Prove that for $f\colon S^1 \to S^1$, $\Vert f(x)-x\Vert<1$ implies $f$ is surjective. I'm working on a topology problem that asks

Let $f:S^1\to S^1$ be continuous where $S^1$ is the unit circle. Suppose that $\Vert f(x)-x\Vert<1$ for all $x$, then $f$ is surjective.

So far, I'm able to show that if $f$ is not surjective, it has degree zero, but from there it gets more complicated. My intuition is that so long as $f$ is continuous, mapping $S_1$ to an interval would mean there is some point that will be "too far" from its image, which is the desired contradiction, but I'm not sure how to make this notion rigorous, or even if its entirely correct. What should I be thinking about in order to figure out how to solve this? Thanks!
 A: Assume it's not surjective, then it has degree zero as you say. So you can create a contradiction by showing $\deg f\neq 0.$
As $\|f(x)-x\|=\|f(x)-\mathrm{id}(x)\|< 1,$ you get a feeling that they are similar.
Hint: Try to find a homotopy $H:f\simeq \mathrm{id}.$ Then use $\deg \mathrm{id}=1.$

Geometry intuition: When $x\in S^1$ circles a round, as $f(x)$ is close to $x,$ $f(x)$ seems to circle a round synchronously. That means $\deg f=1.$ So using the degree to consider doesn't get the question complicated. On the contrary, it gets you closer to the essence of this function.
The homotopy drags $f(x)$ to $x$, emphasizing that the intuition is true, in mathematical meaning.

Actually we can relax the condition to be $\|f(x)-x\|<2,$ i.e. $f(x)\neq -x.$ This means that the chord connecting $f(x)$ and $x$ is not a diameter, it doesn't meet the origin $0.$ That gives you the idea to drag $f(x)$ to $x$ by the minor arc corresponding to the chord. Which is the homotopy
$$
H(t,x)=\frac{tx+(1-t)f(x)}{\|tx+(1-t)f(x)\|}=\frac{f(x)-t(f(x)-x)}{\|f(x)-t(f(x)-x)\|},\quad t\in [0,1]
$$
To prove that it's continuous, we only need to prove $\|tx+(1-t)f(x)\|\neq 0.$ It's obvious once you notice that the line segment connecting them is not a diameter.
If it takes zero value, then $tx=(t-1)f(x),$ so $$\|tx\|=t=1-t=\|(t-1)f(x)\|.$$
It can only happens when $t=\frac{1}{2}.$ However, at this time, $\frac{1}{2}x+\frac{1}{2}f(x)\neq 0$ as $f(x)\neq -x.$ So we always have $\|tx+(1-t)f(x)\|>0.$
