A function on the unit circle that is not homotopic to the identity must be a reflection at some point I have a problem in the theory of homotopy which I have just started learning. The question states:
If we have a function
$$
f:\mathbb{S}^1\rightarrow \mathbb{S}^1 \;\;\; \text{s.t} \;\;\;\text{$f$ is not homotopic to the identity map}
$$
then at some point $x \in \mathbb{S}^1$, $f(x)=-x$.
To try and prove this I tried to use the contrapositive and tried to show that if $f(x)\neq -x \;\;\forall x \in \mathbb{S}^1 $ then $f$ must be homotopic to the identity. However I constructed the homotopy
$$
H(x,t) = x+tf(x)-tx
$$
which seems to be continuos and satisfy the necessary conditions, but I cannot see where to use the assumption, or how this homotopy breaks down in the case $f(x)=-x$.
Any hints or answers appreciated, thanks!
 A: Okay, so using the hint provided by @Zarrax, I think I have a solution. The problem with my homotopy was that its image was not actually a subset of $\mathbb{S}^1$. In other words, for H to be a homotopy, its norm should always be 1 so that for every $(x,t)$ we are still in the unit circle. I then altered by homotopy by normalising it:
$$
H(x,t) = \frac{x+tf(x)-tx}{||x+tf(x)-tx||}
$$
Now the vector has norm 1 at all times but we still have:
$$
H(x,0) = \frac{x}{||x||} = x\;\;\;\;\;H(x,1) = \frac{f(x)}{||f(x)||}=f(x)
$$
However to show that this is a homotopy, we need to show it in continuous and thus that the denominator is never 0:
Suppose that it is, then $x-tf(x)-tx=0$. This can be rearranged to give
$$
\frac{t-1}{t} = \frac{f(x)}{x}
$$
By taking norms we can see,
$$
\frac{|t-1|}{|t|}=\frac{|f(x)|}{|x|}=1
$$
We then have to consider two cases:

*

*$\frac{t-1}{t}=1$ which is of course impossible


*$\frac{t-1}{t} =-1$ which then implies $\frac{f(x)}{x}$=-1 and thus $f(x)=-x$ which we assumed to be false.
As a result, we have found a homotopy and proved the contrapositive.
