So Im reading Munkres Topology, and I've already gone through the Fundamental Group chapter a couple of times, and while I think I kind of grasp the idea that's behind of it all, I don't feel I have deeply understood the matter.
So, if we consider the usual covering $p:\Bbb{R} \to S^1$ and we chose $b_0 = p(0)$ we have the standard isomorphism $\phi:\Pi_{1}(S^{1},b_0) \to \Bbb{Z}$ given by $\phi([f])=\tilde{f}(1)$ where $\tilde{f}$ is the unique lifting of the loop $f$ that starts in $e_0=0$.
However, when I continue reading, there is a step in each of two of the following proofs (The equivalences for continuous maps $f:S^1 \to X$ and the fundamental theorem of algebra) which I don't understand.
I will write each of the statements:
1)Let $p:\Bbb{R} \to S^1$ be the standard covering map, and let $p_0:I \to S^1$ be its restriction to the unit interval. Then $[p_0]$ generates $\Pi_1(S^1,b_0)$ because $p_0$ is a loop in $S^1$ whose lift to $\Bbb{R}$ begins at $0$ and ends at $1$
2)Consider $f(p_0(s))=(e^{2\pi is})^n=(\cos(2\pi ns),\sin(2\pi ns))$. This loop lifts to the path $s \to ns$ in the covering space $\Bbb{R}$. Therefore the loop $f \circ p_0$ corresponds to the integer $n$ under the standard isomorphism.
So what Im not able to understand is, in either case, how to see that the mentioned liftings are effectively the liftings, and how, this implies that each of the loops in question, correspond to each of the integers $1$ and $n$ respectively.
In the first case, the path $\overline f:s \mapsto s$ lifts $p_0$ because $p(\overline f(s)) = p(s) = p_0(s)$ for all $s$ (because $p_0$ is just the restriction of $p$ to the unit interval). This implies that $[p_0]$ corresponds to $1 = \overline f(1)$.
In the second case, the path $\overline f:s \mapsto ns$ lifts $f \circ p_0$ because $$p(\overline f(s)) = p(ns) = e^{2\pi i n s} = (f(p_0(s)).$$ This implies that $[f \circ p_0]$ corresponds to $n = \overline f(1)$.