. For if $\alpha^{\#}$ is any point of $E$, then the lift $\tilde{\alpha}$ of the path $\alpha$ is a path in $E$ from $e_0$ to $\alpha^{\#}$.

$\textbf{Step 7.}$ Finally, $H = p_*(E, e_0)$. Let $\alpha$ be a loop in $B$ at $b_0$. Let $\tilde{\alpha}$ be its lift
to $E$ beginning at $e_0$. Theorem 54.6 tells us that $[\alpha] \in p_{*}(\pi_1(E, e_0))$ if and only if $\tilde{\alpha}$ is a loop in $E$. Now the final point of $\tilde{\alpha}$ is the point $\alpha^{\#}$, and $\alpha^{\#} = e_0$ if and only if $\alpha$ is equivalent to the constant path at $b_0$, i.e., if and only if $[\alpha * \bar{e_{b_0}}] \in H$. This occurs precisely when $[\alpha] \in H$.

$\textbf{Corollary}$
The space $B$ has a universal covering space if and only if $B$ is path connected, locally path connected, and semilocally simply connected.