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This is an exercise from Munkres' topology textbook in chapter $9$. It is said that if we assume , we have known that there is no retraction from $B^{n+1}$ to $S^n$, then show any nowhere vanishing vector field $ F $ on $ B^{n+1}$ has a boundary point $b\in S^n $ at which point the vector is pointing radially inward.
I think we should construct a retraction to arrive at a contradiction. But I don't get the right retraction after trials. Could any one help me?
Let ${\bf F}:B^{n+1}\to \mathbb{R}^{n+1}-\{{\bf 0}\}$ be a nowhere vanishing vector field on $B^{n+1}$, and suppose that ${\bf F}(x)$ does not point radially inward for any $x$ in the boundary $S^n$. Because the restriction ${\bf F}|_{S^n}:S^n\to \mathbb{R}^{n+1} - \{{\bf 0}\}$ extends to the map ${\bf F}:B^{n+1}\to \mathbb{R}^{n+1} - \{{\bf 0}\}$, it follows that ${\bf F}|_{S^n}$ is null-homotopic.
One can show that ${\bf F}|_{S^n}$ is homotopic to the inclusion map $j:S^n\to \mathbb{R}^{n+1} - \{{\bf 0} \}$ as follows. Define $H:S^n\times [0,1]\to \Bbb{R}^{n+1} - \{{\bf 0}\}$ by $$H(x,t) = tx + (1-t){\bf F}(x).$$ As long as $H(x,t)\neq {\bf 0}$, then it defines the necessary homotopy. If $H(x,t) = tx + (1-t){\bf F}(x) = 0$, then $${\bf F}(x) = \frac{t}{t-1}\; x.$$ Since $t\in [0,1]$, the fraction $\frac{t}{t-1}$ is negative, and hence ${\bf F}(x) = \frac{t}{t-1}\; x$ is a negative scalar times $x$, i.e. ${\bf F}$ points radially inward at the point $x\in S^n$. We assumed that cannot happen, and so $H(x,t)\neq {\bf 0}$ for all $x\in S^n$ and $t\in [0,1]$.
Therefore the inclusion $j:S^n\to \mathbb{R}^{n+1} - \{{\bf 0} \}$ is null-homotopic. Since $j$ is null-homotopic, it can be extended to a map $J:B^{n+1}\to \Bbb{R}^{n+1}-\{{\bf 0}\}$. There is a retract $r:\Bbb{R}^{n+1}- \{{\bf 0}\}\to S^n$ defined by $r(x)=\frac{x}{||x||}$, and $r\circ J:B^{n+1}\to S^n$ is a retract from $B^{n+1}$ to $S^n$. This contradicts our assumption that there are no retracts from $B^{n+1}$ to $S^n$, and so ${\bf F}$ must point radially inward for some $x\in S^n$.
