$f:\overline{\mathbb{B^n}}\to\mathbb{R^n}$ and $\forall x\in S^{n-1}$ ($f(x)=0$ or $\angle x0f(x)$ is acute), then $0\in f(\overline{B^n})$

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I struggle with following problem. Let $f:\overline{\mathbb{B^n}}\to\mathbb{R^n}$ be continuous map such that $\forall x\in S^{n-1}$, either $f(x)=0$ or $\angle x0f(x)$ is acute. Show that if mentioned conditions are fulfilled, then $0\in f(\overline{B^n})$.

If there is $x\in S^{n-1}$ such that $f(x)=0$ then we're done. So we can assume that $f|_{S^{n-1}}:S^{n-1}\to\mathbb{R^n_*}$.

I know that if $0\notin f({S^{n-1}})$ and $f|_{S^{n-1}}$ is not null homotopic in $\mathbb{R^n_*}$, then $0\in f(\overline{B^n})$.

However, I do not know how to show that this function is not null homotopic. One way is to show that $f|_{S^{n-1}}$ cannot be extended to continuous function on $\overline{B^n}$. I have no idea how to do it.