Show that maps of degree one from the circle to itself are homotopic to the identity.

Question

Let $f:S^1\rightarrow S^1$ be a map of degree 1. Show that this map is homotopic to the identity. The book gives a hint to consider the exponential map $p:R\rightarrow S^1$ as a covering of the circle and then lift $f\circ p$ to a map $g: R\rightarrow R$. I am not sure how to continue.

Answer 1

Given the lift $g: \mathbb R \to \mathbb R$, also consider the map $id:\mathbb R \to \mathbb R$. Note that since $\frac{g(2 \pi)-g(0)}{2 \pi}=1$, where $0 \in p^{-1}(a)$ (at least up to homotopy) for the pointed space $(S^1,a)$, we also have that the maps $g,id$ can be connected by a homotopy $F(t)=g(t)+(1-t)id$, but the homotopy always lies in the same section of $\mathbb R$ in between $[2 \pi,4 \pi]$ this homotopy can be projected back down to $S^1$ in a continuous fashion via $\mathrm{exp}$.