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Brouwer fixed-point theorem: Difference between revisions

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→Closedness: Fixed some language issues, and removed a paragraph that was wrong and misleading. (The problem with the example is indeed the closedness of the ___domain - the theorem never uses that the *range* of the function is equal to the ___domain, only that the *codomain* is. Indeed, if it did it would be a much weaker theorem. It is true that the *range* of f is (0,1), but we can easily extend the *codomain* to (-1,1).)
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===A proof by Hirsch===
There is also a quick proof, by [[Morris Hirsch]], based on the impossibility of a differentiable retraction. TheLet [[indirect''f'' proof]]denote startsa bycontinuous notingmap from the unit ball D<sup>n</sup> in n-dimensional Euclidean space to itself and assume that ''f'' fixes no point. By continuity and the fact that D<sup>n</sup> is compact, it follows that for some ε > 0, ∥x - ''f''(x)∥ > ε for all x in D<sup>n</sup>. Then the map ''f'' can be approximated by a smooth map retaining the property of not fixing a point; this can be done by using the [[Weierstrass approximation theorem]] or by [[convolution|convolving]] with smooth [[bump function]]s. One then defines a retraction as above which must now be differentiable. Such a retraction must have a non-singular value p ∈ ∂D<sup>n</sup>, by [[Sard's theorem]], which is also non-singular for the restriction to the boundary (which is just the identity). Thus the inverse image ''f''<sup>-1</sup>(p) would be a 1-manifold with boundary. TheIts boundary would have to contain at least two end pointsendpoints, both of which would have to lie on the boundary of the original ball—which is impossible in a retraction.<ref>{{harvnb|Hirsch|1988}}</ref>
 
R. Bruce Kellogg, Tien-Yien Li, and [[James A. Yorke]] turned Hirsch's proof into a [[Computability|computable]] proof by observing that the retract is in fact defined everywhere except at the fixed points.{{sfn|Kellogg|Li|Yorke|1976}} For almost any point, ''q'', on the boundary, (assuming it is not a fixed point) the one manifold with boundary mentioned above does exist and the only possibility is that it leads from ''q'' to a fixed point. It is an easy numerical task to follow such a path from ''q'' to the fixed point so the method is essentially computable.{{sfn|Chow|Mallet-Paret|Yorke|1978}} gave a conceptually similar path-following version of the homotopy proof which extends to a wide variety of related problems.
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