Editing Brouwer fixed-point theorem (section)

===Convexity===
Convexity is not strictly necessary for Brouwer's fixed-point theorem. Because the properties involved (continuity, being a fixed point) are invariant under [[homeomorphism]]s, Brouwer's fixed-point theorem is equivalent to forms in which the domain is required to be a closed unit ball <math>D^n</math>. For the same reason it holds for every set that is homeomorphic to a closed ball (and therefore also [[closed set|closed]], bounded, [[connected space|connected]], [[simply connected|without holes]], etc.).

The following example shows that Brouwer's fixed-point theorem does not work for domains with holes. Consider the  function <math>f(x)=-x</math>, which is a continuous function from the unit circle to itself. Since ''-x≠x'' holds for any point of the unit circle, ''f'' has no fixed point. The analogous example works for the ''n''-dimensional sphere (or any symmetric domain that does not contain the origin). The unit circle is closed and bounded, but it has a hole (and so it is not convex) . The function ''f'' {{em|does}} have a fixed point for the unit disc, since it takes the origin to itself.

A formal generalization of Brouwer's fixed-point theorem for "hole-free" domains can be derived from the [[Lefschetz fixed-point theorem]].<ref>{{cite web | url=https://math.stackexchange.com/q/323841 | title=Why is convexity a requirement for Brouwer fixed points? | publisher=Math StackExchange | access-date=22 May 2015 | author=Belk, Jim}}</ref>