Problem: Let $\alpha:[0,1] \rightarrow \mathbb{C}^*$ be a closed curve $\alpha(0)=\alpha(1)=1$. Then exists a unique $\tilde{\alpha}: [0,1] \rightarrow \mathbb{C}$ such that $\tilde{\alpha}(0)=0$ and $\exp \circ \tilde{\alpha} = \alpha$.
I am having trouble understanding the proof(italics part) - what exactly is meant?
Proof: For any point $a \in \mathbb{C}^*$ exists an open neighborhood $V$, such that $\exp^{-1}(V) = \bigsqcup_n U_n$, where each $U_n$ is homeomoprhically mapped onto $V$.
By Heine Borel, we deduce exictence of division $[0,1]$ such that $\alpha([a_n,a_{n+1}])$ is contained in an open set $V$ as above.
(NOT understand) We can then lift the whole curve, by inductively lifting pieces in the corresponding parallel horizontal strips such that they glue at initial and end points.
So we have the subdivision of $[0,1]$ such that $\alpha([a_n,a_{n+1}])$ is contained in an open $V$ and we have a lift $\tilde{\alpha}_n:[a_n,a_{n+1}] \rightarrow \mathbb{C} $ such that the composition $\exp \circ \tilde{\alpha}_n([a_n,a_{n+1}]) = \alpha([a_n,a_{n+1}])$. Let us start with the partition $[0,a_1]$, then we have an open set, call it $V_0$, where $\alpha([0,a_1])$ has a lift and a collection $\bigsqcup_n U_n^0$ with each $U_n^0 \approx V_0$. Select one of these $U_n^0$ and call it $O_0$. Now, for the partition $[a_1,a_2]$, we have the open set $\alpha([a_1,a_2]) \subset V_1$ and collection $\bigsqcup_n U_n^1$. Notice each $U_n^1$ contains one of $\exp^-1(\alpha(a_1))$ and we choose $O_1=U_n^1$ such that $U_n^1 \cap O_0 \neq \emptyset$. Thus $\alpha([0,a_2])$ has a lift $\tilde{\alpha}:[0,a_2]\rightarrow \mathbb{C}$ defined by $$ \tilde{\alpha}(x) = \left\{ \begin{array}{cc}\tilde{\alpha}_0(x) & x \in [0,a_1] \\ \tilde{\alpha}_1(x) & x \in [a_1,a_2]\end{array}\right.$$
Repeat this process by induction to cover $[0,1]$ and we arrive at the lift $\tilde{\alpha}$ of $\alpha$. At the risk of repeating ad nauseam, that is $\tilde{\alpha}$ is such that $\exp \circ \tilde{\alpha} = \alpha$.