# 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.

## 1 Answer

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!