# Reference
https://math.ucr.edu/~res/math205B-2018/Munkres%20-%20Topology.pdf
# covering space
$\textbf{Definition.}$
Let $p: E \rightarrow B$ be a continuous surjective map. The open set $U$ of $B$ is said to be evenly covered by $p$ if the inverse image $p^{-1}(U)$ can be written as the union of disjoint open sets $V_a$ in $E$ such that for each $\alpha$, the restriction of $p$ to $V_a$ is a homeomorphism onto $U$. The collection $\{V_\alpha\}$ will be called a partition of $p^{-1}(U)$ into slices.


$\textbf{Definition.}$
Let $p: E \rightarrow B$ be continuous and surjective. If every point $b$ of $B$ has a neighborhood $U$ that is evenly covered by $p$, then $p$ is called a covering map, and $E$ is said to be a covering space of $B$.

$\star$ Note that if $p: E \rightarrow B$ is a covering map, then for each $b \in B$, the subspace $p^{-1}(b)$ of $E$ has the discrete topology. For each slice $V_a$, it is open in $E$ and intersects the set $p^{-1}(b)$ in a single point; therefore, this point is open in $p^{-1}(b)$.

$\star$If $p: E \rightarrow B$ is a covering map, then $p$ is a local homeomorphism of $E$ with $B$. That is, each point $e$ of $E$ has a neighborhood that is mapped homeomorphically by $p$ onto an open subset of $B$.

$\textbf{Theorem}$
The map $p: \mathbb{R} \rightarrow {S^1}$ given by the equation $p(x) =( \sin2\pi x ,\cos2\pi x)$ is a covering map.

$Proof$:
The fact that $p$ is a covering map comes from elementary properties of the sine and cosine functions. Consider, for example, the subset $U$ of $S^1$ consisting of those points having a positive first coordinate. The set $p^{-1}(U)$ consists of those points $x$ for which $\cos(2\pi x)$ is positive, that is, it is the union of the intervals:

$V_n = \left(n - \frac{1}{4}, n + \frac{1}{4}\right).$

for all $n \in \mathbb{Z}.$ Now, restricted to any closed interval $\overline{V_n}$, the map $p$ is injective because $\sin(2\pi x)$ is strictly monotonic on such an interval. Furthermore, $p$ carries $\overline{V_n}$ surjectively maps onto $\overline{U}$, and $V_n$ to $U$, by the intermediate value theorem. Since $\overline{V_n}$ is compact, $p | \overline{V_n}$ is a homeomorphism of $\overline{V_n}$ with $\overline{U}$. In particular, $p |_{V_n}$ is a homeomorphism of $V_n$ with $U$.

Similar arguments can be applied to the intersections of $S^1$ with the upper and lower open half-planes, and with the open left-hand half-plane. These open sets cover $S^1$, and each of them is evenly covered by $p$. Hence, $p: \mathbb{R} \rightarrow S^1$ is a covering map.

$\textbf{Theorem}$
Let $p: E \rightarrow B$ be a covering map. If $B_0$ is a subspace of $B$, and if $E_0 = p^{-1}(B_0)$, then the restriction of $p$ is a covering map.

$Proof$
Given $b_0 \in B_0$, let $U$ be an open set in $B$ containing $b_0$ that is evenly covered by $p$; let $\{ V_a \}$ be a partition of $p^{-1}(U)$ into slices. Then $U \cap B_0$ is a neighborhood of $b_0$ in $B_0$, and the sets $v_a \cap E_0$ are disjoint open sets in $E_0$ whose union is $p^{-1}(U \cap B_0)$, and each is mapped homeomorphically onto $U \cap B_0$ by $p$.

$\textbf{theorem}$
If $p: E \rightarrow B$ and $p': E' \rightarrow B'$ are covering maps, then the map
$p \times p': E \times E' \rightarrow B \times B'$
is a covering map.

$Proof$
Given $b \in B$ and $b' \in B'$, let $U$ and $U'$ be neighborhoods of $b$ and $b'$, respectively, that are evenly covered by $p$ and $p'$, respectively. Let $\{ V_a \}$ and $\{ W_b \}$ be partitions of $p^{-1}(U)$ and $(p')^{-1}(U')$, respectively, into slices. Then the inverse image under $p \times p'$ of the open set $U \times U'$ is the union of all the sets $V_a \times W_b$. These are disjoint open sets in $E \times E'$, and each is mapped homeomorphically onto $U \times U'$ by $p \times p'$.

# lifting

$\textbf{Definition.}$
Let $p: E \rightarrow B$ be a covering map. If $f$ is a continuous mapping of some space $X$ into $B$, a lifting of $\tilde{f}$ is a map $\tilde{f}: X \rightarrow E$ such that $p \circ \tilde{f} = f$.

$\textbf{Theorem}$
Let $p: E \rightarrow B$ be a covering map, and let $p(e_0) = b_0$. Any path $f: [0, 1] \rightarrow B$ beginning at $b_0$ has a unique lifting to a path $\tilde{f}: [0, 1]$ in $E$ beginning at $e_0$.

$Proof$
Cover $B$ by open sets $U$ each of which is evenly covered by $p$. Find a subdivision of $[0, 1]$, say $s_0,....s_n$, such that for each $i$, the set $f([s_i, s_{i+1}])$ lies in such an open set $U$. (Here we use the Lebesgue number lemma.)

We define the lifting $\tilde{f}$ step by step.

First, define $\tilde{f}(0) = e_0$. Then, supposing $\tilde{f}(s)$ is defined for $0 \leq s \leq s_i$ , we define $\tilde{f}$ on $[s_i, s_{i+1}]$ as follows: The set $\tilde{f}([s_i, s_{i+1}])$ lies in some open set $U$ that is evenly covered by $p$. Let ${\{V_a\}}$ be a partition of $p^{-1}(U)$ into slices; each set $V_a$ is mapped homeomorphically onto $U$ by $p$. Now $\tilde{f}(s)$ lies in one of these sets, say in $V_0$. Define $\tilde{f}(s)$ for $s \in [s_i, s_{i+1}]$ by the equation:
\begin{equation}
\tilde{f}(s) = (p|_{V_0})^{-1}(f(s)).
\end{equation}
Because $p|_{V_0}: V_0 \rightarrow U$ is a homeomorphism, $\tilde{f}$ will be continuous on $[s_i, s_{i+1}]$.

Continuing in this way, we define $\tilde{f}$ on all of $[0, 1]$. Continuity of $\tilde{f}$ follows from the pasting lemma; the fact that $p \circ \tilde{f} = f$ is immediate from the definition of $f$.

The uniqueness of $\tilde{f}$ is also proved step by step. Suppose that $\widetilde{\widetilde{f}}$ is another lifting of $f$ beginning at $e_0$. Then $\tilde{f}(0) = e_0 = \widetilde{\widetilde{f}}(0)$. Suppose that $\tilde{f}(s) = \widetilde{\widetilde{f}}(s)$ for all $s$ such that $0 \leq s \leq s_i$. Let $V_0$ be as in the preceding paragraph; then for $s \in [s_i, s_{i+1}]$, $\tilde{f}(s)$ is defined as $(p|_{V_0})^{-1}(f(s))$. What can $\widetilde{\widetilde{f}}(s)$ equal? Since $\tilde{f}(s)$ is a lifting of $f$, it must carry the interval $[s_i, s_{i+1}]$ into the set $p^{-1}(U) = \bigcup V_a$. The slices are open and disjoint; because the set $[s_i, s_{i+1}]$ is connected, it must lie entirely in one of the sets $V_a$. Because $\tilde{f}(s_i) = \widetilde{\widetilde{f}}(s_i)$, which is in $V_0$, $\widetilde{\widetilde{f}}$ must carry all of $[s_i, s_{i+1}]$ into the set $V_0$. Thus, for $s$ in $[s_i, s_{i+1}]$, $\widetilde{\widetilde{f}}(s)$ must equal some point $y$ of $V_0$ lying in $p^{-1}(f(s))$. But there is only one such point $y$, namely, $(p|_{V_0})^{-1}(f(s))$. Hence $\tilde{f}(s) = \widetilde{\widetilde{f}}(s)$ for $s \in [s_i, s_{i+1}]$.

$\textbf{theorem}$
Let $p: E \rightarrow B$ be a covering map, and let $p(e_0) = b_0$. Let the map $F: I \times I \rightarrow B$ be continuous, with $F(0, 0) = b_0$. There exists a unique lifting of $F$ to a continuous map
\begin{equation}
\tilde{F}: I \times I \rightarrow E
\end{equation}such that $\tilde{F}(0, 0) = e_0$. If $F$ is a path homotopy, then $\tilde{F}$ is a path homotopy.

$Proof$
Given $F$, we first define $\tilde{F}(0, 0) = e_0$. Next, we use the preceding lemma to extend $\tilde{F}$ to the left-hand edge $0 \times I$ and the bottom edge $I \times 0$ of $I \times 1$. Then we
extend $\tilde{F}$ to all of $I \times I$ as follows:
$\quad$Choose subdivisions
$s_0 < s_1 < \ldots < s_m$
$t_0 < t_1 < \ldots < t_n$

of $I$ fine enough that each rectangle
\begin{equation}
I_i \times J_j = [s_{i-1}, s_i] \times [t_{j-1}, t_j]
\end{equation}
is mapped by $F$ into an open set of $B$ that is evenly covered by $p$. (Use the Lebesgue number lemma.) We define the lifting $\tilde{F}$ step by step, beginning with the rectangle $I_1 \times J_1$, continuing with the other rectangles $I_i \times J_1$ in the "bottom row," then with the rectangles $I_i \times J_2$ in the next row, and so on.

In general, given $i_0$ and $j_0$, assume that $\tilde{F}$ is defined on the set $A$ which is the union of $\{0\} \times I$ and $I \times \{0\}$ and all the rectangles "previous" to $I_{i0} \times J_{j0}$. (those rectangles $I_i \times J_j$ for which $j<j_0$ and those for which $j = j_0$ and $i < i_0.$) Assume also that $\tilde{F}$ is a continuous lifting of $F|_A$. We define $\tilde{F}$ on $I_{i0} \times J_{j0}$. Choose an open set $U$ of $B$ that is evenly covered by $p$ and contains the set $F(I_{i0} \times J_{j0})$. Let $\{ V_a \}$ be a partition of $U$ into slices; each set $V_a$ is mapped homeomorphically onto $U$ by $p$. Now $\tilde{F}$ is already defined on the set $C = A \cap (I_{i0} \times J_{j0})$. This set is the union of the left and bottom edges of the rectangle $I_{i0} \times J_{j0}$, so it is connected. Therefore, $\tilde{F}(C)$ is $connected$ and must lie entirely within one of the sets $V_a$. Suppose it lies in $V_0$. Then, the situation is as pictured in Figure 54.2.


Let $p_0 : V_0 \rightarrow U$ denote the restriction of $p$ to $V_0$. Since $\tilde{F}$ is a lifting of $F|_A$, we
know that for $x \in C$,
$p_0(\tilde{F}(x)) = p(\tilde{F}(x)) = F(x)$, so that $\tilde{F}(x) = p_0^{-1}(F(x))$. Hence we may extend $\tilde{F}$ by defining
\begin{equation}
\tilde{F}(x) = p_0^{-1}(F(x))
\end{equation}
for $x \in I_{i0} \times J_{j0}$. The extended map will be continuous by the pasting lemma.
$\quad$Continuing in this way, we define $\tilde{F}$ on all of $I \times I$.
$\quad$To check uniqueness, note that at each step of the construction of $\tilde{F}$, as we extend $\tilde{F}$ first to the bottom and left edges of $I \times I$, and then to the rectangles $I_{i0} \times J_{j0}$ one by one, there is only one way to extend $\tilde{F}$ continuously. Thus, once the value of $\tilde{F}$ at $(0, 0)$ is specified, $\tilde{F}$ is completely determined.

Now suppose that $F$ is a path homotopy. We wish to show that $\tilde{F}$ is a path homotopy. The map $F$ carries the entire left edge $\{0\} \times I$ of $I \times I$ into a single point $b_0$ of $B$.
Because $\tilde{F}$ is a lifting of $F$, it carries this edge into the set $p^{-1}(b_0)$. But this set has the
discrete topology as a subspace of $E$. Since $0 \times I$ is connected and $\tilde{F}$ is continuous,
$\tilde{F}(0 \times I)$ is connected and thus must equal a one-point set. Similarly, $\tilde{F}(1 \times I)$ must
be a one-point set. Thus, $\tilde{F}$ is a path homotopy.

$\textbf{Theorem}$
Let $p: E \rightarrow B$ be a covering map, and let $p(e_0) = b_0$. Let $f$ and $g$ be two paths in $B$ from $b_0$ to $b_1$, and let $\tilde{f}$ and $\tilde{g}$ be their respective liftings to paths in $E$ beginning at $e_0$. If $f$ and $g$ are path homotopic, then $\tilde{f}$ and $\tilde{g}$ end at the same point of $E$ and are path homotopic.

$Proof$
Let $F: I \times I \rightarrow B$ be the path homotopy between $f$ and $g$. Then $F(0, 0) = b_0$. Let $\tilde{F} : I \times I \rightarrow E$ be the lifting of $F$ to $E$ such that $\tilde{F}(0, 0) = e_0$. By the preceding lemma, $\tilde{F}$ is a path homotopy, so that $\tilde{F}(0 \times I) = \{e_0\}$ and $\tilde{F}(1 \times I)$ is a one-point set $\{e_1\}$.
$\quad$The restriction $\tilde{F}|_{I \times 0}$ of $\tilde{F}$ to the bottom edge of $I \times I$ is a path on $E$ beginning at $e_0$ that is a lifting of $F|_{I \times 0}$. By uniqueness of path liftings, we must have $\tilde{F}(s, 0) = \tilde{f}(s)$. Similarly, $\tilde{F}|_{I \times 1}$ is a path on $E$ that is a lifting of $F|_{I \times 1}$, and it begins at $e_0$ because $\tilde{F}(0 \times I) = \{e_0\}$. By uniqueness of path liftings, $\tilde{F}(s, 1) = \tilde{g}(s)$. Therefore, both $\tilde{f}$ and $\tilde{g}$ end at $e_1$, and $\tilde{F}$ is a path homotopy between them.

# The Fundamental Group of the Circle

$\textbf{Definition}$
Let $X$ be a topological space. Let $\pi_1(X, x_0)$ denote the set of all equivalence classes $[f]$ of $loops$ in $X$ based at $x_0$.

$\textbf{Definition.}$
Let $p: E \rightarrow B$ be a covering map, and let $b_0 \in B$. Choose $e_0$ so that $p(e_0) = b_0$. Given an element $[f]$ of $\pi_1(B, b_0)$, let $\tilde{f}$ be the lifting of $f$ to a path in $E$ that begins at $e_0$. Let $\phi([f])$ denote the endpoint $\tilde{f}(1)$ of $\tilde{f}$. Then $\phi$ is a well-defined set map:

$\phi: \pi_1(B, b_0) \rightarrow p^{-1}(b_0)$

We call $\phi$ the $lifting$ $correspondence$ derived from the covering map $p$. It depends, of course, on the choice of the point $e_0$.

$\textbf{Theorem}$
Let $p: E \rightarrow B$ be a covering map, and let $p(e_0) = b_0$. If $E$ is path connected, then the lifting correspondence $\phi: \pi_1(B, b_0) \rightarrow p^{-1}(b_0)$ is surjective. If $E$ is simply connected, it is bijective.

$Proof$
If $E$ is path-connected,then,given $e_1 \in p^{-1}(b_0)$, there is a path $\tilde{f}$ in $E$ from $e_0$ to $e_1$. Then $f = p \circ \tilde{f}$ is a loop in $B$ at $b_0$, and $\phi([f]) = e_1$ by definition.
Suppose $E$ is simply connected.Let $[f]$ and $[g]$ be two elements of $\pi_1(B, b_0)$ such that $\phi([f]) = \phi([g])$. Let $\tilde{f}$ and $\tilde{g}$ be the liftings of $f$ and $g$, respectively, to paths in $E$ that begin at $e_0$, then $\tilde{f}(1) = \tilde{g}(1)$. Since $E$ is simply connected, there is a path homotopy $\tilde{F}$ in $E$ between $\tilde{f}$ and $\tilde{g}$. Then $p \circ \tilde{F}$ is a path homotopy in $B$ between $f$ and $g$.

$\textbf{Theorem}$
The fundamental group of $S^1$ is isomorphic to the additive group of integers.

${proof}$
Let $p: \mathbb{R} \rightarrow S^1$ be the covering map $p(x) =( \sin2\pi x,\ \cos2\pi x)$ , let $e_0 = 0$, and let $b_0 = p(e_0)$. Then $p^{-1}(b_0)$ is the set $\mathbb{Z}$ of integers. Since $\mathbb{R}$ is simply connected, the lifting correspondence

$\phi: \pi_1(B, b_0) \rightarrow \mathbb{Z}$

is bijective.
We show that $\phi$ is a homomorphism, and the theorem is proved.

Given $[f]$ and $[g]$ in $\pi_1(B, b_0)$, let $\tilde{f}$ and $\tilde{g}$ be their respective liftings to paths on $\mathbb{R}$ beginning at $0$. Let $n = \tilde{f}(1)$ and $m = \tilde{g}(1)$; then $\phi([f]) = n$ and $\phi([m]) = m$ by definition. Let $\widetilde{\widetilde{g}}$ be the path:

$\widetilde{\widetilde{g}}(s) = n + \tilde{g}(s)$

on $\mathbb{R}$. Because $p(n + x) = p(x)$ for all $x \in \mathbb{R}$, the path $\widetilde{\widetilde{g}}$ is a lifting of $g$; it begins at $n$. Then the product $f * \widetilde{\widetilde{g}}$ is defined, and it is the lifting of $f * g$ that begins at $0$, as you can check. The endpoint of this path is $\widetilde{\widetilde{g}}(1)=n + m$. Then, by definition, $\phi([f * g]) = n + m = \phi([f]) * \phi([g])$.

$\textbf{Definition}$
Let $h: X \rightarrow Y$ be a continuous map such that $h(x_0) = y_0$. To denote this, we say $h: (X, x_0) \rightarrow (Y, y_0)$ is a continuous map. We define the homomorphism induced by $h$ as $h_*: \pi_1(X, x_0) \rightarrow \pi_1(Y, y_0)$:

\begin{equation}
h_*([f]) = [h \circ f]
\end{equation}

$\textbf{Theorem}$
Let $p: E \rightarrow B$ be a covering map, and let $p(e_0) = b_0$.
$(a)$The homomorphism $p_*:$ $\pi_1(E, e_0) \rightarrow \pi_1(B, b_0)$ is injective.
$(b)$Let $H = p_*(\pi_1(E, e_0) )$. The lifting correspondence induces an injective map:
\begin{equation}
\phi': \pi_1(B, b_0)/H \rightarrow p^{-1}(b_0)
\end{equation}
which is bijective if $E$ is path connected.
$(c)$If $f$ is a loop in $B$ based at $b_0$, then $[f] \in H$ if and only if $f$ lifts to a loop in $E$ based at $e_0$.

$\textbf{Proof}:$
$(a)$Suppose $\tilde{h}$ is a loop in $E$ at $e_0$, and $p_*([\tilde{h}])$ is the identity element. Let $F$ be a path homotopy between $p \circ \tilde{h}$ and the constant loop. If $\tilde{F}$ is the lifting of $F$ to $E$ such that $\tilde{F}(0, 0) = e_0$, then $\tilde{F}$ is a path homotopy between $\tilde{h}$ and the constant loop at $e_0$.

$(b)$: Given loops $f$ and $g$ in $B$, let $\tilde{f}$ and $\tilde{g}$ be liftings of them to $E$ that begin at $e_0$. Then, $\phi([f]) = \tilde{f}(1)$ and $\phi([g]) = \tilde{g}(1)$. We show that $\phi([f]) = \phi([g])$ if and only if $[f] \in H * [g]$.

First, suppose that $[f] \in H * [g]$. Then $[f] = [h * g]$, where $h = p \circ \tilde{h}$ for some loop $\tilde{h}$ in $E$ based at $e_0$. Now the product $\tilde{h} * \tilde{g}$ is defined, and it is a lifting of $h * g$. Because $[f] = [h * g]$, the liftings $\tilde{f}$ and $\tilde{h} * \tilde{g}$, which begin at $e_0$, must end at the same point of $E$. Then $\tilde{f}$ and $\tilde{g}$ end at the same point of $E$, so that $\phi([f]) = \phi([g])$.
(See Figure 54.3.)


Now suppose that $\phi([f]) = \phi([g])$. Then $\tilde{f}$ and $\tilde{g}$ end at the same point of $E$. The product of $\tilde{f}$ and the reverse of $\tilde{g}$ is defined, and it is a loop $\tilde{h}$ in $E$ based at $e_0$. By direct computation, $[\tilde{h} * \tilde{g}] = [\tilde{f}]$. If $\tilde{F}$ is a path homotopy in $E$ between the loops $\tilde{h} * \tilde{g}$ and $\tilde{f}$, then $p \circ \tilde{F}$ is a path homotopy in $B$ between $h * g$ and $f$, where $h = p \circ \tilde{h}$. Thus, $[f] \in H * [g]$, as desired.

If $E$ is path connected, then $\phi$ is surjective, so that $\phi'$ is surjective as well.

$(c)$ Injectivity of $\phi'$ means that $\phi([f]) = \phi([g])$ if and only if $[f] \in H * [g]$. Applying this result in the case where $g$ is the constant loop, we see that $\phi([f]) = e_0$ if and only if $[f] \in H$. But $[\tilde{f}] = e_0$ precisely when the lift of $f$ that begins at $e_0$ also ends at $e_0$.