# deducing standard Brouwer fixed point theorem

This is a corollary in Dold, Algebraic Topology.(Cor 2.4,2.5,2.6) The goal is to see whether standard brouwer fixed point theorem can be deduced from 2.5 or 2.6.

Cor 2.4 $$S^{n-1}$$ is not a retraction of $$D^n$$ where $$D^n$$ is the closed $$n-$$disk.

Cor 2.5 If $$f:D^n\to R^n$$ is continuous, then either $$f(y)=0$$ for some $$y$$ or $$f(z)=\lambda z$$ for some $$z\in S^{n-1}$$ and some $$\lambda>0$$.

Cor 2.6("Brouwer fixed point Thm": Consider $$g(x)-x$$ as a function in Cor 2.5) If $$g:D^n\to R^n$$ is continuous, then either $$g(y)=y$$ for some $$y\in D^n$$ or $$g(z)=(1+\lambda)z$$ for some $$z\in S^{n-1}$$.

Standard Brouwer fixed point Thm: Continuous function $$f:D^n\to D^n$$ must have at least 1 fixed point.

$$\textbf{Q:}$$ I do not see how to deduce Standard Brouwer fixed point Thm from Cor 2.6 or 2.5 without going through contradiction via constructing retraction.(In other words, I do not know how to eliminate $$g(z)=(1+\lambda) z$$ situation in Cor 2.6 in general without going through standard argument.) I suspect that I can suppose no such fixed point. Then I can pick a sequence of points $$z_i$$ along with a sequence of functions defined as the following.

Start with $$z_1$$. Suppose $$f_1=f$$ does not have fixed point. From Cor 2.6, I can pick out $$z_1$$ s.t $$f(z_1)=\lambda_1 z_1$$ with $$\lambda_1>1$$. Then consider $$f_2=\frac{f}{\lambda_1}$$. If $$f_2$$ has no fixed point, then pick out $$z_2$$. Iterate this procedure. I can hope the sequence converging to $$0$$ as each time $$f_i$$ shrinks. ($$\textbf{How do I see this sequence of function do converge to 0?}$$ Note that $$\lambda_i$$ are varying and I do not have growth estimation of $$\lambda_i$$.) Suppose this holds. I have $$0$$ as limiting point. Then I want to say $$0$$ is my fixed point to get desired contradiction.

Note that $$\{(1+\lambda)z \mid \lambda > 0 \text{ and } z \in \partial D^n\} \cap D^n = \varnothing$$?