The Topological Degree and Applications

Tyson Liddell Supervisor: Dr J´erˆomeDroniou Monash University

February 21, 2015

1 1 Abstract

In many areas of mathematics and the applied sciences, numerous problems boil down to solving equations that are often nonlinear. The Brouwer degree is a power tool used in analysis and topology that, under certain conditions, can guarantee solutions to nonlinear equations exist. This report will present a construction of the Brouwer degree and list a few of its properties. Some applications of the Brouwer degree are then presented; in particular, a new result that pertains to the orientation preserving structure of derivatives of continuous injective maps.

2 Introduction 2.1 Topological Degree n n Consider some open, bounded subset Ω ⊂ R and any continuous map f : Ω → R . In many problems in this general setting we are often posed with a question that reduces to knowing something n about the equation f(x) = y for a given y ∈ R . Specifically, often it would be very useful to know immediately whether or not such an equation has any solutions x ∈ Ω. For exactly this purpose we introduce a tool called the topological degree which, for the sake of brevity, from this point forward we may refer to as the degree. As we shall see, the degree can provide us with knowledge of existence of solutions to the previous equation. Of course, it may not be immediately obvious as to why this is useful. Consider the following problems. Given a square matrix with non-negative entries, does it have n a real eigenvalue? Does a continuous map f : A ⊂ R → A, where A is compact and convex, have a fixed point? Are certain continuous map surjective? As we will show, all of these classical questions can be answered with relative ease using the degree and the fact that it gives a definite answer to the existence of solutions to the earlier equation. As a further example, the reader may also recall the Jordan curve theorem, paraphrased by the intuitively obvious statement that a closed curve in 2 R that does not intersect itself partitions the plane into two parts, an ‘inside’ and ‘outside’. This is surprisingly difficult to show by elementary means; however, using the degree it can be proved it with little effort. The origin of the degree lies in algebraic topology, however we will use principally analytic arguments with some elementary topology.

2.2 Outline We will begin this report by providing further motivation and intuition for the degree by touching on the one-dimensional case. Following this we construct the topological degree for Euclidean space, known as the Brouwer degree, from which we will then demonstrate its use by solving some of the earlier mentioned problems. Up until this point we draw strongly from the work in [Dei85]. We then present an original result pertaining to the orientation preserving structure of derivatives of continuous injective maps.

2.3 Notation A full notation reference is provided in Appendix A, however we take this opportunity to mention some of what will be covered. Unless otherwise explicitly stated, Ω will denote an open and bounded n subset of R . The set of all continuous (or k-times diﬀerentiable with continuous derivatives up to

2 n order k respectively) functions f :Ω → R will be denoted by C(Ω) = C0(Ω) (or Ck(Ω) respectively). ∞ k Similarly, the notation C (Ω) will be used to denote the set ∩k∈NC (Ω). For differentiable functions k on compact sets we define for any k ∈ N ∪ {0} the set of all restricted functions C (Ω) = {f| Ω : n n f ∈ Ck(R )}. We will use | · | to denote the Euclidean norm and for any f ∈ C(S) where S ⊂ R is compact we use the notation |f|0 = max{|f(x)| : x ∈ S}. The open and closed balls of radius r centered at x will be denoted by Br(x) and Br(x) respectively. The notation ‘id’ will be used n n n for the identity map id : R → R which may also be restricted to a domain of a subset of R n n dependent on the context. For any y ∈ R and A ⊂ R the distance of y to A will be denoted %(y, A) = inf{|y − a| : a ∈ A}. For a function f : A → B and some y ∈ B, the inverse image f −1({y}) = {x ∈ A : f(a) = y} will be denoted by the obvious shorthand f −1(y). n A function f ∈ Ω → R is differentiable at a ∈ Ω if and only if there exists a linear map n n n n n f 0(a): R → R and remainder r : R → R such that for all h ∈ R satisfying a + h ∈ Ω we have f(a + h) = f(a) + f 0(a)h + r(h) 0 with limh→0 |r(h)|/|h| = 0. In this case f (a) is unique and we denote the Jacobian at a by 0 −1 n Jf (a) = det(f (a)). Furthermore, if f (y) = {x1, . . . , xm}, we call y ∈ R a regular value of f if Jf (xi) 6= 0 for i = 1, 2, . . . , m. However, if Jf (xi) = 0 for some i, then y is called singular value −1 and the corresponding xi a critical point of f. This definition implies that if f (y) = ∅ then y is a regular value of f.

3 The degree in one dimension

Let us motivate the basic idea of the topological degree with the following simple example.

0.2 0.4 0.6 0.8 1

Figure 1: Graph of an arbitrary function satisfying the assumptions of Proposition 3.1 .

Proposition 3.1. Let f : [0, 1] → R be a continuously diﬀerentiable function with f(0) 6= 0, f(1) 6= 0 and let A = f −1(0) be the set of zeroes of f. If for all x ∈ A we have f 0(x) 6= 0, that is 0 is a regular value of f, then A is a ﬁnite set and 1 X (sgn f(1) − sgn f(0)) = sgn f 0(x) (1) 2 x∈A

3 where sgn : R → {−1, 0, 1} denotes the sign function deﬁned by  −1, x < 0  sgn(x) = 0, x = 0 . 1, x > 0

Proof. To show that A is finite, let us assume for a contradiction that A is an infinite set. Those familiar with basic analysis will immediately recognise that [0, 1] is a compact set and as such A ⊂ [0, 1] must have an accumulation point γ ∈ [0, 1]. That is there exists a sequence (xn) in A \{γ} such that limn→∞ xn = γ. As f(xn) = 0 for all n ∈ N, the continuity of f implies that 0 f(γ) = f(limn→∞ xn) = limn→∞ f(xn) = 0 and so γ ∈ A. Furthermore, by assumption f (γ) 6= 0 and so, by the Inverse Function Theorem (B.4), there exists a neighbourhood Uγ of γ such that the restriction f|Uγ : Uγ → f(Uγ ) is a diffeomorphism. Therefore, as f|Uγ is injective we must have f(x) 6= 0 for all x ∈ Uγ \{γ}. However, this is a contradiction as γ is an accumulation point of A and thus at least one xn belongs to Uγ \(γ). Therefore, A is a finite set as stated. We now show that (1) holds. If A = ∅ then we must have sgn f(α) = sgn f(β) ∈ {−1, 1} for all α, β ∈ [0, 1], otherwise, by the intermediate value theorem, we would have A non-empty. Therefore, clearly (1) is true using the empty sum convention. Now for the case where A is non-empty, let us order the elements of A = {a1, . . . , an} so that a1 < a2 < . . . < an. By the continuity of f, for each disjoint interval I ∈ {[0, a1), (a1, a2),..., (an, 1]} there exists cI ∈ {−1, 1} such that for all x ∈ I, sgn f(x) = cI by the same argument we used for the case of the empty set. Furthermore, 0 0 by differentiability, f(ai + h) = f(ai) + hf (ai) + r(h) where for small h, |hf (ai)| > |r(h)| because 0 0 f (ai) 6= 0. Therefore, f(ai + h) = hf (ai) + r(h) and for small h, by the previous inequality, 0 0 sgn f(ai ± h) = sgn[±hf (ai)] = ± sgn[hf (ai)]. In other words, sgn f(ai + h) = − sgn f(ai − h). What we have shown is the obvious fact that the sign of f must alternate at each zero and remain 0 constant everywhere else as our intuition tells us. It is also obvious that sgn f (ai) indicates how the sign of f changes across the zero ai. Using this fact, if we let α1 = 0, α2 ∈ (a1, a2), α3 ∈ (a2, a3), . . . , αn ∈ (an−1, an), αn+1 = 1 and apply telescopic summation we obtain

n n X X 0 sgn f(1) − sgn f(0) = (sgn f(αi+1) − sgn f(αi)) = 2 sgn f (ai) i=1 i=1 and rearranging gives the result at (1). This completes the proof. On ﬁrst glance Proposition 3.1 may seem obvious and its proof a waste of mathematical machin- ery. However, we will shortly generalise this idea to continuous functions on arbitrary Euclidean n n spaces (R → R ) where we are not so well served by our intuition alone. The reader may now query: what does this have to do with the existence of solutions to an equation? We note that in this case if we have X sgn f 0(x) 6= 0 (2) x∈A we can then deduce that the equation f(x) = 0 has at least one solution x ∈ (0, 1). This is actually a simple application of the topological degree. Indeed, it turns out that the simplest of case of the degree in R is given by precisely the sum at (2) and, contrary to the general case, the left hand side of equation (1) provides a simple expression for it. Furthermore, in the subsequent construction of the degree (2) will serve as our foundation and we will reuse some of the methods seen in the previous proof.

4 4 Construction of the degree 4.1 The degree is unique Before we give the explicit construction of the topological (Brouwer) degree, it is useful to discuss its uniqueness. This will aid in building some intuition as well as simplifying and further motivating its construction/definition. The impatient reader may skip ahead to Section 4.2 for the complete n definition of the degree at their own risk. Now, let Ω ⊂ R be open and bounded, f ∈ C(Ω) and n y ∈ R \ f(∂Ω). If we assume that we can associate with each such triple (f, Ω, y) an integer called the degree, denoted by d(f, Ω, y), with the three properties (D1)–(D3) below, then these properties uniquely determine that integer. (D1) d(id, Ω, y) = 1 for y ∈ Ω. P (D2) d(f, Ω, y) = i d(f, Ωi, y) for a finite number of disjoint open subsets Ωi ⊂ Ω such that y 6∈ f(Ω \(∪i Ωi)).

n n (D3) For any two continuous functions h : [0, 1] × Ω → R and y : [0, 1] → R satisfying y(t) 6∈ h(t, ∂Ω) for all t ∈ [0, 1], the integer d(h(t, ·), Ω, y(t)) is constant on [0, 1]. As we shall soon see, such a degree does in fact exist.

4.1.1 The trivial case Before we proceed with showing that the degree is unique in general, let us get a trivial case out of the way. Proposition 4.1. If f ∈ C(Ω), y 6∈ f(∂Ω) and f −1(y) = ∅ then d(f, Ω, y) = 0. Proof. This is just a simple consequence of (D2). Indeed, from (D2) we have d(f, Ω, y) = d(f, Ω, y)+ d(f, ∅, y) which implies that d(f, ∅, y) = 0. Therefore, if Ω1 ⊂ Ω is open and y 6∈ f(Ω \ Ω1) = f(Ω \(Ω1 ∪ ∅)) it must follow again from (D2) that d(f, Ω, y) = d(f, Ω1, y)+d(f, ∅, y) = d(f, Ω1, y). Thus, setting Ω1 = ∅ we obtain the result required. The property that we derived in coming to this result is so useful that we will state it explicitly for future reference along with a simple corollary:

(D4) d(f, Ω, y) = d(f, Ω1, y) if Ω1 ⊂ Ω is open and y 6∈ f(Ω \ Ω1). (D5) d(f, Ω, y) 6= 0 implies that there exists x ∈ Ω such that f(x) = y. As we have already shown that d(f, Ω, y) = 0 if f −1(y) = ∅, we will take f −1(y) to be nonempty from this point forward unless otherwise stated.

4.1.2 Reducing to C2 maps In an axiomatic fashion we will now show that, in general, properties (D1), (D2) and (D3) of the degree uniquely determine some integer. The approach we will take comes from [Dei85] and involves a series of reductions beginning with arbitrary continuous maps and ending with the simplest cases of linear maps, where we can apply the tools of linear algebra. This reduction will be made in several stages beginning with a reduction to C2 maps. We make the following proposition.

5 n Proposition 4.2. Let A ⊂ R be compact (that is, closed and bounded) and f ∈ C(A). Then for any n > 0 there exists a twice continuously diﬀerentiable function g ∈ C2(R ) such that |f(x)−g(x)| ≤ for all x ∈ A.

n + Proof. Let us deﬁne the molliﬁer function ϕα : R → R by

( 2 4 f1(x) := c(|x| − 1) , |x| < 1 ϕ1(x) = (4.3) f2(x) := 0, otherwise −n ϕα(x) = α ϕ1(x/α) (4.4)

R 2 n where c ∈ R is chosen such that n ϕ1(x) dx = 1. We now show that ϕ1 ∈ C (R ), the case of ϕα R n follows by the chain rule. Clearly, ϕ1 is smooth for all x ∈ R satisfying |x|= 6 1 by the composition and product of smooth functions. We also observe that from the interior of the ball B1(0) the ﬁrst n order partial derivatives of ϕ1 approach zero at ∂B1(0) and on R \ B1(0) they are identically zero. Therefore, if we can show that these partial derivatives of ϕ1 are zero for all x such that |x| = 1, we 1 n 2 n n can deduce that ϕ1 ∈ C (R ). Indeed, as f1, f2 ∈ C (R ), for any a ∈ S1(0) = {x ∈ R : |x| = 1} we have

( ∂f1 f1(a) + (a)h + hr1(h), if a + hej ∈ B1(0) ϕ (a + he ) = ∂xi 1 i ∂f2 f2(a) + (a)h + hr2(h), otherwise ∂xi ( 0 + 0h + hr (h), if a + he ∈ B (0) = 1 j 1 0 + 0h + hr2(h), otherwise

= ϕ1(a) + 0h + hR(h) where limh→0 R(h) = limh→0 r1(h) = limh→0 r2(h) = 0. Therefore, by the definition of differentiability, all first order partial derivatives of ϕ1 exist and are equal to zero for all points a ∈ S1(0) and 2 1 n ∂ ϕ1 so ϕ1 ∈ C ( ). We can clearly obtain (a) = 0 for all a ∈ S1(0) using the same idea as that R xixj 2 n above and conclude that ϕ1 ∈ C (R ) as required. Now, define the convolution Z ˜ ˜ fα(x) = f ∗ ϕα(x) = f(t)ϕα(x − t) dt, n R ˜ 2 n where f is the continuous extension defined by Proposition B.2. As ϕα ∈ C (R ) is compactly ˜ n 2 n supported and f ∈ C(R ), by the properties of the convolution operator we have fα ∈ C (R ). ˜ Finally, we show that fα converges uniformly to f on A as α → 0 and then we are done because fα with α > 0 chosen small enough satisfies the property required in the proposition. To achieve this we will make use of the following identity. Z Z ˜ ˜ ˜ ˜ ˜ fα(x) − f(x) = (f(t) − f(x))ϕα(x − t) dt = (f(t) − f(x))ϕα(x − t) dt. n R Bα(x) ˜ Now, let A˜ = A + B1(0) = {y : %(y, A) ≤ 1} and let > 0 be given. As f is uniformly continuous on the compact set A˜, there exists γ > 0 such that |f˜(t) − f˜(x)| ≤ for all t, x ∈ A˜ satisfying

6 |x − t| ≤ γ. Now, for δ ≤ min{γ, 1}, we have Bδ(x) ⊂ A˜ for all x ∈ A and so for any such x

Z ˜ ˜ ˜ fδ(x) − f(x) = (f(t) − f(x))ϕδ(x − t) dt Bδ (x) Z ˜ ˜ ≤ (f(t) − f(x))ϕδ(x − t) dt Bδ (x) Z ≤ |ϕδ(x − t)| dt Bδ (x) Z = ϕδ(t) dt Bδ (0) = .

˜ Therefore, fα converges uniformly on A to f as required. This completes the proof. This statement is quite useful as it tells us that we can uniformly approximate any continuous function on a compact set with a C2 function to any level of accuracy. Using this fact, for any n f ∈ C(Ω) and y ∈ R \ f(∂Ω) we have α = %(y, f(∂Ω)) > 0 and so there exists g ∈ C2(Ω) such that n |f − g|0 < α. The continuous homotopy h : [0, 1] × Ω → R given by h(t, x) = (1 − t)f(x) + t g(x) satisﬁes for all x ∈ ∂Ω and all t ∈ [0, 1]

|h(x, t) − y| = |f(x) − y + t(g(x) − f(x))| ≥ |f(x) − y| − t|g(x) − f(x)|

≥ |f(x) − y| − |f − g|0 > α − α = 0.

In other words, y 6∈ h(t, ∂Ω) for all t ∈ [0, 1] and therefore, by the third property of the degree, we have d(f, Ω, y) = d(g, Ω, y). This is a very important result as it means that the degree of any continuous function is equal to the degree of some C2 function. Thus, it will suﬃce to show that the degree of any C2 function in unique.

4.1.3 Reducing to regular values We now show that the degree of any f ∈ C2(Ω) at y 6∈ f(∂Ω) is equal to the degree of f at some regular value. That is d(f, Ω, y) = d(f, Ω, yˆ), wherey ˆ is a regular value of f. This will be crucial for our reduction to linear maps later. The reasoning here is straightforward. Sard’s lemma (Lemma B.3) says the set of all singular values of any continuously differentiable function defined on an open set has a Lebesgue measure of zero. From this we conclude the set of regular values of n f is dense in R . Therefore, for α = %(y, f(∂Ω)) > 0 we choose a regular valuey ˆ ∈ Bα(y), which is n guaranteed to exist, and define γ(t) := ty+(1−t)ˆy as well as the trivial homotopy h : [0, 1]×Ω → R by h(t, x) := f(x). By the convexity of Bα(y) we have γ(t) ∈ Bα(y) for all t ∈ [0, 1] and so by (D3) we have d(f, Ω, y) = d(f, Ω, γ(t)) = d(f, Ω, yˆ). Thus we have shown that the degree at any singular value is equal to the degree of some regular value. Therefore, it will suffice to show uniqueness of the degree for regular values alone.

7 4.1.4 Some further simpliﬁcation Before we progress with our reduction to the linear case, let us note the following.

n Proposition 4.3. Let Ω ⊂ R be open and bounded, f ∈ C2(Ω) and y 6∈ f(∂Ω). If y is a regular value of f then the set f −1(y) is ﬁnite.

Proof. This is a generalisation of a result we proved in Proposition 3.1 for the one-dimensional case of the degree. Let us assume for a contradiction that S = f −1(y) is infinite. Then, as Ω is compact and S ⊂ Ω, there exists an accumulation point γ ∈ Ω of S. Furthermore, as f(x) = y for all x ∈ S and f is continuous, we must have f(γ) = y and so γ ∈ S. Therefore, it follows 0 that Jf (γ) = det(f (γ)) 6= 0 and so by the inverse function theorem (Theorem B.4) there exists a 2 neighbourhood Uγ of γ such that the restriction f|Uγ : Uγ → f(Uγ ) is a C diffeomorphism. Note that as y 6∈ f(∂Ω) we have S ⊂ Ω, with Ω open, and so our use of the inverse function theorem here is justified by restricting f to Ω. Now, as f|Uγ is one-to-one we have f|Uγ (x) 6= y for all x ∈ Uγ \{γ} which means that S ∩ Uγ \{γ} = ∅. This is a contradiction as γ is an accumulation point of S. Therefore, f −1(y) must be finite as stated. This competes the proof.

Let us look at this result in light of where we are currently situated. At the moment we are trying to show that if f ∈ C2(Ω) and y 6∈ f(∂Ω) is a regular value of f then d(f, Ω, y) is −1 uniquely determined by our properties (D1)–(D3). As the set f (y) = {x1, x2, . . . , xm} is ﬁnite by Proposition 4.3 and y is a regular value of f, the inverse function theorem tells us that there −1 exist disjoint open neighbourhoods Uxi for i ∈ {1, 2, . . . , m} of each xi that cover the set f (y) and such that f is a diﬀeomorphism. Therefore, by (D2) we have |Uxi

m X d(f, Ω, y) = d(f, Uxi , y) i=1 and this leads to our ﬁnal reduction.

4.1.5 Reduction to linear maps

We now show that each term d(f, Uxi , y) in the previous sum can be reduced to the degree of a linear map. As f ∈ C2(Ω), by diﬀerentiability of f and regularity of y we have

0 f(x) = y + f (xi)(x − xi) + r(x − xi)

0 n n 0 |r(h)| for f (xi) ∈ L(R , R ) with det(f (xi)) 6= 0 and limh→0 |h| = 0. Let us denote A = f(xi) and n note that as the set S1(0) = {z ∈ R : |z| = 1} is compact, the set S = {|Az| : |z| = 1} is compact. Therefore, as |Az| > 0 for all z ∈ S1(0), it must follow that |Az| ≥ c for all such z for some c > 0.

This means that for any z ∈ n \{0} we have A z ≥ c and so |Az| ≥ c|z| by linearity. We deﬁne R |z| n the homotopy h : [0, 1] × Ω → R by h(t, x) = tf(x) + (1 − t)A(x − xi) and y(t) := yt. From the limiting behavior of r(h) as h → 0, there exists some δ > 0 such at for all x ∈ ∂Bδ(xi) ⊂ Uxi we

8 have c|x − xi| > |r(x − xi)|. So, by the previous fact, for all such x and all t ∈ [0, 1] we obtain

|h(t, x) − y(t)| = |tf(x) + (1 − t)A(x − xi) − ty|

= |t(y + A(x − xi) + r(x − xi)) + (1 − t)A(x − xi) − ty|

= |A(x − xi) + t · r(x − xi)|

≥ c|x − xi| − |r(x − xi)| > 0.

Therefore, as f(x) 6= y for x ∈ Uxi \ Bδ(xi) since f is a diﬀeomorphism on Uxi , by (D3) and (D4) we deduce that

d(f, Uxi , y) = d(f, Bδ(xi), y) = d(h(0, x),Bδ(xi), y(0)) = d(A − Axi,Bδ(xi), 0).

n Furthermore, as A is non-singular, Ax − Axi = 0 has the unique solution x = xi on R and so by (D4) for any r > 0 such that Br(0) ⊃ Bδ(xi) we have d(A − Axi,Bδ(xi), 0) = d(A − Axi,Br(0), 0). n If we now define another homotopy g : [0, 1] × Br(0) → R by g(t, x) = Ax − t(Axi) we note that for a fixed t ∈ [0, 1] the unique solution xt = txi to g(t, x) = 0 satisfies |xt| = |txi| ≤ |xi|. It is then clear that g(t, x) 6= 0 for all (t, x) ∈ [0, 1] × ∂Br(0) and so finally, by (D3), we obtain

d(f, Uxi , y) = d(A − Axi,Br(0), 0) = d(A, Br(0), 0). In fact, as we have det(A) 6= 0 and so Ax = 0 which is equivalent to x = 0, (D4) tells us that the radius r > 0 may now be made arbitrary. This completes our reduction to the linear case. All that remains now is to show that for an invertible matrix A and r > 0 the integer d(A, Br(0), 0) is uniquely determined by (D1)–(D3) and our demonstration of uniqueness of the degree will be complete.

4.1.6 Degree of an invertible linear map is unique It is reasonably straightforward to show that for an invertible map A the degree at zero is given by d(A, Br(0), 0) = sgn det(A) which completes our discussion on the uniqueness of the degree. Firstly, consider the case where det(A) > 0. Then by Proposition B.6 there exists a continuous n×n f : [0, 1] → R such that f(0) = A, f(1) = id and det(f(t)) 6= 0 for t ∈ [0, 1]. Therefore, if we n deﬁne h : [0, 1] × Br(0) → R by h(t, x) = [f(t)]x we have 0 6∈ h([0, 1] × ∂Br(0)) and so

d(A, Br(0), 0) = d(id,Br(0), 0) = 1 = sgn(det(A)).

In the second case, where det(A) < 0, we will need a little more ingenuity. Let us deﬁne the diagonal n × n matrix −1 0 0 ... 0  0 1 0 ... 0    0 0 1 ... 0 Ae =    . . . ..   . . . . 0 0 0 0 ... 1 From Proposition B.6, by the same argument as that above, we know that

d(A, Br(0), 0) = d(A,e Br(0), 0).

9 Thus, all we need to do is show that d(A,e Br(0), 0) = −1 and we will be done. To do this we first fix Ω = B2(0) and define Ω1, Ω2 ⊂ Ω by Ω1 = {x ∈ Ω: x1 ∈ (−2, 0)} and Ω2 = {x ∈ Ω: x1 ∈ (0, 2)}. n n We also define f : Ω → R by f(x) = (|x1| − 1, x2, . . . , xn), the constant function g : Ω → R by g(x) := (1, 0,..., 0) and the homotopy

h(t, x) = tf(x) + (1 − t)g(x) = t(|x1| − 2, x2, . . . , xn) + (1, 0,..., 0)

As d(g, Ω, 0) = 0 by (D5) and we observe that 0 6∈ h([0, 1] × ∂Ω), from (D3) we have d(f, Ω, 0) = d(g, Ω, 0) = 0. Furthermore, by applying (D2) we observe that d(f, Ω, 0) = d(f, Ω1, 0) + d(f, Ω2, 0).

Restricting f leads to f| Ω1 = Ae − (1, 0,..., 0) and f| Ω2 = id −(1, 0,....0). Clearly, the only zeros of f| Ω1 and f| Ω2 occur at (−1, 0,..., 0) ∈ Ω1 and (1, 0,..., 0) ∈ Ω2 respectively and so

d(f, Ω1, 0) = d(Ae − (1, 0,..., 0), Ω, 0) = d(A,e Ω, 0), where we have used the homotopy h˜(t, ·) = A˜−t(1, 0,..., 0) and applied (D3) once more. Similarly, d(f, Ω2, 0) = d(id, Ω, 0). Therefore, we deduce that

d(A,e B2(0), 0) = d(f, Ω1, 0) = 0 − d(f, Ω2, 0) = − d(id, Ω, 0) = −1 = sgn(det(A)) as required. This completes our discussion on the uniqueness of the degree.

10 4.2 The Brouwer degree Given that the reader is now familiar with the previous arguments – that is the uniqueness of a function with properties (D1)–(D3) which we called the degree – the construction of such a function is straightforward. We now give a definition of the Brouwer degree. Definition 4.1. If Ω is open and bounded, f ∈ C2(Ω) and y 6∈ f(∂Ω) is a regular value of f then we define X d(f, Ω, y) := sgn det(Jf (x)), x∈f −1(y) P where we use the empty sum convention ∅ = 0. Definition 4.2. If Ω is open and bounded, f ∈ C2(Ω) and y 6∈ f(∂Ω) (not necessarily a regular value of f) then we define d(f, Ω, y) := d(f, Ω, yˆ) where yˆ is a regular value of f with |y −yˆ| < %(y, f(∂Ω)), where on the right hand side we can apply Definition 4.1. Definition 4.3. If Ω is open and bounded, f ∈ C(Ω) and y 6∈ f(∂Ω) then we define

d(f, Ω, y) := d(g, Ω, y)

2 where g ∈ C (Ω) with |f −g|0 < %(y, f(∂Ω)) and on the right hand side we can apply Definition 4.2. Note that Proposition 4.2 guarantees that such a function g exists in Definition 4.3 and our reasoning in Section 4.1.3 with Sard’s lemma tells us that such ay ˆ exists in Definition 4.2.

4.2.1 The degree is well defined We have given a definition of the degree, although it remains to be shown that it is in fact well defined. We know from Proposition 4.3 that the sum in Definition 4.1 is finite; nevertheless, two main questions arise from Definitions 4.2 and 4.3: How do we know that all sufficiently close regular valuesy ˆ and sufficiently close functions g ∈ C2(Ω) will result in the same degree as per Definitions 4.1 and 4.2 respectively? We will show this to be the case in the style of [Dei85] and will first need the following result.

n n Proposition 4.4. Let Ω ⊂ R be open and bounded, f ∈ C2(Ω) and y ∈ R \(f(∂Ω)) be a regular value of f. Then there exists an y,f > 0 such that for all 0 < < y,f we have d(f, Ω, y) = R Ω ϕ(f(x) − y)Jf (x) dx, where ϕ is the approximation of identity deﬁned by (4.3). −1 Proof. In the case where f (y) = ∅ we have d(f, Ω, y) = 0. If we choose y,f = %(y, f(∂Ω)), for any 0 < < y,f we have |f(x) − y| > %(y, f(∂Ω)) = y,f > for all x ∈ Ω and so Z Z ϕ(f(x) − y)Jf (x) dx = 0 · Jf (x) dx = 0 = d(f, Ω, y) Ω Ω as required. Now, let us assume that f −1(y) 6= ∅. From our discussion at the end of Section 4.1.4 we −1 have f (y) = {x1, . . . , xn} ⊂ Ω for some n ∈ N and there exist disjoint neighbourhoods Uxi ,

11 each connected, of each xi such that f : Ux → Vi is a diﬀeomorphism by the inverse function |Uxi i theorem (Theorem B.4). Note that sgn Jf (·) is constant on each Uxi by the continuity of each partial derivative and the fact that Jf (·) 6= 0 on Uxi (see the discussion in Section 6.2 if this is n −1 unfamiliar). We let γ > 0 such that Bγ (y) ⊂ ∩i=1Vi and we let Ui = Uxi ∩ f (Bγ (y)). As n n f(x) 6= y for all x ∈ Ω \ ∪i=1Ui, and Ω \ ∪i=1Ui is closed, we have |f(x) − y| ≥ A > 0 for some A ∈ R. Therefore, for 0 < < A we have

n Z X Z ϕ(f(x) − y)Jf (x) dx = sgn Jf (xi) ϕ(f(x) − y)|Jf (x)| dx Ω i=1 Ui n X Z = sgn Jf (xi) ϕ(f(x) − y)|Jf−y(x)| dx i=1 Ui n X Z = sgn Jf (xi) ϕ(x) dx. (5) i=1 Bγ (0)

Where at (5) we used integration by substitution and the fact that Jf (x) = Jf−y(x) and f(Ui)−y = Bγ (0). Therefore, we obtain the result by choosing y,f = min{A, γ} to ensure the final integral is equal to 1 and using the definition of the degree at Definition 4.1. This completes the proof. We are now ready to show that all regular values of f ∈ C2(Ω) which are sufficiently close yield the same degree. This gets us half way to showing that our inductive definition of the degree is well defined.

n Proposition 4.5. Let Ω ⊂ R be open and bounded, f ∈ C2(Ω), y 6∈ f(∂Ω) and α = %(y, f(∂Ω)), then d(f, Ω, y1) = d(f, Ω, y2) for all regular values y1, y2 of f satisfying |y − yi| < α.

Proof. Let δ = α −max{|yi −y| : i = 1, 2}. As y1 and y2 are regular values of f, by Proposition 4.4, we can express the degrees by the following integrals for some satisfying 0 < < min{δ, y1,f , y2,f }: Z Z d(f, Ω, y1) = ϕ(f(x) − y1)Jf (x) dx d(f, Ω, y2) = ϕ(f(x) − y2)Jf (x) dx Ω Ω

n n where y1,f , y2,f are given in the previous proposition. The function ϕ : R → R is again the mollifier defined at (4.3). Now, using this alternative representation of the degree we will show that Z Z d(f, Ω, y1) − d(f, Ω, y2) = ϕ(f(x) − y1)Jf (x) dx − ϕ(f(x) − y2)Jf (x) dx = 0. Ω Ω The way that we achieve this is somewhat technical. Our approach will be to construct a function 1 n R v ∈ C (R ) that satisfies d(f, Ω, y1) − d(f, Ω, y2) = Ω div v dx and use the divergence theorem n n to reduce the integral on the right hand side to zero. We begin by defining w : R → R by R 1 w(x) = (y1 − y2) 0 ϕ(x − y1 + t(y1 − y2)) dt and we note that if we set a = x − y1, x¯ = x − y2 we

12 y2

y1 y

∂Bα−δ(y)

f(∂Ω)

Figure 2: All points in the convex combination (t − 1)y1 + ty2 are contained in Bα−δ(y). obtain

ϕ(x − y2) − ϕ(x − y1) = ϕ(¯x) − ϕ(a) n X Z 1 ∂ = (¯xi − ai) [ϕ (a + t(¯x − a))] dt ∂x i=1 0 i n X Z 1 ∂ = (yi − yi ) [ϕ (x − y + t(y − y ))] dt 1 2 ∂x 1 1 2 i=1 0 i n X ∂ Z 1 = (yi − yi ) ϕ (x − y + t(y − y )) dt ∂x 1 2 1 1 2 i=1 i 0 n X ∂wi = (x) ∂x i=1 i = div w(x). Here we have used Taylor’s theorem with the remainder formula, diﬀerentiation under the integral sign and a simple change of variables. Now, by convexity we have |(1−t)y1 +ty2 −y| ≤ max{|yi −y| : i = 1, 2} = α − δ as seen in Figure 2. Building on this we note that if w(x) 6= 0, then there exists t ∈ [0, 1] such that ϕ(x−y1 +t(y1 − y2)) 6= 0 and thus |x − y1 + t(y1 − y2)| < . Therefore, for all such x we observe that

|x − y| = |x − y1 − t(y2 − y1) + y1 + t(y2 − y1) − y|

≤ |x − y1 − t(y2 − y1)| + |y1 + t(y2 − y1) − y| ≤ + α − δ. n Furthermore, if ` is a limit point of S := {x ∈ R : w(x) 6= 0} and (xn) ⊂ S satisﬁes xn → ` as n → ∞, then by the continuity of the norm function we have

|` − y| = lim xn − y = lim |xn − y| ≤ + α − δ n→∞ n→∞

13 by the order limit theorem and so we have |x − y| ≤ α − (δ − ) for all x ∈ S = supp w. In other n n words, supp w ⊂ Br(y) for r = α − (δ − ) < α. Now, we deﬁne v : R → R by

(Pn j=1 wj(f(x))dij(x), x ∈ Ω vi(x) = , for i = 1, 2, . . . , n 0, otherwise where dij(x) = Cji(x) is the cofactor of the i-th column and j-th row of Jf (x). From the product rule and chain rule we obtain for x ∈ Ω

n n n ∂vi(x) X X ∂wj(f(x)) ∂fk(x) X ∂dij(x) = dij + wj(f(x)). ∂xi ∂xk ∂xi ∂xi j=1 k=1 j=1 Pn We note that i=1 dij ∂ifk(x) = δjkJf (x) with Kronecker’s δjk by proper/improper cofactor expansion (expansion along a row that is the same/diﬀers from its cofactors) and so, using Proposition B.7 from Appendix B, for all x ∈ Ω we obtain

n X ∂vi(x) div v(x) = ∂x i=1 i n n n n n X X X ∂wj(f(x)) ∂fk(x) X X ∂dij(x) = dij(x) + wj(f(x)) ∂xk ∂xi ∂xi i=1 j=1 k=1 i=1 j=1 n n " n # n " n # X X X ∂wj(f(x)) ∂fk(x) X X ∂dij(x) = dij(x) + wj(f(x)) ∂xk ∂xi ∂xi j=1 k=1 i=1 j=1 i=1 n n " n # X X X ∂fk(x) ∂wj(f(x)) = dij(x) · + 0 ∂xi ∂xk j=1 k=1 i=1 n n X X ∂wj(f(x)) = δjkJf (x) ∂xk j=1 k=1 n X ∂wj(f(x)) = J (x) ∂x f j=1 j

= [div w(f(x))] · Jf (x).

1 n Since supp w ⊂ Br(y) ⊂ Bα(y), we have supp v ⊂ Ω, which implies that v ∈ C (R ). Therefore, from our alternative representation of the degree given in Proposition 4.4, for some cube Q =

14 n [−a, a]n ⊂ R containing Ω we obtain Z Z d(f, Ω, y1) − d(f, Ω, y2) = ϕ(f(x) − y1)Jf (x) dx − ϕ(f(x) − y2)Jf (x) dx Ω Ω Z = [ϕ(f(x) − y1) − ϕ(f(x) − y2)] Jf (x) dx Ω Z = [div w(f(x))] Jf (x) dx Ω Z = div v(x) dx Ω Z = div v(x) dx Q = 0,

n where we have used the fact that v ∈ C1(R ) and supp v ⊂ Ω ⊂ Q which implies that v = 0 on ∂Q with the result following from the divergence theorem. This completes the proof. Now, all that remains is to show that the degree does not depend on the choice of g ∈ C2(Ω) such that |f − g|0 < %(y, f(∂Ω)) and we will have shown that Deﬁnition 4.3 stated earlier is well deﬁned. To do this we need to prove the following result using the implicit function theorem (Theorem B.5) given in the Appendix.

n Proposition 4.6. Let f ∈ C2(Ω), y ∈ R \ f(∂Ω). Then for any g ∈ C2(Ω) there exists δ = δf,g,y > 0 such that d(f, Ω, y) = d(f + tg, Ω, y) for all |t| < δ. Proof. Firstly, we consider the trivial case were f −1(y) = ∅, that is f(x) 6= y for all x ∈ Ω, and let α α = %(y, f(Ω)) > 0. As g is continuous, g(Ω) is bounded by |g|0 and if we set δ = > 0, for 2|g|0+1 all x ∈ Ω and |t| < δ we obtain α |f(x) + tg(x) − y| ≥ |f(x) − y| − |tg(x)| ≥ α − |t||g| > > 0 0 2 and so clearly (f + tg)−1(y) = ∅ which gives us d(g + tg, Ω, y) = 0 = d(f, Ω, y) for all |t| < δ by (D5). Thus we are done for the trivial case. Let us now consider the case where f −1(y) 6= ∅ and y is a regular value of f. We know from −1 Proposition 4.3 that f (y) = {x1, x2, . . . , xn} for some x ∈ N. Let us deﬁne ft = f + tg and h(t, x) = ft(x) − y. We observe that h(0, xi) = f(xi) − y = 0 and Jh(0,·) = Jf (xi) 6= 0. Therefore, by Theorem B.5 and shrinking r > 0 and ρ > 0 as necessary, there exists an interval I = (−r, r), n disjoint sets Bρ(xi) and continuous functions wi : I → Bρ(xi) such that for V = ∪i=1Bρ(xi) we −1 have ft (y) ∩ V = {w1(t), w2(t), . . . , wn(t)}. This ﬁnal set equality follows from the fact that on each ball Bρ(xi) we have

ft(x) = y ⇐⇒ h(x, t) = 0 ⇐⇒ wi(t) = x from Theorem B.5. Now, we want to show that there are no other solutions outside V to ft(x) = y so that we can apply Deﬁnition 4.1 of the degree. Let us shrink ρ (and thus r) as necessary so that sgn Jf (x) = sgn Jf (xi) on Bρ(xi) for all i = 1, 2, . . . , n. As continuous functions preserve compactness, f(Ω \ V ) is compact and as y 6∈ f(Ω \ V ) it then follows that %(y, f(Ω \ V )) > 0. This

15 means that there exists γ > 0 such that |f(x) − y| > γ for all x ∈ Ω \ V . For all such x and γ |t| < δ0 = we obtain |g|0+1

|ft(x) − y| = |f(x) − y + tg(x)| ≥ |f(x) − y| − |tg(x)|

≥ |f(x) − y| − |t||g|0 γ ≥ |f(x) − y| − |g|0 |g|0 + 1 > |f(x) − y| − γ > 0 and so ft(x) 6= y on Ω \ V for |t| < δ0. Therefore, we have found that

−1 −1 −1 ft (y) = (ft (y) ∩ V ) ∪ (ft (y) ∩ (Ω \ V )) = {w1(t), w2(t), . . . , wn(t)}.

Furthermore, as (t, x) 7→ Jft (x) is uniformly continuous on [−δ0, δ0] × V , for 0 < < min{|Jf (z)| : z ∈ V } there exists 0 < δ ≤ δ0 such that |Jft (x) − Jf (x)| ≤ for |(t, x) − (0, x)| = |t| ≤ δ with x ∈ V . Therefore, we deduce that sgn Jft (wi(t)) = sgn Jf (wi(t)) = sgn Jf (xi) for |t| ≤ δ, where the last equality holds because wi : I → Bρ(xi) and sgn Jf (x) is constant on Bρ(xi) by our choice of ρ. This means that d(ft, Ω, y) = d(f, Ω, y) for |t| < δ by Deﬁnition 4.1. Note that we have also used −1 the fact that ft (y) = {w1(t), w2(t), . . . , wn(t)} and Jft (wi(t)) 6= 0 and so y is a regular value of ft. n Finally, if y ∈ R \ f(∂Ω) is not a regular value of f, let us choose a regular valuey ˆ ∈ Bα/3(y) with α = %(y, f(∂Ω)). From our previous reasoning, there exists δ0 > 0 such that d(ft, Ω, yˆ) = d(f, Ω, yˆ) for |t| < δ0 and by Deﬁnition 4.2 we have

d(ft, Ω, yˆ) = d(f, Ω, yˆ) = d(f, Ω, y)

α n α o 2α as |yˆ − y| < < α. Now, if we let δ = min δ0, we have |yˆ − f(x)| ≥ for x ∈ ∂Ω and 3 3|g|0+1 3 α α |tg(x)| = |t||g(x)| ≤ |t||g|0 ≤ |g0| < 3|g|0 + 1 3 which gives us for x ∈ ∂Ω and |t| < δ 2α α α |yˆ − f (x)| = |yˆ − f(x) − tg(x)| ≥ |yˆ − f(x)| − |tg(x)| > − = . t 3 3 3

In particular, this implies that |yˆ − y| < %(ˆy, ft(∂Ω)) for |t| < δ and so by Definition 4.2 we have d(ft, Ω, y) = d(ft, Ω, yˆ) which means that d(ft, Ω, y) = d(f, Ω, y) for all t < δ as required. This completes the proof. We can now use this result to complete our argument that Definition 4.3 is well defined. However, let us first give the following lemma which will prove useful now and in some later arguments.

n m Lemma 4.7. If X ⊂ R is connected and f : X → R is a locally constant function, then f is constant on X.

16 Proof. Let a ∈ X be arbitrary and S = {x ∈ X : f(x) = f(a)}. To show that f is constant on X we will show that S = X. To achieve this it suﬃces to show that S is both open and closed n in the induced topology on X from R . Indeed, if S is both open and closed then Sc is open and X = S ∪ Sc is the union of two open sets. However, as S is non-empty and X is a connected space, it must be that Sc = ∅ and so S = X as required. To see why S is open all we need to observe is that, by assumption, for any x ∈ S there exists a neighbourhood Ux ⊂ X such that f(Ux) = {f(x)} = {f(a)} which implies that Ux ⊂ S and so each point of S is an interior point of S. To show that S is closed let ` ∈ X be a limit point of S,(xn) ⊂ S a sequence satisfying xn → ` as n → ∞, and U` ⊂ X be the neighbourhood of ` where f is locally constant. Then there exists an M ∈ N such that for all n ≥ M we have xn ∈ U` by convergence. This means that for any such xn ∈ S we have f(`) = f(xn) = f(a) by the locally constant property and so ` ∈ S. Therefore, S contains all its limit points and S = S (in the relative topology) as required. This completes the proof.

n 2 Now, let f ∈ C(Ω), y ∈ R \ f(∂Ω), α = %(y, f(∂Ω)) and g1, g2 ∈ C (Ω) with |f − gi|0 < α for i = 1, 2 as per the assumptions of Definition 4.3. Let us define h : [0, 1]×Ω by h(t, ·) = g1 +t(g2 −g1) and φ(t) = d(h(t, ·), Ω, y) for t ∈ [0, 1]. We have h ∈ C2([0, 1]×Ω) and simple algebra yields h(t, ·) = h(t0, ·) + (t − t0)(g2 − g1). Proposition 4.6 thus implies that φ(t) = d(h(t0, ·) + (t − t0)(g2 − g1), Ω, y) is constant on some neighbourhood of t0 for all t0 ∈ [0, 1]. Therefore, Lemma 4.7 implies that φ is constant on [0, 1] and so d(g1, Ω, y) = d(g2, Ω, y) as required. We have now shown that the Brouwer degree given in Definitions 4.1-4.3 is well defined. Al- though this process was quite tedious, what we shall see in just a moment is how useful this mathematical tool can be and how easy it is to use. However, before we show some applications of the degree, let us show that the properties (D1), (D2) and (D3) stated earlier are satisfied. This will then confirm that the degree is the unique function that satisfies these properties.

4.2.2 Properties of the degree Recall we had the following properties which we mentioned are satisfied by the degree upon their introduction: (D1) d(id, Ω, y) = 1 for y ∈ Ω. P (D2) d(f, Ω, y) = i d(f, Ωi, y) for a finite number of disjoint open subsets Ωi ⊂ Ω such that y 6∈ f(Ω \(∪i Ωi)). n n (D3) For any two continuous functions h : [0, 1] × Ω → R and y : [0, 1] → R satisfying y(t) 6∈ h(t, ∂Ω) for all t ∈ [0, 1] we have d(h(t, ·), Ω, y(t)) is constant on [0, 1]. n We first show that (D1) is satisfied. Consider f = id, Ω ⊂ R open and y ∈ Ω. Clearly Jf (x) = 1 6= 0 on Ω and so y is a regular value of f. Therefore, by Definition 4.1 we have X d(id, Ω, y) = d(f, Ω, y) = sgn Jf (x) = sgn Jf (y) = sgn(1) = 1 x∈f −1(x) as required. Now, let us show (D2). Let f ∈ C(Ω), the sets Ω1, Ω2,..., Ωm ⊂ Ω be disjoint and open, y 6∈ f(Ω \(∪i Ωi)) and

γ = min{%(y, f(∂Ω)),%(y, f(∂Ω1)),%(y, f(∂Ω2)),...,%(y, f(∂Ωm))}.

17 2 γ As γ > 0, by Proposition 4.2 we can ﬁnd a function g ∈ C (Ω) with |f − g|0 < 3 . By Sard’s lemma γ we can choose a regular valuey ˆ of g with |y − yˆ| < 3 . This gives us d(f, Ω, y) = d(g, Ω, y) = d(g, Ω, yˆ) X = sgn Jg(x) x∈g−1(ˆy) X X X = sgn Jg(x) + ... + sgn Jg(x) + sgn Jg(x) x∈g−1(ˆy) x∈g−1(ˆy) x∈g−1(ˆy) m x∈Ω1 x∈Ωm x∈Ω \(∪i=1 Ωi) X = d(g, Ω1, yˆ) + ... + d(g, Ωm, yˆ) + sgn Jg(x) x∈∅ m X = d(g, Ωi, y) i=1

2γ where we have used Definitions 4.1,4.2, and 4.3 and the fact that %(y, g(∂Ωi)) > 3 and so |y − yˆ| < γ γ 3 < %(y, g(∂Ωi)). As we also have |f − g|0 < 3 < %(y, g(∂Ωi)) we can reduce this further. Indeed, m m X X d(f, Ω, y) = d(g, Ωi, y) = d(f, Ωi, y) i=1 i=1 as required. Thus, we have shown (D2) is satisfied. n n Let h : [0, 1] × Ω → R and y : [0, 1] → R both be continuous with y(t) 6∈ h(t, ∂Ω) for each t ∈ [0, 1]. To show that (D3) is satisfied we will approximate the homotopy h by a sufficiently close C2 map and reduce to the regular case once again. In doing so we will show that d(h(t, ·), Ω, y(t)) is constant on a neighbourhood of each t ∈ [0, 1] and hence constant on [0, 1]. We begin by showing that the set S = {%(y(t), h(t, ∂Ω)) : t ∈ [0, 1]} has an infimum greater than zero. Since y(t) 6∈ h(t, ∂Ω) for all t ∈ [0, 1], the continuous function F (t, x) = h(t, x) − y(t) satisfies 0 6∈ F ([0, 1] × ∂Ω) =: X. By the continuity of F and the compactness of [0, 1] × ∂Ω we deduce that X is compact. Thus, the fact that 0 6∈ X implies that %(0,X) = δ0 > 0 and so α = inf(S) ≥ δ0 > 0. n From this point forward we fix t ∈ [0, 1]. Applying Proposition 4.2, let eh : [0, 1] × Ω → R α ˜ n satisfy |eh − h|0 < 10 with h twice continuously differentiable and choose a regular valuey ˆt ∈ R ˜ α of h(t, ·) satisfying |y(t) − yˆt| < 10 by using Lemma B.3 (Sard’s). By the continuity of eh and y we α α choose δ > 0 satisfying |eh(t0, ·) − eh(t, ·)|0 < 10 and |y(t0) − y(t)| < 10 for all |t0 − t| ≤ δ. Due to our conservative choices of approximation it is not hard to see that by applying Definition 4.3 and 4.2 for all such t0 we have

18 we obtain

d(eh(t, ·), Ω, yˆt) = d(eh(t, ·), Ω, y(t0)) = d(eh(t0, ·), Ω, y(t0)) = d(h(t0, ·), Ω, y(t0)). by Deﬁnitions 4.2 and 4.3 again and recalling that %(y(t0), h(t0, ∂Ω)) ≥ α. Therefore, d(h(t, ·), Ω, y(t)) = d(h(t0, ·), Ω, y(t0)) which means that d(h(t, ·), Ω, y(t)) is locally constant on [0, 1] and so by Lemma 4.7 it is constant on [0, 1]. Thus, we have shown that (D3) is satisﬁed.

5 Applications

The reader may be delighted to ﬁnd just how easy the following results are to prove using the degree.

5.1 Brouwer’s ﬁxed point theorem

Theorem 5.1. Let f : B1(0) → B1(0) be continuous. Then f has a ﬁxed point, that is there exists some x ∈ B1(0) with f(x) = x.

Proof. Firstly, if f(x) = x for some x ∈ ∂B1(0) we are done, so let us assume that this is not the case. We define F (x) = x − f(x) and note that f has a fixed point if and only if F (x) = 0 has a solution on B1(0). This immediately allows us to apply the degree. Indeed, if we can show that d(F,B1(0), 0) 6= 0 then, by applying (D5), we have shown that a fixed point exists. To achieve this we seek a homotopy from F to a simpler function, where the degree is known, and a reduction to id : B1(0) → B1(0) seems ideal. We define h(t, ·) = tF + (1 − t) id and note that for all x ∈ ∂B1(0) and t ∈ [0, 1) we have

|h(t, x) − 0| = |x − tf(x)| ≥ |x| − t|f(x)| = 1 − t|f(x)| ≥ 1 − t > 0.

Furthermore, for x ∈ ∂B1(0) and t = 1 we note that |h(t, x) − 0| = |x − f(x)| 6= 0 by assumption. Therefore, by (D3) and (D1) we have

d(F,B1(0), 0) = d(h(1, ·),B1(0), 0) = d(h(0, ·),B1(0), 0) = d(id,B1(0), 0) = 1.

This means that the equation F (x) = 0, or equivalently f(x) = x, has a solution on B1(0) by applying (D5). This completes the proof.

It should be noted that Theorem 5.1 can be extended to (any set that is homeomorphic to) a general convex compact set with little eﬀort. This was not done here because the details are purely technical and do not apply the degree any diﬀerently.

5.2 Perron-Frobenius Theorem

Theorem 5.2. Let A = [aij] be an n × n matrix such that aij ≥ 0 for all 1 ≤ i, j ≤ n. Then there exists a non-negative eigenvale λ and corresponding eigenvector x = (x1, . . . , xn) of A such that xi ≥ 0 for all 1 ≤ i ≤ n.

19 Proof. Let us define ( n ) n X D = x ∈ R : xi ≥ 0 for all i and xi = 1 . i=1 We note that for any x, y ∈ D we have n n n X X X [tx + (1 − t)y]i = t xi + (1 − t) yi = t + 1 − t = 1 i=1 i=1 i=1 Pn from which it follows that D is convex. Furthermore, as x 7→ i=1 xi is continuous, it is easy to see that any convergent sequence in D converges in D and so D is closed. Now, if Ax = 0 for some x ∈ D we are done, so let us assume the contrary. From the compactness of {Ax : x ∈ D} and the Pn fact that aij ≥ 0, it then follows that there exists an α > 0 such that i=1[Ax]i ≥ α for all x ∈ D. n n Pn −1 Let us define the normalising function f : R → R by f(x) = ( i=1[Ax]i) · Ax. Clearly, f is continuous by composition and f(D) ⊂ D. Therefore, by the extension of Theorem 5.1, f has a fixed point. That is

n !−1 n ! X X f(x) = x =⇒ [Ax]i · Ax =⇒ Ax = [Ax]i · x =⇒ Ax = λx. i=1 i=1 This completes the proof.

5.3 Surjective maps result n Proposition 5.3. Let f ∈ C(R ) with hf(x), xi /|x| → ∞ as |x| → ∞, where h·, ·i denotes the n n euclidean inner product. Then f : R → R is surjective. n Proof. Our approach here will be to show that for any y ∈ R there exists some r > 0 such that d(f, Br(0), y) 6= 0 using a homotopy to the identity map. This implies that f(x) = y has at least one n n solution on Br(0) which is sufficient. Indeed, let us fix an arbitrary y ∈ R and define h : [0, 1]×R by h(t, x) = tx + (1 − t)f(x) − y. We note that (1 − t) 1 hh(t, x), xi = t hx, xi + (1 − t) hf(x), xi − hy, xi = |x| · t|x| + hf(x), xi − hy, xi . |x| |x| Thus, using the growth property in the proposition, we choose r > |y| such that hf(x), xi /|x| > |y| for all |x| ≥ r and by application of the Cauchy-Schwartz inequality we obtain for all |x| = r (1 − t) 1 hh(t, x), xi = r · tr + hf(x), xi − hy, xi |x| |x| (1 − t) 1 ≥ r · tr + hf(x), xi − |y||x| |x| |x| (1 − t) = r · tr + hf(x), xi − |y| |x| > r · [t · |y| + (1 − t)|y| − |y|] = 0.

Therefore, for all |x| = r it follows that |h(t, x) − 0| > 0 and so d(f, Br(0), y) = d(id,Br(0), y) = 1 as required. This completes the proof.

20 6 Investigating continuous injective maps 6.1 Orientation preserving maps

Recall that any ordered basis of Rn has a particular orientation. In R orientation is the idea of 2 direction along the real line, in R it is the idea of clockwise and anti-clockwise when rotating from 3 the ﬁrst basis vector to the next and in R it is the idea of left and right handedness. In general, a basis B of Rn is said to be positively oriented if the transition matrix from B to the standard basis {e1, . . . , en} has positive determinant and negatively oriented otherwise. A rotation is the classic example of an orientation preserving linear map and a reﬂection an orientation reversing linear map. In general, a linear map A : Rn → Rn preserves orientation if det(A) > 0 and reverses it if det(A) < 0. As derivatives are linear maps we can consider whether or not a derivative f 0(x) is orientation preserving by inspecting the the sign of the Jacobian determinant Jf (x).

6.2 Preservation of orientation on C1 diffeomorphisms Now, let us consider an arbitrary C1 diffeomorphism f ∈ C1(Ω). This means that f is bijective, 3 differentiable and f −1 is differentiable. Intuitively, we can visualise such a map in R by considering the transformation of a malleable closed unit ball. Under such a, possibly non-linear, transformation the ball can be stretched and deformed in any way as long as we do not cut or crease it sharply. What insight can we gain into whether or not the derivative of such a map f at some point x ∈ Ω preserves orientation when Ω is connected? It is rather easy to see that if f 0(x) perseveres orientation for some x ∈ Ω, then for allx ˆ ∈ Ω, f 0(ˆx) preserves orientation. Indeed, as f −1 ◦ f = id, by differentiability we observe that det (f −1 ◦ f)0(x) = det (f −1)0(f(x)) ◦ f 0(x) = det (f −1)0(f(x)) · det [f 0(x)] = 1.

0 It then follows that det(f (x)) 6= 0 for all x ∈ Ω. Furthermore, as Ω is connected and Jf is continuous on Ω, Jf (Ω) ⊂ R must be connected and so I = Jf (Ω) is an interval. The fact that det(f 0(x)) > 0 and 0 6∈ I then implies that det(f 0(ˆx)) > 0 for anyx ˆ ∈ Ω as required. This is equivalent to saying that if f 0(x) preserves orientation for a single point x ∈ Ω, it must do so for all xˆ ∈ Ω. Similarly, the same reasoning tells us that if the derivative reverses orientation at one point, it must do so for all points in the domain. This result generalises to each connected component of Ω if Ω is not connected. Results like this are often of interest in analysis, as knowing that global behaviour is entirely determined locally can often simplify ones mathematical reasoning.

6.3 Generalising to continuous injective maps Under the strong assumption of a C1 diffeomorphism, it was very easy to deduce the global behavior of the orientation of the derivatives by inspecting local behavior as we just saw. This begs the question, can we do the same thing with weaker assumptions? What about if f is only continuous and injective? The previous approach will not suffice here as we may have det f 0(x) = 0 for some x ∈ Ω or it may not even be defined. Furthermore, there is nothing to suggest that Jf need be continuous or Jf (Ω) a connected set. Indeed, on first glance it does not appear that the local behavior of the derivatives of f should influence the global behavior and this is where the power of the degree really becomes obvious. We find that, surprisingly, we can deduce the same global behavior of the derivatives of a continuous injective map defined on a connected set with only the same local knowledge.

21 n Proposition 6.1. Let Ω ⊂ R be open, connected and bounded and f ∈ C(Ω) injective. Then 0 (a) For all a ∈ Ω such that f is diﬀerentiable at a and f (a) is invertible we have sgn Jf (a) = d(f, Ω, f(a))

0 (b) sgn Jf is constant on Ωf = {x : f is diﬀerentiable at x and det(f (x)) 6= 0} Part (a). The proof here is very similar to our methods in Section 4.1.5 and our goal is to construct a homotopy from f to a linear map where we can evaluate the degree by elementary means after applying (D3). The linear map we will use is the obvious choice of f 0(a) and we proceed as follows. 0 As f is diﬀerentiable at a we have f(x) = f(a) + f (a)(x − a) + r(x − a) with limh→0 |r(h)|/|h| = 0. n Furthermore, as f 0(a) is invertible we can deduce that |f 0(a)z| ≥ c|z| for some c > 0 for all z ∈ R .

From diﬀerentiability, let us choose δ0 > 0 such that c|x − a| > |r(x − a)| for all x ∈ ∂Bδ0 (a) and 1 let δ = min{δ0, 2 %(a, ∂Ω)}. Note that as a 6∈ ∂Ω and we have δ > 0. As f is injective we have f(x) 6= f(a) for all x ∈ Ω \ Bδ(a) and so by (D4) we deduce that d(f, Ω, f(a)) = d(f, Bδ(a), f(a)). n 0 n Let us deﬁne h : [0, 1] × Bδ(a) → R by h(t, x) = tf(x) + (1 − t)f (a)(x − a) and γ : [0, 1] → R by γ(t) = tf(a). We note that for all x ∈ ∂Bδ(a) and all t ∈ [0, 1] we have |h(t, x) − γ(t)| = |tf(x) + (1 − t)f 0(a)(x − a) − tf(a)| = |t(f(a) + f 0(a)(x − a) + r(x − a)) + (1 − t)f 0(a)(x − a) − tf(a)| = |f 0(a)(x − a) + t · r(x − a)| ≥ |f 0(a)(x − a)| − t|r(x − a)| ≥ |f 0(a)(x − a)| − |r(x − a)| ≥ c|x − a| − |r(x − a)| > 0.

0 0 This allows us to conclude that d(f, Bδ(a), f(a)) = d(f (a) − f (a)a, Bδ(a), 0) by (D3). The fact that det(f 0(a)) 6= 0 implies that x = a is the only solution to f 0(a)x − f 0(a)a = 0 and so (f 0(a)−f 0(a)a)−1(f(a)) = {a}. As linear functions are smooth and all values of an invertible linear transformation are regular we have 0 0 0 d(f, Ω, f(a)) = d(f (a) − f (a)a, Bδ(a), 0) = sgn det(f (a)) = sgn Jf (a). This completes the proof. Part (b). As f is continuous and Ω is connected it follows that f(Ω) is connected. From (D3) we also observe that d(f, Ω, ·) is locally constant and so by applying Lemma 4.7 we deduce that 0 d(f, Ω, ·) is constant on f(Ω). Thus, for all x, x ∈ Ωf we have 0 0 sgn Jf (x) = d(f, Ω, f(x)) = d(f, Ω, f(x )) = sgn Jf (x ). This completes the proof. What we have confirmed in the previous proposition is that once we know that a single invertible derivative f 0(x), for some x ∈ Ω, preserves orientation, we know that all invertible derivatives of f on Ω must preserve orientation regardless of how complicated the derivatives may be. Thus, using the degree we have deduced a subtle layer of structure of any continuous, injective map f ∈ C(Ω) defined on a connected set. As a final note, we observe that from our assumptions of continuity and injectivity on the compact set Ω, we have a sufficient condition for f −1 being continuous and so f is actually a homeomorphism in this case.

22 A Symbols

• |·| (Euclidean norm)

• |·|0 (maximum norm) • h·, ·i (Euclidean inner product)

• Br(x) (open ball of radius r centered at x)

• Br(x) (closure of Br(x))

n n •C (Ω) (set of all continuous functions f : R → R restricted to Ω) n n •C k(Ω) (set of all k times continuously diﬀerentiable functions f : R → R restricted to Ω) n n •C ∞(Ω) (set of all smooth functions f : R → R restricted to Ω) • d(·, ·, ·) (topological degree)

∂fi(x) • ∂jfi(x) = (partial derivative of the ith component of f with respect to the jth compo- ∂xj nent of x) • f 0(·) (derivative of f) • id (identity map)

• Jf (·) (Jacobian determinant of f)

n n n • L(R , R ) (space of linear operators on R ) n • Ω (open, bounded subset of R ) • Ω (closure of Ω)

• R (set of all read numbers) n • R (n-dimensional Euclidean space) • %(·, ·) (distance from a point to a set)

• sgn (sign function)

23 B Useful results

n Lemma B.1. If A ⊂ R is compact, then there exists an at most countable subset {a1, a2,...} ⊂ A that is dense in A.

Proof. Consider the open cover of A given by Cn = {Br(x): r = 1/n, x ∈ A}. As A is compact, for each n there exists an integer m and a set Pn = {xn1 , xn2 , . . . , xnm } ⊂ A such that Cfn =

{Br(xn1 ),Br(xn2 ),...,Br(xnm ): r = 1/n} is a finite subcover of A. Now, if we take S = P1 ∪ P2 ∪ ..., we have a countable union of finite sets and so S is at most countable. Clearly S ⊂ A so all that remains is to show that S is dense in A. Indeed, for any x ∈ A \ S let Bα(x) be an arbitrary neighbourhood (open ball) of x. Clearly some element of Pdαe ⊂ S must be in Bα(x) by definition and so x is a limit point of S. This completes the proof.

n n Proposition B.2. Let A ⊂ R be compact and f : A → R be a continuous. Then there exists a n n continuous function f˜ : R → R such that f˜(x) = f(x) on for all x ∈ A. We call f˜ a continuous n extension of f on R .

Proof. By Lemma B.1 there exists an at most countable set S = {a1, a2,...} ⊂ A that is dense in A. Let us define the distance %(x, A) = inf{|x − a| : a ∈ A} and |x − a | ϕ (x) = max 2 − i , 0 for x 6∈ A. i %(x, A) Then, we will show that the function given by  f(x), x ∈ A f˜ = −1 P −i P −i  i≥1 2 ϕi(x) i≥1 2 ϕi(x)f(ai) , x 6∈ A satisfies the requirements. To see why the denominator in f˜ is non-zero consider the following. As n S is dense in A, for any > 0 and x ∈ R \ A there existsa ˜ ∈ S such that |x − a˜| ≤ %(x, A) + and %(x,A) so setting = 2 we find that for some i ∈ N |x − a | |x − a˜| (3/2) %(x, A) 1 2 − i = 2 − ≥ 2 − = > 0. %(x, A) %(x, A) %(x, A) 2

n Therefore, it follows that for each x ∈ R \ A, the denominator in f˜ is non-zero as stated. We note −i 1−i P 1 n that for each i ∈ N we have |2 ϕi(x)| ≤ 2 for all x ∈ A. As the series n≥0( 2 ) converges, by P −i the Weierstrass M-test, the series of functions i≥1 2 ϕi(x) converges uniformly. Furthermore, as n each term in the series is a continuous function on R \ A the series must converge to a continuous n n function on R \ A. Therefore, on R \ A, the ﬁrst factor in the deﬁnition of f˜ is continuous and similarly the second factor is too. We have used the fact that we can similarly bound the second factor due to the fact that A is compact and so {f(ai): ai ∈ S} ⊂ f(A) is bounded. Therefore, it n follows that f˜ is continuous on R \ ∂A. All that remains is to show that f˜ is continuous on ∂A. Indeed, for γ ∈ ∂A ⊂ A, by the continuity of f on A there exists δ > 0 such that for all a ∈ A ∗ δ satisfying |a − γ| ≤ δ we have |f(a) − f(γ)| ≤ . Furthermore, for any 0 < δ < 3 , x ∈ Bδ∗ (γ) \ A and ai ∈ A \ Bδ(γ) we have |x − a | δ − δ∗ (2/3)δ 2 − i ≤ 2 − < 2 − = 0. %(x, A) δ∗ (1/3)δ

24 Therefore, for all x ∈ Bδ∗ (γ) \ A we have

P −i 2 ϕi(x)f(ai) |f˜(x) − f˜(γ)| = i≥1 − f(γ) P −i i≥1 2 ϕi(x) P −i P −i 2 ϕi(x)f(ai) − f(γ) 2 ϕi(x) = i≥1 i≥1 P −i i≥1 2 ϕi(x) P −i 2 ϕi(x)[f(ai) − f(γ)] = i≥1 P −i i≥1 2 ϕi(x) P −i 2 ϕi(x) ≤ i≥1 · P −i i≥1 2 ϕi(x) = and so we can conclude that f˜ is continuous on ∂A as we wanted to show. This completes the proof.

n Lemma B.3 (Sard’s lemma). Let Ω ⊂ R be open and f ∈ C1(Ω), then the regular values of f are dense in the range of f. Lemma B.3 is a special case of Sard’s Lemma which states that the set of singular values of f has zero measure. For a proof see [Dei85, Chap. 1].

n n Theorem B.4 (Inverse function Theorem). Let U ⊂ R be open and F : U → R with F ∈ Ck(U). If DF (a) is invertible at a ∈ U then there exists neighbourhoods U0 of a and V0 of f(a), each k connected, such that F|U0 : U0 → V0 is a C diffeomorphism. n Theorem B.5 (Implicit function Theorem). Let h : R × Ω → R be a continuously differentiable function with h(t0, x0) = 0 and Jh(t0,·) 6= 0 for some (t0, x0) ∈ R × Ω. Then there exists an interval I = (t0 − r, t0 + r), an open ball Bδ(x0) ⊂ Ω and a continuous function f : I → Bδ(x0) such that −1 f(t0) = x0 and h (0) ∩ (I × Bδ(x0)) is the graph of f. That is for (t, x) ∈ I × Bδ(x0) we have h(x, t) = 0 if and only if f(t) = x. For a proof of Theorem B.4 and Theorem B.5 see [Lee13, Appendix C]. Lemma B.6. If we define the diagonal n × n matrices

1 0 0 ... 0 −1 0 0 ... 0 0 1 0 ... 0  0 1 0 ... 0     0 0 1 ... 0 ˜  0 0 1 ... 0 I =   and I =   . . . ..   . . . ..  . . . . 0  . . . . 0 0 0 0 ... 1 0 0 0 ... 1 and A is an n × n matrix satisfying det(A) > 0 (or det(A) < 0 respectively) then there exists a n×n continuous function f : [0, 1] → R such that f(0) = A and f(1) = I (or f(1) = I˜ respectively) with det(f(t)) 6= 0 for all t ∈ [0, 1].

25 Proof. We show the case for det(A) > 0, the case for det(A) < 0 is similar. The approach taken here will be rather informal and we will explain the construction of such an f in words. We begin by noting that the adding of one row to another in A does not affect the value of the determinant and that this operation can be done in a continuous fashion while maintaining a constant determinant. n×n n×n For example, if 1 ≤ k ≤ n and 1 ≤ l ≤ n with k 6= l, then g : [0, 1] × R → R defined by ( Aij + tAkj, if i = l g(t, Aij) = Aij, otherwise is an explicit illustration of how this can be done continuously. Therefore, from this point forward all row operations of this form will be taken to be performed continuously. Now, as det(A) > 0, in the last column of A there is a non-zero entry in some row. Let us add a multiple of this particular row to the last row so that the last entry in the main diagonal is non-zero. Following this, we add a sufficient multiple of the final row to every other row to introduce zeroes to every other entry in the last column. We then repeat this process column-wise. What we then have is a matrix of the form  0   .   .  A1 =  B   0  0 ... 0 α for some n − 1 × n − 1 matrix B and α 6= 0 satisfying det(A1) = det(A). By then repeating the process on the submatrix B and its subsequent submatrices we will eventually obtain a diagonal matrix   λ1  λ2 0  D =    ..   .  0 λn n with det(D) = det(A). Now, as det(D) = Πi=1λi > 0, the negative λi must occur in pairs. We note that for any such pair λi, λj we can perform the following row and column operations continuously without changing the value of the determinant throughout:

Rj = Rj + Ri,Ri = Ri − Rj,Rj = Rj + Ri,Ci = Ci + Cj,Cj = Cj − Ci,Ci = Ci + Cj. If we let D˜ be the matrix that results from applying all of these operations, in the listed order to D we can easily check that D˜ ii = −λj and D˜ jj = −λi. Therefore, by repeating this process on all negative pairs we obtain a diagonal matrix Db with positive diagonal entries that satisfies det(Db) = det(A). We can then multiply each row k of Db by a suitable scalar γk > 0 and obtain the matrix I. Note that such row operations can also be performed in a continuous fashion without n×n n×n changing the sign of the determinant. For example, if 1 ≤ k ≤ n then g : [0, 1] × R → R defined by ( [tγk + (1 − t)]Aij, if i = l g(t, Aij) = Aij, otherwise gives the continuous approach to multiplying row k by the scalar γk and satisfies det(g(t, Db)) > 0 for all t ∈ [0, 1]. Therefore, by putting all of the previous continuous operations together, we can define a piecewise continuous function f that satisfies our requirements. This completes the proof.

26 n 2 Lemma B.7. Let Ω be an open subset of R , f ∈ C (Ω) and dij(x) the cofactor of ∂fj(x)/ ∂xi in Jf (x). Then n X ∂dij(x) = 0 for j = 1, 2, . . . , n ∂x i=1 i

2 n Proof. We present an example of the R case only. For a complete proof in R see [Dei85, chap. 1]. T Note that dij(x) = [C ]ij(x), where C is the matrix of cofactors of Jf , and so     ∂f1 ∂f ∂f2 ∂f 1 − 1  ∂x1 ∂x2  T  ∂x2 ∂x2  Jf =   =⇒ C =    ∂f2 ∂f   ∂f2 ∂f   2  − 1  ∂x1 ∂x2 ∂x1 ∂x1 from which we obtain

2 2  ∂ f2 ∂ f2 n 2 T  − , for j = 1 X ∂dij X ∂[C ]ij ∂x ∂x ∂x ∂x = = 2 1 1 2 = 0 2 2 xi xi  ∂ f1 ∂ f1 i=1 i=1  − , for j = 2 ∂x2 ∂x1 ∂x1 ∂x2 where we have used the fact that f ∈ C2(Ω) and applied Clairaut’s theorem. This completes the demonstration of the 2 × 2 case.

27 References

[Dei85] K. Deimling, Nonlinear Functional Analysis. Springer-Verlag, Berlin, 1985. [HK71] K. Hoﬀman, R. Kunze, Linear Algebra. Prentice-Hall, New Jersey, 2nd edition, 1971. [Lee13] J. M. Lee, Introduction to Smooth Manifolds. Springer, New York, 2nd edition, 2013. [Mun00] J. R. Munkres, Topology. Prentice-Hall, New Jersey, 2nd edition, 2000.

[OCC06] D. O’Regan, Y. J. Cho, Y. Chen, Topological degree theory and applications. Chapman and Hall/CRC, 2006. [Rud76] W. Rudin, Principles of Mathematical Analysis. McGraw-Hill, 3rd edition, 1976.

Acknowledgements

I would like to thank my supervisor Dr. J´erˆomeDroniou for his tireless assistance and guidance as well as AMSI for the generous funding I received to undertake this research project. I would also like to mention Hagen von Eitzen of http://math.stackexchange.com/q/315356 for the idea of the rather elementary proof for Lemma B.6.