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I want to prove the following fact without using topological degree theory or related algebraic topology

Let $h:\overline{B}(0,1)\to \mathbb{R}^n$ be a continuous map such that $|h(x)-x|\leq \delta$ when $|x|=1$. Then $B(0,1-\delta)\subset h(B(0,1))$.

The result is obvious if one uses topological degree theory, since $h$ is homotopic to the indentity map when restricted to $B(0,1-\delta)$. Or with more effort, one can use Brouwer fixed point theorem to prove it. Here I want an elementary proof of this fact, which does not involve any algebraic topological concept.

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up vote 7 down vote accepted

Note that the result you mention implies in particular that any $h\in C^0(\overline{B},\overline{B})$ with $h_{|\partial B}={\operatorname {id}}_{\partial B}$is surjective. This immediately implies the non-retraction theorem, from which traditionally the Brouwer FPT easily follows. In conclusion, all these results are quite close to each other in terms of proof length. Note also that there are elementary proofs of the non-retraction theorem; check e.g. the Wiki article, that lists a number of them.

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