# Math4202 Topology II (Lecture 16)

## Algebraic Topology

### Fundamental group of the circle

Recall from previous lecture, we have unique lift for covering map.

#### Lemma for unique lifting for covering map

Let $p: E\to B$ be a covering map, and $e_0\in E$ and $p(e_0)=b_0$. Any path $f:I\to B$ beginning at $b_0$, has a unique lifting to a path starting at $e_0$.

Back to the circle example, it means that there exists a unique correspondence between a loop starting at $(1,0)$ in $S^1$ and a path in $\mathbb{R}$ starting at $0$, ending in $\mathbb{Z}$.

<details>
<summary>Proof</summary>

$\forall t\in I$, by the covering map, partition into slices property, there exist some open neighborhood $U_{f(t)}$ of $f(t)$ such that $p^{-1}(U_{f(t)})\subseteq E$ is a neighborhood of $e_0$. And $p^{-1}(U_{f(t)})$ is a disjoint union of $\{V_{f(t),\alpha}\}$

Since $f:I\to B$ is continuous, then $V_t$ of $t\in I$ is open in $I$ and we can find some small open neighborhood $f^{-1}(V_t)\subseteq U_{f(t)}$.

Note that $\{V_t\}$ is an open cover of $[0,1]$. As $[0,1]$ is compact, $\{V_t\}$ has a finite subcover, $\{V_{t_i}\}_{i=1}^k$.

Then we can use $\{V_{t_i}\}_{i=1}^k$ to partition $I$ into $0<s_1<s_2<\cdots<s_m=1$, such that $[s_i,s_{i+1}]$ is contained in one of $V_{t_i}$.

We can do so by consider the $[0,t_1)\subseteq V_{t_1}$, therefore $[t_1,1]\subseteq \bigcup_{i=2}^k V_{t_i}$. (These open sets should intersect with $[t_1,1]$ non trivially, no useless choice shall be made.)

Since $\bigcup_{i=2}^k V_{t_i}$ is open, there is $z_1<t_1$ and $(z_1,1]\subseteq \bigcup_{i=2}^k V_{t_i}$.

Then we can choose $z_1<s_1<t_1$, since $s_1\in \bigcup_{i=2}^k V_{t_i}$, there exists some $V_{t_2}$ such that $s_1\in V_{t_2}$.

Note that we can find $[0,t_2)\subseteq V_{t_1}\cup V_{t_2}$, and $t_2>t_1$.

Continue this process, we can find our partition $0<s_1<s_2<\cdots<s_m=1$, such that $[s_i,s_{i+1}]$ is contained in one of $V_{t_i}$.

In conclusion, we find the above partition of $I$ such that

$$
f([s_i,s_{i+1}])\subset f(V_{t_i})\subseteq U_i
$$

Define $\tilde{f}:I\to E$, on $[0,s_1]$, $f([0,s_1])\subseteq U_0$, $p^{-1}(U_0)=\bigcup_\alpha W_{0,\alpha}$, $\exists \alpha_0$ such that $e_0\in W_{0,\alpha_0}$.

$$
\tilde{f}=(p^{-1}|_{W_{s_i,\alpha_i}})\circ f
$$
Repeat the same construction for each $[s_i,s_{i+1}]$.

---

The uniqueness for lifting map is guaranteed by the following:

- For $e_0$, $W_{0,\alpha_0}\to U_0$ is a homeomorphism, suppose $\tilde{f}$ and $\tilde{f}'$ are liftings of $f$, then they must agree on $\tilde{f}([0,s_1])=\tilde{f}'([0,s_1])=p^{-1}(U_0)\circ f$.
- Repeat the same construction for each $[s_i,s_{i+1}]$.

</details>

#### Lemma for unique lifting homotopy for covering map

Let $p: E\to B$ be a covering map, and $e_0\in E$ and $p(e_0)=b_0$. Let $F:I\times I\to B$ be continuous with $F(0,0)=b_0$. There is a unique lifting of $F$ to a continuous map $\tilde{F}:T\times I\to E$, such that $\tilde{F}(0,0)=e_0$.

Further more, if $F$ is a path homotopy, then $\tilde{F}$ is a path homotopy.

Discuss on Friday.
