Editing Brouwer fixed-point theorem (section)

==Importance of the pre-conditions==
The theorem holds only for functions that are ''endomorphisms'' (functions that have the same set as the domain and codomain) and for nonempty sets that are ''compact'' (thus, in particular, bounded and closed) and ''convex'' (or [[Homeomorphism|homeomorphic]] to convex). The following examples show why the pre-conditions are important.

===The function ''f'' as an endomorphism===
Consider the function
:<math>f(x) = x+1</math>
with domain [-1,1]. The range of the function is [0,2]. Thus, f is not an endomorphism.

===Boundedness===
Consider the function
:<math>f(x) = x+1,</math>
which is a continuous function from <math>\mathbb{R}</math> to itself. As it shifts every point to the right, it cannot have a fixed point. The space <math>\mathbb{R}</math> is convex and closed, but not bounded.

===Closedness===
Consider the function
:<math>f(x) = \frac{x+1}{2},</math>
which is a continuous function from the open interval <math>(-1,1)</math> to itself. Since the point <math>x=1</math> is not part of the interval, there is no point in the domain such that <math>f(x) = x</math>. The set <math>(-1,1)</math> is convex and bounded, but not closed. On the other hand, the function <math>f</math>  does have a fixed point in the ''closed'' interval <math>[-1,1]</math>, namely <math>x=1</math>. The closed interval <math>[-1,1]</math> is compact, the open interval <math>(-1,1)</math> is not.

===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>

===Notes===
The continuous function in this theorem is not required to be [[bijective]] or [[surjective]].