If $f:S^1\to S^1$ doesn't have any fixed point then it is homotopic to the identity How to show that every continuous function $f:S^1\to S^1$ without fixed points is homotopic to the identity? (without using homology nor the concept of degree). 
 A: Here is a slightly longer argument that doesn't go through the antipodal map:
Suppose $f:S^1\rightarrow S^1$ has no fixed points. Identify $S_1$
with $\mathbb{R}/\mathbb{Z}$ and let $\pi:\mathbb{R}\rightarrow S_1$
be the quotient map. Let
$\gamma:[0,1] \rightarrow S^1$ be the loop $t \mapsto f(t+\mathbb{Z})$,
starting at $f(\bar{0})$.
Choosing $x_0 \in \pi^{-1}(f(\bar{0}))\cap [0,1)$, we have a
unique lift $\overline{\gamma}:[0,1]\rightarrow \mathbb{R}$
which starts at $x_0$.  Now we can define the function
\begin{align*}
        F:S^1 \times [0,1] &\longrightarrow S^1 \\ 
        (t+\mathbb{Z}, \lambda) &\longmapsto \pi \left( \lambda \cdot \overline{\gamma}
        (t) + (1-\lambda)\cdot t \right) 
    \end{align*}
where $t$ is chosen such that $t\in [0,1)$. This is clearly
continuous if $t\neq 0$. Remains to check continuity at $\bar{0}$.
By the definition of a lift, we know that $\overline{\gamma}(1) \in
    \{x_0+n\;|\; n\in \mathbb{Z}\}$. Claim for all $t$, we have $t < \overline{\gamma}(t) <
    t + 1 $. If not, suppose (wlog) the lower bound doesn't hold. But we
chose $f(0)=x_0>0$, so by the
intermediate value theorem we have $\overline{\gamma}(t)=t$ for
some $t$ -- we have $\pi\circ \overline{\gamma}(t) =
    \pi(t)$ i.e. $f(\bar{t}) = \bar{t}$, giving us a fixed point of
$f$ [a contradiction].
This shows $\overline{\gamma}(1) = x_0 + 1$, hence
$\lim_{t\rightarrow 1} F(t+\mathbb{Z}, \lambda) = \lim_{t\rightarrow 0}
    F(t+\mathbb{Z},\lambda)$ as required. Thus (F) is a homotopy $f\simeq \text{id}$.
A: $f : S^1 \to S^1$ be a map with no fixed points. Consider the projection of the straightline homotopy $$H(s, t) = \frac{(1-t)f(s) - ts}{\left \lVert(1-t)f(s)-ts\right \rVert}$$ between $f$ and the antipodal map $-\text{id}$, which is well-defined since $f(x) \neq x$ for all $x$. Compose this with the homotopy $$H(s, t) = e^{i\pi (1-t)} s$$ which rotates $-\text{id}$ to the identity map $\text{id}$. Thus, by transitivity, $f \sim \text{id}$.
