Jae Choon Cha
POSTECH
Covering maps
Definition. A map $p\colon E\to B$ is a covering map if $p$ is onto and every $x\in B$ has an open neighborhood $U$ such that $p^{1}(U) =\bigcup_\alpha V_\alpha$ where $\{V_\alpha\}_ \alpha$ is a collection of mutually disjoint open subsets $V_\alpha$ in $E$ for each of which the restriction $p_{V_\alpha}\colon V_\alpha \to U$ is a homeomorphism.
If it is the case, we say that $B$ is the base space of the covering, and $E$ is a covering space of $B$. Also we say that an open set $U$ in $B$ satisfying the above condition is evenly covered, and $V_\alpha$ is a sheet over $U$. For brevity, we also say that $p$ is a covering, and $E$ is a covering of $B$.
Examples.

A homeomorphism $E\to B$ is a covering.

If $A$ is a discrete space (namely each onepoint subset is open), then the projection $A\times B \to B$ is a covering. Indeed, the whole base space $B$ is evenly covered, where each $\{a\}\times B \subset A\times B$ is a sheet over $B$.

A special case of the above is when $A=\{1,2,\ldots,n\}$ with discrete topology. In this case, there are $n$ sheets $\{i\}\times B$ ($i=1,\ldots,n$).
Infinite cyclic covering of the circle
The following is one of the most important examples of coverings.
Let $\exp\colon \R\to S^1$ be the map $\exp(s) = e^{2\pi i s}$, where $S^1$ is regarded as the unit circle in the complex plane $\C$. This is a covering. Indeed, for every $z_0\in S^1$, writing $z=e^{2\pi i s_0}$ for some $s_0\in \R$, let $U=S^1\{z_0\}$ and $V_k = (s_0+k, s_0+k+1)\subset \R$ for $k\in \Z$. Then $p^{1}(U) = \R\sm p^{1}(z_0)$ is the union of the $V_k$. Furthermore, $p_{V_k}\colon V_k \to U$ is a homeomorphism for each $k$; we leave it as an exercise for the readers. Perhaps the only possibly less straightforward part is the continuity of the inverse $(p_ {V_k})^{1}$. A direct $\epsilon$$\delta$ argument can be used for this. (Or, you may appeal to the fact that there is a complex logarithm for an appropriate choice of a branch cut; then you still need to show that the logarithm is continuous.) From the above it follows that our $U$ is evenly covered. By varying $z_0$, we obtain evenly covered open sets which cover the base space $B$. (Indeed, using $z_0=1,1\in S^1$, we obtain two evenly covered open sets $S^1\sm\{1\}$ and $S^1\sm\{1\}$, which cover $S^1$.) This proves that $\exp\colon \R \to S^1$ is a covering map.
Observe that $\exp^{1}(b)$ is an infinite discrete subset for each point $b$ in the base space $S^1$. This is why the word "infinite" is used in the term "infinite cyclic covering." The word "cyclic" is justified below in what follows.
Covering transformations
Defintion. Suppose $p\colon E\to B$ is a covering. A covering transformation is a homeomorphism $f\colon E\to E$ satisfying $pf=p$. That is, the following diagram is commutative:
Let $G(EB)$ be the set of all covering transformations for the covering $p$. It is straightforward to see that $G(EB)$ is a group under composition. (The identity and inverse are the function identity and inverse respectively.) The group $G(EB)$ is called the covering transformation group for the covering $p\colon E\to B$. Sometimes we say deck transformation instead of covering transformation.
In the case of $\exp\colon \R \to S^1$, it is easily seen that the translation map $f_k\colon \R\to \R$ defined by $f_k(s)=s+k$ is a covering transformation for each integer $k\in\Z$. Furthermore, the function $\Phi\colon \Z\to G(\RS^1)$ given by $\Phi(k)=f_k$ is a group homomorphism, since $f_{k+\ell} =f_k\circ f_\ell$. It is obvious that $\Phi$ is 11, since $f_k$ and $f_\ell$ are different maps whenever $k\ne \ell$. Moreover, we will show later that $\Phi$ is an isomorphism, that is, the covering transformation group is an infinite cyclic group. This is why $\exp\colon\R^1\to S^1$ is called an infinite cyclic covering.
Finite cyclic cover of the circle
The following is another covering of $S^1$ which is similar to $\exp\colon \R\to S^1$. Fix a positive integer $d$. Let $p_d\colon S^1\to S^1$ be the map $p_d(z)=z^d$. We claim that $U=S^1\sm\{z_0\}$ is evenly covered for every $z_0\in S^1$. From the claim it follows that $p_d$ is a covering map. To show the claim, observe $p_d^{1}(U)=S^1\sm\{w_0,\ldots,w_{d1}\}$ where $w_0,\ldots, w_{d1}$ are the roots of the equation $x^n=z_0$. Namely, writing $z=e^{2\pi i s_0}$ with $0\le s_0 < 1$, we have $w_k = e^{2\pi i (s_0+k)/d}$. Let $V_k = \{e^{2\pi i s} \in S^1 \mid \frac{s_0+k}{d} < s <\frac{s_0+k+1}{d}\}$ for $k=0,\ldots,d1$. Then it can be shown that $p_d _{V_k}\colon V_k \to U$ is a homeomorphism. That is, $U$ is evenly covered and the $V_k$ are sheets.
Note that $p_d^{1}(z_0)$ consists of $d$ points. The map $f_k\colon S^1\to S^1$ defined by $f_k(z)=e^{2\pi i k/d}\cdot z$ is a covering transformation for the covering $p_d$. Let $\Z_d$ be the quotient group of $\Z$ by the subgroup $d\Z$. Namely, $\Z_d$ is a finite cyclic group of order $d$ consisting of the cosets $k+d\Z$ which we will often denote by $k\in \Z_d$ as an abuse of notation. Then $\Phi\colon \Z_d \to G(S^1S^1)$ given by $\Phi(k) = f_k$ ($k=0,\ldots,d1$) is an injective group homomorphism (check this!). Indeed it turns out that this $\Phi$ is an isomorphism too. We will prove this later.
Product covering
The following observation is useful in constructing coverings.
Lemma. If $p_1\colon E_1\to B_1$ and $p_2\colon E_2\to B_2$ are coverings, then the map $p_1\times p_2\colon E_1\times E_2 \to B_1\times B_2$ defined by $(e_1,e_2) \mapsto (p_1(e_1),p_2(e_2))$ is a covering.
Proof. If $U_1\subset B_1$ and $U_2\subset B_2$ are evenly covered, then $U_1\times U_2 \subset B_1\times B_2$ is evenly covered. The conclusion follows immediately from this. $\quad\square$
Example. Let $T^2 = S^1\times S^1$ be the torus. The map $\R^2\to T^2$ defined by $(s,t)\mapsto (e^{2\pi i s}, e^{2\pi i t})$ is a covering map, using the above lemma. Also, $T^2\to T^2$ defined by $(z,w) \to (z^n, z^m)$ is a covering for any nonzero integers $n$ and $m$. (Prove this!)
Graphs as topological spaces
To discuss more examples of coverings, it is useful to use graphs. Formally, define an oriented graph $X$ to be a triple $X=(V,E,\partial)$ of sets $V$ and $E$ together with a function $\partial\colon E\to V\times V$. We will often call it a graph, omitting the word "oriented." An element $v\in V$ is called a vertex, and $e\in E$ is called an edge. In our mind, vertices are points, and edges are arcs joining vertices. The function $\partial$ is regarded as the indication of the two endpoints of each edge.
More precisely, we construct a topological space from a graph as follows. Let $\partial_$, $\partial_+\colon E\to V$ be the functions given by $\partial(e)=(\partial_ e, \partial_+ e)$. Regard $V$ and $E$ as spaces with discrete topology, take the disjoint union of $V$ and $E\times I$, and define an equivalence relation $\sim$ generated by $(e,0)\sim\partial_e$, $(e,1)\sim\partial_+e$ for $e\in E$. As an abuse of notation, denote the quotient space by the same symbol $X$. That is,
Note that each edge $e\in E$ is associated to a path $\gamma_e \colon I\to X$ defined by $\gamma_e(s)= [e,s]$, where $[e,s]\in X$ designates the equivalence class of the point $(e,s)\in E\times I$ in the quotient space $X$. From the definition it follows that the equivalence class $[\partial_e]\in X$ of the point $\partial_e\in V$ is the starting point of this path, and $[\partial_+e]\in X$ is the endpoint.
Examples.

Let $V=\{*\}$ and $E=\{e\}$ be singletons. Let $X=(V,E,\partial)$ be the graph given by $V$, $E$ and the function $\partial\colon E\to V\times V$ defined by $\partial(e)=(*,*)$. What is $X$ as a space? We have one vertex $*$ and one edge $e$ whose both endpoints are attached to $*$. So, the space $X$ is homeomorphic to the circle $S^1$.

Let $V=\{t,u,v\}$ and $E=\{a,b,c,d,e\}$. Define $\partial\colon E\to V\times V$ by $\partial(a)=(v,u)$, $\partial(b)=(u,v)$, $\partial(c)=(u,t)$, $\partial(d)=(v,t)$, $\partial(e)=(t,t)$. Can you draw a picture of the space $X$? See below.
Words and paths on a graph
In general, a word $w$ on a set $S$ is defined to be a finite sequence $\{s_i\}_ {i=1}^n$ consisting of elements $s_i\in S$. Elements in $S$ are called letters. For brevity, we denote by $w=s_1\cdots s_n$.
For a graph $X=(V,E)$, define $\overline E=\{e^{1}\mid e\in E\}$. Here, $e^{1}$ is just a formal symbol, which does not have any algebraic meaning for now. Let $S=E\cup \overline E$. For convenience we often denote $e\in S$ by $e^{+1}$. Then a word on $S$ is of the form $w=e_1^{\epsilon_1}e_2^{\epsilon_2}\cdots e_n^{\epsilon_n}$ where $e_i\in E$, $\epsilon_i \in \{1,+1\}$. For instance, in the second example given above, $w=abde^{1}$ is such a word. As a convention, we often denote $s^n=\underbrace{s\cdot s \cdots s}_n$ and $s^{n}=\underbrace{s^{1}\cdot s^{1}\cdots s^{1}}_n$ for $n>0$.
Now, for a word $w$ on $E\cup \overline E$, we will associate a path $\gamma_w$ in the graph $X$ (viewed as a space). Intuitively, $\gamma_w$ will be the path travelling $X$ along the edges which appears in $w$ in the given order. For instance, for the case of $w=abde^{1}$, start from the starting point of the edge $a$, which is the vertex $\partial_(a)=v$ in the above example, and proceed along $a$ to reach $u$. Then, continue to proceed along $b$ and $d$, and finally proceed along $e$ in the reversed direction (since we have $e^{1}$, not $e$, at the end of $w$). So, the endpoint of $\gamma_w$ is the vertex $t$ in this case. Observe that we need endpoint matching conditions to take a product of paths as in the above paragraph. For instance, in case of $w=abde^{1}$, the endpoint of $a$ is equal to the starting point $b$, and so on.
In general, given $w=e_1^{\epsilon_1} \cdots e_n^{\epsilon_n}$, suppose
Since $\gamma_{e_i}^{\epsilon_i}$ starts from the vertex $\partial_{\epsilon_i}(e_i)$ and finishes at the vertex $\partial_{\epsilon_i}(e_i)$, this hypothesis guarantees that the product
is defined. We take this as the path associated to the word $w$. For example, in case of the above example, $\gamma_w$ for $w=abde^{1}$ is illustrated below.
Maps between graphs specified by words
Suppose $X=(V(X),E(X),\partial^X)$ and $Y=(V(Y),E(Y),\partial^Y)$ are graphs. Here we will discuss a class of maps $X \to Y$ which can be described using words, where $X$ and $Y$ are viewed as spaces. Suppose $f_0\colon V(X)\to V(Y)$ and $f_1\colon E(X) \to \{$words on $E(Y)\cup \overline{E(Y)}$ satisfying $(*)\}$ are functions. The idea is to send vertices of $X$ to vertices of $Y$ via $f_0$, and edges of $X$ to paths in $Y$ via $f_1$. For this purpose, suppose that $\gamma_{f_1(e)}(0)=[f_0(\partial^X_(e))]$ and $\gamma_{f_1(e)}(1)=[f_0(\partial^X_+(e))]$. Intuitively, this condition says that if an endpoints of an edge $e\in E(X)$ is attached to a vertex $v\in E(X)$, then $e$ must be sent to a path with corresponding endpoint $f_0(v)$. In fact, it is straightforward to verify that $f\colon X\to Y$ defined by $f([v])=[f_0(v)]$ for $v\in V(X)$ and $f([e,s])=\gamma_{f_1(e)}(s)$ for $e\in E(X)$, $s\in I$ is a welldefined continuous map on the space $X$. (Here, recall from the above that $\gamma_{f_1(e)}$ is a path in $Y$ associated to the word $f_1(e)$.)
Example. Let $X$ be the graph with $V=\{*\}$, $E=\{e\}$ which was described in the above example. Recall that $X\cong S^1$ as spaces. We will define a map $X\to X$ by assigning words to edges. First, define $f_0\colon V\to V$ by $f_0(*)=*$. For the edge, fix an integer $d>0$, and define $f_1$ by $f_1(e)=e^d$. Let $f\colon X\to X$ be the map determined by $f_0$ and $f_1$. What is this map $f$? Observe that the path $\gamma_{e^d}$ wraps the circle $S^1$ $d$ times. Indeed, by our definition of $f$ given above, it turns out that $f$ is equal to the covering map $p_d\colon S^1\to S^1$ given by $p_d(z)=z^d$.
We often describe a map $f\colon X\to Y$ between two graphs by drawing a diagram as follows. On the right hand side, we draw the graph $Y$ which is the codomain, and label vertices and edges with their names. On the left hand side, we draw the graph $X$ which is the domain, but now lables for $X$ are the images of vertices and edges under $f_0$ and $f_1$. Namely, in this example, the vertex on the left hand side is labeled by $*$ because its image under $f_0$ is $*$. The edge is labeled by $e^d$, which is the value of $f_1$ on the edge.
More Examples.
 Let $f$ be the map given as follows. We leave it as an exercise to prove that $f$ is a covering map.
 The following is another example of a covering map between graphs:
Here is a bit challenging exercise: what is the covering transformation group of these coverings? At this moment, finding a proof may be difficult, but it is doable to make a guess.