Definition 3.1
Suppose $p$ is a continuous map between two topological spaces
$$p: E \longrightarrow B$$
Let $U$ be an open set in $B$ and let $\{V_{\alpha}\}$ be a set (finite or infinite) of disjoint open sets in $E$.
$U$ is said to be evenly covered by $p$ if
$$p^{-1} (U) = \bigcup V_{\alpha}$$
and, for each $\alpha$, the map
$$P \Big|_{V_{\alpha}}: V_{\alpha} \longrightarrow U$$ is a homeomorphism.
The collection $\{V_{\alpha}\}$ is called a partition of
$p^{-1}(U)$
into slices.
Note that if $U$ is evenly covered by $p$, and $W$ is an open subset of $\bigcup V_{\alpha}$, then $W$ is also evenly covered by $p$.
In other words, each $V_{\alpha}$, when plugged into $p$, creates an identical copy of $U$. Note that this isn't the same as mapping $E$ to $B^n$. Each copy of $U$ lies directly on top of the rest. If we think of $U$ as a flat disk, $\{V_{\alpha}\}$ slices $U$ horizontally into infinitesimally thin slices.
To illustrate this with some examples, we need one more definition.
Definition 3.2
Let $p: E \longrightarrow B$ be continuous and surjective. If every point $b \in B$ has a neighborhood $U_b$ that is evenly covered by $p$, then $p$ is called a covering map}, and $E$ is called a \textbf{covering space of $B$.
Example 3.3
A simple example of a covering map is as follows: Let $S^1$ denote the unit circle. The map
$$p: S^1 \longrightarrow S^1$$ given by
$$p(s) = s^n$$
where $n$ is a positive integer, and the notation $s^n$ represents a path that travels around the circle $n$ times. $p$ is an example of a covering map. This map creates $n$ copies of $S^1$, but is still itself contained in $S^1$.
If $n$ was not an integer, say $s^{3/2}$, where we wrap around the circle 1$\frac{1}{2}$ times, this would not be a covering map, since each point on the circle needs the same amount of covers. Hence the term evenly
covered.
Example 3.4
We also have the trivial cover, which is given by the identity map
$$i: E \longrightarrow E$$
which maps every element in $E$ to itself. Every space covers itself, so this is a covering map
Let $X$ be a space and let $i: X \longrightarrow X$ be the identity map. $i$ is a covering map.
More generally, let
$$E = X \times \{1,2,...,n \}$$
The map $P: E \longrightarrow X$ given by
$$p(x,i) = x \quad \quad \forall i = 1,2,...,n$$
is a covering map.
Theorem 3.5
Let $p:E \longrightarrow B$ be a covering map. Then:
Proof
:
a. To show that $p^{-1}(b)$ has the discrete topology, we need to show that every subspace of $p^{-1}(b)$ is open in the subspace topology of $p^{-1}(b)$.
Choose any $b \in B$. By the definition of a covering map, there exists an open neighborhood $U_b$ of $b$ that is evenly covered by $p$. Therefore we have a set $\{V_{\alpha}\}$ of disjoint open sets in $E$ such that
$$p^{-1} (U_b) = \bigcup V_{\alpha}$$
where
$$P \big|_{V_{\alpha}}: V_{\alpha} \longrightarrow U_b$$
is a homeomorphism for each $V_{\alpha}$.
Since $p$ is a homeomorphism, we can say that $\left(P \big|_{V_{\alpha}} \right)^{-1}(b)$ is open in $V_{\alpha}$.
Now we can rewrite $p^{-1}(b)$ as
$$p^{-1}(b) = \bigcup \Big( p^{-1}(b) \cap V_{\alpha} \Big)$$
Since $p$ is bijective, $p^{-1}(b) \cap V_{\alpha}$ is a single point set for each $\alpha$, and is open in $p^{-1}(b)$.
We've shown that every individual point $p^{-1}(b) \cap V_{\alpha}$ is open in $p^{-1}(b)$.
Therefore, $p^{-1}(b)$ has the discrete topology. \\
b. To show that $p$ is an open map, we need to show that for any open set $A $ in $ E$, $p(A)$ is open in $B$.
So, let $A$ be an open set in $E$. Given $x \in p(A)$, choose a neighborhood $U$ of $x$ that is evenly covered by $p$. Let $\{V_{\alpha} \}$ be a partition of $p^{-1}(U)$ into slices.
Then, there is a point $y \in A$ such that $p(y) = x$.
Let $V_{\beta}$ be the set containing $y$.
Since the set $V_{\beta} \cap A$ is open in $E$, we know that it is also open in $V_{\beta}$. Furthermore, since $p$ maps $V_{\beta}$ homeomorphically onto $U$, the set $p\Big( V_{\beta} \cap A \Big)$ is open in $U$ and therefore open in $B$.
Finally, since $p\Big( V_{\beta} \cap A \Big)$ is a neighborhood of $x$ contained in $p(A)$, we can conclude that $P(A)$ is open in $B$.
Therefore, $p$ is an open map.
Theorem 3.6
The map $p: \mathbb{R} \longrightarrow S^1$ given by
$$p(x) = (\cos x, \sin x)$$
is a covering map.
Proof : This is the infinite version of the covering map in Example 3.3, which wraps the entire real number line around the unit circle. We will treat this example more rigorously.
We need to show that for every point on the circle $S^1$, there exists a neighborhood $U_b$ around it that is evenly covered by $p$. Note that the definition of covering map does not require that that the sets $U_b$ be disjoint. We will break $S^1$ into the following open sets:
$$U^+_x = S^1 \cap \left\{(x,y) \in \mathbb{R}^2 | x>0\right \}$$
$$U^-_x = S^1 \cap \left\{(x,y) \in \mathbb{R}^2 | x< 0\right\}$$
$$U^+_y = S^1 \cap \left\{(x,y) \in \mathbb{R}^2 | y>0\right \}$$
$$U^-_y = S^1 \cap \left\{(x,y) \in \mathbb{R}^2 | y< 0\right\}$$
So
$$S^1 = U_x^+ \cup U_x^- \cup U_y^+ \cup U_y^- $$
We will show that each of these sets is evenly covered by $p$.
Looking just at $U_x^+$, this narrows the angle $x$ down to $(-\frac{\pi}{2} + 2\pi k, \frac{\pi}{2} + 2\pi k)$ where $k$ is an integer. In other words, we have
$$p^{-1}\left( U_x^+\right) = \bigcup_{k \in \mathbb{Z} } \left(2 \pi k-\frac{\pi}{2}, 2 \pi k-\frac{\pi}{2} \right) $$
For instance, if $k=1$, this is the interval $(\frac{3\pi}{2}, \frac{5 \pi}{2})$
Now we can define our open sets $\{V_k\}$ to be each of these intervals, giving us the map
$$p^{-1}\left( U_x^+\right) = \bigcup_{n \in \mathbb{Z}} V_k$$
and $$p \big|_{V_k} : V_k \longrightarrow U_x^+$$ for all $k\in \mathbb{Z}$. To show that this is a homeomorphism, we need to show that, given any $k$, $p \big|_{V_k}$ is bijective, continuous, and has a continuous inverse.
$p(x) = \big( \cos x, \sin x \big) $ is injective on each $(-\frac{\pi}{2} + 2\pi k, \frac{\pi}{2} + 2\pi k)$ because, when restricted to that interval, $\sin x$ is strictly increasing. Furthermore it is surjective simply by the definition of $U_x^+$.
We know that $p$ is continuous by virtue of sine and cosine being continuous on any continuous domain. Given this, and given that each $V_k$ is a compact set, its image much also be compact. Thus the inverse map $\left(p \big|_{V_k} \right)^{-1}: U^+_x \longrightarrow V_k$ is continuous.
Therefore, each $$p \big|_{V_k}: V_k \longrightarrow U_x^+$$
is a homeomorphism.
The same argument follows for $U_x^-$, $U_y^+$, and $U_y^-$, proving that $p$ is a covering map.
Example 3.7
The map $p: \mathbb{R}^+ \longrightarrow S^1$ given by
$$p(x) = (\cos x, \sin x)$$
is surjective and locally homeomorphic. However, it isn't a covering map.
Hang on a second, why not? This is the map that wraps only the positive number line around the unit circle. To show that it is not a covering map, we need to find a point that it does not evenly cover.
The problem here lies with $x=0$. Let $U_0$ be the very small interval $$U_0 = \left\{ (\cos x, \sin x) \Big| \, |x| < \epsilon \right\}$$ There is no interval in $\mathbb{R}^+$ that maps surjectively onto $U_0$, therefore $p$ cannot be a covering map.
However, the interval $U_0$ is vanishingly small enough for $p$ to still be locally homeomorphic.
Theorem 3.8
Let
$$p:E \longrightarrow B$$
be a covering map.
If $B_0$ is a subspace of $B$ such that $E_0 = p^{-1}(B_0)$, then the map
$$p \big|_{E_0}:E_0 \longrightarrow B_0$$
is a covering map.
Proof
: We need to show that for every $b_0 \in B_0$, there exists a neighborhood around it that is evenly covered by $p \big|_{E_0}$.
Given we know that $p$ is a covering map, we can let $b_0 \in B_0$ and let $U$ be an open subset of $B$ which contains $b_0$ and is openly covered by $p$.
Let $ \{V_{\alpha} \}$ be a partition of
$$p^{-1}(U) = \cup V_{\alpha}$$ into slices.
Then we have:
$$p^{-1}(U \cap B_0) = p^{-1}(U) \cap p^{-1} (B_0) = \left( \cup V_{\alpha}\right) \cap E_0$$
$$= \cup \left( V_{\alpha} \cap E_0\right)$$
with each $ V_{\alpha} \cap E_0$ is mapped homoeomorphically onto $U \cap B_0$ by $p$.
Therefore, $p \big|_{E_0}$ is a covering map.
Theorem 3.9
If
$$p: E \longrightarrow B \quad \text{ and } \quad p': E' \longrightarrow B' $$
are covering maps, then
$$p' \times p: E \times E' \longrightarrow B \times B'$$
is a covering map.
Proof : Given $b \in B$ and $b' \in B'$, let $U$ and $U'$ be open neighborhoods of $b$ and $b'$ respectively ,which are evenly covered by $p$ and $p'$ respectively.
Now, let $\{V_{\alpha} \}$ and $\{V'_{\beta} \}$ be partitions of $p^{-1}(U)$ and $(p')^{-1}(U')$, respectively, into slices. Then,
$$(p \times p')^{-1} (U \times U') = \cup \left( V_{\alpha} \times V'_{\beta} \right) $$
where the sets $V_{\alpha} \times V'_{\beta}$ are disjoint open subsets of $E \times E'$ and mapped homeomorphically onto $U \times U'$ by $p \times p'$.
Example 3.10
Consider the space
$$T = S^1 \times S^1$$
This space is called the torus .
The product map
$$p \times p: \mathbb{R} \times \mathbb{R} \longrightarrow S^1 \times S^1$$
is a covering map of the torus by the plane $\mathbb{R}^2$, where $p$ denotes the covering map
$$p:\mathbb{R} \longrightarrow S'$$
given by
$$p(x) = (\cos 2 \pi x, \sin 2 \pi x)$$
This map is a covering map by the same argument from Theorem 2.
$p$ on its own takes each unit interval in $\mathbb{R}$ and wraps it around the unit circle. $p \times p$ takes each unit square
$[n, n+1] \times [m, m+1]$ where $n,m \in \mathbb{R}$ and wraps it around the torus.
In reality this would be a surface in $\mathbb{R}^4$, but to make it easier to visualize we have dropped down to to $\mathbb{R}^3$
Let's look at a specific example. Let $D$ be the donut-shaped surface that is obtained from rotating the circle $C^1$ centered at $(1,0,0)$ and with radius $\frac{1}{3}$ in the $xz$ plane about the $z$ axis.
Let $C_2$ be the circle of radius $1$ centered at the origin. So we have an image that looks like this:
We can show that $S^1 \times S^1$ is homeomorphic to $D$. Let $a$ be a point on the circle $C^1$ and let $b$ be a point on $C^2$. Define
$$f : C^1 \times C^2 \longrightarrow D$$ by
defining $f(a \times b)$ to be the the point where $a$ ends up after rotating $C^1$ around the $z$ axis until its center hits point $b$.
We can show that this is homeomorphism of $C_1 \times C_2$ with $D$. Firstly, we can see that $f$ is continuous since it involves only rotating continuously about the $z$ axis. The inverse will also be continuous since it simply rotates backwards.
To show that it $f$ is bijective, we will employ two different approaches. First, we will show that $f$ is surjective.
Let $x_0$ be any point that lies on our torus lite. We can draw a circle which lies on the torus that contains it. This circle has a center $b_0$. Furthermore, we can measure some angle from the axis of rotation. That angle, when places on the original $C^1$, will correspond to $a$. Therefore, every point is mapped to, and $f$ is surjective.
Now, to show that $f$ is injective, we can come up with the formula that for $f(a,b)$
For clarity, we can first express $C^1$ and $C^2$ in Cartesian coordinates
$$C^2 = x^2 + y^2 = 1$$
$C^1 = (x-1)^2 + z^2 = \frac{1}{9}$
Now we can see that the formula
$f(a\times b)$
$$b = (\cos \varphi, \sin \varphi, 0) \quad \quad 0 \leq \varphi \leq 2 \pi$$
gives us any point on $C^2$ and
$$a = \left(\frac{1}{3}\cos \theta +1, 0, \frac{1}{3} \sin \theta \right) \quad \quad 0 \leq \theta \leq 2 \pi$$
gives us any point on $C^1$. The $x$ component will correspond to the radius of the partial circle drawn by $f(a \times b)$ while the $z$ terms corresponds to the height of that partial circle
This gives us the map
$$f(a \times b) = \left( \left( \frac{1}{3}\cos \theta +1 \right) \cos \varphi , \left( \frac{1}{3}\cos \theta +1 \right) \sin \varphi, \frac{1}{3} \sin \theta \right)$$
For $0 \leq \theta, \varphi \leq 2 \pi$...
If $$\left( \frac{1}{3}\cos \theta +1 \right) \cos \varphi = \left( \frac{1}{3}\cos \theta' +1 \right) \cos \varphi' $$
$$\left( \frac{1}{3}\cos \theta +1 \right) \sin \varphi = \left( \frac{1}{3}\cos \theta' +1 \right) \sin \varphi' $$
$$ \frac{1}{3} \sin \theta = \frac{1}{3} \sin \theta' $$
only if $\theta = \theta'$ and $\varphi = \varphi'$
Just looking at the first equation, assume that $\theta \neq \theta'$. In order for the first equality to be true, we need either $$\theta = \frac{\pi}{4}, \quad \theta' = \frac{7 \pi}{4}$$ (or vice versa), or $$\theta = \frac{3\pi}{4}, \quad \theta' = \frac{5\pi}{4}$$
But if we look at the third equality, both of these pairs will give us
$$ \frac{1}{3} \sin \theta = - \frac{1}{3} \sin \theta' $$
Hence, there is no way for all three equalities to hold unless we have $\theta = \theta'$ and, using the same reasoning, $\varphi = \varphi'$, proving injectivity.
We will end this section with a few more examples of covering maps which involve Cartesian products.
Example 3.11
Consider the covering map
$$p \times p : \mathbb{R} \times \mathbb{R} \longrightarrow S^1 \times S^1$$
Let $b_0$ be the point $p(0) \in S^1$ and let $B_0$ be the subspace
$$B_0 = (S^1 \times b_0) \cup (S^1 \times b_0)$$
In other words, $B_0$ is the union of two circles that have a single point $b_0$ in common. We call this the figure eight space .
The space
$$E_0 = p^{-1}(B_0)$$
is the infinite grid
$$E_0 = (\mathbb{R} \times \mathbb{Z}) \times (\mathbb{R} \times \mathbb{Z})$$
Thus, the map
$$P_0 : E_0 \longrightarrow B_0$$
obtained by restricting $p \times p$ is a covering map.
Example 3.12
Consider the covering map
$$p \times i: \mathbb{R} \times \mathbb{R^+} \longrightarrow S^1 \times \mathbb{R}^+$$
where $i$ is the identity map on $\mathbb{R}^+$ and $p$ is the map
$$p(x) = (\cos x, \sin x)$$
Take the standard homeomorphism of $S^1 \times \mathbb{R}^+$ with $\mathbb{R}^2 \mathbin{/} \{0\} $. That is,
$$f\Big((\cos x, \sin x), r\Big) = (r \cos x, r \sin x)$$
This gives us a covering map
$$\mathbb{R} \times \mathbb{R^+} \longrightarrow \mathbb{R}^2 \mathbin{/} \{0\}$$
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