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Abstract and Applied Analysis
Volume 2012, Article ID 370946, 18 pages
http://dx.doi.org/10.1155/2012/370946
Research Article

A Note on Coincidence Degree Theory

Department of Mathematics, Yüzüncü Yıl University, 65080 Van, Turkey

Received 13 March 2012; Accepted 17 June 2012

Academic Editor: Yong Zhou

Copyright © 2012 Ali Sırma and Sebaheddin Ṣevgin. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

The background of definition of coincidence degree is explained, and some of its basic properties are given.

1. Introduction

Gaines and Mawhin introduced coincidence degree theory in 1970s in analyzing functional and differential equations [1, 2]. Mawhin has continued studies on this theory later on and has made so important contributions on this subject since then this theory is also known as Mahwin's coincidence degree theory. Coincidence theory is very powerful technique especially in existence of solutions problems in nonlinear equations. It has especially so broad applications in the existence of periodic solutions of nonlinear differential equations so that many researchers have used it for their investigations (see [332] and references therein). The main goal in the coincidence degree theory is to search the existence of a solutions of the operator equation 𝐿𝑥=𝑁𝑥(1.1) in some bounded and open set Ω in some Banach space for 𝐿 being a linear operator and 𝑁 nonlinear operator using Leray-Schauder degree theory. As it is known that, in finite dimensional case, for Ω𝑛, 𝑓𝐶(Ω), and  𝑝𝑛𝑓(𝜕Ω), the degree of 𝑓 on Ω with respect to 𝑝, 𝑑(𝑓,Ω,𝑝) is well defined. But unfortunately this is not the case in infinite dimension for 𝑓𝐶(Ω) (see [33], page 172). Luckily, in an arbitrary Banach space 𝑋, Leray and Schauder proved that for Ω𝑋 open, bounded set, 𝑀Ω𝑋 compact operator and for  𝑝𝑋(𝐼𝑀)(𝜕Ω) the degree of compact perturbation of identity 𝐼𝑀 in Ω with respect to 𝑝, 𝑑(𝐼𝑀,Ω,𝑝) is well defined [34]. One of the main useful properties of degree theory is that if 𝑑(𝐼𝑀,Ω,𝑝)0 then (𝐼𝑀)𝑥=𝑝 has at least one solution in Ω. In particular if we take 𝑝=0 and 𝑑(𝐼𝑀,Ω,𝑝)0 then the compact operator 𝑀 has at least one fixed point in Ω. In [1], Gaines and Mawhin studied existence of a solution of an operator equation (1.1) defined on a Banach space 𝑋 in an open bounded set Ω using the Leray-Schauder degree theory. But since the operator 𝐼(𝐿𝑁) is not compact in general the need to define a compact operator 𝑀 such that its set of fixed points in Ω would be equal to a solution set of (1.1) in Ω aroused. In [1], the compact operator 𝑀 is given and the coincidence degree for the couple (𝐿,𝑁) in Ω is defined by 𝑑[(𝐿,𝑁),Ω]=𝑑(𝐼𝑀,Ω,0).

The aim of this paper is to make an effort to understand the theoretical background of the definition of coincidence degree which has similar properties with the Leray-Schauder degree for an operator couple (𝐿,𝑁) satisfying some special conditions, to analyze the dependence of coincidence degree to the components of the compact operator 𝑀 and in this way to prepare good resource for one who wants to study and to improve the coincidence degree theory.

The paper is basically prepared using [1]. In this study, we tried to explain the theory that was given densely in [1]. Besides we give proofs of some results that their proofs not given in [1]. Namely, we give proofs of Lemmas 2.1, 2.2, and 3.19 and Theorems 3.3, 4.1, and 4.2. We state and prove Lemma 3.17 which is essential for Proposition 3.18. In Proposition 3.6 we show that the operator Π𝑄 is an isomorphism and explain important details, and, in Proposition 3.20, we show that 𝐴 is an automorphism and make necessary explanations. Also in each proof we tried to make important contributions to make the proofs much more understandable and so that it can be improved by interested researchers.

In summary, in Section 2, some preliminaries which are used in the definition of coincidence degree are used. In Section 3, definition of coincidence degree for some linear perturbations of Fredholm mappings on normed spaces is given. In Section 4, some basic properties of coincidence degree are given.

2. Algebraic Preliminaries

In this section, we will give some facts that will be used throughout the paper.

Let 𝑋 and 𝑍 be two vector spaces, the domain of operator 𝐿, Dom𝐿 is a linear subspace of 𝑋, and 𝐿Dom𝐿𝑍 is a linear operator. Assume that the operators 𝑃𝑋𝑋 and 𝑄𝑍𝑍 linear projection operators such that the chain 𝑋𝑃Dom𝐿𝐿𝑍𝑄𝑍(2.1) is exact, that is, Im𝑃=ker𝐿 and Im𝐿=ker𝑄. Let us define the restriction of 𝐿 to Dom𝐿ker𝑃 as 𝐿𝑃Dom𝐿ker𝑃Im𝐿.

Now, let us give the following lemma about 𝐿𝑃.

Lemma 2.1. 𝐿𝑃 is an algebraic isomorphism.

Proof. Firstly, let us show that 𝐿𝑃 is one-to-one mapping. For this let us take 𝑥ker𝐿𝑃ker𝐿=Im𝑃, so that there exists 𝑦Dom𝑃 such that 𝑥=𝑃𝑦. Since 𝑃 is a projection operator we get 𝑥=𝑃𝑦=𝑃2𝑦=𝑃(𝑃𝑦)=𝑃𝑥=0. Therefore, 𝑥=0, so that we obtain that ker𝐿𝑃={0}. This means that 𝐿𝑃 is one-to-one.
Now let us show that 𝐿𝑃 is onto. Since 𝑃𝑋𝑋 is a projection operator, we can write the vector space 𝑋 as direct sums 𝑋=ker𝑃Im𝑃. From the exactness of the chain above, we get 𝑋=ker𝑃ker𝐿. Take 𝑧Im𝐿, so that there exists 𝑥Dom𝐿𝑋 with 𝐿𝑥=𝑧. Since 𝑋=ker𝑃ker𝐿, there exists unique elements 𝑒ker𝑃 and 𝑓ker𝐿 such that we can write 𝑥=𝑒+𝑓. From here, we can obtain 𝑧=𝐿𝑥=𝐿(𝑒+𝑓)=𝐿𝑒+𝐿𝑓=𝐿𝑒+0=𝐿𝑒. This means that 𝑒Dom𝐿. So we get 𝑒Dom𝐿 and 𝑒ker𝑃 and 𝐿𝑃𝑒=𝑧. So the result follows.

Now, let us define 𝐾𝑃=𝐿𝑃1. It is clear that 𝐾𝑃Im𝐿𝑍Dom𝐿ker𝑃 is one-to-one, onto, and 𝑃𝐾𝑃=0.

Lemma 2.2. (1) On Im𝐿, we have 𝐿𝐾𝑃=𝐼. (2) On Dom𝐿, we have 𝐾𝑃𝐿=𝐼𝑃.

Proof. (1) Take 𝑥Im𝐿. Therefore, 𝐿𝐾𝑃𝑥=𝐿(𝐾𝑃(𝑥))=𝐿𝑃(𝐾𝑃(𝑥))=𝐼𝑥.
(2) Since Im𝑃=ker𝐿, then we have 𝐿𝑃=0, so we obtain 𝐾𝑃𝐿=𝐾𝑃𝐿(𝐼𝑃). So that in order to prove (2), we need to show the equality 𝐾𝑃𝐿(𝐼𝑃)=𝐾𝑃𝐿𝑃(𝐼𝑃). If we can have Im(𝐼𝑃)Dom(𝐿𝑃)=Dom𝐿ker𝑃, then the result follows. Take 𝑥Dom𝐿. Since 𝑃(𝑥)ker𝐿Dom𝐿 and Dom𝐿 is a vector subspace of 𝑋, we have (𝑥𝑃𝑥)Dom𝐿.
Since 𝑃(𝑥𝑃𝑥)=𝑃𝑥𝑃2𝑥=𝑃𝑥𝑃𝑥=0, then (𝑥𝑃𝑥)ker𝑃; therefore, we have (𝑥𝑃𝑥)Dom𝐿ker𝑃. From here, we obtain Im(𝐼𝑃)Dom𝐿ker𝑃. So using (1), the result 𝐾𝑃𝐿(𝐼𝑃)=𝐾𝑃𝐿𝑃(𝐼𝑃)=𝐼𝑃 follows.

Now, let us define the canonic surjection Π as Π𝑍Coker𝐿𝑧𝑧+Im𝐿.(2.2) Here, Coker𝐿=𝑍/Im𝐿 is the quotient space of 𝑍 under the equivalence relation 𝑧𝑧𝑧𝑧Im𝐿. Thus, Coker𝐿={𝑧=𝑧+Im𝐿𝑧𝑍}. It is clear that the canonic surjection operator Π is linear and kerΠ=ker𝑄.

Proposition 2.3. If there exists an one-to-one operator ΛCoker𝐿ker𝐿, then 𝐿𝑥=𝑦,𝑦𝑍(2.3) will be equivalent to (𝐼𝑃)𝑥=ΛΠ+𝐾𝑃,𝑄𝑦.(2.4) Here, the operator 𝐾𝑃,𝑄𝑍𝑋 is defined as 𝐾𝑃,𝑄=𝐾𝑃(𝐼𝑄).

Proof. Since Im𝐿=ker𝑄=kerΠ, then for 𝑦Im𝐿 we have 𝑄𝑦=0 and ΛΠ𝑦=ΛIm𝐿=Λ0=0. From here, it is seen that 𝐿𝑥=𝑦𝐿𝑥=𝑦𝑄𝑦𝐾𝑃𝐿𝑥=𝐾𝑃(𝑦𝑄y)(𝐼𝑃)𝑥=𝐾𝑃(𝐼𝑄)𝑦(𝐼𝑃)𝑥=ΛΠ+𝐾𝑃(𝐼𝑄)𝑦.(2.5)

Now, let us consider another projection operator couple (𝑃,𝑄) that will make the chain 𝑋𝑃Dom𝐿𝐿𝑍𝑄𝑍(2.6) exact, and let us search the relation of this operator couple with (𝑃,𝑄).

From Lemma 2.2, since 𝐿𝐾𝑃=𝐼, and 𝐿𝐾𝑃=𝐼 then we have 𝐿(𝐾𝑃𝐾𝑃)=0. So for any 𝑧Im𝐿, we have (𝐾𝑃𝐾𝑃)𝑧ker𝐿. Therefore, we can write 𝐾𝑃𝐾𝑃Im𝐿ker𝐿=Im𝑃=Im𝑃. Since the projection operator 𝑃 behaves on Im𝑃 as an identity operator, we have 𝐾𝑃𝐾𝑃=𝑃(𝐾𝑃𝐾𝑃). As a result, the equality 𝐾𝑃𝐾𝑃𝐾=𝑃𝑃𝐾𝑃=𝑃𝐾𝑃𝐾𝑃(2.7) follows.

Lemma 2.4. The following relations hold.

(i) 𝑃𝐾𝑃+𝑃𝐾𝑃=0,   (ii) 𝐾𝑃=(𝐼𝑃)𝐾𝑃.

Proof. (i) Using 𝑃𝐾𝑃=0, 𝑃𝐾𝑃=0, and (2.7), the result 𝑃𝐾𝑝𝐾𝑃=𝑃𝐾𝑃𝐾𝑃𝑃𝐾𝑃=𝑃𝐾𝑃(2.8) follows.
(ii) Again using (2.7) and (i), we obtain 𝐾𝑃𝐾𝑃=𝑃𝐾𝑃𝐾𝑃𝐾𝑃𝐾𝑃=𝑃𝐾𝑃𝑃𝐾𝑃𝐾𝑃=𝑃𝐾𝑃+𝑃𝐾𝑃𝐾𝑃=𝐼𝑃𝐾𝑃.(2.9) In a similar manner, the equality 𝐾𝑃=(𝐼𝑃𝐾𝑃) can be obtained.

3. Definition of Coincidence Degree for Some Linear Perturbations of Fredholm Mappings on Normed Spaces

In this section, definition of coincidence degree for some linear perturbations of Fredholm mappings on normed spaces is given.

Let 𝑋 and 𝑍 be two real norm spaces, Ω𝑋 an open, bounded subset of 𝑋 and Ω an closure of Ω. Let us assume that the operators 𝐿Dom𝐿𝑋𝑍,𝑁Ω𝑋𝑍(3.1) satisfy the following conditions:(i)𝐿is linear and Im𝐿 is an closed subset of 𝑍,(ii)ker𝐿 and Coker𝐿=𝑍/Im𝐿 are finite dimensional spaces and dimker𝐿=dimCoker𝐿,(iii)the operator 𝑁Ω𝑋𝑍 is continuous and Π𝑁(Ω) is bounded,(iv)the operator 𝐾𝑃,𝑄𝑁Ω𝑍 is compact on Ω.

Definition 3.1. The operator 𝐿 which satisfies the conditions (i) and (ii) will be called as Fredholm operator of index zero.

Definition 3.2. The operator 𝑁Ω𝑍 which satisfies the conditions (iii) and (iv) will be called 𝐿-compact operator.

It is clear that if we take 𝑋=𝑍 and 𝐿=𝐼 the operator Π reduced to zero operator and the operator 𝐾𝑃,𝑄 turns to an identity operator then 𝐿-compactness of 𝑁 on Ω reduced to usual compactness for operators.

Theorem 3.3. Let 𝑍 be a Banach space. If the operator 𝐿 is a Fredholm operator of index zero then there exist continuous projections 𝑃𝑋𝑋 and 𝑄𝑍𝑍 such that the chain 𝑋𝑃Dom𝐿𝐿𝑍𝑄𝑍(3.2) will be exact.

Proof. Assume that ker𝐿 is finite dimensional, and the set {𝑦1,𝑦2,,𝑦𝑛} is a basis for ker𝐿. Define the vector subspaces as 𝑋𝑘=span{𝑦1,𝑦2,,𝑦𝑘1,𝑦k+1,,𝑦𝑛}. Since 𝑋𝑘 is finite dimensional so is a closed subspace of 𝑋 (see [35, Theorem  2.4-3] and 𝑦𝑘𝑋𝑘. Let 𝐵 be a basis of 𝑋 such that {𝑦1,𝑦2,,𝑦𝑛}𝐵. Now, let us define the linear operators which satisfy the conditions 𝐹𝑘𝐹𝑋,𝑘(𝑦)=1,if𝑦=𝑦𝑘𝑦0,if𝑦𝐵𝑘.(3.3) Therefore, the operator 𝑃 defined by 𝑃𝑥=𝑛𝑘=1𝐹𝑘(𝑥)𝑦𝑘(3.4) is a continuous projection operator (see [36, Remark  2.1.19]).
Now, let us prove the existence of continuous projection operators 𝑄 on 𝑍 with ker𝑄=Im𝐿. We know that there exists a subspace 𝑍 such that 𝑍𝑍=Im𝐿 (see [37, Proposition  I]). The projection operator 𝑄 defined on 𝑍 with the rule 𝑄𝑧(𝑧)=𝑄ker𝑄+𝑧𝑍=𝑧𝑍(3.5) satisfies the relations ker𝑄=Im𝐿 and 𝑍Im𝑄=. Since dimCoker𝐿 is finite dimensional so is Im𝑄, therefore it is closed in 𝑍. Since 𝑍 is a Banach space, ker𝑄 and Im𝑄 are closed subsets of 𝑍, therefore the projection operator 𝑄 is continuous (see [38, Theorem  6.12.6]).

Moreover, the canonical surjection Π𝑍Coker𝐿 is continuous with the quotient topology on Coker𝐿. Now, let us state two theorems that will be used in the proof of following proposition.

Theorem 3.4 (see [35]). Assume 𝑋 and 𝑌 are normed spaces and the operator 𝑇𝑋𝑌 is linear. Therefore,(a)If 𝑇 is bounded and dim(Im𝑇)< then 𝑇 is compact.(b)If dim𝑋< then 𝑇 is continuous and compact.

Theorem 3.5 (see [35], Lemma  8.3-2). Let 𝑋 be normed space, 𝑇𝑋𝑋 be a linear compact operator, and 𝑆𝑋𝑋 a linear bounded (continuous) operator. So the operators 𝑇𝑆 and 𝑆𝑇 are also compact.

The following proposition states that the condition (iv) does not depend on the choice of the projection operators 𝑃 and 𝑄.

Proposition 3.6. Assume that the conditions (i), (ii), (iii) are all satisfied. If the condition (iv) is satisfied for the projection operator couple (𝑃,𝑄) that makes the chain exact then for any projection operator couple (𝑃,𝑄) that makes the chain exact is satisfied.

Proof. Let us denote the restriction of Π to Im𝑄 with Π𝑄, and let us show that the linear operator Π𝑄Im𝑄𝑍/𝐼𝑚𝐿 is one-to-one and onto. Since ker(Π𝑄)Im𝐿=ker𝑄 and ker𝑄Im𝑄={0}, then we have ker(Π𝑄)={0}. Therefore Π𝑄 is one-to-one. To show surjection, let us take an arbitrary element 𝑧𝑍/Im𝐿. So there exists 𝑧𝑍 such that Π𝑧=𝑧 holds. Since the space 𝑍 can be written as 𝑍=Im𝑄ker𝑄=Im𝑄Im𝐿 there exist unique elements 𝑧Im𝑄Im𝑄 and 𝑧Im𝐿Im𝐿 such that the relation 𝑧=𝑧Im𝑄+𝑧Im𝐿 is satisfied. Since we have 𝑧𝑧=ΠIm𝑄+𝑧Im𝐿𝑧=ΠIm𝑄𝑧+ΠIm𝐿𝑧=ΠIm𝑄,(3.6) then the surjectivity of Π𝑄 follows.
Since we have dim(Im𝑄)dim(ImΠ𝑄)dim(Coker𝐿)=𝑛<, then Im𝑄 is a finite dimensional linear subspace of 𝑍. Similarly, Im𝑄 is also a finite dimensional subspace of 𝑍. Therefore, since we have dim(Im(𝑄𝑄))dim(Im𝑄)+dim(Im𝑄), then Im(𝑄𝑄) is also a finite dimensional subspace of 𝑍.
Now, let us show that for an arbitrary 𝛼𝑍 the relation Π𝑄1Π(𝛼)=Q(𝛼) holds. Since we can write 𝛼=𝑄𝛼+(𝐼𝑄)𝛼), then Π𝑄1Π(𝛼)=Π𝑄1Π((𝑄𝛼+(𝐼𝑄)𝛼))=Π𝑄1Π(𝑄𝛼)+Π𝑄1Π((𝐼𝑄)𝛼)=Π𝑄1Π(𝑄𝛼)=Π𝑄1Π𝑄(𝑄𝛼)=𝑄𝛼(3.7) is obtained.
Let 𝐾𝑃 denote the restriction of the operator 𝐾𝑃 to the finite dimensional space Im(𝑄𝑄). Using the results obtained until here in this proof and using the equality 𝐾𝑃=(𝐼𝑃)𝐾𝑃, 𝐾𝑃,𝑄𝑁=𝐾𝑃𝐼𝑄𝑁=𝐼𝑃𝐾𝑃𝐼𝑄𝑁=𝐼𝑃𝐾𝑃𝐼𝑄+𝑄+𝑄𝑁=𝐼𝑃𝐾𝑃(𝐼𝑄)𝑁+𝐼𝑃𝐾𝑃𝑄𝑄𝑁=𝐼𝑃𝐾𝑃,𝑄𝑁+𝐼𝑃𝐾𝑃𝑄𝑄𝑁=𝐼𝑃𝐾𝑃,𝑄𝑁+𝐼𝑃𝐾𝑃𝑄𝑄𝑁=𝐼𝑃𝐾𝑃,𝑄𝑁+𝐼𝑃𝐾𝑃Π𝑄1ΠΠ𝑄1Π𝑁=𝐼𝑃𝐾𝑃,𝑄𝑁+𝐼𝑃𝐾𝑃Π𝑄1Π𝑄1Π𝑁(3.8) is achieved. Now, let us explain the operator 𝐾𝑃,𝑄𝑁 is compact. Since the operator 𝐾𝑃,𝑄𝑁 is compact and 𝐼𝑃 is continuous, then the operator (𝐼𝑃)𝐾𝑃,𝑄𝑁 is compact. Since dim(Coker𝐿)=𝑛<, then the operator Π𝑄1Coker𝐿Im𝑄 is compact. From the same reason, the operator Π𝑄1 is also compact. Since the operators 𝑁, Π, and 𝐼𝑃 are all continuous, the compactness of 𝐾𝑃,𝑄𝑁 follows.

Proposition 3.7. The element 𝑥Dom𝐿Ω is a solution of the operator equation (1.1) if and only if it satisfies (𝐼𝑃)𝑥=ΛΠ+𝐾𝑃,𝑄𝑁𝑥.(3.9) In other words, the set of solutions of (1.1) is equal to the set of fixed points of the operator 𝑀Ω𝑋 defined by 𝑀=𝑃+ΛΠ+𝐾𝑃,𝑄𝑁.(3.10) Here, ΛCoker𝐿ker𝐿 is any isomorphism.

Proof. Clear from Proposition 2.3.

Remark 3.8. Note that since Im𝑃=ker𝐿Dom𝐿, ImΛ=ker𝐿Dom𝐿, and Im𝐾𝑃,𝑄Dom𝐿ker𝑃Dom𝐿, then by definition 𝑀ΩDom𝐿. That is, any fixed points of 𝑀, if they exist, should be in the set ΩDom𝐿. Therefore, if (1.1) has a solution in Ω, then the solution should be in the set ΩDom𝐿.

Proposition 3.9. Assume that the conditions (i)–(iv) hold. Then, the operator 𝑀 is compact on Ω.

Proof. The projection operator 𝑃 is bounded and Im𝑃=ker𝐿 then Im𝑃 is a finite dimensional therefore, from Theorem 3.4 (a), 𝑃 is compact. By assumption (iv), 𝐾𝑃,𝑄𝑁 is compact. Beside these the operator ΛCoker𝐿ker𝐿 is linear isomorphism and dim(Coker𝐿)=𝑛<, therefore Λ is compact. Since ΛΠ is continuous then ΛΠ𝑁 is compact. As a result, we obtained the compactness of the operator 𝑀 on a set Ω.

Let 𝜕Ω denote the boundary of a set Ω.

(v) If 0(𝐿𝑁)(Dom𝐿𝜕Ω), then the Leray-Schauder degree 𝑑(𝐼𝑀,Ω,0) is well defined [34], since this condition by Proposition 3.7 gives us 0(𝐼𝑀)(𝜕Ω).

Now, let us search how much the degree depends upon the choice of the operators 𝑃, 𝑄, and Λ. To show this, we will need the following definition and results.

Let Λ𝐿 will be the set of all linear isomorphism from Coker𝐿 to ker𝐿.

Definition 3.10. If there exists a continuous Λ, Λ[]ΛCoker𝐿×0,1ker𝐿(,0)=Λ,Λ(,1)=Λ(3.11) such that for any 𝜆[0,1] the operator Λ(,Λ)Λ𝐿 then the operator Λ,ΛΛ𝐿 is called homotopic in Λ𝐿.

Being homotopic is an equivalence relation in the set Λ𝐿. Therefore, this equivalence relation divides the set Λ𝐿 into homotopy classes.

Proposition 3.11. The operators Λ and Λ are homotopic in Λ𝐿 if and only if det(ΛΛ)>0.

Proof. Assume that Λ and Λ are homotopic in Λ𝐿. From the condition (ii) we know that we have dimker𝐿=dimCoker𝐿=𝑛. Let Λ be the operator defined in Definition 3.10, [𝑎1,𝑎2,,𝑎𝑛] and [𝑏1,𝑏2,,𝑏𝑛] be bases of the spaces Coker𝐿 and ker𝐿, respectively. If for any 𝜆[0,1], Δ(𝜆) denotes the determinant of the matrix corresponding to Λ(,𝜆) with respect to these bases, then, for any 𝜆[0,1], Δ(𝜆)0 since Λ(,𝜆) is an isomorphism for any 𝜆[0,1]. Beside this, since Λ is continuous, then Δ is also continuous with respect to 𝜆. Using continuity and the fact that for any 𝜆[0,1], Δ(𝜆)0 we have the number Δ(𝜆) is always positive or negative, that is it has always same sign. In particular Δ(0) and Δ(1) have the same signs, therefore we haveΛdetΛ1Λ=detΛdet1=Λdet=det(Λ)Δ(1)Δ(0)>0.(3.12) Conversely assume that det(ΛΛ1)>0. With respect to bases of Coker𝐿 and ker𝐿, let Λ and Λ denote the matrix representations of the operators Λ and Λ, respectively. By assumption det(Λ) and Λdet() have the same sign. Therefore, they belong to same connected component of the topological group GL(𝑛,𝑟). Since GL(𝑛,𝑟) is locally arcwise connected then the corresponding component is also path connected. Therefore, there exists a continuous operator Λ[]ΛCoker𝐿×0,1ker𝐿(Λ,0)=Λ,(Λ,1)=.(3.13) Therefore, for any 𝜆[0,1], if we take Λ(,𝜆) as a family of isomorphisms corresponding to continuous matrices defined from Coker𝐿 to ker𝐿, then the proof will be completed.

Corollary 3.12. Λ𝐿 is separated into two homotopy classes.

Therefore, the set of all isomorphisms ΛCoker𝐿ker𝐿 with the same sign of determinant will be in the same classes. So one class will be with positive determinant and the other one will be with negative determinant.

Note the following: let ΛCoker𝐿ker𝐿 be any isomorphism from the set Λ𝐿. The sign of determinant of the matrix corresponding to Λ depends upon not only the basis chosen for Coker𝐿 and ker𝐿 but also the order of the elements in these basis. If the operators Λ and Λ are homotopic with respect to chosen bases for Coker𝐿 and ker𝐿, then they are homotopic with respect to any basis chosen for these spaces.

Now, let us fix an orientation on Coker𝐿 and ker𝐿, and let [𝑎1,𝑎2,,𝑎𝑛] be a basis for Coker𝐿 for the chosen orientation.

Definition 3.13. Let the operator ΛCoker𝐿ker𝐿 be given. If [Λ𝑎1,Λ𝑎2,,Λ𝑎𝑛] has the same orientation with basis chosen in ker𝐿, then the operator Λ is said to be an orientation preserving transformation. Otherwise, it is said to be an orientation reversing transformation.

Proposition 3.14. If Coker𝐿 and ker𝐿 are oriented, then the operators Λ and Λ are homotopic in Λ𝐿 if and only if they are both orientation preserving or both orientation reversing transformations.

Proof. Assume that [𝑎1,𝑎2,,𝑎𝑛] and [𝑏1,𝑏2,,𝑏𝑛] are, respectively, bases of Coker𝐿 and ker𝐿 with respect to chosen the orientation. The basis [Λ𝑎1,Λ𝑎2,,Λ𝑎𝑛] on ker𝐿 has the same orientation with [𝑏1,𝑏2,,𝑏𝑛] if and only if the determinant of the matrix 𝑆=(𝑠𝑖𝑗) defined by Λ𝑎𝑗=𝑛𝑗=1𝑠𝑖𝑗𝑏𝑗(3.14) will be positive. Namely, let 𝑀1 be the transition matrix from the basis [𝑎1,𝑎2,,𝑎𝑛] to the basis [𝑏1,𝑏2,,𝑏𝑛] and 𝑀2 be the transition matrix from the basis [𝑎1,𝑎2,,𝑎𝑛] to the basis [Λ𝑎1,Λ𝑎2,,Λ𝑎𝑛], 𝑎1,𝑎2,,𝑎𝑛𝑀1𝑏1,𝑏2,,𝑏𝑛,𝑎1,𝑎2,,𝑎𝑛𝑀2Λ𝑎1,Λ𝑎2,,Λ𝑎𝑛,Λ𝑎1,Λ𝑎2,,Λ𝑎𝑛𝑆𝑏1,𝑏2,,𝑏𝑛,(3.15) then we have 𝑀1=𝑆𝑀2 and det(𝑀1)=det(𝑆)det(𝑀2). [Λ𝑎1,Λ𝑎2,,Λ𝑎𝑛] has the same orientation with [𝑏1,𝑏2,,𝑏𝑛] if and only if the determinants of the matrixes 𝑀1 and 𝑀2 have the same sign. This is only possible in the case the determinant of 𝑆 is positive. Therefore, since the determinant of the matrixes 𝑀1 and 𝑀2 have the same sign, using the relation det(𝑀1)=det(𝑆)det(𝑀2), we obtain that det(𝑆)>0.
Let us assume that 𝑆 is a matrix related to a basis [Λ𝑎1,Λ𝑎2,,Λ𝑎𝑛]. In this case if the matrix 𝐺=(𝑔𝑖𝑗) is the matrix represent the operator ΛΛ1 with respect to basis [𝑏1,𝑏2,,𝑏𝑛], then we have 𝑛𝑗=1𝑠𝑖𝑗𝑏𝑗=Λ𝑎𝑖=ΛΛΛ1𝑎𝑖=ΛΛ1Λ𝑎𝑖,ΛΛ1𝑛𝑘=1𝑠𝑘𝑖𝑏𝑘=𝑛𝑘=1𝑠𝑘𝑖ΛΛ1𝑏𝑘,𝑛𝑘=1𝑠𝑛𝑘𝑖𝑗=1𝑔𝑗𝑘𝑏𝑗.(3.16) Therefore, (𝑆)𝑇=𝑆𝑇𝐺𝑇 and 𝑆=𝐺𝑆 are obtained. Since det(𝑆)>0 and det(𝑆)>0, then det(𝐺)>0, that is det(ΛΛ1)>0. This means that Λ and Λ1 have the same orientation.
Conversely, if the operators Λ and Λ1 have the same orientation, then det(ΛΛ1)>0. Therefore, from the Proposition 3.11, Λ and Λ1 are homotopic.

Lemma 3.15. Let 𝑌 be a vector space and 𝑆,𝑆𝑌𝑌 be two projection operators with Im𝑆=Im𝑆0. Therefore the operator 𝑆 defined by 𝑆=𝑎𝑆+𝑏𝑆, 𝑎,𝑏, is a projection operator with the property Im𝑆=Im𝑆 if and only if 𝑎+𝑏=1.

Proof. Let 𝑎 and 𝑏 are real numbers and assume that the operator 𝑆 defined 𝑆=𝑎𝑆+𝑏𝑆 is a projection operator with its image is equal to Im𝑆=Im𝑆. Since for any 𝑥Im𝑆 we have 𝑆𝑥=𝑥 and for any 𝑦𝑌, 𝑆𝑦Im𝑆=Im𝑆 then, for any 𝑦𝑌 we have 𝑆𝑆𝑦=𝑆(𝑆𝑦)=𝑆𝑦. Therefore, we get the relation 𝑆𝑆=𝑆. In a similar manner, the equality 𝑆𝑆=𝑆 can be shown. So 𝑎𝑆+𝑏𝑆=𝑆=𝑆2=𝑎𝑆+𝑏𝑆𝑎𝑆+𝑏𝑆=𝑎2𝑆2+𝑎𝑏𝑆𝑆+𝑎𝑏𝑆𝑆+𝑏2𝑆2=𝑎2𝑆+𝑎𝑏𝑆𝑆+𝑎𝑏𝑆𝑆+𝑏2𝑆=𝑎2𝑆+𝑎𝑏𝑆+𝑎𝑏𝑆+𝑏2𝑆=𝑎𝑎𝑆+𝑏𝑆+𝑏𝑎𝑆+𝑏𝑆=(𝑎+𝑏)𝑎𝑆+𝑏𝑆(3.17) is obtained. From here, we get the result (𝑎+𝑏1)(𝑎𝑆+𝑏𝑆)=0, that is (𝑎+𝑏1)𝑆=0. The assumption Im𝑆0 forces the fact that 𝑎+𝑏=1.
Conversely, if 𝑎+𝑏=1, then 𝑆=𝑆2=𝑎𝑆+𝑏𝑆𝑎𝑆+𝑏𝑆=𝑎2𝑆2+𝑎𝑏𝑆𝑆+𝑎𝑏𝑆𝑆+𝑏2𝑆2=𝑎2𝑆+𝑎𝑏𝑆𝑆+𝑎𝑏𝑆𝑆+𝑏2𝑆=𝑎2𝑆+𝑎𝑏𝑆+𝑎𝑏𝑆+𝑏2𝑆=(𝑎+𝑏)𝑎𝑆+𝑏𝑆=𝑎𝑆+𝑏𝑆=𝑆(3.18) is obtained. Therefore, 𝑆 is a projection operator. Since Im𝑆=Im𝑆 is a vector space and 𝑆=𝑎𝑆+𝑏𝑆, then we have Im𝑆Im𝑆. Now, let us take an arbitrary element 𝑥Im𝑆=Im𝑆{0}. Therefore, 𝑆𝑥=𝑎𝑆𝑥+𝑏𝑆𝑥=𝑎𝑥+𝑏𝑥=(𝑎+𝑏)𝑥=𝑥(3.19) and from here we obtain 𝑥Im𝑆 and Im𝑆Im𝑆. So the result Im𝑆=Im𝑆 follows.

Lemma 3.16. If 𝑃 and 𝑃 are projection operators onto ker𝐿, 𝑎+𝑏=1 and 𝑃=𝑎𝑃+𝑏𝑃, then 𝐾𝑃=𝑎𝐾𝑃+𝑏𝐾𝑃.

Proof. In the case ker𝐿={0}, the proof is clear. Assume that Im𝑃=Im𝑃=ker𝐿{0}. Since 𝑎+𝑏=1 by Lemma 3.15, 𝑃 is a projection operator and Im𝑃=Im𝑃. Since 𝐾𝑃=(𝐼𝑃)𝐾𝑃 and 𝑃𝐾𝑃=0, then the relation 𝐾𝑃=𝐼𝑃𝐾𝑃=𝐼𝑎𝑃𝑏𝑃𝐾𝑃=𝐾𝑃𝑎𝑃𝐾𝑃𝑏𝑃𝐾𝑃=1𝐾𝑃𝑏𝑃𝐾𝑃=(𝑎+𝑏)𝐾𝑃𝑏𝑃𝐾𝑃=𝑎𝐾𝑃+𝑏𝐾𝑃𝑏𝑃𝐾𝑃=𝑎𝐾𝑃+𝑏𝐼𝑃𝐾𝑃=𝑎𝐾𝑃+𝑏𝐾𝑃(3.20) is obtained.

Lemma 3.17. Let 𝑍 be a vector space, 𝑆,𝑆𝑍𝑍 two projection operators with ker𝑆=ker𝑆, then for 𝑎,𝑏, 𝑎+𝑏=1 the operator 𝑆 defined by 𝑆=𝑎𝑆+𝑏𝑆 is a projection operator with ker𝑆=ker𝑆.

Proof. First of all, let us show that 𝑆 is a projection operator. Since 𝑆𝑍𝑍𝑆𝑍𝑍𝑍=ker𝑆Im𝑆,𝑍=ker𝑆Im𝑆,(3.21) then for any 𝑧𝑍 there exist unique elements 𝑧0ker𝑆, 𝑧1Im𝑆, 𝑧0ker𝑆, and 𝑧1Im𝑆 such that 𝑧=𝑧0+𝑧1 and 𝑧=𝑧0+𝑧1 hold. Therefore, 𝑆2(𝑧)=𝑎2𝑆(𝑧)+𝑎𝑏𝑆𝑆(𝑧)+𝑎𝑏𝑆𝑆(𝑧)+𝑏2𝑆(𝑧)=𝑎2𝑆𝑧0+𝑧1+𝑎𝑏𝑆𝑆𝑧0+𝑧1+𝑎𝑏𝑆𝑆𝑧0+𝑧1+𝑏2𝑆𝑧0+𝑧1=𝑎2𝑆𝑧1+𝑎𝑏𝑆𝑆𝑧1+𝑎𝑏𝑆𝑆𝑧1+𝑏2𝑆𝑧1=𝑎2𝑆𝑧1𝑧+𝑎𝑏𝑆1+𝑎𝑏𝑆𝑧1+𝑏2𝑆𝑧1𝑧=(𝑎+𝑏)𝑎𝑆1+𝑏𝑆𝑧1𝑧=𝑎𝑆1+𝑏𝑆𝑧1𝑧=𝑎𝑆0𝑧+𝑎𝑆1+𝑏𝑆𝑧0+𝑏𝑆𝑧1=𝑎𝑆(𝑧)+𝑏𝑆(𝑧)=𝑆(𝑧)(3.22) is obtained.
Now, let us show that ker𝑆=ker𝑆=ker𝑆. For this, take an arbitrary element 𝑥ker𝑆=ker𝑆. Therefore, 𝑆(𝑥)=𝑎𝑆(𝑥)+𝑏𝑆(𝑥)=𝑎.0+𝑏.0=0.(3.23) This means that ker𝑆=ker𝑆ker𝑆. Now, take 𝑥ker𝑆. Since 𝑍=ker𝑆Im𝑆, then there exist unique elements 𝑒ker𝑆 and 𝑓Im𝑆 such that 𝑥=𝑒+𝑓 holds. Therefore, we obtain 0=𝑆(𝑥)=𝑎𝑆(𝑥)+𝑏𝑆(𝑥)=𝑎𝑆(𝑒+𝑓)+𝑏𝑆(𝑒+𝑓)=𝑎𝑆(𝑓)+𝑏𝑆(𝑓)=𝑎𝑓+𝑏𝑆(𝑓).(3.24) That is 𝑎𝑓+𝑏𝑆(𝑓)=0. If 𝑏=0, since 𝑎+𝑏=1 then 𝑎=1 and then 𝑓=0. So that 𝑥ker𝑆=ker𝑆. If 𝑏0, then 𝑆(𝑓)=(a/b)𝑓. Then, 𝑆(𝑓)=𝑆2(𝑓)=𝑆𝑆(𝑓)=𝑆𝑎𝑏𝑓𝑎=𝑏𝑆(𝑓)(3.25) is obtained. In this case, we get 𝑆(𝑓)=(𝑎/𝑏)𝑆(𝑓). Since 𝑎+𝑏=1, this gives us 𝑆(𝑓)=0. From here, we get that ker𝑆ker𝑆=ker𝑆 is obtained. So in any case we showed that ker𝑆=ker𝑆=ker𝑆.

Proposition 3.18. If the assumptions (i)−(v) hold, then Leray-Schauder degree 𝑑[𝐼𝑀,Ω,0] depends on only 𝐿, 𝑁, Ω and homotopy class of Λ in Λ𝐿.

Proof. Let the operators 𝑃, 𝑃, 𝑄, 𝑄 be the projection operators with the properties Im𝑃=Im𝑃=ker𝐿, ker𝑄=ker𝑄=Im𝐿 and Λ, Λ two isomorphisms from Coker𝐿 to ker𝐿 in the same homotopy class. From Lemma 3.15 and Lemma 3.16, it is clear that for each 𝜆[0,1] the operators 𝑃(𝜆)=(1𝜆)𝑃+𝜆𝑃,𝑄(𝜆)=(1𝜆)𝑄+𝜆𝑄(3.26) are the projection operators with the property of for each 𝜆[0,1], Im𝑃(𝜆)=ker𝐿 and ker𝑄(𝜆)=Im𝐿. Beside this, from Lemma 3.16, we have 𝐾𝑃(𝜆)=(1𝜆)𝐾𝑃+𝜆𝐾𝑃. Let the operator ΛCoker𝐿×[0,1]ker𝐿 be the operator given in Definition 3.1. Using Proposition 3.7, we see that for each 𝜆[0,1] the fixed points of the operator 𝑀(,𝜆)𝑀Ω𝑋(,𝜆)=𝑃(𝜆)+ΛΠ𝑁(,𝜆)+𝐾𝑃(𝜆),𝑄(𝜆)𝑁(3.27) coincide with the solutions of the operator equation (1.1). From the condition (v), we have 0(𝐿𝑁)(Dom𝐿𝜕Ω) and 𝑥𝑀[].(𝑥,𝜆),𝑥𝜕Ω,𝜆0,1(3.28) Clearly, we have 𝑀(,0)=𝑀=𝑃+ΛΠ𝑁+𝐾𝑃,𝑄𝑁,𝑀(Λ,1)=𝑀=𝑃+Π𝑁+𝐾𝑃,𝑄𝑁.(3.29) Now let us show that 𝑀 is compact on Ω×[0,1]. From the open form 𝑀(,𝜆)=(1𝜆)𝑃𝑥+𝜆𝑃𝑥+ΛΠ𝑁(,𝜆)+(1𝜆)𝐾𝑃+𝜆𝐾𝑃𝐼(1𝜆)𝑄𝜆𝑄𝑁,(3.30) it is clear that 𝑀 is continuous. So, in order to show that the set 𝑀(Ω×[0,1]) is relatively compact the only delicate point is the last term. Using the fact that 𝐾𝑃=(𝐼𝑃)𝐾𝑃, we obtain the last term as 𝐼𝜆𝑃𝐾𝑃(𝐼𝑄)𝑁+𝜆𝐼𝜆𝑃𝐾𝑃𝑄𝑄𝑁.(3.31) Therefore, compactness can be proven like in the proof of Proposition 3.9. Using the invariance of Leray-Schauder degree with respect to compact homotopy, we obtain that 𝑑(𝐼𝑀(,0),Ω,0)=𝑑(𝐼𝑀(,1),Ω,0),(3.32) that is, 𝑑(𝐼𝑀,Ω,0)=𝑑𝐼𝑀,Ω,0.(3.33)

Now, let us indicate how the degree 𝑑(𝐼𝑀,Ω,0) depends on homotopy class of Λ. For this, let us prove the following lemma.

Lemma 3.19. If 𝐺ker𝐿ker𝐿 is any automorphism and 𝑀=𝑃+𝐺ΛΠ+𝐾𝑃,𝑄𝑁(3.34) then the relation 𝐼𝑀=(𝐼𝑃+𝐺𝑃)(𝐼𝑀)(3.35) is satisfied.

Proof. Since 𝑃2=𝑃, 𝑃𝐾𝑃,𝑄=0, and 𝑃Λ=Λ, then (𝐼𝑃+𝐺𝑃)(𝐼𝑀)=𝐼𝑀𝑃+𝑃𝑀+𝐺𝑃𝐺𝑃𝑀=𝐼𝑃ΛΠ𝑁𝐾𝑃,𝑄𝑁𝑃+𝑃2+𝑃ΛΠ𝑁+𝑃𝐾𝑃,𝑄𝑁+𝐺𝑃𝐺𝑃𝑀=𝐼𝑃ΛΠ𝑁𝐾𝑃,𝑄𝑁+𝑃ΛΠ𝑁+𝐺𝑃𝐺𝑃𝑀=𝐼𝑃ΛΠ𝑁𝐾𝑃,𝑄𝑁+𝑃ΛΠ𝑁+𝐺𝑃𝐺𝑃𝑃+ΛΠ𝑁+𝐾𝑃,𝑄𝑁=𝐼𝑃ΛΠ𝑁𝐾𝑃,𝑄𝑁+𝑃ΛΠ𝑁+𝐺𝑃𝐺𝑃2𝐺𝑃ΛΠ𝑁𝐺𝑃𝐾𝑃,𝑄𝑁=𝐼𝑃ΛΠ𝑁𝐾𝑃,𝑄𝑁+𝑃ΛΠ𝑁𝐺𝑃ΛΠ𝑁=𝐼𝑃+𝐺𝑃ΛΠ+𝐾𝑃,𝑄𝑁=𝐼𝑀.(3.36)

Proposition 3.20. If Λ,ΛΛ𝐿 and Λ𝑀=𝑃+Π+𝐾𝑃,𝑄𝑁(3.37) then we have 𝑑𝐼𝑀Λ,Ω,0=sgndetΛ1𝑑(𝐼𝑀,Ω,0).(3.38)

Proof. In Lemma 3.19, if we take 𝐺=ΛΛ1, then 𝐼𝑀=𝐼𝑃+ΛΛ1𝑃(𝐼𝑀)(3.39) is obtained. Now let us show that the operator 𝐴=𝐼𝑃+ΛΛ1𝑃 is an automorphism on 𝑋. For this take, 𝑥ker𝐿, then we have 𝑥𝑃𝑥+ΛΛ1𝑃𝑥=0.(3.40) If we apply the operator 𝑃 to both sides, we get ΛΛ1𝑃𝑥=0. Since ΛΛ1 is one-to-one, this result gives 𝑃𝑥=0. If we substitute this result in (3.40) we obtain that 𝑥=0, and therefore 𝐴 is one-to-one. For surjectivity, take 𝑦𝑋. Therefore, there exists unique elements 𝑘ker𝑃, 𝑗Im𝑃 such that 𝑦=𝑘+𝑗. Now, we are looking for 𝑥𝑋, 𝑘ker𝑃, 𝑗Im𝑃 such that 𝑥=𝑘+𝑗 and 𝐴𝑥=𝑦. So Λ𝐴𝑥=𝑦𝐼𝑃+Λ1𝑃Λ𝑥=𝑦𝐼𝑃+Λ1𝑃𝑘+𝑗𝑘=𝑘+𝑗+ΛΛ1𝑗=𝑘+𝑗.(3.41) Using uniqueness in direct sum and the fact that ΛΛ1 is an automorphism on ker𝐿=Im𝑃, we get 𝑘=𝑘 and ΛΛ1𝑗=𝑗. Therefore, taking 𝑥=𝑘+ΛΛ1𝑗 ontoness of the operator 𝐴 is proved. Therefore, 𝐴 is an automorphism on 𝑋. So using the identity 𝐼𝑀=(𝐼𝑃+ΛΛ1𝑃)(𝐼𝑀) and Leray Product Theorem, we have 𝑑𝐼𝑀,Ω,0=𝑑𝐴,𝐵𝜖(0),0𝑑(𝐼𝑀,Ω,0).(3.42) Therefore the result 𝑑𝐼𝑀,Ω,0=𝑑𝐼𝑃+ΛΛ1𝑃,𝐵𝜖(0),0𝑑(𝐼𝑀,Ω,0)=𝑑𝐼𝑃+ΛΛ1𝑃ker𝐿,𝐵𝜖Λ(0)ker𝐿,0𝑑(𝐼𝑀,Ω,0)=𝑑Λ1,𝐵𝜖Λ(0)ker𝐿,0𝑑(𝐼𝑀,Ω,0)=sgndetΛ1𝑑(𝐼𝑀,Ω,0)(3.43) is achieved.

Corollary 3.21. Under the assumptions of Proposition 3.18, Leray-Schauder degree |𝑑(𝐼𝑀,Ω,0)| only depends upon 𝐿, 𝑁, and Ω.

Now, if the orientation on the spaces ker𝐿 and Coker𝐿 is fixed, then we can give the following beautiful and fruitful definition.

Definition 3.22. If the operators 𝐿 and 𝑁 satisfy the conditions (i)–(v) then the coincidence degree of 𝐿 and 𝑁 in Ω defined by 𝑑[](𝐿,𝑁),Ω=𝑑(𝐼𝑀,Ω,0).(3.44) Here, Λ in 𝑀 is an orientation preserving isomorphism.

This definition is supported with all the arguments given in this paper.

4. Basic Properties of Coincidence Degree

In this section, we will see that the coincidence degree satisfies all the basic properties of the Leray-Schauder degree. First, let us consider the simplest case where 𝑋=𝑍 and 𝐿=𝐼. In this situation, ker𝐿={0} and Im𝐿=𝑍=𝑋, so that Coker𝐿=𝑍/Im𝐿={0}. Therefore dimker𝐿=dimCoker𝐿=0 and then the assumptions (i) and (ii) are clearly satisfied. Since Im𝑃=ker𝐿=0 and ker𝑄=Im𝐿=𝑍, then 𝑃=0, and 𝑄=0. Thus 𝐾𝑃,𝑄=𝐾𝑃(𝐼𝑄)=𝐼(𝐼0)=𝐼, and Π=0. Therefore, the conditions (iii) and (iv) reduced to the compactness of 𝑁 on Ω. Since 𝐿=𝐼 and Dom𝐿=𝑋, then the condition (v) in this case means that 𝑁 has no fixed point on 𝜕Ω. Since 𝑃=0, Π=0, and 𝐾𝑃,𝑄=0, then 𝑀=𝑃+(ΛΠ+𝐾𝑃,𝑄)𝑁=𝑁. Therefore, 𝑑[][](𝐿,𝑁),Ω=𝑑(𝐼,𝑁),Ω=𝑑(𝐼𝑁,Ω,0).(4.1) That is the coincidence degree of 𝐿 and 𝑁 in this case is nothing but the Leray-Schauder degree of 𝐼𝑁.

Now, we will give the basic properties of coincidence degree.

Theorem 4.1. Assume that the conditions (i) to (v) are satisfied. Then coincidence degree satisfies the following basic properties. (1)Existence theorem: if 𝑑[(𝐿,𝑁),Ω]0, then 0(𝐿𝑁)(dom𝐿Ω). (2)Excision property: if Ω0Ω is an open set such that (𝐿𝑁)1(0)Ω0, then 𝑑[](𝐿,𝑁),Ω=𝑑(𝐿,𝑁),Ω0.(4.2)(3) Additivity property: if Ω=Ω1Ω2 with Ω1 and Ω2 are open, bounded, disjoint subsets of 𝑋, then 𝑑[](𝐿,𝑁),Ω=𝑑(𝐿,𝑁),Ω1+𝑑(𝐿,𝑁),Ω2.(4.3)(4)Invariance under homotopy property: if the operator 𝑁[](Ω×0,1𝑍𝑥,𝜆)𝑁(𝑥,𝜆)(4.4) is 𝐿-compact in Ω×[0,1] and such that for each 𝜆[0,1], 0[𝐿𝑁(,𝜆)](dom𝐿𝜕Ω), then coincidence degree 𝑑[(𝐿,𝑁(,𝜆)),Ω], is independent of 𝜆 in [0,1]. In particular 𝑑[][].(𝐿,𝑁(,1)),Ω=𝑑(𝐿,𝑁(,0)),Ω(4.5)

Proof. (1) If 𝑑(𝐼𝑀,Ω,0)=𝑑[(𝐿,𝑁),Ω]0, then 𝑥Ω such that (𝐼𝑀)𝑥=0. But in fact, we know that 𝑥Dom𝐿Ω. Also, by Proposition 3.7, 𝐿𝑥=𝑁𝑥. That is 0(𝐿𝑁)(dom𝐿Ω).
(2) Assume that Ω0Ω is an open set such that (𝐿𝑁)1(0)Ω0, then by Proposition 3.7,  (𝐼𝑀)1(0)Ω0. Therefore, by the excision property of the Leray-Schauder degree, 𝑑(𝐼𝑀,Ω,0)=𝑑(𝐼𝑀,Ω0,0). So, by the definition of coincidence degree, we have 𝑑[(𝐿,𝑁),Ω]=𝑑[(𝐿,𝑁),Ω0].
(3) If Ω=Ω1Ω2 with Ω1 and Ω2 are open, bounded, disjoint subsets of 𝑋, then additive property of Leray-Schauder degree we have 𝑑(𝐼𝑀,Ω,0)=𝑑(𝐼𝑀,Ω1,0)+𝑑(𝐼𝑀,Ω2,0). So the result follows from the definition of coincidence degree.
(4) Since the operator 𝑁(,𝜆) is 𝐿-compact for each 𝜆[0,1] and for each 𝜆[0,1],  0[𝐿𝑁(,𝜆)](dom𝐿𝜕Ω), then for each 𝜆[0,1] the coincidence degree 𝑑[(𝐿,𝑁(,𝜆)),Ω] is well defined. Since the operator 𝑁 is 𝐿-compact on Ω×[0,1] then it is a homotopy of compact operators on Ω. Therefore, by invariance of the Leray-Schauder degree under homotopy property the result follows.

The famous Borsuck theorem for degree theory is also valid for coincidence degree.

Theorem 4.2. If Ω is symmetric with respect to 0 and contains it and if 𝑁(𝑥)=𝑁(𝑥) in Ω, then coincidence degree 𝑑[(𝐿,𝑁),Ω] is an odd integer.

Proof. We proved that the operator 𝑀 is compact on Ω. Since a projection operator 𝑃 is linear then it is odd in Ω and 𝑁 is odd in Ω then the operator 𝑀=𝑃+(ΛΠ+𝐾𝑃,𝑄)𝑁 is odd in Ω. Therefore, the result follows from the validity of Borsuck theorem in the Leray-Schauder degree.

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