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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.

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

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  • $\begingroup$ That was dumb of me. The last conclusion was to rule out range. Thanks. $\endgroup$ – user45765 Dec 21 '18 at 23:30

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