I have a doubt on what seems to be a simple step on a proof I've seen, but I just can't seem to get it.
Theorem. Suppose $\pi:E\rightarrow B$ is a fiber bundle with fiber $F$. For every path $a:[s_0,s_1]\rightarrow B$ and point $x_0\in E$ such that $\pi(x_0)=a(s_0)$, there exists a lift $\tilde a:[s_0,s_1]\rightarrow E$ such that $\tilde a(s_0)=x_0$.
The first part of the proof I was shown goes as follows:
Proof. Initially, suppose $a([s_0,s_1])$ is contained in a trivializing neighbourhood $U$ (that is, an open set $U$ such that there exists a homeomorphism $\varphi_U:U\times F\rightarrow\pi^{-1}(U)$ satisfying $\pi\circ\varphi_U={proj}_1$). Then, fixing any point $y_0\in F$, defining $\tilde a$ by $\tilde a(s)=\varphi_U(a(s),y_0)$ gives the desired lift.
Now, I can't see why $\tilde a(s_0)=x_0$ as desired. I was thinking of the bundle $B=S^1$, $E=\mathbb R$, $F=\mathbb Z$ and can't see how the choice of $y_0$ would be arbitrary. You'd have to choose $y_0\in\mathbb Z$ corresponding to the "floor" in which $x_0$ is when looking at $\mathbb R$ as a coil, right?
I hope my question was clear enough and thank you in advance for your attention and any help provided.