Injective self-map of a circle

Every continuous injective map $g:S^1\to S^1$ is a homeomorphism. After normalizing $g(1)=1$ by a rotation, it is homotopic either to the identity or to complex conjugation, and has degree $1$ or $-1$.

To see this without assuming surjectivity, lift $t\mapsto g(e^{2\pi it})$ on $[0,1]$ through the covering map $u\mapsto e^{2\pi iu}$ to a path $G$ with $G(0)=0$. Injectivity makes $G$ injective on $[0,1)$, hence strictly monotone. Its endpoint $d=G(1)$ is an integer. Strict monotonicity excludes $d=0$; if $|d|>1$, the intermediate value theorem gives an interior integer value, contradicting injectivity of $g$. Thus $d=\pm 1$, and $\exp (2\pi i((1-s)G(t)+sdt))$ descends to a homotopy from $g$ to $z\mapsto z^d$. If $g$ omitted a point, a continuous branch of the complex argument on the punctured circle would make it null-homotopic, contradicting this nonzero degree. Hence it is surjective. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism.