I am in the process of trying to use the general lifting lemma (Lemma 79.1) Munkres to show that every continuous map $f$ from $S^n$ to $S^1$ is null-homotopic for $n>1$. Now I know this question has been asked before on this site but my question is related to a small detail in the proof of it.
Because the fundamental group of $\pi_1(S^n)= 0$ I know that my $f$ can be lifted to a map $\tilde{f}$ from $S^n$ to $\Bbb{R}$ that is a covering space for $S^1$. This map is defined as follows: Suppose the basepoint of $S^n$ is $y_0$, the basepoint of $S^1$ is $x_0$ and the basepoint of $\Bbb{R}$ is $z_0$. For any $y_1 \in S^n$, choose a path $\alpha$ from $y_0 $ to $y_1$. Then $f \circ \alpha$ is a path in $S^1$ from $x_0$ to $f(y_1)$. I now lift this to a path $\gamma \in \Bbb{R}$ with startpoint $z_0$ and endpoint
$p^{-1}(f(y_1)).$
I now say that $\tilde{f}(y_1) = p^{-1}(f(y_1))$. $p$ is the usual covering map from $\Bbb{R}$ to $S^1$ given by the exponential function: $p(x) = e^{ix}$.
I want to define a null homotopy between $\tilde{f}$ and the constant map at $z_0$. I want it to be a straight line homotopy and I try with $\begin{eqnarray*} F :&S^n \times I& \longrightarrow \Bbb{R} \\ &(y_1,t)& \longmapsto z_0(1-t) + tp^{-1}(f(y_1)). \end{eqnarray*}$ The problem now is that when I take the inverse image under $p$ I don't have a well defined map? How do I get around this? Can I just say we pick a point in the inverse image and it will do the job?