# Open map which “almost fixes” the boundary of an open ball

We have a continuous function $f:\bar{B}\to\mathbb{R}^n$, where $\bar{B}=\{x\in\mathbb{R}^n:\|x\|\le 1\}$, such that if $\|x\|=1$ then $\|f(x)-x\|<\epsilon$, for a fixed $\epsilon\in(0,1)$. We have to prove that $B(0,1-\epsilon)\subseteq f(\bar{B})$.

This appears as a lemma in Rudin's Real and Complex Analysis. The author claims that it is possible to prove it without Brouwer's fixed point theorem, under the additional hypothesis that $f$ is open.

So far I've only observed that the problem reduces to showing that $f(\bar{B})\cap B(0,1-\epsilon)$ is not empty. Any ideas?

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Can you give us a more precise reference? Which lemma? –  Siminore Aug 21 '12 at 15:19
Lemma 7.23, page 151 (third edition). It is used in the proof of the change-of-variables theorem. –  Mizar Aug 21 '12 at 15:38
I see, I was reading the first edition. –  Siminore Aug 21 '12 at 15:39
If $f(\overline{B}) \subset \mathbb{R}^n \setminus B(0,1-\varepsilon)$, then you are essentially mapping a ball into a domain with a hole. I guess this is impossible, by comparing homologies... –  Siminore Aug 22 '12 at 8:20
Thank you, but I was looking for a solution as elementary as possible, just because Rudin seems to suggest its existence.. –  Mizar Aug 22 '12 at 19:31