The Brouwer’s Fixed Point Theorem  is one of the most well known and important existence principles in mathematics. Since, the theorem and its many equivalent formulations or extensions are powerful tools in showing the existence of solutions for many problems in pure and applied mathematics, many scholars have been studying its further extensions and applications. The Brouwer Theorem itself gives no information about the location of fixed points. However, effective ways have been developed to calculate or approximate the fixed points. Such techniques are important in various applications including calculation of economic equilibria. Because Brouwer Fixed Point Theorem has a significant role in mathematics, there are many generalizations and proofs of this theorem. In this paper, we will try to show several proves of Brouwer Fixed Point Theorem. First, let’s take a look at Brouwer Theorem from real world illustrations. There are several real world examples, and we will take in consideration few of them.
1. Brouwer Fixed Point Theorem on R1
Take two sheets of paper with identical images one lying directly above the other. If you fold the top paper and change its shape completely, and you put on the top of the other sheet. The Brouwer Theorem says there is at least one point on the top paper that is directly above the corresponding point on the bottom sheet. For instance, suppose we have two identical deck’s cards. Let’s keep one of the cards unchanged, and let the other rotate and stretch (fold) it. However, we do not cut or tore. Furthermore, let’s take the deformed card and put on top of the unchanged card. According to Brouwer’s Theorem, there must be a point on the deformed deck’s card that will map exactly on the same point of the unchanged deck’s card.
Another example on Brouwer’s Theorem that is similar to previous examples is if we take two identical same disks. We place one on the top of the other. Then we change the shape and the form of the disk on the top, and on the other hand, we do not make any change to the bottom disk. Moreover, we put the deformed disk on the top of the disk that is on the bottom. Brouwer Theorem tells us there is at least one point of the top disk that is mapped exactly on the same point of the bottom disk.
As we mention early, we are going to present several proves of Brouwer’s Theorem by starting from the simplest one on R.
Theorem 1. If f is continuous function and it is mapping f : R-> Ris a compact convex set in to itself there is a point x0such that f(x0) = x0.1Proof. The domain [a, b] is mapping in codomain [a, b] then f(a) ≥ a and f(b) ≤ b. Let define a continuous function g(x) in the closed interval [a, b] such that [a, b] E R where g(x) = f (x0) = x0. By applying Intermediate Value Theorem, we can write as follows, f(a) < g(x) < f(b). Then it followsg(a) < 0 and g(b) > 0 where g(a) and g(b) are with opposite signs. There exists a x0 E [a, b] such that g(x) = f(x0)− x0 we will get f(x0)−x0 = 0 Hence f(x0) = x0.
We will present two short lemmas, which plays a crucial role on implying Brouwer’s Theorem. First, we will show Lemma 2 the mapping of a function to itself does not have to be a bijection in order to have fixed point. The other Lemma contributes on proving Brouwer’s Theorem by retraction.
Lemma 2. If X has fixed point property that is any continuous function, f : X -> X, then any space homomorphic to X has fixed point property.
Proof. Let Y be homomorphic to X, and function h:X -> Y homomorphic, g :Y -> Y ↑h h −1 ↓ f : X -> X defined by f =h−1g ◦h is contained inX,and X has fixed point property. There is x0E X with f (x0) = x0. That means h−1 (g (h ( x0))) = x0 ⇒ g (h (x 0)) = h (x0). So, g has a fixed point. Thus Y has fixed point property.
Lemma 3. There is no retrtraction from S1 to D2. 2
2. Combinatorial: Sperner’s Lemma
Emanuel Sperner is known for two theorems from 1928. Sperner’s Lemma states that every Sperner coloring of a triangulation of an n - dimensional simplex contains a cell with a complete set of colors. The lemma was proven by Sperner. Later, it was noticed this Lemma implies Brouwer Fixed Point Theorem without explicit use of homology.
A Sperner’s Labeling  is a labeling of a triangulation with the numbers 1, 2,and 3 such that
- The three corners of the triangle are labeled 1, 2, and 3.
- Every vertex on the line connected vertex i and vertex j is labeled i or j
The simplest proof of Sperner’s Lemma is by considering 1-dimensional space. Here we have a line segment (a, b) subdivide into smaller segments, and let’s color vertices of subdivision with two colors. There is required that a, and b receive different colors. The number of small segments that receives two colors is odd.
Corollary 4. If [a, b] is a 1 − 2 segment subdivided in to N subintervals labeled by Sperner’s labeling, the number of the 1 − 2 segment in the subdivision is odd.
Proof. We show by induction the number of the [1 − 2] segment is odd. First, we will check subintervals with the 1 − 2 segment, when the number of subdivision N = 0 and N = 1 Assume that for a finite number of subdivision is N subintervals with an odd 1−2 segment. If the subdivision has N +1 subintervals it is going to yield an odd numbers of the 1 − 2 segment.
If we subdivide any subinterval of the finite subdivision of N subintervals,there will appear three cases. Case #1: If the subinterval is labeled with 1 − 2 segment and we subdivide into two subintervals, there will be two choices. There is a choice 1 − 1 − 2 segment, and the other choice is 1 − 2 − 2 segment. Both choices will contribute zero 1−2 segment (See Figure 2). Case #2: If the subinterval is labeled with 1−1 segment and we subdivide into two subintervals, there will be two choices. There is a choice 1−1 −1 segment that will contribute zero 1−2 segment, and the other choice is 1−2−1 segment will contribute with two 1 −2 segment (See Figure 2). Case #3: If the subinterval is labeled with 2 − 2 segment and we subdivide into two subintervals, there will be two choices. There is a choice 2−2 −2 segment that will contribute zero 1 − 2 segment, and the other is 2 − 1 − 2 segment will contribute with two 1 − 2 segment (See Figure 2). In three cases the number of contributions is even. Adding an even number and an odd number results into an odd number, and it completes the proof that N + 1 subintervals will have an odd number of subintervals with 1 − 2 segment.
Let’s start with Sperner’s Lemma  in Dimension−2 by letting P be a polygon of the plane, and consider a triangulation of the polygon. Color each vertex of triangulation by one of three colors 1, 2,and 3. An edge is called 1-2 edge if its endpoints are colored 1, and 2, and triangle is said to be complete if each of its vertices are colored using different colors as is shown in Figure 2.
Lemma 5 (Sperner’s Lemma 1928). If we have a polygon whose vertices are colored by three colors and triangulation is given by for this polygon, the number of complete triangles (with three different colors) is equal to the numbers of 1-2 edges on the boundary of the polygon ( mod 2).
number of dots inside the triangle in two different ways. Firstly, each interior segment contributes whether 0 or 2 dots (depends weather it is a 1-2 edge), each boundary segment contributes 0 or 1 dots. The number of dots is equal to the number of 1-2 edges on the boundary of polygon ( mod 2). Now we count the number of dots in each triangle in the Figure 2. Complete triangles contain one dot (check this), while other triangles an even number of dots. Therefore, the number of dots is equal to the number of complete triangles ( mod 2). This completes the proof.
Note from the editor: This extract doesn’t contain any figures.
- Quote paper
- Duli Pllana (Author), 2018, Reflection on Brouwer's Fixed Point Theorem, Munich, GRIN Verlag, https://www.grin.com/document/428448