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Suppose $f:\mathbb{B}^n \to \mathbb{R}^n$ is continuous (here $\mathbb{B}^n$ refers to the $n$-dimensional unit ball). Suppose also that its restriction $g := f|_{\mathbb{S}^{n-1}}$ does not have $0$ in its image, and $g:\mathbb{S}^{n-1} \to \mathbb{R}^n \setminus\{0\}$ is homotopic to the identity map.

Is it necessarily true that $f(x) = 0$ for some $x\in \mathbb{B}^n$?

I have proved it in the affirmative below. I'm hoping there may be a proof that uses only elementary results (i.e. avoids Borsuk-Ulam). I used this result in the corrected proof of this question about Chebyshev sets.

Eli Johnson
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2 Answers2

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If, by contradiction, $f$ has no zero, then $f|_{\mathbb S^{n-1}}$ is homotopic to a constant in $\mathbb R^n\setminus\{0\}$ by $$ H(x,t)=f(tx), \quad x\in S^{n-1}, \ t\in[0,1]. $$ Therefore, by transitivity, the identity map on $\mathbb S^{n-1}$ is homotopic to a constant. However the inclusion of $\mathbb S^{n-1}$ in $\mathbb R^n\setminus\{0\}$ is easily seen to be a homotopy equivalence so we deduce that $\mathbb S^{n-1}$ is contractible which it obviously isn't.

Ruy
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Write $c\mathbf{v}\in\mathbb{B}^n$, with $c \in [0,1]$ and $\mathbf{v}\in \mathbb{S}^{n-1}$. If $H:\mathbb{S}^{n-1} \times I \to \mathbb{R}^n \setminus \{0\}$ is the assumed homotopy taking $g\to\mathrm{Id}_{\mathbb{S}^{n-1}}$, then let $h: \mathbb{B}^n \to \mathbb{R}^n$ be given by

$$ h(c \mathbf{v}) := \left\{\begin{array}{ll}f(2c\mathbf{v}) &: c\leq 1/2 \\H(\mathbf{v},2c-1) &: c>1/2\end{array}\right.$$

Observe that $h$ restricts to the identity on $\mathbb{S}^{n-1}$, which is in particular an odd function, so $h$ has a zero by the Borsuk-Ulam theorem. By construction, points in the image of $h$ are either in the image of $f$ or $H$. And $H$ was presumed to avoid zero, so $f$ must have a zero.

Eli Johnson
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