A Note on Coincidence Degree Theory

and Applied Analysis 3 Now let us show that LP is onto. Since P : X → X is a projection operator, we canwrite the vector spaceX as direct sumsX kerP ⊕ ImP . From the exactness of the chain above, we get X kerP ⊕ kerL. Take z ∈ ImL, so that there exists x ∈ Dom L ⊂ X with Lx z. Since X kerP ⊕kerL, there exists unique elements e ∈ kerP and f ∈ kerL 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 e ∈ Dom L. So we get e ∈ Dom L and e ∈ kerP and LPe z. So the result follows. Now, let us define KP : L−1 P . It is clear that KP : ImL ⊂ Z → DomL ∩ kerP is one-to-one, onto, and PKP 0. Lemma 2.2. (1) On ImL, we have LKP I. (2) On Dom L, we have KPL I − P . Proof. 1 Take x ∈ ImL. Therefore, LKPx L KP x LP KP x Ix. 2 Since ImP kerL, then we have LP 0, so we obtain KPL KPL I − P . So that in order to prove 2 , we need to show the equality KPL I − P KPLP I − P . If we can have Im I − P ⊆ Dom LP DomL ∩ kerP , then the result follows. Take x ∈ DomL. Since P x ∈ kerL ⊂ DomL and DomL is a vector subspace of X, we have x − Px ∈ DomL. Since P x − Px Px − P 2x Px − Px 0, then x − Px ∈ kerP ; therefore, we have x − Px ∈ DomL ∩ kerP . From here, we obtain Im I − P ⊂ DomL ∩ kerP . So using 1 , the result KPL I − P KPLP I − P I − P follows. Now, let us define the canonic surjection Π as Π : Z −→ CokerL z −→ z ImL. 2.2 Here, CokerL Z/ ImL is the quotient space of Z under the equivalence relation z ∼ z′ ⇔ z − z′ ∈ ImL. Thus, CokerL {z z ImL : z ∈ Z}. 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 Λ : CokerL → kerL, then Lx y, y ∈ Z 2.3 will be equivalent to I − P x ΛΠ KP,Q ) y. 2.4 Here, the operator KP,Q : Z → X is defined as KP,Q KP I −Q . Proof. Since ImL kerQ kerΠ, then for y ∈ ImL we have Qy 0 and ΛΠy Λ ImL Λ0 0. From here, it is seen that Lx y ⇐⇒ Lx y −Qy ⇐⇒ KPLx KP ( y −Qy ⇐⇒ I − P x KP I −Q y ⇐⇒ I − P x ΛΠ KP I −Q y. 2.5 4 Abstract and Applied Analysis Now, let us consider another projection operator couple P ′, Q′ that will make the chain X P ′ −→ Dom L L −→ Z Q ′ −→ Z 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 KP − KP ′ 0. So for any z ∈ ImL, we have KP − KP ′ z ∈ kerL. Therefore, we can write KP − KP ′ : ImL → kerL ImP ImP ′. Since the projection operator P behaves on ImP as an identity operator, we have KP −KP ′ P KP −KP ′ . As a result, the equality KP −KP ′ P KP −KP ′ P ′ KP −KP ′ 2.7


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 3-32 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 Ω ⊂ R n , f ∈ C Ω , and p ∈ R n \ f ∂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 f ∈ C Ω see 33 , page 172 .Luckily, in an arbitrary Banach space X, Leray and Schauder proved that for Ω ∈ X open, bounded set, M : Ω → X compact operator and for p ∈ X \ I −M ∂Ω the degree of compact perturbation of identity I −M in Ω with respect

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, Dom L is a linear subspace of X, and L : Dom L → Z is a linear operator.Assume that the operators P : X → X and Q : Z → Z linear projection operators such that the chain is exact, that is, Im P ker L and Im L ker Q.Let us define the restriction of L to Dom L ∩ ker P as L P : Dom L ∩ ker P → Im L. Now, let us give the following lemma about L P .
Proof.Firstly, let us show that L P is one-to-one mapping.For this let us take x ∈ ker L P ⊂ ker L Im P , so that there exists y ∈ Dom P such that x Py.Since P is a projection operator we get x Py P 2 y P Py Px 0. Therefore, x 0, so that we obtain that ker L P {0}.This means that L P is one-to-one.Now let us show that L P is onto.Since P : X → X is a projection operator, we can write the vector space X as direct sums X ker P ⊕ Im P .From the exactness of the chain above, we get X ker P ⊕ ker L. Take z ∈ Im L, so that there exists x ∈ Dom L ⊂ X with Lx z.Since X ker P ⊕ ker L, there exists unique elements e ∈ ker P and f ∈ ker L 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 e ∈ Dom L. So we get e ∈ Dom L and e ∈ ker P and L P e z.So the result follows.Now, let us define K P : L −1 P .It is clear that K P : Im L ⊂ Z → Dom L ∩ ker P is one-to-one, onto, and PK P 0.

Lemma 2.2. (1) On
Im L, we have LK P I. (2) On Dom L, we have K P L I − P .
Proof. 1 Take x ∈ Im L. Therefore, LK P x L K P x L P K P x Ix. 2 Since Im P ker L, then we have LP 0, so we obtain K P L K P L I − P .So that in order to prove 2 , we need to show the equality K P L I − P K P L P I − P .If we can have Im I − P ⊆ Dom L P Dom L ∩ ker P , then the result follows.Take x ∈ Dom L. Since P x ∈ ker L ⊂ Dom L and Dom L is a vector subspace of X, we have x − Px ∈ Dom L.
Since P x − Px Px − P 2 x Px − Px 0, then x − Px ∈ ker P ; therefore, we have x − Px ∈ Dom L ∩ ker P .From here, we obtain Im I − P ⊂ Dom L ∩ ker P .So using 1 , the result K P L I − P K P L P I − P I − P follows.
Now, let us define the canonic surjection Π as will be equivalent to Here, the operator K P,Q : Z → X is defined as K P,Q K P I − Q .
Proof.Since Im L ker Q ker Π, then for y ∈ Im L we have Qy 0 and ΛΠy Λ Im L Λ0 0. From here, it is seen that Lx y ⇐⇒ Lx y − Qy ⇐⇒ K P Lx K P y − Qy Now, let us consider another projection operator couple P , Q that will make the chain exact, and let us search the relation of this operator couple with P, Q .From Lemma 2.2, since LK P I, and LK P I then we have L K P − K P 0. So for any z ∈ Im L, we have K P − K P z ∈ ker L. Therefore, we can write K P − K P : Im L → ker L Im P Im P .Since the projection operator P behaves on Im P as an identity operator, we have K P − K P P K P − K P .As a result, the equality follows.
i PK P P K P 0, ii K P I − P K P .
Proof.i Using PK P 0, P K P 0, and 2.7 , the result follows.
ii Again using 2.7 and i , we obtain K P − K P P K P − K P ⇒ K P − K P P K P − P K P ⇒ K P P K P −PK P ⇒ K P I − PK P .

2.9
In a similar manner, the equality K P I − P K P can be obtained.

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, Ω ⊂ X an open, bounded subset of X and Ω an closure of Ω.Let us assume that the operators iii the operator N : Ω ⊂ X → Z is continuous and ΠN Ω is bounded, iv the operator K P,Q N : Ω → Z 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 N : Ω → Z 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 K P,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 P : X → X and Q : Z → Z such that the chain will be exact.
Proof.Assume that ker L is finite dimensional, and the set {y 1 , y 2 , . . ., y n } is a basis for ker L. Define the vector subspaces as X k span{y 1 , y 2 , . . ., y k−1 , y k 1 , . . ., y n }.Since X k is finite dimensional so is a closed subspace of X see 35, Theorem 2.4-3 and y k / ∈ X k .Let B be a basis of X such that {y 1 , y 2 , . . ., y n } ⊆ B. Now, let us define the linear operators which satisfy the conditions

3.3
Therefore, the operator P defined by 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 ker Q Im L. We know that there exists a subspace Z such that Z Im L ⊕ Z see 37, Proposition I .The projection operator Q defined on Z with the rule Since Z is a Banach space, ker Q and Im Q are closed subsets of Z, therefore the projection operator Q is continuous see 38, Theorem 6.12.6 .
Moreover, the canonical surjection Π : Z → Coker L is continuous with the quotient topology on Coker L. Now, let us state two theorems that will be used in the proof of following proposition.
Let X be normed space, T : X → X be a linear compact operator, and S : X → X 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 Im Q with Π Q , and let us show that the linear operator Therefore Π Q is one-to-one.To show surjection, let us take an arbitrary element z ∈ Z/ Im L. So there exists z ∈ Z such that Πz z holds.Since the space Z can be written as Abstract and Applied Analysis 7 Now, let us show that for an arbitrary α ∈ Z the relation Let K P denote the restriction of the operator K P to the finite dimensional space Im Q − Q .Using the results obtained until here in this proof and using the equality K P I −P K P , is achieved.Now, let us explain the operator K P ,Q N is compact.Since the operator K P,Q N is compact and I − P is continuous, then the operator From the same reason, the operator Π −1 Q is also compact.Since the operators N, Π, and I − P are all continuous, the compactness of K P ,Q N follows.
Proposition 3.7.The element x ∈ Dom L ∩ Ω is a solution of the operator equation 1.1 if and only if it satisfies In other words, the set of solutions of 1.1 is equal to the set of fixed points of the operator M : Ω → X defined by
Remark 3.8.Note that since Im P ker L ⊂ Dom L, Im Λ ker L ⊂ Dom L, and Im K P,Q ⊂ Dom L ∩ ker P ⊂ Dom L, then by definition M : Ω → Dom L. That is, any fixed points of M, if they exist, should be in the set Ω ∩ Dom L. Therefore, if 1.1 has a solution in Ω, then the solution should be in the set Ω ∩ Dom L. 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 Im P ker L then Im P is a finite dimensional therefore, from Theorem 3.4 a , P is compact.By assumption iv , K P,Q N is compact.Beside these the operator Λ : Coker L → ker L is linear isomorphism and dim Coker L 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 / ∈ L − N Dom L ∩ ∂Ω , then the Leray-Schauder degree d I − M, Ω, 0 is well defined 34 , since this condition by Proposition 3.7 gives us 0 / ∈ I − M ∂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 Λ L will be the set of all linear isomorphism from Coker L to ker L.
Definition 3.10.If there exists a continuous Λ, Being homotopic is an equivalence relation in the set Λ L .Therefore, this equivalence relation divides the set Λ L into homotopy classes.
Proposition 3.11.The operators Λ and Λ are homotopic in Λ L if and only if det ΛΛ > 0.
Proof.Assume that Λ and Λ are homotopic in Λ L .From the condition ii we know that we have dimker L dimCoker L n.Let Λ be the operator defined in Definition 3.10, a 1 , a 2 , . . ., a n and b 1 , b 2 , . . ., b n be bases of the spaces Coker L and ker L, 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 have Conversely assume that det Λ Λ −1 > 0. With respect to bases of Coker L and ker L, let Λ and Λ denote the matrix representations of the operators Λ and Λ , respectively.By assumption det Λ and det Λ 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

3.13
Therefore, for any λ ∈ 0, 1 , if we take Λ •, λ as a family of isomorphisms corresponding to continuous matrices defined from Coker L to ker L, then the proof will be completed.
Therefore, the set of all isomorphisms Λ : Coker L → ker L 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 Λ : Coker L → ker L be any isomorphism from the set Λ L .The sign of determinant of the matrix corresponding to Λ depends upon not only the basis chosen for Coker L and ker L but also the order of the elements in these basis.If the operators Λ and Λ are homotopic with respect to chosen bases for Coker L and ker L, then they are homotopic with respect to any basis chosen for these spaces.Now, let us fix an orientation on Coker L and ker L, and let a 1 , a 2 , . . ., a n be a basis for Coker L for the chosen orientation.Definition 3.13.Let the operator Λ : Coker L → ker L be given.If Λa 1 , Λa 2 , . . ., Λa n has the same orientation with basis chosen in ker L, then the operator Λ is said to be an orientation preserving transformation.Otherwise, it is said to be an orientation reversing transformation.have the same sign.This is only possible in the case the determinant of S is positive.Therefore, since the determinant of the matrixes M 1 and M 2 have the same sign, using the relation det M 1 det S det M 2 , we obtain that det S > 0. Let us assume that S is a matrix related to a basis Λ a 1 , Λ a 2 , . . ., Λ a n .In this case if the matrix G g ij is the matrix represent the operator Λ Λ −1 with respect to basis b 1 , b 2 , . . ., b n , then we have g jk b j .

3.16
Therefore, S T S T G T and S GS are obtained.Since det S > 0 and det S > 0, then det G > 0, that is det Λ Λ −1 > 0. This means that Λ and Λ −1 have the same orientation.
Conversely, if the operators Λ and Λ −1 have the same orientation, then det Λ Λ −1 > 0. Therefore, from the Proposition 3.11, Λ and Λ −1 are homotopic.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 Im S Im S .Since for any x ∈ Im S we have Sx x and for any y ∈ Y , S y ∈ Im S Im S then, for any y ∈ Y we have SS y S S y S y.Therefore, we get the relation SS S .In a similar manner, the equality S S S can be shown.So Proof.First of all, let us show that S is a projection operator.Since

3.21
then for any z ∈ Z there exist unique elements z 0 ∈ ker S, z 1 ∈ Im S, z 0 ∈ ker S , and z 1 ∈ Im S such that z z 0 z 1 and z z 0 z 1 hold.Therefore, is obtained.Now, let us show that ker S ker S ker S .For this, take an arbitrary element x ∈ ker S ker S .Therefore, S x aS x bS x a.0 b.0 0.

3.23
This means that ker S ker S ⊆ ker S .Now, take x ∈ ker S .Since Z ker S ⊕ Im S, then there exist unique elements e ∈ ker S and f ∈ Im S 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 x ∈ ker S ker S .If b / 0, then S f − a/b f.Then, 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 ker S ⊆ ker S ker S is obtained.So in any case we showed that ker S ker S ker S .Proof.Let the operators P , P , Q, Q be the projection operators with the properties Im P Im P ker L, ker Q ker Q Im L and Λ, Λ two isomorphisms from Coker L to ker L in the same homotopy class.From Lemma 3.15 and Lemma 3.16, it is clear that for each λ ∈ 0, 1 the operators are the projection operators with the property of for each λ ∈ 0, 1 , Im P λ ker L and ker Q λ Im L. Beside this, from Lemma 3.16, we have K P λ 1 − λ K P λK P .Let the operator Λ : Coker L × 0, 1 → ker L 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 coincide with the solutions of the operator equation 1.1 .From the condition v , we have 0 / ∈ L − N Dom L ∩ ∂Ω and x / ∈ M x, λ , ∀x ∈ ∂Ω, ∀λ ∈ 0, 1 .

3.28
Clearly, we have 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 K P I −P K P , we obtain the last term as 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  For this take, x ∈ ker L, then we have x − Px Λ Λ −1 Px 0.

3.40
If we apply the operator P to both sides, we get Λ Λ −1 Px 0. Since Λ Λ −1 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 y ∈ X.Therefore, there exists unique elements k ∈ ker P , j ∈ ImP such that y k j.Now, we are looking for x ∈ X, k ∈ ker P , j ∈ Im P such that x k j and Ax y.So

3.41
Using uniqueness in direct sum and the fact that Λ Λ −1 is an automorphism on ker L Im P , we get k k and Λ Λ −1 j j.Therefore, taking x k Λ Λ −1 j ontoness of the operator A is proved.Therefore, A is an automorphism on X.So using the identity I − M I − P Λ Λ −1 P I − M and Leray Product Theorem, we have Therefore the result This definition is supported with all the arguments given in this paper.

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, ker L {0} and Im L Z X, so that Coker L Z/ Im L {0}.Therefore dimkerL dimCoker L 0 and then the assumptions i and ii are clearly satisfied.Since Im P ker L 0 and ker Q Im L Z, then P 0, and Q 0. Thus K P,Q K P I − Q I I − 0 I, and Π 0. Therefore, the conditions iii and iv reduced to the compactness of N on Ω.Since L I and Dom L X, then the condition v in this case means that N has no fixed point on ∂Ω.Since P 0, Π 0, and K P,Q 0, then M P That is the coincidence degree of L and N in this case is nothing but the Leray-Schauder degree of I − N. Now, we will give the basic properties of coincidence degree.4 Since the operator N •, λ is L-compact for each λ ∈ 0, 1 and for each λ ∈ 0, 1 , 0 / ∈ L − N •, λ dom L ∩ ∂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.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 K P,Q N is odd in Ω.Therefore, the result follows from the validity of Borsuck theorem in the Leray-Schauder degree.

2 . 2 Proposition 2 . 3 .
Here, Coker L Z/ Im L is the quotient space of Z under the equivalence relation z ∼ z ⇔ z − z ∈ Im L. Thus, Coker L {z z Im L : z ∈ Z}.It is clear that the canonic surjection operator Π is linear and ker Π ker Q.If there exists an one-to-one operator Λ : Coker L → ker L, then Lx y, y ∈ Z 2.3

Theorem 3 . 4
see35 .Assume X and Y are normed spaces and the operator T : X → Y is linear.Therefore,

Lemma 3 . 15 .
Let Y be a vector space and S, S : Y → Y be two projection operators with Im S Im S / 0. Therefore the operator S defined by S aS bS , a, b ∈ R, is a projection operator with the property Im S Im S if and only if a b 1.

2 aS bS aS bS a 2 S 2 abSS abS S b 2 S 2 a 2 S 2 aS bS aS bS a 2 S 2 2 a 2 SLemma 3 . 16 .Lemma 3 . 17 .
abSS abS S b 2 S a 2 S abS abS b 2 S a aS bS b aS bS a b aS bS 3.17 is obtained.From here, we get the result a b − 1 aS bS 0, that is a b − 1 S 0. The assumption Im S / 0 forces the fact that a b 1.Conversely, if a b 1, then S S abSS abS S b 2 S abSS abS S b 2 S a 2 S abS abS b 2 S a b aS bS aS bS S 3.18 is obtained.Therefore, S is a projection operator.Since Im S Im S is a vector space and S aS bS , then we have Im S ⊆ Im S. Now, let us take an arbitrary element x ∈ Im S Im S / {0}.Therefore, S x aSx bS x ax bx a b x x 3.19 and from here we obtain x ∈ Im S and Im S ⊆ Im S .So the result Im S Im S follows.If P and P are projection operators onto ker L, a b 1 and P aP bP , then K P aK P bK P .Proof.In the case ker L {0}, the proof is clear.Assume that Im P Im P ker L / {0}.Since a b 1 by Lemma 3.15, P is a projection operator and Im P Im P .Since K P I − P K P and PK P 0, then the relation K P I − P K P I − aP − bP K P K P − aP K P − bP K P 1 • K P − bP K P a b K P − bP K P aK P bK P − bP K P aK P b I − P K P aK P bK P Let Z be a vector space, S, S : Z → Z two projection operators with ker S ker S , then for a, b ∈ R, a b 1 the operator S defined by S aS bS is a projection operator with ker S ker S.

3 . 33 Now
, let us indicate how the degree d I − M, Ω, 0 depends on homotopy class of Λ.For this, let us prove the following lemma.

Lemma 3 . 19 . 36 Proposition 3 . 20 .
If G : ker L → ker L is any automorphism and Since P 2 P , PK P,Q 0, and P Λ Λ, thenI −P GP I −M I −M−P PM GP −GP M I −P −ΛΠN −K P,Q N −P P 2 P ΛΠN PK P,Q N GP −GP M I −P −ΛΠN −K P,Q N P ΛΠN GP −GP M I −P −ΛΠN −K P,Q N P ΛΠN GP −GP P ΛΠN K P,Q N I −P −ΛΠN −K P,Q N P ΛΠN GP −GP 2 −GP ΛΠN −GP K P,Q N I −P −ΛΠN −K P,Q N P ΛΠN −GP ΛΠN I − P GP ΛΠ K P,Q N I −M .3.If Λ, Λ ∈ Λ L and M P Λ Π K P,Q N 3.37then we haved I − M , Ω, 0 sgn det Λ Λ −1 d I − M, Ω, 0 .3.38Proof.In Lemma 3.19, if we take G Λ Λ −1 , thenI − M I − P Λ Λ −1 P I − M 3.39is obtained.Now let us show that the operator A I − P Λ Λ −1 P is an automorphism on X.

Corollary 3 . 21 .
Under the assumptions of Proposition 3.18, Leray-Schauder degree |d I − M, Ω, 0 | only depends upon L, N, and Ω.Now, if the orientation on the spaces ker L and Coker L 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 I − M, Ω, 0 .3.44Here, Λ in M is an orientation preserving isomorphism.

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.

Proposition 3.14.
If Coker L and ker L are oriented, then the operators Λ and Λ are homotopic in Λ L if and only if they are both orientation preserving or both orientation reversing transformations.Proof.Assume that a 1 , a 2 , . . ., a n and b 1 , b 2 , . . ., b n are, respectively, bases of Coker L and ker L with respect to chosen the orientation.The basis Λa 1 , Λa 2 , . . ., Λa n on ker L has the same orientation with b 1 , b 2 , . . ., b n if and only if the determinant of the matrix S Namely, let M 1 be the transition matrix from the basis a 1 , a 2 , . . ., a n to the basis b 1 , b 2 , . . ., b n and M 2 be the transition matrix from the basis a 1 , a 2 , . . ., a n to the basis Λa 1 , Λa 2 , . . ., Λa n , SM 2 and det M 1 det S det M 2 .Λa 1 , Λa 2 , . . ., Λa n has the same orientation with b 1 , b 2 , . . ., b n if and only if the determinants of the matrixes M 1 and M 2 a 1 , a 2 , . . ., a n M 2 − −− → Λa 1 , Λa 2 , . . ., Λa n , Λa 1 , Λa 2 , . . ., Λa n S − → b 1 , b 2 , . . ., b n , 3.15 then we have M 1