Let $\Bbb D^n$ be n-dimensional ball and $S^{n-1}$ the $n-1$ dimensional sphere realized as boundary of $\Bbb D^n$. Prove that following are equivalent.
There is no retraction $\Bbb D^n\to S^{n-1}$
Every Continuous map $\Bbb D^n \to \Bbb D^n$ has a fixed point.
I can do for the case of $n=2$. For this case, I don't even need the help of 2. to prove 1. just i will show there is induced homomorphism between them such that $\pi _*(S^{1}) \to \pi _*(\Bbb D^2) \to \pi_*(S^1)$ does not commute. Next direction is also easy for the case of n=2
But I can't do for the case of $n>2$. could u please do for me for higher dimension.
1) implies 2):
Let $f$ be a continuous mapping of $D^n$ into itself. Suppose $f$ does not have any fixed points. For every $x\in D^n$, let $g(x)$ be the unique point of intersection of the ray starting from $f(x)$ and passing through $x$ with $\partial D^n$ (its boundary as a subspace of $\mathbb{R}^n$). Verify that $g$ is a continuous function and that $g:$ $D^n\rightarrow S^{n-1}$ is a retraction. Ttherefore contradiction, thus $f$ must have a fixed point.
2) implies 1) :
Suppose there is a retraction $r:D^n\rightarrow S^{n-1}$. Let $i:S^{n-1}\rightarrow D^n$ be the natural embedding and $R:D^n\rightarrow D^n$ be the map that sends a vector $x$ to $-x$. Consider the map $Rir:D^n\rightarrow D^n$. By (2) this map must have a fixed point. Show that this fixed point can not be in the boundary of $D^n$ or in the interior of $D^n$. Therefore contradiction.