= Solution
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> $p:\widetilde X\to X$ 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> $\widetilde f:Y\to\widetilde X$ satisfying $p\circ\widetilde f=f$.
The <path lifting theorem> says that a <path> $\gamma:[0,1]\to X$ and a point $\widetilde x_0\in p^{-1}(\gamma(0))$ determine a unique <path> $\widetilde\gamma:[0,1]\to\widetilde X$ with $p\widetilde\gamma=\gamma$ and $\widetilde\gamma(0)=\widetilde x_0$.
On $S^1\setminus\{1\}$ there is a continuous branch $\operatorname{Arg}_{(0,2\pi)}$ of the <complex argument>, taking values in $(0,2\pi)$. If $f$ avoids $1$, set $\theta(z)=\operatorname{Arg}_{(0,2\pi)}f(z)$. Then the explicit <homotopy>
$$
\boxed{H(z,s)=\exp\big(i((1-s)\theta(z)+s\pi)\big)}
$$
starts 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 $E(t)\neq1$. Thus $d=1$. If $G$ is decreasing, the same reasoning gives $d=-1$. Now define
$$
\boxed{K(E(t),s)=\exp\big(2\pi i((1-s)G(t)+sdt)\big).}
$$
At $t=0$ 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 \b[$g$ 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 \b[every continuous injective <circle> self-map is surjective]. The lifting argument did not assume surjectivity, so the reasoning is not circular.
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