Editing Brouwer fixed-point theorem (section)

===One-dimensional case===
[[File:Théorème-de-Brouwer-dim-1.svg|200px|right]]
In one dimension, the result is intuitive and easy to prove.  The continuous function ''f'' is defined on a closed interval [''a'',&nbsp;''b''] and takes values in the same interval.  Saying that this function has a fixed point amounts to saying that its graph (dark green in the figure on the right) intersects that of the function defined on the same interval [''a'',&nbsp;''b''] which maps ''x'' to ''x'' (light green).

Intuitively, any continuous line from the left edge of the square to the right edge must necessarily intersect the green diagonal.  To prove this, consider the function ''g'' which maps ''x'' to ''f''(''x'')&nbsp;−&nbsp;''x''. It is ≥&nbsp;0 on ''a'' and ≤&nbsp;0 on&nbsp;''b''.  By the [[intermediate value theorem]], ''g'' has a [[Root of a function|zero]] in [''a'',&nbsp;''b'']; this zero is a fixed point.

Brouwer is said to have expressed this as follows: "Instead of examining a surface, we will prove the theorem about a piece of string. Let us begin with the string in an unfolded state, then refold it. Let us flatten the refolded string. Again a point of the string has not changed its position with respect to its original position on the unfolded string."<ref name=Arte />