# Math4202 Topology II (Lecture 15)

## Algebraic Topology

### Fundamental group of the circle

Recall from previous lecture, we have $p:\mathbb{R}\to S^1$ by $x\mapsto e^{2\pi ix}$.

We want to study the relation between the paths in $\mathbb{R}$ starting at $0$ and the loops in $S^1$ at $1$.

#### Definition for lift

Let $p:E\to B$ be a map. If $f$ is a continuous map from $X\to B$, a lifting of $f$ is a map $\tilde{f}:X\to E$ such that $p\circ \tilde{f}=f$

> A natural question is, whether lifting always exists? and how many of them (up to homotopy)?

Back to the circle example, we have $f:I\to S^1$, representing a loop, and $p:\mathbb{R}\to S^1$, by $p(x)=e^{2\pi ix}$.

#### 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>Idea for Proof</summary>

Starting at $b_0$, by the covering map property, there exist some open neighborhood $U_0$ of $b_0$ such that $V_0=p^{-1}(U_0)$ is a neighborhood of $e_0$. And $p|_{V_0}$ is a homeomorphism on to $U_0$. 

Since $f$ is continuous, then $f^{-1}(U_0)$ is open in $I$ and we can find some small open neighborhood $[0,s_1]$, such that $f^{-1}([0,s_1])\subset V_0$.

Then we define $\tilde{f}:[0,s_1]\to E$, by $\tilde {f}(t)=(p|_{V_0})^{-1}\circ f$.

Continue with compactness property... Continue on Wednesday.

</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.

