Brouwer fixed-point theorem: Difference between revisions

Revision as of 05:34, 15 December 2023

Revision as of 13:41, 22 December 2023

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A variation of the preceding proof does not employ the Sard's theorem, and goes as follows. If <math>r\colon B\to \partial B</math> is a smooth retraction, one considers the smooth deformation <math>g^t(x):=t r(x)+(1-t)x,</math> and the smooth function

:<math>\varphi(t):=\int_B \det D g^t(x) \, dx.</math>

Differentiating under the sign of integral it is not difficult to check that ''{{prime|φ}}''(''t'') = 0 for all ''t'', so ''φ'' is a constant function, which is a contradiction because ''φ''(0) is the ''n''-dimensional volume of the ball, while ''φ''(1) is zero. The geometric idea is that ''φ''(''t'') is the oriented area of ''g''^''t''(''B'') (that is, the Lebesgue measure of the image of the ball via ''g''^''t'', taking into account multiplicity and orientation), and should remain constant (as it is very clear in the one-dimensional case). On the other hand, as the parameter ''t'' passes formfrom 0 to 1 the map ''g''^''t'' transforms continuously from the identity map of the ball, to the retraction ''r'', which is a contradiction since the oriented area of the identity coincides with the volume of the ball, while the oriented area of ''r'' is necessarily 0, as its image is the boundary of the ball, a set of null measure.<ref>{{harvnb|Kulpa|1989}}</ref>

===A proof using the game Hex===