How do I prove this winding number is not zero? Let $\alpha:[0,1]\rightarrow S^1:t\mapsto e^{2\pi it}$ be a path.
Let $f:S^1\rightarrow S^1$ be a continuous map such that $-f(x)=f(-x)$ on $S^1$.
How do I show that the winding number $Wnd(f\circ \alpha,0)$ is not zero?
I have proven this by proving $f$ is not null-homotopic, but I'm now trying to show to prove this directly.
Here's how I tried.
Note that $-(f\circ \alpha)(t)=(f\circ \alpha)(t+1/2)$.
Set $f\circ \alpha = r e^{i\theta}$
If you draw a circle, then it's intuitively clear that $\theta(0)\neq \theta(1)$.
I think this is due to connectedness of $[0,1]$, but I don't know how to prove this rigorously.
Please help :)

Divide [0,1] into two parts [0,1/2] and [1/2,1] and note that $r=1$
Now, let's decompose $\theta$ into two parts so that $\theta=\theta_1 \ast \theta_2$ and set $g_1=e^{i\theta_1} , g_2=e^{i\theta_2}$
Note that $Wnd(f\circ \alpha,0)=Wnd(g_1,0) + Wnd(g_2,0)$ and $g_2=-g_1$
Hence, $e^{\theta_1 - \theta_2} = -1$. Since the range of this is discrete, $\theta_1 = \theta_2 + (2n+1)\pi$
Of course $Wnd(g_1,0)$ is not 0 since $\theta_1(0)\neq \theta_1(1)$
Hence $Wnd(f\circ \alpha,0)= 2 Wnd(g_1,0)$ and thus not 0.
