We learned in our class about the map $deg: \pi_1(S^1,1) \to \mathbb{Z}$. Which works as follows: take $\alpha$ lookat $1$ in $S^1$ take the unique lifting beginning at $0$ with respect to the standart covering map $\mathbb{R} \to S^1$ $x\mapsto e^{2\pi ix}$. then the Degree is the endpoint of the path.
Now we have the following lemma which says that $f:S^1\to S^1$ $z\to z^n $ has degree $n$. Now Im confused what is meant by this. $f$ is no loop at all as it is not defined on $I$. So how can we lift that? In the proof than he dont proof that $f$ has degree $n$ but that $g:I \to S^1$ $t\mapsto e^{2\pi nit}$ has degree $n$ which is clear but this are just different maps or what is meant here?
Can someone tell me how to understand this?
For any topological space $X$ and any continuous function $f : S^1 \to X$ you get a loop $g : [0,1] \to X$ based at the point $f(1)$ by using the formula $g(t) = f(e^{2\pi i t})$.