There is a proof of the Brouwer Fixed Point Theorem via the Jordan Curve Theorem ?

The Brouwer Fixed Point Theorem. Let $B=\{x\in \mathbb R^2 :∥x∥≤1\}$ be the closed unit ball in $\mathbb R^2$ . Any continuous function $f:B\rightarrow B$ has a fixed point.

The Jordan curve theorem. The image of a continuous injective mapping $J:S^1\rightarrow \mathbb R^2$ divides the plane into exactly two components, one of which is unbounded and the other bounded. Moreover, both of these components have the image of the mapping $J$ as their boundary.

The Jordan Curve Theorem via the Brouwer Fixed Point Theorem

$\partial B=S^{1}$

Any hints would be appreciated.


You might enjoy David Gale's paper The Game of Hex and the Brouwer Fixed-Point Theorem, The Monthly, Dec 1979, pp. 818-827. He gives direct proofs in both directions between the Brouwer Theorem and the:

Hex Theorem: The game of Hex cannot end in a tie.

As Gale explains, the Hex Theorem is completely elementary (Gale gives an algorithm that takes in a completed Hex board and finds a win for one of the two players). But the Hex Theorem can be thought of as a sort of "finite" or "discrete" version of the Jordan theorem (the weak form, needed for Brouwer, only says that at least one player wins; the strong form, that both cannot win, is closer to Jordan), and the former does follow from the latter (Beck, Bleicher, and Crow, Excursion into Mathematics, 1969, pp. 327-339). For further discussion relationing the Hex and Jordan Theorems, see the three or so paragraphs after "Hex Theorem" in op. cit., p. 820.

  • $\begingroup$ As a side-note: The Jan 2014 AMM has an article with the following abstract: "A short, inductive proof is presented of the fact that a Hex board cannot be colored such that winning conditions are satisfied for both players." Appropriately titled, the article has as its citation: Huneke, S. C. An Inductive Proof of Hex Uniqueness. *The American Mathematical Monthly, 121*(1), 78-80. $\endgroup$ – Benjamin Dickman Dec 17 '13 at 8:14

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