Let $g_k:S^1 \rightarrow S^1$ be defined by $z \mapsto z^k$, sending a complex number $z$ of norm 1 to its $k$-th power, where $k$ is an integer.
Question: For which $k$ is the map $g_k$ homotopic to a homeomorphism?
My immediate thought is that the only homeomorphism will be where $k=2n\pi$, since any other $k$ will make it such that at least a point is hit more than once, so it won't be surjective. I do not, however, know how to show if this is true for homotopy to homeomorphism. Any help is appreciated. Thank you.
Here is an outline of an approach; some details I leave to you.
$1)$ Show that the degree of the map $g_{k}$ is $k$.
$2)$ Show that the degree of a homeomorphism is $\pm 1$ (there are many approaches here, but one you might use is that degree is multiplicative across composition, and the identity map has degree $1$).
$3)$ Recall that the degrees of two homotopic maps are equal; conclude that we must have $k = \pm 1$.
$4)$ Note that if $k = 1$ or $k = -1$, then $g_{k}$ is actually a homeomorphism; if $k = 1$, this is immediate. If $k = -1$, you can observe that $g_{k}$ is an involution.