Path lifting lemma - proof clarification

Question

So i was studying the proof of the 'path lifting lemma' on my book and I got stuck. Here $ S^1 $ is identified with complex numbers of norm 1 and $ E: \mathbb{R} \rightarrow S^1 $ is defined by $ E(x)= e^{2\pi ix}, x\in \mathbb{R}$.

Lemma: If $ \alpha $ is a path in $ S^1 $ with starting point 1, there exist a unique path $\alpha '$ in $ \mathbb{R} $ with starting point $0$ such that $ E \alpha ' = \alpha $.

Proof: Because I is compact and $ \alpha $ is continuous, we can choose subintervals of I, $ [t_0,t_1],...[t_{n-1},t_n],\ 0=t_0 < t_1$ $ <...< t_n=1$ such that $$|\alpha (t) - \alpha (t')| < 1 \; \forall t,t' \in [t_{i-1},t_i] \;\; \forall i=1...n \; \; \; \; \; \; \;\; \; \; \; \; \; \;\; \; \; \; \; \; \; \textbf{(1)} $$ and so such that $$ \alpha ([t_{i-1},t_i])\mbox{ is contained in } S^1 \backslash \{ - \alpha (t_{i-1}) \} \;\; \forall i=1...n \; \; \; \; \; \; \;\; \; \; \; \ \textbf{ (2) } $$...(the proof goes on)

I'm ok with the rest of the proof (that I didn't write) but I don't understand this beginning: why can we choose such intervals? Where the compactness of I and the continuity of $ \alpha $ are used? Why (1) implies (2) ?

Thanks to everyone.

Answer 1

Fix a path $\alpha: I\to S^1$. Note that you can reformulate the statement as follows: "It is possible to subdivide the interval into pieces $[0,1] = [0,t_1]\cup [t_1,t_2]\cup...\cup[t_{n-1},t_n]$ such that the image $\alpha([t_i,t_{i+1}])$ of any piece is contained in an open ball of radius say $1/4$. " My statement is stronger than the one from the book.

Here is how to prove this. Let $\mathcal U$ be the open cover of $S^1$ consisting of all open balls of radius $1/4$. Set $\mathcal V = \{\gamma^{-1}(U)\,|\, U\in \mathcal U\}$. Note that $\mathcal V$ is an open cover of the interval $I$. The result now follows from the Lebegue-number lemma: Explanations of Lebesgue number lemma. There is a number $1/n >0$ such that each set of diameter at most $1/n$ is contained in a set from the cover $\mathcal V$. The partition $[0,1/n] \cup [1/n,2/n]\cup...$ will do the job. $[t_i,t_{i+1}] \subset \gamma^{-1}(U)$ for some $U \in \mathcal U$, hence $\gamma([t_i,t_{i+1}])\subset U$ and $U$ is an open ball of radius $1/4$.

Answer 2

(1)

It is well-known that any continuous functon from a compact metric space to a metric space is uniformly continuous. This implies that there exists $\delta > 0$ such that $|\alpha (t) - \alpha (t')| < 1$ for all $t, t'\in I$ with $|t - t'| < \delta$. Let $n \in \mathbb N$ such that $1/n \le \delta$. Then $t_i = i/n$ ($i=0,\dots,n$) will do.

(2)

We have $|\alpha (t_{i-1}) - (-\alpha (t_{i-1}))| = 2$, thus for $t \in [t_{i-1},t_i]$ we get $$|\alpha (t) - (-\alpha (t_{i-1}))| = |\alpha (t_{i-1}) - (-\alpha (t_{i+1})) + \alpha (t) - \alpha (t_{i-1})| \\ \ge |\alpha (t_{i-1}) - (-\alpha (t_{i+1}))| - | \alpha (t) - \alpha (t_{i-1})| > 2 -1 = 1 ,$$ hence $\alpha (t) \ne -\alpha (t_{i-1})$.