I'm stuck on a past paper question:
Note: $a=1$
I thought that perhaps as $g$ injective and a map from $S^1$ we could perhaps view it as a path from $[0,1]$ and thus it would have a lift $\tilde{g}$. Then considering the induced maps on the fundamental groups, we get that the induced map of g must be trivial. I'm not sure if that helps or what to do next! Any help much appreciated :)
(Also, I think there must be a flaw in this idea as there is a statement in my notes saying that the identity map on $S^1$ doesn't lift with the standard covering of the real line. Not sure what the error in my thinking is. Either way I am stuck!)
The "standard result" from real analysis that they probably have in mind is that an injective (continuous) function $f:[0,1] \to \Bbb R$ must either be strictly increasing or strictly decreasing.
One approach is as follows: let $p:\Bbb R \to S^1$ dentoe the covering map $p(x) = e^{2 \pi i x}$. Let $h:[0,1] \to S^1$ be the map $h = g \circ p|_{[0,1]}$. Because $g$ is injective, $h$ is a path in $S^1$ satisfying $h(0) = h(1)$, and $h$ must be injective over $[0,1)$.
Consider the lift $\tilde h : [0,1] \to \Bbb R$ with $\tilde h(0) = 0$. Since $\tilde h$ is injective over $[0,1)$, $\tilde h$ is either strictly increasing or decreasing. Since $h = p\circ \tilde h$ is injective over $[0,1)$, it must be that $\tilde h([0,1)) \subset [-1,1]$. Since $p(\tilde h(0)) = p(\tilde h(1))$ and $\tilde h$ is either increasing or decreasing, we cannot have $\tilde h(1) = 0$ and we must have $\tilde h(1) = 1$ or $\tilde h(1) = -1$.
Since $g \circ p|_{[0,1]}$ is surjective, $g$ is necessarily surjective. Since $g$ is a continuous bijection from a compact space to a Hausdorff space, it is a homeomorphism. Thus, the induced map $g_*:\pi_1(S^1) \to \pi_1(S^1)$ is a group isomorphism. So, either $g_*(1) = 1$, which implies that $g$ is homotopic to the identity, or $g_*(1) = -1$, which means that $g$ is homotopic to the conjugation map.
Alternatively, we can complete this proof without group theory.
Suppose that $\tilde h(1) = 1$. In order to show that $g$ is homotopic to the identity map, it suffices to find a path homotopy between $\tilde h:[0,1] \to [0,1]$ and the identity map over $[0,1]$. Define $F:[0,1]\times [0,1] \to [0,1]$ as follows: $$ F(s,t) = (1-t) \tilde h(s) + st. $$ Verify that we have $f_0(s) = F(s,0) = \tilde h(s)$ and $f_1(s) = F(s,1) = s$.
Suppose that $\tilde h(1) = -1$. In order to show that $g$ is homotopic to the conjugation map, it suffices to find a path homotopy between $\tilde h$ and the map $s \mapsto -s$ over $[0,1]$. With our work from the $\tilde h(1) = 1$ case, we have a homotopy $F$ between $-\tilde h$ and the identity map, which means that $-F$ is a homotopy between $\tilde h$ and $s \mapsto -s$, which is what we need.