# The Fundamental Group of the Circle, Part 6

/
__ Part 1:__ Set-up/observations

__Part 2:__ Show $\Phi$ is well defined

__Part 3:__ Show $\Phi$ is a group homomorphism

__Part 4:__ Show $\Phi$ is surjective

__Part 5:__ Show $\Phi$ is injective

__Part 6:__ Prove the two lemmas used in parts 4 and 5

**Lemma 1:**For each path $f:I\to S^1$, with $f(0)=x_0$ and for each $\tilde{x_0}\in p^{-1}(x_0)$, there is a unique lift $\tilde{f}:I\to\mathbb{R}$ such that $\tilde{f}(0)=x_0$.

**Lemma 2:**For each homotopy $f_t:I\to S^1$ of paths starting at $f_t(0)=x_0$ and for each $\tilde{x}_0\in p^{-1}(x_0)$, there is a unique lift $\tilde{f}_t:I\to\mathbb{R}$ such that $\tilde{f}_t(0)=\tilde{x}_0$.

**Proposition:**Given any space $Y$, a map $F:Y\times I\to S^1$, and a map $\tilde{F}:Y\times\{0\}\to\mathbb{R}$ which lifts $F\big|_{Y\times\{0\}}$, there exists a unique map $G:Y\times I\to \mathbb{R}$ which lifts $F$ and is an extension of $\tilde{F}$.

## Proving the Proposition*

*Observation:* There is an open cover $\{U_\alpha\}$ of $S^1$ such that for each $\alpha$, the set $p^{-1}(U_\alpha)$ is a disjoint union of open sets each of which is mapped homeomorphically to $U_\alpha$ by $p$. (For an example, see the footnote ** below)

*is*true for $i=0$ since $N\times [0,0]$ is a subset of $Y\times\{0\}$ and by assumption we already have the map $\tilde{F}:Y\times\{0\}\to\mathbb{R}$.) Now since $F(N\times[t_i,t_{i+1}])\subset U_i$ we know by our opening Observation that $p^{-1}(U_i)$ is a disjoint union of open sets in $\mathbb{R}$, each of which is mapped homeomorphically onto $U_i$. Hence there exists a $\tilde{U}_i\in p^{-1}(U_i)$ which contains the point $G(y_0,t_i)$ (this point exists by the induction hypothesis) since $(y_0,t_i)$ sits inside $N\times [0,t_i]$, and the latter maps into $U_i$ by $F$.

This is well-defined since $p\big|_{\tilde{u}_i}$ is a homeomorphism, so we can repeat this inductive step finitely many times to obtain the desired lift $G:N\times I\to\mathbb{R}$ for some neighborhood $N$ of $y_0$. And since $y_0\in Y$ was arbitrary, we have our lift $G$ on all of $Y\times I$.

QED!

*Footnotes:*

* This proof is directly from Hathcer's *Algebraic Topology* chapter 1.1, but I've included the pictures to help make sense of things.

** For example, take $\{U_\alpha\}=\{U_1,U_2\}$ where $U_1=S^1\smallsetminus \{(1,0)\}$ and $U_2=S^1\smallsetminus \{(-1,0)\}$. Then $p^{-1}(U_1)=\{(n,n+1)\}_{n\in\mathbb{Z}}$ is a disjoint collection and $p(n,n+1)\cong U_1$ for all $n$. Similarly $p^{-1}(U_2)=\{(n-1/2,n+1/2)\}_{n\in\mathbb{Z}}$ is disjoint and $p(n-1/2,n+1/2)\cong U_2$ for all $n$.

*** Since we can replace $N\times {t_i}$ with the intersection $(N\times {t_i})\cap \tilde{G}\big|_{N\times\{t_i\}}^{-1}(\tilde{U}_i)$.