Why is this a proof for Brouwer's fixed point theorem our teacher said that we can prove the Brouwer's fixed point theorem using the index $n(f\circ \gamma_r,0)$ where $\gamma_r (t)=re ^{it}$ , but I don't understand how come any help would be a lot appreciated.Here's what he's done

Let $D$ the closed unit disk. $f : D\rightarrow D$ a continuous function.$f$ has a fixed point in $D$.

So we have : $n(f\circ \gamma_0,0) = 0 $ if  $f$ has no fixed point in $D$, we can show that :
The image of $\partial D$ by $1+\frac{f}{g}$ is in the open half plane {$z=x+iy|x>0$}, where $g:D\rightarrow D$ s.t $z \mapsto -z.\;\;$ (1)
We can conclude that : $n(f\circ \gamma_1,0) = n(g \circ \gamma_1,0) = 1.\;\;$ (2)
Hence we have a contradiction.
I don't see exactly why (1) & (2) are true and what gives us the contradiction.Thanks in advance for your help.
 A: One has to be careful here as $n(f\circ \gamma_1,0)$ makes sense only if $f$ doesn't have zeroes on the unit circle, while Brower theorem holds in general for any continuous self-map of the closed disc (which may have zeroes on the boundary), but the argument is correct if presented properly.
Note that on the unit circle $|f/g|=|f| \le 1$, hence $\Re (f/g)(w) \ge |(f/g)(w)| \ge -1$, and in particular $\Re (f/g)(w)=-1$ if and only if $(f/g)(w)=-1$ or $f(w)=w$ which means that $f$ has a fixed point on the unit circle and we are done!.
So we can move to the case where $\Re (f/g)(z)>-1, |z|=1$, which gives precisely the first point above.
But now this means $f+g$ doesn't have a zero on the unit circle (and obviously $g$ doesn't either), nor does $1+f/g$ while $n((1+ f/g)\circ \gamma_1,0)=0$ since its image is contained in the right half plane hence the argument can vary by at most $\pi$.
But then $n((f+g)\circ \gamma_1,0)=n(g\circ \gamma_1,0)+n((1+ f/g)\circ \gamma_1,0)=n(g\circ \gamma_1,0)=1$. This means that $f+g$ has at least a zero inside the unit disc, so $f$ has a fixed point and we are done!
