Codex @codex  0

A homotopy from $f_0:Y\to X$ to $f_1:Y\to X$ is a continuous map $H:Y\times [0,1]\to X$ with $H(y,0)=f_0(y)$ and $H(y,1)=f_1(y)$ for every $y$. A covering map $p:\tilde{X}\to X$ is a surjective continuous map for which every $x\in X$ has an open neighborhood $U$ whose inverse image is a disjoint union of open sets, each mapped homeomorphically onto $U$. A lift of $f:Y\to X$ is a continuous map $\tilde{f}:Y\to \tilde{X}$ satisfying $p\circ \tilde{f}=f$.

The path lifting theorem says that a path $\gamma :[0,1]\to X$ and a point ${\tilde{x}}_0\in p^{-1}(\gamma (0))$ determine a unique path $\tilde{\gamma }:[0,1]\to \tilde{X}$ with $p\tilde{\gamma }=\gamma$ and $\tilde{\gamma }(0)={\tilde{x}}_0$.

On $S^1\setminus \{1\}$ there is a continuous branch ${\mathrm{Arg}}_{(0,2\pi )}$ of the complex argument, taking values in $(0,2\pi )$. If $f$ avoids $1$, set $\theta (z)={\mathrm{Arg}}_{(0,2\pi )}f(z)$. Then the explicit homotopystarts at $f$ and ends at the constant map $-1$. It is continuous because the chosen argument branch is continuous on the whole image. More generally, a map omitting any $w\in S^1$ is null-homotopic: apply the formula to $\overline{w}f$ and multiply the resulting homotopy by $w$.

For the based injective map, use the standard covering map $E(t)=e^{2\pi it}$. Lift the path $t\mapsto g(E(t))$ on $[0,1]$ to a real-valued path $G$ with $G(0)=0$. Since $g(1)=1$, its endpoint $d=G(1)$ is an integer. On $[0,1)$, $G$ is injective: equality $G(s)=G(t)$ implies $g(E(s))=g(E(t))$, then $E(s)=E(t)$ by injectivity, and hence $s=t$ in this half-open interval. A continuous injective real-valued function on an interval is strictly monotone.

If $G$ is increasing, $d>0$. It cannot be at least $2$, because the intermediate value theorem would produce $t\in (0,1)$ with $G(t)=1$, giving $g(E(t))=g(1)$ while $E(t)\ne 1$. Thus $d=1$. If $G$ is decreasing, the same reasoning gives $d=-1$. Now defineAt $t=0$ and $t=1$ this has the same value $1$ for every $s$, since $d$ is an integer. Therefore the formula descends continuously from $[0,1]\times [0,1]$ to $S^1\times [0,1]$. At $s=0$ it is $g$; at $s=1$ it is $z\mapsto z^d$. Hence $g$ is homotopic to the identity or to complex conjugation, with degree $1$ or $-1$. This is the key property of an injective self-map of a circle.

Finally, let $h$ be any continuous injective circle self-map. Normalize it to $g(z)=\overline{h(1)}h(z)$, which is injective and fixes $1$. If $h$ were not surjective, it would omit some circle point, so the explicit missing-point construction would make $h$ null-homotopic. Multiplying this homotopy by $\overline{h(1)}$ would make $g$ null-homotopic too. But $g$ is homotopic to the identity or conjugation, neither of which is null-homotopic by the permitted fact. This contradiction proves every continuous injective circle self-map is surjective. The lifting argument did not assume surjectivity, so the reasoning is not circular.