Fredholm Theory Relative to Any Algebra Homomorphisms

Abstract

In this paper, we give another definition of Ruston elements and almost Ruston elements, which is equivalent to the definitions given by Mouton and Raubenheimer in the case that the homomorphism has a closed range and Riesz property. For two homomorphisms, we consider the preserver problems of Fredholm theory and Fredholm spectrum theory. In addition, we study the spectral mapping theorems of Fredholm (Weyl, Browder, Ruston, and almost Ruston) elements relative to a homomorphism. Last but not least, the dependence of Fredholm theory on three homomorphisms is considered, and meanwhile, the transitivity of Fredholm theory relative to three homomorphisms is illustrated. Furthermore, we consider the Fredholm theory relative to more homomorphisms.

1. Introduction

Fredholm theory originates from the problem of solving integral equations and plays a significant role in operator theory. The classical Fredholm theory can be built in compact manifolds as the study of Fredholm sections of some Banach bundle over the Banach manifold [1]. As we have known, Atkinson established a connection between Fredholm operators and the invertible elements in Calkin algebra [2], which indicates that the Fredholm operators can be regarded as “weakened invertible operators”. Since then, the research on classical Fredholm theory has received widespread attention and flourishing development, which has promoted the development of spectral theory [3,4,5].

On the basis of classical Fredholm theory, many research studies extend Fredholm operator theory to the abstract Fredholm theory in rings and algebras from some novel perspectives. Barnes [6] characterized the Fredholm elements in a ring. After that, Fredholm theories in a semisimple Banach algebra and primitive Banach algebra were studied by Smyth, Aiena, Berkani, etc. [7,8,9,10,11]. Since then, the abstract Fredholm theory has attracted much attention as a hot topic in operator theory and operator algebra, a typical manifestation of which is that Fredholm theory relative to homomorphisms has emerged in the last three decades. Indeed, Harte [12] proposed Fredholm, Weyl and Browder theory relative to a unital homomorphism ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ between general unital Banach algebras ${\displaystyle \mathcal{A}}$ and ${\displaystyle \mathcal{B}}$ as the following notions: an element ${\displaystyle a\in \mathcal{A}}$ is called Fredholm if ${\displaystyle 0}$ is not in the spectrum of ${\displaystyle \mathcal{T}a}$, while ${\displaystyle a}$ is Weyl (Browder) if there exist (commuting) elements ${\displaystyle b}$ and ${\displaystyle c}$ in ${\displaystyle \mathcal{A}}$ with ${\displaystyle a-b+c}$ such that ${\displaystyle 0}$ is not in the spectrum of ${\displaystyle b}$ and ${\displaystyle c}$ is in the null space of ${\displaystyle \mathcal{T}}$. As a continuation of Harte’s work, T. Mouton and H. Raubenheimer focused on Fredholm theory relative to two Banach algebra homomorphisms. And furthermore, H.T. Mouton, S. Mouton, and H. Raubenheimer considered the Ruston elements and Fredholm theory relative to arbitrary homomorphisms. T. Mouton and H. Raubenheimer defined Ruston elements and almost Ruston elements relative to a homomorphism and studied their properties. After that, many wonderful studies for the Fredholm theory relative to homomorphisms have sprung up. Other outstanding works that you can refer are the references [13,14,15,16,17,18,19,20].

Although the above literature has conducted in-depth and systematic research on the Fredholm theory based on Banach algebraic homomorphisms, the spectral theory of Fredholm-type elements relative to a homomorphism still needs further development. And the degree to which Fredholm’s theory relies on homomorphisms is not clear. In order to resolve the above problems, continue H.T. Mouton, S. Mouton and H. Raubenheimer’s work, and extend Harte’s work, this paper focuses on Fredholm theory relative to any algebra homomorphisms. We give the new definitions of Ruston elements and almost Ruston elements and consider the preserver problems of Fredholm theory and Fredholm spectrum theory for two homomorphisms. In addition, we study the spectral theory of Fredholm (Weyl, Browder, Ruston, and almost Ruston) elements relative to a homomorphism. Last but not least, the dependence of Fredholm theory on three homomorphisms is considered, and meanwhile, the transitivity of Fredholm theory relative to three homomorphisms is illustrated.

The result will improve and extend the Fredholm theory relative to a homomorphism and solve the above two problems. The proposal of another concept of Ruston elements and almost Ruston elements is not only necessary and novel, but also it is equivalent to T. Mouton and H. Raubenheimer’s definitions. The paper will extend and enrich the existing Fredholm theory relative to a homomorphism, and promote the development of Fredholm theory.

Throughout this paper, let ${\displaystyle L\left(H\right)}$ denote the algebra of bounded linear operators on an infinite dimensional Hilbert space ${\displaystyle H}$, ${\displaystyle K\left(H\right)}$ the closed ideal of compact operators on ${\displaystyle H}$, and ${\displaystyle L\left(H\right)/K\left(H\right)}$ the Calkin algebra. As we have known, Atkinson in [2] characterized the set of Fredholm operators on a Banach space ${\displaystyle X}$ as those bounded linear operators, which are invertible modulo the two-sided ideal of compact operators on ${\displaystyle X}$. Harte [21] proved that ${\displaystyle T\in L\left(H\right)}$ is a Fredholm operator if and only if ${\displaystyle n\left(T\right)<\infty }$, ${\displaystyle d\left(T\right)<\infty }$ and the range of ${\displaystyle T}$ is closed. Here, ${\displaystyle n\left(T\right)}$, the nullity of ${\displaystyle T}$, is the dimension of the null space ${\displaystyle N\left(T\right)}$, and ${\displaystyle d\left(T\right)}$, the defect of ${\displaystyle T}$, is the co-dimension of the range of ${\displaystyle T}$. Define the index of a Fredholm operator ${\displaystyle T}$ by ${\displaystyle \mathrm{ind}\left(T\right)=n\left(T\right)-d\left(T\right)}$. If ${\displaystyle T\in L\left(H\right)}$ is a Fredholm operator with index 0, we call ${\displaystyle T}$ a Weyl operator. Also, ${\displaystyle T\in B\left(H\right)}$ is called a Browder operator if it is a Fredholm operator with finite ascent ${\displaystyle \alpha \left(T\right)}$ and finite descent ${\displaystyle \beta \left(T\right)}$.

2. Preliminaries

In this paper, all Banach algebras will be complex and unital. We denote the set of all invertible elements of a Banach algebra ${\displaystyle \mathcal{A}}$ by ${\displaystyle {\mathcal{A}}^{-1}}$ and the radical of ${\displaystyle \mathcal{A}}$ by ${\displaystyle Rad\left(\mathcal{A}\right)}$. Suppose ${\displaystyle \mathcal{A}}$ and ${\displaystyle \mathcal{B}}$ are Banach algebras; if ${\displaystyle \mathcal{T}ab=\mathcal{T}a\mathcal{T}b}$ and ${\displaystyle \mathcal{T}}$ maps the unit of ${\displaystyle \mathcal{A}}$ onto the unit of ${\displaystyle \mathcal{B}}$, then we call ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ a homomorphism. Clearly, ${\displaystyle \mathcal{T}{\mathcal{A}}^{-1}\subseteq {\mathcal{B}}^{-1}}$.

We denote the spectrum of an element ${\displaystyle a}$ in a Banach algebra ${\displaystyle \mathcal{A}}$ by ${\displaystyle \sigma \left(a\right)}$, the sets of isolated and accumulation points of ${\displaystyle \sigma \left(a\right)}$ by ${\displaystyle iso\sigma \left(a\right)}$ and ${\displaystyle acc\sigma \left(a\right)}$, respectively, and the spectral radius of ${\displaystyle a}$ by ${\displaystyle r\left(a\right)}$. For ${\displaystyle \lambda \in C\setminus acc\sigma \left(a\right)}$, we write ${\displaystyle p(a,\lambda )}$ for the spectral idempotent of ${\displaystyle a\in \mathcal{A}}$ relative to ${\displaystyle \lambda }$. Suppose ${\displaystyle a\in \mathcal{A}}$; if ${\displaystyle 0\notin acc\sigma \left(a\right)}$, then we call ${\displaystyle a}$ almost invertible. The set of all almost invertible elements of ${\displaystyle \mathcal{A}}$ is denoted by ${\displaystyle {\mathcal{A}}^D}$. An ideal ${\displaystyle I}$ in ${\displaystyle \mathcal{A}}$ will be called inessential whenever ${\displaystyle acc\sigma \left(a\right)\subseteq \left\{0\right\}}$ for all ${\displaystyle a\in I}$. If ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ is a homomorphism and ${\displaystyle N\left(\mathcal{T}\right)}$ is an inessential ideal of ${\displaystyle \mathcal{A}}$, then ${\displaystyle \mathcal{T}}$ is said to have the Riesz property. We denote the set of quasinilpotent (nilpotent) elements in ${\displaystyle \mathcal{A}}$ by ${\displaystyle QN\left(\mathcal{A}\right)\left(N\right(\mathcal{A}\left)\right)}$. The Fredholm theory relative to a homomorphism ${\displaystyle \mathcal{T}}$ is dedicated to dealing with the relationship between the spectra ${\displaystyle \sigma (a,\mathcal{A})}$ and ${\displaystyle \sigma (\mathcal{T}a,\mathcal{B})}$. It is evident that if ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ is a homomorphism and ${\displaystyle a\in \mathcal{A}}$, then ${\displaystyle \sigma (\mathcal{T}a,\mathcal{B})\subset \sigma (a,\mathcal{A})}$.

Proposition 1

([22], Proposition 2.1). Let ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ be a homomorphism. If ${\displaystyle a\in \mathcal{A}}$ and ${\displaystyle \Gamma }$ is a contour in ${\displaystyle C\setminus \sigma \left(a\right)}$ having winding number 0 or 1 around each point of ${\displaystyle \sigma \left(a\right)}$, then ${\displaystyle \mathcal{T}(\frac{1}{2\pi i}{\int }_{\Gamma }{(\lambda -a)}^{-1}d\lambda )=\frac{1}{2\pi i}{\int }_{\Gamma }{(\lambda -\mathcal{T}a)}^{-1}d\lambda }$.

We now recall Harte’s definitions of Fredholm, Weyl, and Browder elements in Banach algebras.

Definition 1

([12], pp. 431–432). Let ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ be a homomorphism. An element ${\displaystyle a\in \mathcal{A}}$ is called the following:

1. Fredholm if ${\displaystyle \mathcal{T}a\in {\mathcal{B}}^{-1}}$;

2. Weyl if there exist elements ${\displaystyle b\in {\mathcal{A}}^{-1}}$ and ${\displaystyle c\in N\left(\mathcal{T}\right)}$ such that ${\displaystyle a=b+c}$;

3. Browder if there exist commuting elements ${\displaystyle b\in {\mathcal{A}}^{-1}}$ and ${\displaystyle c\in N\left(\mathcal{T}\right)}$ such that ${\displaystyle a=b+c}$;

4. Almost invertible Fredholm if it is Fredholm and almost invertible.

Denote by ${\displaystyle F_{\mathcal{T}}, W_{\mathcal{T}}, B_{\mathcal{T}}}$ and ${\displaystyle {\mathcal{A}}^D\bigcap F_{\mathcal{T}}}$ the sets of Fredholm, Weyl, Browder and almost invertible Fredholm elements of ${\displaystyle \mathcal{A}}$ relative to ${\displaystyle \mathcal{T}}$, respectively. From Harte’s results, if ${\displaystyle \mathcal{T}}$ has the Riesz property, then ${\displaystyle a}$ is a Browder element if and only if it is a Fredholm element and an almost invertible element.

If the spectrum of the element ${\displaystyle a+{\mathcal{T}}^{-1}\left(0\right)}$ in the quotient algebra ${\displaystyle \mathcal{A}/{\mathcal{T}}^{-1}\left(0\right)}$ consists of zero, then we call ${\displaystyle a\in \mathcal{A}}$ a Riesz element relative to the closed two-sided ideal ${\displaystyle {\mathcal{T}}^{-1}\left(0\right)}$. We denote the set of Riesz elements in ${\displaystyle \mathcal{A}}$ by ${\displaystyle R_{\mathcal{T}}}$. A two-sided ideal ${\displaystyle I}$ of ${\displaystyle \mathcal{A}}$ is called inessential if for every ${\displaystyle a\in I}$, the spectrum of ${\displaystyle a}$ is either finite or a sequence converging to 0. If ${\displaystyle a\in \mathcal{A}}$, we shall write ${\displaystyle comm\left(a\right)=\{b\in \mathcal{A}:ab=ba\}}$ for the commutant of ${\displaystyle a}$ and ${\displaystyle com{m}^2\left(a\right)=\{c\in \mathcal{A}:ab=ba\Rightarrow bc=cb\}}$ for the double commutant of ${\displaystyle a}$. The closure of a subset ${\displaystyle E}$ of ${\displaystyle \mathcal{A}}$ will be denoted by ${\displaystyle \overline{E}}$. T. Mouton and H. Raubenheimer have given the definitions of Ruston elements and almost Ruston elements as follows.

Definition 2

([18], Definition 3.1). Let ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ be a bounded homomorphism. With respect to ${\displaystyle \mathcal{T}}$, we shall describe an element ${\displaystyle a\in \mathcal{A}}$ as follows:

1. Ruston if ${\displaystyle a\in {\mathcal{A}}^{-1}⨁R_{\mathcal{T}}=\{b+c:b\in {\mathcal{A}}^{-1},c\in R_{\mathcal{T}}andbc=cb\}}$, the commuting sum of an invertible element and a Riesz element relative to ${\displaystyle {\mathcal{T}}^{-1}\left(0\right)}$;

2. Almost Ruston if ${\displaystyle a\in {\mathcal{A}}^{-1}⨁R_{\mathcal{T}}=\{b+c:b\in {\mathcal{A}}^{-1},c\in R_{\mathcal{T}}and\mathcal{T}b\mathcal{T}c=\mathcal{T}c\mathcal{T}b\}}$, the sum of an invertible element and a Riesz element whose images commute.

Evidently, if ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ is a homomorphism, then we have the relation that invertible ${\displaystyle \Rightarrow }$ almost invertible Fredholm ${\displaystyle \Rightarrow }$ Browder ${\displaystyle \Rightarrow }$ Weyl element ${\displaystyle \Rightarrow }$ Fredholm element. And the following relation holds: Browder ${\displaystyle \Rightarrow }$ Ruston ${\displaystyle \Rightarrow }$ almost Ruston. In the following, we first study the preserver problem of Fredholm theory and Fredholm spectrum theory relative to two homomorphisms.

3. The Dependence of Fredholm Theory on Homomorphisms

In this section, we will be concerned with the Fredholm theory relative to two homomorphisms ${\displaystyle \mathcal{T}}$ and ${\displaystyle \mathcal{S}}$. We will consider the condition that ${\displaystyle a\in \mathcal{A}}$ is an ${\displaystyle \mathcal{S}-}$Fredholm (Weyl, Browder, Ruston, and almost Ruston) element when ${\displaystyle a}$ is a ${\displaystyle \mathcal{T}-}$Fredholm type element. And in addition, the spectrum preservation problem relative to the homomorphisms is studied. If there are no special instructions, we assume that ${\displaystyle \mathcal{T}}$ and ${\displaystyle \mathcal{S}}$ are bounded linear.

Theorem 1.

Let ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ and ${\displaystyle \mathcal{S}:\mathcal{A}\to \mathcal{D}}$ be homomorphisms. Suppose ${\displaystyle \mathcal{T}}$ is a one-to-one homomorphism and ${\displaystyle a\in \mathcal{A}}$ satisfies the condition that ${\displaystyle \sigma (a,\mathcal{A})}$ is totally disconnected. Then, if ${\displaystyle a}$ is a ${\displaystyle \mathcal{T}-}$Fredholm element, then ${\displaystyle a}$ is an ${\displaystyle \mathcal{S}-}$Fredholm element.

Proof. 

From ([16], Corollary 2.5), it can be seen that ${\displaystyle \sigma (a,\mathcal{A})=\sigma (\mathcal{T}a,\mathcal{B})}$, from which we obtain that ${\displaystyle a\in {\mathcal{A}}^{-1}}$. Since ${\displaystyle \mathcal{S}}$ is a homomorphism, we obtain that ${\displaystyle \mathcal{S}a}$ is invertible, which indicates that ${\displaystyle a}$ is an ${\displaystyle \mathcal{S}-}$Fredholm element. ${\displaystyle \square }$

Proposition 2.

Let ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ and ${\displaystyle \mathcal{S}:\mathcal{A}\to \mathcal{D}}$ be homomorphisms. Suppose ${\displaystyle \mathcal{S}}$ has the Riesz property and ${\displaystyle a\in \mathcal{A}}$ is a ${\displaystyle \mathcal{T}-}$Fredholm element. Then, we have the following:

(1) If ${\displaystyle a}$ is ${\displaystyle \mathcal{S}-}$Ruston, then ${\displaystyle a}$ is ${\displaystyle \mathcal{T}-}$Ruston.

(2) if ${\displaystyle a}$ is ${\displaystyle \mathcal{S}-}$almost Ruston, then ${\displaystyle a}$ is ${\displaystyle \mathcal{T}-}$almost Ruston.

Proof. 

1. If ${\displaystyle a}$ is ${\displaystyle \mathcal{S}-}$Ruston, then we can see that ${\displaystyle a}$ is ${\displaystyle \mathcal{S}-}$Browder owing to the fact that ${\displaystyle \mathcal{S}}$ has the Riesz property and the homomorphism is bounded linear from ([16], Remark 2.8). According to ([16], Theorem 3.1), it is demonstrated that ${\displaystyle a}$ is ${\displaystyle \mathcal{T}-}$Browder, from which we obtain that ${\displaystyle a}$ is ${\displaystyle \mathcal{T}-}$Ruston.

2. If ${\displaystyle a}$ is ${\displaystyle \mathcal{S}-}$almost Ruston, from ([16], Remark 2.8), it can be seen that ${\displaystyle a}$ is an ${\displaystyle \mathcal{S}-}$Weyl element since ${\displaystyle \mathcal{S}}$ has the Riesz property and the homomorphism is bounded linear. Combining with ([16], Theorem 3.1), we obtain that ${\displaystyle a}$ is a ${\displaystyle \mathcal{T}-}$Weyl element, which indicates that it is a ${\displaystyle \mathcal{T}-}$almost Ruston element. ${\displaystyle \square }$

In the following, we represent the condition such that ${\displaystyle a\in \mathcal{A}}$ is ${\displaystyle \mathcal{T}-}$Fredholm type elements if and only if ${\displaystyle a}$ is ${\displaystyle \mathcal{S}-}$Fredholm type elements.

Corollary 1.

Let ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ and ${\displaystyle \mathcal{S}:\mathcal{A}\to \mathcal{D}}$ be homomorphisms. Suppose ${\displaystyle \mathcal{T},\mathcal{S}}$ are one-to-one homomorphisms and ${\displaystyle a\in \mathcal{A}}$ satisfies the condition that ${\displaystyle \sigma (a,\mathcal{A})}$ is totally disconnected. Then, ${\displaystyle a}$ is a ${\displaystyle \mathcal{T}-}$Fredholm element if and only if ${\displaystyle a}$ is a ${\displaystyle \mathcal{S}-}$Fredholm element.

Proof. 

Note that ${\displaystyle \mathcal{T}}$ is one to one and ${\displaystyle \sigma (a,\mathcal{A})}$ is totally disconnected. Combined with Theorem 1, we obtain that if ${\displaystyle a}$ is a ${\displaystyle \mathcal{T}-}$Fredholm element, then ${\displaystyle a}$ is an ${\displaystyle \mathcal{S}-}$Fredholm element. Similarly, associated with the fact that ${\displaystyle \mathcal{S}}$ is one to one, it follows that ${\displaystyle a}$ is a ${\displaystyle \mathcal{T}-}$Fredholm element if ${\displaystyle a}$ is an ${\displaystyle \mathcal{S}-}$Fredholm element from Theorem 1. ${\displaystyle \square }$

Corollary 2.

Let ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ and ${\displaystyle \mathcal{S}:\mathcal{A}\to \mathcal{D}}$ be homomorphisms. Suppose ${\displaystyle \mathcal{S},\mathcal{T}}$ have the Riesz property and ${\displaystyle a\in \mathcal{A}}$ is ${\displaystyle \mathcal{T}-}$Fredholm and ${\displaystyle \mathcal{S}-}$Fredholm. Then, we have the following:

(1) ${\displaystyle a}$ is ${\displaystyle \mathcal{S}-}$Ruston if and only if ${\displaystyle a}$ is ${\displaystyle \mathcal{T}-}$Ruston.

(2) ${\displaystyle a}$ is ${\displaystyle \mathcal{S}-}$almost Ruston if and only if ${\displaystyle a}$ is ${\displaystyle \mathcal{T}-}$almost Ruston.

Proof. 

(1) If ${\displaystyle a}$ is an ${\displaystyle \mathcal{S}-}$Ruston element, then one can obtain that ${\displaystyle a}$ is ${\displaystyle \mathcal{T}-}$Ruston since ${\displaystyle \mathcal{S}}$ has the Riesz property and ${\displaystyle a\in \mathcal{A}}$ is a ${\displaystyle \mathcal{T}-}$Fredholm element according to Proposition 2. Similarly, it can be obtained in the other direction.

(2) If ${\displaystyle a}$ is an ${\displaystyle \mathcal{S}-}$almost Ruston element, associated with the fact that ${\displaystyle \mathcal{S}}$ has the Riesz property and ${\displaystyle a\in \mathcal{A}}$ is a ${\displaystyle \mathcal{T}-}$Fredholm element, it can be obtained that ${\displaystyle a}$ is a ${\displaystyle \mathcal{T}-}$almost Ruston by means of Proposition 2. Similarly, the other direction can be proved by Proposition 2. ${\displaystyle \square }$

In the following, the spectrum preservation problem relative to homomorphisms will be discussed. Recall that ([16], Definition 2.9) we only use the same notation.

Definition 3

([16], Definition 2.9). Letting ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ be a homomorphism and let ${\displaystyle a\in \mathcal{A}}$ with respect to ${\displaystyle \mathcal{T}}$, we have the following:

(1) The spectrum ${\displaystyle \sigma (\mathcal{T}a,\mathcal{B})}$ is the Fredholm spectrum of ${\displaystyle a}$.

(2) The set ${\displaystyle {\omega }_{\mathcal{T}}\left(a\right)=\{\lambda \in C:\lambda -aisnotWeyl\}=\bigcap \{\sigma (a-c,\mathcal{A}):\mathcal{T}c=0\}}$ is the Weyl spectrum of ${\displaystyle a}$.

(3) The set ${\displaystyle {\omega }_{\mathcal{T}}^{comm}\left(a\right)=\{\lambda \in C:\lambda -aisnotBrowder\}=\bigcap \{\sigma (a-c,\mathcal{A}):\mathcal{T}c=0, ac=ca\}}$ is the Browder spectrum of ${\displaystyle a}$.

(4) The set ${\displaystyle {\nu }_{\mathcal{T}}\left(a\right)=\{\lambda \in C:\lambda -aisnotalmostRuston\}}$ is the almost Ruston spectrum of ${\displaystyle a}$.

(5) The set ${\displaystyle {\nu }_{\mathcal{T}}^{comm}\left(a\right)=\{\lambda \in C:\lambda -aisnotRuston\}}$ is the Ruston spectrum of ${\displaystyle a}$.

Next, we give the spectrum preservation properties of Ruston and almost Ruston spectra under the homomorphisms.

Corollary 3.

Let ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ and ${\displaystyle \mathcal{S}:\mathcal{A}\to \mathcal{D}}$ be homomorphisms. Suppose ${\displaystyle \mathcal{S}}$ has the Riesz property and ${\displaystyle a\in \mathcal{A}}$. If ${\displaystyle \sigma (\mathcal{T}a,\mathcal{B})\subset \sigma (\mathcal{S}a,\mathcal{D})}$, then we have the following.

(1) ${\displaystyle {\nu }_{\mathcal{T}}\left(a\right)\subset {\nu }_{\mathcal{S}}\left(a\right)}$;

(2) ${\displaystyle {\nu }_{\mathcal{T}}^{comm}\left(a\right)\subset {\nu }_{\mathcal{S}}^{comm}\left(a\right)}$.

Proof. 

(1) Suppose ${\displaystyle \lambda \notin {\nu }_S\left(a\right)}$, then it can be seen that ${\displaystyle a-\lambda }$ is an ${\displaystyle \mathcal{S}-}$almost Ruston element, from which we obtain that ${\displaystyle a-\lambda }$ is an ${\displaystyle \mathcal{S}-}$Fredholm element. This shows that ${\displaystyle \mathcal{S}(a-\lambda )}$ is invertible. In other words, ${\displaystyle \lambda \notin \sigma (\mathcal{S}a,D)}$, which induces that ${\displaystyle \lambda \notin \sigma (\mathcal{T}a,B)}$. That is to say, ${\displaystyle \mathcal{T}a-\lambda }$ is invertible, and hence ${\displaystyle a-\lambda }$ is ${\displaystyle \mathcal{T}-}$Fredholm. According to Proposition 2, we obtain that ${\displaystyle a-\lambda }$ is a ${\displaystyle \mathcal{T}-}$almost Ruston element, which is equivalent to ${\displaystyle \lambda \notin {\nu }_T\left(a\right)}$.

(2) Suppose ${\displaystyle \lambda \notin {\nu }_S^{comm}\left(a\right)}$, then we obtain that ${\displaystyle a-\lambda }$ is an ${\displaystyle \mathcal{S}-}$Ruston element, which indicates that ${\displaystyle a-\lambda }$ is an ${\displaystyle \mathcal{S}-}$Fredholm element, and hence ${\displaystyle \mathcal{S}(a-\lambda )}$ is invertible. In other words, ${\displaystyle \lambda \notin \sigma (\mathcal{S}a,D)}$, from which we obtain that ${\displaystyle \lambda \notin \sigma (\mathcal{T}a,B)}$. In other words, ${\displaystyle \mathcal{T}a-\lambda }$ is invertible. Hence, ${\displaystyle a-\lambda }$ is ${\displaystyle \mathcal{T}-}$Fredholm. According to Proposition 2, we know that ${\displaystyle a-\lambda }$ is a ${\displaystyle \mathcal{T}-}$Ruston element. Therefore, it follows that ${\displaystyle \lambda \notin {\nu }_T^{comm}\left(a\right)}$. ${\displaystyle \square }$

Corollary 4.

Let ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ and ${\displaystyle \mathcal{S}:\mathcal{A}\to \mathcal{D}}$ be homomorphisms. Suppose ${\displaystyle \mathcal{T},\mathcal{S}}$ have the Riesz property and ${\displaystyle a\in \mathcal{A}}$. If ${\displaystyle \sigma (\mathcal{T}a,\mathcal{B})=\sigma (\mathcal{S}a,\mathcal{D})}$, then

(1) ${\displaystyle {\nu }_{\mathcal{T}}\left(a\right)={\nu }_{\mathcal{S}}\left(a\right)}$;

(2) ${\displaystyle {\nu }_{\mathcal{T}}^{comm}\left(a\right)={\nu }_{\mathcal{S}}^{comm}\left(a\right)}$.

Proof. 

Since ${\displaystyle \mathcal{S}}$ has the Riesz property and ${\displaystyle \sigma (\mathcal{T}a,B)\subset \sigma (\mathcal{S}a,D)}$, then one can obtain that ${\displaystyle {\nu }_T\left(a\right)\subset {\nu }_S\left(a\right)}$ and ${\displaystyle {\nu }_T^{comm}\left(a\right)\subset {\nu }_S^{comm}\left(a\right)}$ from Corollary 3. Conversely, associated with the fact that ${\displaystyle \mathcal{T}}$ has the Riesz property and ${\displaystyle \sigma (\mathcal{S}a,D)\subset \sigma (\mathcal{T}a,B)}$, it can be obtained that ${\displaystyle {\nu }_S\left(a\right)\subset {\nu }_T\left(a\right)}$ and ${\displaystyle {\nu }_S^{comm}\left(a\right)\subset {\nu }_T^{comm}\left(a\right)}$, which leads to the conclusion that ${\displaystyle {\nu }_{\mathcal{T}}\left(a\right)={\nu }_{\mathcal{S}}\left(a\right)}$ and ${\displaystyle {\nu }_{\mathcal{T}}^{comm}\left(a\right)={\nu }_{\mathcal{S}}^{comm}\left(a\right)}$. ${\displaystyle \square }$

In order to research the preserving problem, in detail, if ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ and ${\displaystyle \mathcal{S}:\mathcal{A}\to \mathcal{D}}$ are homomorphisms, is ${\displaystyle a}$ a ${\displaystyle \mathcal{T}-}$Fredholm (Weyl, Browder, Ruston, almost Ruston) element if and only if ${\displaystyle a}$ is an ${\displaystyle \mathcal{S}-}$Fredholm (Weyl, Browder, Ruston, almost Ruston) element? Next, we provide another characterization method.

Theorem 2.

Let ${\displaystyle \mathcal{A},\mathcal{B}}$ be Banach algebras and let ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ and ${\displaystyle \mathcal{S}:\mathcal{A}\to \mathcal{D}}$ be homomorphisms such that ${\displaystyle \mathcal{S}}$ has the Riesz property. Suppose that ${\displaystyle \mathcal{T}}$ has the property that whenever ${\displaystyle p}$ is a spectral idempotent in ${\displaystyle {\mathcal{S}}^{-1}\left(0\right)}$, then ${\displaystyle p\in {\mathcal{T}}^{-1}\left(0\right)}$. Then, we have the following:

(1) If ${\displaystyle a\in \mathcal{A}}$ is ${\displaystyle \mathcal{S}-}$Ruston, then it is ${\displaystyle \mathcal{T}-}$Ruston;

(2) If ${\displaystyle a\in \mathcal{A}}$ is ${\displaystyle \mathcal{S}-}$almost Ruston, then it is ${\displaystyle \mathcal{T}-}$almost Ruston.

Proof. 

1. If ${\displaystyle a\in \mathcal{A}}$ is ${\displaystyle \mathcal{S}-}$Ruston, then ${\displaystyle a}$ is an ${\displaystyle \mathcal{S}-}$Browder element since ${\displaystyle \mathcal{S}}$ has the Riesz property and the homomorphism is bounded linear according to ([16], Remark 2.8). From ([17], Theorem 5.4), it can be obtained that ${\displaystyle a}$ is ${\displaystyle \mathcal{T}-}$Browder, which implies that ${\displaystyle a}$ is a ${\displaystyle \mathcal{T}-}$Ruston element.

2. If ${\displaystyle a}$ is ${\displaystyle \mathcal{S}-}$almost Ruston, then from ([16], Remark 2.8), we obtain that ${\displaystyle a}$ is an ${\displaystyle \mathcal{S}-}$Weyl element since ${\displaystyle \mathcal{S}}$ has the Riesz property and the homomorphism is bounded linear. By means of ([17], Theorem 5.4), it can be obtained that ${\displaystyle a}$ is a ${\displaystyle \mathcal{T}-}$Weyl element, which indicates that ${\displaystyle a}$ is a ${\displaystyle \mathcal{T}-}$almost Ruston element. ${\displaystyle \square }$

Proposition 3.

Let ${\displaystyle \mathcal{A},\mathcal{B}}$ be Banach algebras and let ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ and ${\displaystyle \mathcal{S}:\mathcal{A}\to \mathcal{D}}$ be homomorphisms such that ${\displaystyle \mathcal{S},\mathcal{T}}$ have the Riesz property. Suppose that ${\displaystyle p}$ is a spectral idempotent in ${\displaystyle {\mathcal{S}}^{-1}\left(0\right)}$ if and only if ${\displaystyle p\in {\mathcal{T}}^{-1}\left(0\right)}$. Then, we have the following:

(1) ${\displaystyle a\in \mathcal{A}}$ is ${\displaystyle \mathcal{S}-}$Ruston if and only if it is ${\displaystyle \mathcal{T}-}$Ruston;

(2) ${\displaystyle a\in \mathcal{A}}$ is ${\displaystyle \mathcal{S}-}$almost Ruston if and only if ${\displaystyle a}$ is a ${\displaystyle \mathcal{T}-}$almost Ruston element.

Proof. 

(1) Since ${\displaystyle \mathcal{S}}$ has the Riesz property, and ${\displaystyle p\in {\mathcal{T}}^{-1}\left(0\right)}$ if ${\displaystyle p}$ is a spectral idempotent in ${\displaystyle {\mathcal{S}}^{-1}\left(0\right)}$, we obtain that ${\displaystyle a}$ is a ${\displaystyle \mathcal{T}-}$Ruston element if it is a ${\displaystyle \mathcal{S}-}$Ruston from Theorem 2. Similarly, from the fact that ${\displaystyle \mathcal{T}}$ has the Riesz property and ${\displaystyle p\in {\mathcal{S}}^{-1}\left(0\right)}$ if ${\displaystyle p}$ is a spectral idempotent in ${\displaystyle {\mathcal{T}}^{-1}\left(0\right)}$, it can be obtained in the other direction.

(2) According to Theorem 2, one can prove the result similar to (1). ${\displaystyle \square }$

One can obtain the corresponding results about the Ruston spectrum and almost Ruston spectrum.

Corollary 5.

Let ${\displaystyle \mathcal{A},\mathcal{B}}$ be Banach algebras and let ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ and ${\displaystyle \mathcal{S}:\mathcal{A}\to \mathcal{D}}$ be homomorphisms such that ${\displaystyle \mathcal{S}}$ has the Riesz property. Suppose ${\displaystyle p\in {\mathcal{S}}^{-1}\left(0\right)}$ is a spectral idempotent, then ${\displaystyle p\in {\mathcal{T}}^{-1}\left(0\right)}$. Then, we have the following:

(1) ${\displaystyle {\nu }_{\mathcal{T}}\left(a\right)\subset {\nu }_{\mathcal{S}}\left(a\right)}$;

(2) ${\displaystyle {\nu }_{\mathcal{T}}^{comm}\left(a\right)\subset {\nu }_{\mathcal{S}}^{comm}\left(a\right)}$.

Proof. 

(1) Suppose ${\displaystyle \lambda \notin {\nu }_S\left(a\right)}$, that is to say, ${\displaystyle a-\lambda }$ is a ${\displaystyle \mathcal{S}-}$almost Ruston element, which induces that ${\displaystyle a-\lambda }$ is a ${\displaystyle \mathcal{T}-}$almost Ruston element according to Theorem 2. In other words, ${\displaystyle \lambda \notin {\nu }_T\left(a\right)}$.

(2) Suppose ${\displaystyle \lambda \notin {\nu }_S^{comm}\left(a\right)}$, that is to say, ${\displaystyle a-\lambda }$ is an ${\displaystyle \mathcal{S}-}$Ruston element. This shows that ${\displaystyle a-\lambda }$ is a ${\displaystyle \mathcal{T}-}$Ruston element according to Theorem 2. In other words, ${\displaystyle \lambda \notin {\nu }_T^{comm}\left(a\right)}$. ${\displaystyle \square }$

Corollary 6.

Let ${\displaystyle \mathcal{A},\mathcal{B}}$ be Banach algebras and let ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ and ${\displaystyle \mathcal{S}:\mathcal{A}\to \mathcal{D}}$ be homomorphisms such that ${\displaystyle \mathcal{S},\mathcal{T}}$ has the Riesz property. Suppose ${\displaystyle p}$ is a spectral idempotent, then ${\displaystyle p\in {\mathcal{T}}^{-1}\left(0\right)}$ if and only if ${\displaystyle p\in {\mathcal{S}}^{-1}\left(0\right)}$. Then, we have the following:

(1) ${\displaystyle {\nu }_{\mathcal{T}}\left(a\right)={\nu }_{\mathcal{S}}\left(a\right)}$;

(2) ${\displaystyle {\nu }_{\mathcal{T}}^{comm}\left(a\right)={\nu }_{\mathcal{S}}^{comm}\left(a\right)}$.

Proof. 

According to Corollary 5, it suffices to prove that ${\displaystyle {\nu }_T\left(a\right)\supset {\nu }_S\left(a\right)}$ and ${\displaystyle {\nu }_T^{comm}\left(a\right)\supset {\nu }_S^{comm}\left(a\right)}$. Since ${\displaystyle \mathcal{T}}$ has the Riesz property and ${\displaystyle p\in {\mathcal{S}}^{-1}\left(0\right)}$ if ${\displaystyle p}$ is a spectral idempotent in ${\displaystyle {\mathcal{T}}^{-1}\left(0\right)}$, we can see that ${\displaystyle {\nu }_S\left(a\right)\subset {\nu }_T\left(a\right)}$ and ${\displaystyle {\nu }_S^{comm}\left(a\right)\subset {\nu }_T^{comm}\left(a\right)}$ from Corollary 5. ${\displaystyle \square }$

Except the above determination methods, we will provide the third method such that the preserving problems hold. At first, a necessary lemma is listed as follows.

Lemma 1

([17], Theorem 6.6). Let ${\displaystyle \mathcal{A},\mathcal{B}}$ be Banach algebras and let ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ be a homomorphism with closed range which satisfies the Riesz property. Then, we have the following:

(1) ${\displaystyle a}$ is a Browder element if and only if ${\displaystyle a}$ is a Ruston element.

(2) ${\displaystyle a}$ is a Weyl element if and only if ${\displaystyle a}$ is an almost Ruston element.

Proposition 4.

Suppose that ${\displaystyle \mathcal{A},\mathcal{B},\mathcal{D}}$ are Banach algebras. Let ${\displaystyle \mathcal{T},\mathcal{S}:\mathcal{A}\to \mathcal{B}}$ be bounded linear homomorphisms.

(1) Let ${\displaystyle \mathcal{S}:\mathcal{A}\to \mathcal{D}}$ be a homomorphism with closed range which satisfies the Riesz property and ${\displaystyle a\in \mathcal{A}}$ is ${\displaystyle \mathcal{T}-}$Fredholm. Then, we have the following:

(i) If ${\displaystyle a}$ is ${\displaystyle \mathcal{S}-}$Ruston, then ${\displaystyle a}$ is ${\displaystyle \mathcal{T}-}$Ruston;

(ii) If ${\displaystyle a}$ is ${\displaystyle \mathcal{S}-}$almost Ruston, then ${\displaystyle a}$ is ${\displaystyle \mathcal{T}-}$almost Ruston.

(2) Suppose ${\displaystyle \mathcal{S},\mathcal{T}}$ has the Riesz property with closed range, and ${\displaystyle a\in \mathcal{A}}$ is ${\displaystyle \mathcal{T}-}$Fredholm and ${\displaystyle \mathcal{S}-}$Fredholm. Then, we have the following:

(i) ${\displaystyle a}$ is ${\displaystyle \mathcal{S}-}$Ruston if and only if ${\displaystyle a}$ is ${\displaystyle \mathcal{T}-}$Ruston;

(ii) ${\displaystyle a}$ is ${\displaystyle \mathcal{S}-}$almost Ruston if and only if ${\displaystyle a}$ is ${\displaystyle \mathcal{T}-}$almost Ruston element.

(3) Let ${\displaystyle \mathcal{S}:\mathcal{A}\to \mathcal{D}}$ be homomorphism with closed range satisfying the Riesz property. Suppose ${\displaystyle \mathcal{T}}$ has the property that whenever ${\displaystyle p}$ is a spectral idempotent in ${\displaystyle {\mathcal{S}}^{-1}\left(0\right)}$, then ${\displaystyle p\in {\mathcal{T}}^{-1}\left(0\right)}$. Then, we have the following:

(i) If ${\displaystyle a\in \mathcal{A}}$ is ${\displaystyle \mathcal{S}-}$Ruston, then ${\displaystyle a}$ is ${\displaystyle \mathcal{T}-}$Ruston;

(ii) If ${\displaystyle a\in \mathcal{A}}$ is ${\displaystyle \mathcal{S}-}$almost Ruston, then ${\displaystyle a}$ is a ${\displaystyle \mathcal{T}-}$almost Ruston element.

(4) Let ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ and ${\displaystyle \mathcal{S}:\mathcal{A}\to \mathcal{D}}$ be homomorphisms with closed range satisfying the Riesz property. Suppose that whenever ${\displaystyle p}$ is a spectral idempotent in ${\displaystyle {\mathcal{S}}^{-1}\left(0\right)}$ if and only if ${\displaystyle p\in {\mathcal{T}}^{-1}\left(0\right)}$. Then, we have the following:

(i) ${\displaystyle a\in \mathcal{A}}$ is ${\displaystyle \mathcal{S}-}$Ruston if and only if ${\displaystyle a}$ is ${\displaystyle \mathcal{T}-}$Ruston;

(ii) ${\displaystyle a\in \mathcal{A}}$ is ${\displaystyle \mathcal{S}-}$almost Ruston if and only if ${\displaystyle a}$ is a ${\displaystyle \mathcal{T}-}$almost Ruston element.

Proof. 

(1) Suppose ${\displaystyle a}$ is an ${\displaystyle \mathcal{S}-}$Ruston element, then one can obtain that ${\displaystyle a}$ is an ${\displaystyle \mathcal{S}-}$Browder element from Lemma 3. According to ([16], Theorem 3.1), it is demonstrated that ${\displaystyle a}$ is ${\displaystyle \mathcal{T}-}$Browder, which indicates that ${\displaystyle a}$ is a ${\displaystyle \mathcal{T}-}$Ruston element. Similarly, if ${\displaystyle a}$ is an ${\displaystyle \mathcal{S}-}$almost Ruston element, then we obtain that ${\displaystyle a}$ is an ${\displaystyle \mathcal{S}-}$Weyl element by Lemma 3, which implies that ${\displaystyle a}$ is a ${\displaystyle \mathcal{T}-}$Weyl element from ([16], Theorem 3.1). Consequently, it follows that ${\displaystyle a}$ is a ${\displaystyle \mathcal{T}-}$almost Ruston element.

(2) It suffices to prove that if ${\displaystyle a}$ is a ${\displaystyle \mathcal{T}-}$Ruston (${\displaystyle \mathcal{T}-}$almost Ruston) element, then ${\displaystyle a}$ is an ${\displaystyle \mathcal{S}-}$Ruston (${\displaystyle \mathcal{S}-}$ almost Ruston) element. Indeed, from the fact that ${\displaystyle \mathcal{T}}$ has the Riesz property with closed range and ${\displaystyle a}$ is an ${\displaystyle \mathcal{S}-}$Fredholm element, one can obtain the conclusion by means of the above (1).

(3) According to Theorem 2, one can obtain the conclusion directly.

(4) From the above result (3), it suffices to prove that if ${\displaystyle a}$ is a ${\displaystyle \mathcal{T}-}$Ruston (${\displaystyle \mathcal{T}-}$almost Ruston) element, then ${\displaystyle a}$ is an ${\displaystyle \mathcal{S}-}$Ruston (almost Ruston) element, which holds by means of the fact that ${\displaystyle \mathcal{T}}$ has the Riesz property and ${\displaystyle p\in {\mathcal{S}}^{-1}\left(0\right)}$ if ${\displaystyle p}$ is a spectral idempotent in ${\displaystyle {\mathcal{T}}^{-1}\left(0\right)}$. ${\displaystyle \square }$

Similarly, one can obtain the corresponding results about the spectra.

Remark 1.

Let ${\displaystyle \mathcal{A},\mathcal{B}}$ be Banach algebras and let ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ and ${\displaystyle \mathcal{S}:\mathcal{A}\to \mathcal{D}}$ be homomorphisms such that ${\displaystyle \mathcal{T},\mathcal{S}}$ have the Riesz property. Suppose ${\displaystyle \mathcal{T}}$ and ${\displaystyle \mathcal{S}}$ have the same spectral idempotent set. Then, we have the following:

(1) ${\displaystyle a\in \mathcal{A}}$ is ${\displaystyle \mathcal{S}-}$Browder if and only if ${\displaystyle a}$ is ${\displaystyle \mathcal{T}-}$Browder;

(2) ${\displaystyle a\in \mathcal{A}}$ is ${\displaystyle \mathcal{S}-}$Weyl if and only if ${\displaystyle a}$ is ${\displaystyle \mathcal{T}-}$Weyl.

Indeed, the definitions of Ruston elements and almost Ruston elements in [17] are equivalent to the definitions of [16]. We can provide a new definition of Ruston elements and almost Ruston elements denoted by K-Ruston elements and almost K-Ruston elements, which is equivalent to the definitions of [16,17] in the case where the homomorphism has a closed range and satisfies the Riesz property. It is a reasonable definition, which coincides with the other two definitions given by [16,17].

Definition 4.

Let ${\displaystyle \mathcal{A}}$ and ${\displaystyle \mathcal{B}}$ be Banach algebras and let ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ be a homomorphism with respect to ${\displaystyle \mathcal{T}}$. We shall describe an element ${\displaystyle a\in \mathcal{A}}$ as follows:

1. K-Ruston if ${\displaystyle a\in {\mathcal{A}}^{-1}⨁{\mathcal{T}}^{-1}\left(N\left(\mathcal{B}\right)\right)=\{b+c,b\in {\mathcal{A}}^{-1},\mathcal{T}c\in N\left(\mathcal{B}\right)andbc=cb\}}$, the commuting sum of an invertible element and an element whose image is nilpotent in ${\displaystyle \mathcal{B}}$;

2. Almost K-Ruston if ${\displaystyle a\in {\mathcal{A}}^{-1}⨁{\mathcal{T}}^{-1}\left(N\left(\mathcal{B}\right)\right)=\{b+c,b\in {\mathcal{A}}^{-1},\mathcal{T}c\in N\left(\mathcal{B}\right)and\mathcal{T}b\mathcal{T}c=\mathcal{T}c\mathcal{T}b\}}$, the commuting sum of an invertible element and an element whose image is nilpotent in ${\displaystyle \mathcal{B}}$.

The above definitions are equivalent to the definitions of Ruston elements and almost Ruston elements in [16,17] when the homomorphism has a closed range and the Riesz property. Hence, similar to ([17], Theorems 6.2 and 6.6), one can obtain the following conclusion.

Theorem 3.

(1) If ${\displaystyle \mathcal{A}}$ and ${\displaystyle \mathcal{B}}$ are Banach algebras and ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ is a homomorphism, then every (almost) K-Ruston element is Fredholm.

(2) Let ${\displaystyle \mathcal{A}}$ and ${\displaystyle \mathcal{B}}$ be Banach algebras and let ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ be a homomorphism with closed range which satisfies the Riesz property. Then, every K-Ruston element is a Browder element, and every almost K-Ruston element is a Weyl element.

Proof. 

(1) Suppose ${\displaystyle a=b+c}$ with ${\displaystyle b\in {\mathcal{A}}^{-1}}$, ${\displaystyle \mathcal{T}c\in N\left(B\right)}$, ${\displaystyle \mathcal{T}b\mathcal{T}c=\mathcal{T}c\mathcal{T}b}$. Since ${\displaystyle \mathcal{T}b\in {\mathcal{B}}^{-1}}$, we obtain that ${\displaystyle 0\notin \sigma (\mathcal{T}b,B)}$. Associated with the fact that ${\displaystyle \mathcal{T}b}$ and ${\displaystyle \mathcal{T}c}$ commutes, it can be seen that ${\displaystyle \sigma (\mathcal{T}a,B)=\sigma (\mathcal{T}b+\mathcal{T}c,B)\subseteq \sigma (\mathcal{T}b,B)+\sigma (\mathcal{T}c,B)=\sigma (\mathcal{T}b,B)+\left\{0\right\}=\sigma (\mathcal{T}b,B)}$. This shows that ${\displaystyle 0\notin \sigma (\mathcal{T}a,B)}$, from which we obtain that ${\displaystyle a}$ is a Fredholm element.

(2) Suppose ${\displaystyle a=b+c}$, where ${\displaystyle b\in {\mathcal{A}}^{-1}}$, ${\displaystyle \mathcal{T}c\in N\left(B\right)}$ and ${\displaystyle bc=cb}$. It is evident that ${\displaystyle \mathcal{T}c\in QN\left(B\right)}$. It follows that ${\displaystyle a}$ is almost invertible according to the proof of ([17], Theorem 6.6). From the above result (1), we obtain that ${\displaystyle a}$ is a Browder element owing to the fact that ${\displaystyle \mathcal{T}}$ has the Riesz property.

Suppose ${\displaystyle a=b+c}$, where ${\displaystyle b\in {\mathcal{A}}^{-1}}$, ${\displaystyle \mathcal{T}c\in N\left(B\right)}$ and ${\displaystyle \mathcal{T}b\mathcal{T}c=\mathcal{T}c\mathcal{T}b}$. One can check that ${\displaystyle \mathcal{T}\left(b^{-1}c\right)\in N\left(B\right)}$, and hence ${\displaystyle b^{-1}a=1+b^{-1}c}$ is K-Ruston. It follows that ${\displaystyle b^{-1}a}$ is Browder, which implies that ${\displaystyle a}$ is a Weyl element as ${\displaystyle a=b\left(b^{-1}a\right)}$. ${\displaystyle \square }$

Similarly, one can obtain the results about the Ruston spectrum and almost Ruston spectrum.

4. The Spectral Mapping Theorem Relative to a Homomorphism

In this section, we focus on the spectral mapping theorem relative to a homomorphism. At first, some necessary lemmas are listed.

Lemma 2.

Let ${\displaystyle \mathcal{A}}$ and ${\displaystyle \mathcal{B}}$ be Banach algebras, ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ be a bounded linear homomorphism satisfying the Riesz property and ${\displaystyle a_1,a_2\in \mathcal{A}}$ such that ${\displaystyle a_1{a}_2=a_2{a}_1}$. If ${\displaystyle a_1{a}_2}$ is a Ruston element, then both ${\displaystyle a_1}$ and ${\displaystyle a_2}$ are Ruston elements.

Proof. 

If ${\displaystyle a_1{a}_2}$ is a Ruston element, then ${\displaystyle a_1{a}_2}$ is a Browder element from ([16], Remark 2.8). One can obtain that ${\displaystyle a_1}$ and ${\displaystyle a_2}$ are Browder elements from ([17], Theorem 8.10), which induces that ${\displaystyle a_1}$ and ${\displaystyle a_2}$ are Ruston elements according to ([18], Corollary 3.6). ${\displaystyle \square }$

Lemma 3.

Let ${\displaystyle \mathcal{A}}$ and ${\displaystyle \mathcal{B}}$ be Banach algebras, ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ be a homomorphism, ${\displaystyle a\in \mathcal{A}}$ and ${\displaystyle n\in N}$. If ${\displaystyle a}$ is a Fredholm element, then ${\displaystyle a^n}$ is a Fredholm element.

Proof. 

It suffices to prove that ${\displaystyle a^2}$ is a Fredholm element if ${\displaystyle a}$ is a Fredholm element. If ${\displaystyle a}$ is a Fredholm element, then ${\displaystyle \mathcal{T}a}$ is invertible in ${\displaystyle \mathcal{B}}$. Note that ${\displaystyle \mathcal{T}{a}^2=\mathcal{T}a\mathcal{T}a}$ is invertible in ${\displaystyle \mathcal{B}}$, and we obtain that ${\displaystyle a^2}$ is a Fredholm element. It is not difficult to see that ${\displaystyle a^n}$ is a Fredholm element. ${\displaystyle \square }$

In the following, for the Fredholm elements, Weyl elements, and Browder elements relative to a homomorphism, we consider the spectral mapping theorems of the Fredholm spectrum, Weyl spectrum, and Browder spectrum. One can find that the Fredholm spectrum and Browder spectrum relative to a homomorphism satisfy the spectral mapping theorem, but the Weyl spectrum does not satisfy it, which coincides with the situation of the spectra of Fredholm, Weyl, and Browder operators. The results extend the classical spectrum theory.

Theorem 4.

Let ${\displaystyle \mathcal{A}}$ and ${\displaystyle \mathcal{B}}$ be Banach algebras and ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ be a homomorphism. Then, ${\displaystyle {\sigma }_e(f\left(a\right),\mathcal{A})=f\left({\sigma }_e(a,\mathcal{A})\right)}$ for all ${\displaystyle a\in \mathcal{A}}$ and for all ${\displaystyle f}$ analytic and one to one on a neighborhood of ${\displaystyle \sigma (a,\mathcal{A})}$, where ${\displaystyle {\sigma }_e\left(a\right)=\sigma (\mathcal{T}a,\mathcal{B})}$ is the Fredholm spectrum.

Proof. 

If the function is analytic and one to one on a neighborhood of ${\displaystyle \sigma (a,\mathcal{A})}$, then it is not difficult to show the equality from ([23], Lemma 3.11). So let ${\displaystyle \lambda \notin f\left({\sigma }_e(a,\mathcal{A})\right)}$. We may suppose that ${\displaystyle \lambda \in f\left(\sigma \right(a,\mathcal{A}\left)\right)}$ since ${\displaystyle {\sigma }_e(f\left(a\right),\mathcal{A})\subseteq \sigma (f\left(a\right),\mathcal{A})}$. It follows from ([24], Corollaries 3.8 and 3.9) that there exist ${\displaystyle {\lambda }_1,{\lambda }_2,\cdots ,{\lambda }_n\in \sigma (a,\mathcal{A})}$, ${\displaystyle k_1,\cdots ,k_n\in N}$ and a function ${\displaystyle g}$ which is analytic on a neighborhood ${\displaystyle G}$ of ${\displaystyle \sigma (a,\mathcal{A})}$ and has no zeroes in ${\displaystyle G}$ such that ${\displaystyle f\left(z\right)-\lambda ={(z-{\lambda }_1)}^{k_1}\cdots {(z-{\lambda }_n)}^{k_n}g\left(z\right)}$ for all ${\displaystyle z\in G}$. Since ${\displaystyle f}$ is one to one on ${\displaystyle G}$, from which we deduce that ${\displaystyle n=1}$, ${\displaystyle f\left(z\right)-\lambda ={(z-{\lambda }_1)}^{k_1}g\left(z\right)}$, ${\displaystyle f\left(a\right)-\lambda ={(a-{\lambda }_1)}^{k_1}g\left(a\right)}$ with ${\displaystyle g\left(a\right)\in {\mathcal{A}}^{-1}}$ and ${\displaystyle a-{\lambda }_1}$ is thus a Fredholm element. We obtain that ${\displaystyle {(a-{\lambda }_1)}^k}$ is a Fredholm element. Therefore, ${\displaystyle f\left(a\right)-\lambda }$ is Fredholm, which indicates that ${\displaystyle \lambda \notin {\sigma }_e(f\left(a\right),\mathcal{A})}$. ${\displaystyle \square }$

Remark 2.

Let ${\displaystyle \mathcal{A}}$ and ${\displaystyle \mathcal{B}}$ be Banach algebras and ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ be a bounded homomorphism. If either ${\displaystyle \mathcal{B}}$ is commutative or ${\displaystyle \mathcal{T}}$ has a closed range and satisfies the Riesz property, then ${\displaystyle {\omega }_{\mathcal{T}}(f\left(a\right),\mathcal{A})\subseteq f\left({\omega }_{\mathcal{T}}(a,\mathcal{A})\right)}$ for all ${\displaystyle a\in \mathcal{A}}$ and every function ${\displaystyle f}$ is analytic on a neighborhood of ${\displaystyle \sigma (a,\mathcal{A})}$, which is non-constant on each component of its domain of definition.

Next, the spectral mapping theorem of the Browder spectrum relative to a homomorphism is illustrated.

Lemma 4.

Let ${\displaystyle \mathcal{A}}$ and ${\displaystyle \mathcal{B}}$ be Banach algebras, ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ be a homomorphism, ${\displaystyle a\in \mathcal{A}}$, and ${\displaystyle n\in N}$. If ${\displaystyle a}$ is a Browder element, then ${\displaystyle a^n}$ is a Browder element.

Proof. 

It suffices to prove that ${\displaystyle a^2}$ is a Browder element. Since ${\displaystyle a}$ is a Browder element, then ${\displaystyle a=b_1+c_1}$, where ${\displaystyle b_1\in {\mathcal{A}}^{-1},c_1\in {\mathcal{T}}^{-1}\left(0\right),b_1{c}_1=c_1{b}_1}$. One can calculate that ${\displaystyle a^2=(b_1+c_1)(b_1+c_1)=b_1{b}_1+b_1{c}_1+c_1{b}_1+c_1{c}_1}$. It is not difficult to prove that ${\displaystyle b_1{b}_1\in {\mathcal{A}}^{-1}}$. It is clear that ${\displaystyle \mathcal{T}(b_1{c}_1+c_1{b}_1+c_1{c}_1)=0}$, and ${\displaystyle b_1{b}_1}$ commutes with ${\displaystyle b_1{c}_1+c_1{b}_1+c_1{c}_1}$. Hence, we obtain that ${\displaystyle a^2}$ is a Browder element. Similarly, one can obtain that ${\displaystyle a^n}$ is a Browder element. ${\displaystyle \square }$

Lemma 5.

Let ${\displaystyle \mathcal{A}}$ and ${\displaystyle \mathcal{B}}$ be Banach algebras, ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ be a homomorphism and ${\displaystyle d\in {\mathcal{A}}^{-1}\bigcap com{m}^2\left(a\right)}$. If ${\displaystyle a}$ is a Browder element, then ${\displaystyle ad}$ is a Browder element.

Proof. 

Suppose ${\displaystyle a=b_1+c_1}$, where ${\displaystyle b_1\in {\mathcal{A}}^{-1},c_1\in {\mathcal{T}}^{-1}\left(0\right),b_1{c}_1=c_1{b}_1}$, then ${\displaystyle ad=b_{1}d+c_{1}d}$. It is not difficult to calculate that ${\displaystyle b_{1}d\in {\mathcal{A}}^{-1}}$, ${\displaystyle \mathcal{T}\left(c_{1}d\right)=\mathcal{T}\left(c_1\right)\mathcal{T}d=0}$, which indicates that ${\displaystyle c_{1}d\in {\mathcal{T}}^{-1}\left(0\right)}$. Since ${\displaystyle d\in {\mathcal{A}}^{-1}\bigcap com{m}^2\left(a\right)}$, it follows that ${\displaystyle d{b}_1=b_{1}d,d{c}_1=c_{1}d}$. Hence, one can calculate that ${\displaystyle b_{1}d{c}_{1}d=c_{1}d{b}_{1}d}$. Consequently, it follows that ${\displaystyle ad}$ is a Browder element. ${\displaystyle \square }$

Theorem 5.

Let ${\displaystyle \mathcal{A}}$ and ${\displaystyle \mathcal{B}}$ be Banach algebras and ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ be a homomorphism. Then, ${\displaystyle {\omega }_{\mathcal{T}}^{comm}(f\left(a\right),\mathcal{A})=f\left({\omega }_{\mathcal{T}}^{comm}(a,\mathcal{A})\right)}$ for all ${\displaystyle a\in \mathcal{A}}$ and for all ${\displaystyle f}$ analytic and one to one on a neighborhood of ${\displaystyle \sigma (a,\mathcal{A})}$.

Proof. 

If the function is analytic and one to one on a neighborhood of ${\displaystyle \sigma (a,\mathcal{A})}$, then the equality is not difficult to check from ([23], Lemma 3.11). Let ${\displaystyle \lambda \notin f\left({\omega }_{\mathcal{T}}^{comm}(a,\mathcal{A})\right)}$. We may suppose that ${\displaystyle \lambda \in f\left(\sigma \right(a,\mathcal{A}\left)\right)}$ owing to ${\displaystyle {\omega }_{\mathcal{T}}^{comm}(f\left(a\right),\mathcal{A})\subseteq \sigma (f\left(a\right),\mathcal{A})}$. It is demonstrated that from ([24], Corollaries 3.8 and 3.9), there exist ${\displaystyle {\lambda }_1,{\lambda }_2,\cdots ,{\lambda }_n\in \sigma (a,\mathcal{A})}$, ${\displaystyle k_1,\cdots ,k_n\in N}$ and a function ${\displaystyle g}$ which is analytic on a neighborhood ${\displaystyle G}$ of ${\displaystyle \sigma (a,\mathcal{A})}$ and has no zeroes in ${\displaystyle G}$ such that ${\displaystyle f\left(z\right)-\lambda ={(z-{\lambda }_1)}^{k_1}\cdots {(z-{\lambda }_n)}^{k_n}g\left(z\right)}$ for all ${\displaystyle z\in G}$. Note that ${\displaystyle f}$ is one to one on ${\displaystyle G}$; we obtain that ${\displaystyle n=1}$, and hence ${\displaystyle f\left(z\right)-\lambda ={(z-{\lambda }_1)}^{k_1}g\left(z\right)}$, ${\displaystyle f\left(a\right)-\lambda ={(a-{\lambda }_1)}^{k_1}g\left(a\right)}$ with ${\displaystyle g\left(a\right)\in {\mathcal{A}}^{-1}}$ and ${\displaystyle a-{\lambda }_1}$ is a Browder element. According to the above Lemmas 4 and 5, it follows that ${\displaystyle f\left(a\right)-\lambda }$ is a Browder element, which indicates that ${\displaystyle \lambda \notin {\omega }_{\mathcal{T}}^{comm}(f\left(a\right),\mathcal{A})}$. ${\displaystyle \square }$

Remark 3.

Let ${\displaystyle \mathcal{A}}$ and ${\displaystyle \mathcal{B}}$ be Banach algebras and ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ be a homomorphism satisfying the Riesz property. Then, from ([17], Theorem 8.8), one can directly obtain that ${\displaystyle {\omega }_{\mathcal{T}}^{comm}(f\left(a\right),\mathcal{A})=f\left({\omega }_{\mathcal{T}}^{comm}(a,\mathcal{A})\right)}$ for all ${\displaystyle a\in \mathcal{A}}$ and for all ${\displaystyle f}$ analytic and one to one on a neighborhood of ${\displaystyle \sigma (a,\mathcal{A})}$.

5. Fredholm Theory Relative to Three Homomorphisms or More Homomorphisms

In this section, we consider the Fredholm theory relative to three homomorphisms or more homomorphisms. And the dependence on homomorphisms of the Fredholm theory is revealed. At first, we study the Fredholm theory relative to three homomorphisms.

Proposition 5.

(1) Suppose ${\displaystyle \mathcal{A},\mathcal{B},\mathcal{D}}$ are complex Banach algebras with identity and ${\displaystyle \mathcal{B}\bigcap \mathcal{D}\ne \varnothing }$. If ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B},\mathcal{S}:\mathcal{A}\to \mathcal{D},\nabla :\mathcal{A}\to \mathcal{B}\bigcap \mathcal{D}}$ are homomorphisms and ${\displaystyle \mathcal{T}}$(or ${\displaystyle \mathcal{S}}$) are one-to-one homomorphisms, ${\displaystyle a\in \mathcal{A}}$ is a ${\displaystyle \mathcal{T}-}$Fredholm element (or an ${\displaystyle \mathcal{S}-}$Fredholm element) and ${\displaystyle \sigma (a,\mathcal{A})}$ is totally disconnected, then ${\displaystyle a}$ is an ${\displaystyle \nabla -}$Fredholm element.

(2) Suppose ${\displaystyle \mathcal{A},\mathcal{B},\mathcal{D}}$ are complex Banach algebra with identity and ${\displaystyle \mathcal{B}\bigcap \mathcal{D}\ne \varnothing }$. If ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B},\mathcal{S}:\mathcal{A}\to \mathcal{D},\nabla :\mathcal{A}\to \mathcal{B}\bigcap \mathcal{D}}$ are homomorphisms and ${\displaystyle \mathcal{T}}$ (or ${\displaystyle \mathcal{S}}$) are bounded below (or one to one), ${\displaystyle a\in \mathcal{A}}$ is a ${\displaystyle \mathcal{T}-}$Weyl (Browder) element (or an ${\displaystyle \mathcal{S}-}$Weyl (Browder) element), then ${\displaystyle a}$ is a ${\displaystyle \nabla -}$Weyl (Browder) element.

Proof. 

Since ${\displaystyle \mathcal{T}}$ is a one-to-one homomorphism and ${\displaystyle \sigma (a,\mathcal{A})}$ is totally disconnected, one can obtain that ${\displaystyle \sigma (a,\mathcal{A})=\sigma (\mathcal{T}a,B)}$ from ([16], Corollary 2.5). It follows that ${\displaystyle \mathcal{T}a}$ is invertible in ${\displaystyle \mathcal{B}}$ from the fact that ${\displaystyle a}$ is a ${\displaystyle \mathcal{T}-}$Fredholm element. In other words, ${\displaystyle a}$ is invertible in ${\displaystyle \mathcal{A}}$. Thus, ${\displaystyle \nabla a}$ is invertible, which implies that ${\displaystyle a}$ is a ${\displaystyle \nabla -}$Fredholm element.

Suppose ${\displaystyle a\in \mathcal{A}}$ is a ${\displaystyle \mathcal{T}-}$Weyl element, then ${\displaystyle a=b+c}$, where ${\displaystyle b\in {\mathcal{A}}^{-1}}$, ${\displaystyle c\in N\left(\mathcal{T}\right)}$. It follows that ${\displaystyle c=0}$, as ${\displaystyle \mathcal{T}}$ is bounded below. Hence, it is evident that ${\displaystyle a=b}$ is invertible, which shows that ${\displaystyle a}$ is a ${\displaystyle \nabla -}$Weyl element. ${\displaystyle \square }$

Corollary 7.

Let ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B},\mathcal{S}:\mathcal{A}\to \mathcal{D},\nabla :\mathcal{A}\to \mathcal{B}\bigcap \mathcal{D}}$ be homomorphisms. Suppose ${\displaystyle \mathcal{T}}$ (or ${\displaystyle \mathcal{S}}$) has the Riesz property and ${\displaystyle a\in \mathcal{A}}$ is ${\displaystyle \nabla -}$Fredholm. Then, we have the following:

(1) If ${\displaystyle a}$ is ${\displaystyle \mathcal{T}-}$Browder (or ${\displaystyle \mathcal{S}-}$Browder), then ${\displaystyle a}$ is ${\displaystyle \nabla -}$Browder;

(2) If ${\displaystyle a}$ is ${\displaystyle \mathcal{T}-}$Weyl (or ${\displaystyle \mathcal{S}-}$Weyl), then ${\displaystyle a}$ is a ${\displaystyle \nabla -}$Weyl element.

Proof. 

Suppose ${\displaystyle a}$ is ${\displaystyle \mathcal{T}-}$Browder, then ${\displaystyle a}$ must be ${\displaystyle \nabla -}$Browder by means of ([16], Theorem 3.1). Similarly, since ${\displaystyle \mathcal{T}}$ has the Riesz property, then we obtain that ${\displaystyle a}$ is a ${\displaystyle \nabla -}$Weyl element if ${\displaystyle a}$ is ${\displaystyle \mathcal{T}-}$Weyl according to ([16], Theorem 3.1). ${\displaystyle \square }$

Corollary 8.

Let ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B},\mathcal{S}:\mathcal{A}\to \mathcal{D},\nabla :\mathcal{A}\to \mathcal{B}\bigcap \mathcal{D}}$ be homomorphisms and ${\displaystyle \mathcal{T}}$ (or ${\displaystyle \mathcal{S}}$) have the Riesz property. Suppose the kernels of ${\displaystyle \mathcal{T},\mathcal{S},\nabla }$ have the same spectral idempotent sets. Then, we have the following:

(1) If ${\displaystyle a}$ is ${\displaystyle \mathcal{T}-}$Browder (or ${\displaystyle \mathcal{S}-}$Browder), then ${\displaystyle a}$ is ${\displaystyle \nabla -}$Browder;

(2) If ${\displaystyle a}$ is ${\displaystyle \mathcal{T}-}$Weyl (or ${\displaystyle \mathcal{S}-}$Weyl), then ${\displaystyle a}$ is a ${\displaystyle \nabla -}$Weyl element.

Proof. 

Supposing ${\displaystyle a}$ is ${\displaystyle \mathcal{T}-}$Browder and ${\displaystyle \mathcal{T}}$ has the Riesz property, one can obtain that ${\displaystyle a}$ is ${\displaystyle \nabla -}$Browder according to ([17], Theorem 5.4). Similarly, one can obtain that ${\displaystyle a}$ is a ${\displaystyle \nabla -}$Weyl element if ${\displaystyle a}$ is ${\displaystyle \mathcal{T}-}$Weyl (or ${\displaystyle \mathcal{S}-}$Weyl) by means of ([17], Theorem 5.4). ${\displaystyle \square }$

Corollary 9.

Let ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B},\mathcal{S}:\mathcal{A}\to \mathcal{D},\nabla :\mathcal{A}\to \mathcal{B}\bigcap \mathcal{D}}$ be homomorphisms and ${\displaystyle \mathcal{T}}$ (or ${\displaystyle \mathcal{S}}$) have the Riesz property and ${\displaystyle a\in \mathcal{A}}$ be ${\displaystyle \nabla -}$Fredholm. Then, we have the following:

(1) If ${\displaystyle a\in \mathcal{A}}$ is ${\displaystyle \mathcal{S}-}$Ruston (or ${\displaystyle \mathcal{T}-}$Ruston), then ${\displaystyle a}$ is an ${\displaystyle \nabla -}$Ruston element;

(2) If ${\displaystyle a}$ is ${\displaystyle \mathcal{T}-}$almost Ruston (or ${\displaystyle \mathcal{S}-}$almost Ruston), then ${\displaystyle a}$ is an ${\displaystyle \nabla -}$almost Ruston element.

Proof. 

Suppose ${\displaystyle a}$ is ${\displaystyle \mathcal{T}-}$Ruston and ${\displaystyle \mathcal{T}}$ has the Riesz property, from Proposition 2, one can obtain that ${\displaystyle a}$ is a ${\displaystyle \nabla -}$Ruston element. Similarly, it follows that ${\displaystyle a}$ is a ${\displaystyle \nabla -}$almost Ruston element if ${\displaystyle a}$ is ${\displaystyle \mathcal{T}-}$almost Ruston from Proposition 2. ${\displaystyle \square }$

For the Fredholm theory relative to three homomorphisms, next, we consider the transitivity of Fredholm theory.

Theorem 6.

Suppose ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B},\mathcal{S}:\mathcal{B}\to \mathcal{D},\mathcal{S}\circ \mathcal{T}:\mathcal{A}\to \mathcal{D}}$ and ${\displaystyle {\mathcal{S}}^{-1}\left({\mathcal{D}}^{-1}\right)\subset {\mathcal{T}}^{-1}\left({\mathcal{B}}^{-1}\right)}$, where ${\displaystyle \mathcal{A},\mathcal{B},\mathcal{D}}$ are Banach algebras with identity and ${\displaystyle \mathcal{B}\subset \mathcal{A}}$. If ${\displaystyle a\in \mathcal{A}}$ is a ${\displaystyle \mathcal{T}-}$Fredholm element, ${\displaystyle b\in \mathcal{B}}$ is an ${\displaystyle \mathcal{S}-}$Fredholm element, then ${\displaystyle ab}$ is an ${\displaystyle \mathcal{S}\circ \mathcal{T}-}$Fredholm element.

Proof. 

Since ${\displaystyle a\in \mathcal{A}}$ is ${\displaystyle \mathcal{T}-}$Fredholm, ${\displaystyle b\in \mathcal{B}}$ is an ${\displaystyle \mathcal{S}-}$Fredholm element, then it can be seen that ${\displaystyle \mathcal{T}a\in {\mathcal{B}}^{-1},\mathcal{S}b\in {\mathcal{D}}^{-1}}$. One can calculate that ${\displaystyle \mathcal{S}\circ \mathcal{T}\left(ab\right)=\mathcal{S}\circ \left[\mathcal{T}\right(ab\left)\right]=\mathcal{S}\circ [\mathcal{T}a·\mathcal{T}b]=\mathcal{ST}a·\mathcal{ST}b}$. It follows that ${\displaystyle \mathcal{ST}a\in {\mathcal{D}}^{-1}}$ from ${\displaystyle \mathcal{T}a\in {\mathcal{B}}^{-1}}$. Associated with ${\displaystyle \mathcal{S}b\in {\mathcal{D}}^{-1}}$, one can obtain that ${\displaystyle \mathcal{T}b\in {\mathcal{B}}^{-1}}$, which implies that ${\displaystyle \mathcal{S}\circ \mathcal{T}b\in {\mathcal{D}}^{-1}}$. Hence, it is evident that ${\displaystyle \mathcal{ST}a·\mathcal{ST}b\in {\mathcal{D}}^{-1}}$, from which we deduce that ${\displaystyle ab}$ is an ${\displaystyle \mathcal{S}\circ \mathcal{T}-}$Fredholm element. ${\displaystyle \square }$

Proposition 6.

Suppose ${\displaystyle \mathcal{A},\mathcal{B},\mathcal{D}}$ are Banach algebras and ${\displaystyle \mathcal{B}\subset \mathcal{A}}$, ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B},\mathcal{S}:\mathcal{B}\to \mathcal{D},\mathcal{S}\circ \mathcal{T}:\mathcal{A}\to \mathcal{D}}$ are homomorphisms and ${\displaystyle ker\mathcal{S}\subset ker(\mathcal{S}\circ \mathcal{T})}$. If ${\displaystyle a\in \mathcal{A}}$ is a ${\displaystyle \mathcal{T}-}$Weyl element, ${\displaystyle b\in \mathcal{B}}$ is a ${\displaystyle \mathcal{S}-}$Weyl element, then ${\displaystyle ab\in \mathcal{B}}$ is an ${\displaystyle \mathcal{S}\circ \mathcal{T}-}$Weyl element.

Proof. 

Since ${\displaystyle a\in \mathcal{A}}$ is a ${\displaystyle \mathcal{T}-}$Weyl element, then there exist ${\displaystyle b_1\in {\mathcal{A}}^{-1},c_1\in ker\mathcal{T}}$ such that ${\displaystyle a=b_1+c_1}$. Since ${\displaystyle b\in \mathcal{B}}$ is an ${\displaystyle \mathcal{S}-}$Weyl element, then there exist ${\displaystyle b_2\in {\mathcal{B}}^{-1},c_2\in ker\mathcal{S}}$ such that ${\displaystyle a=b_2+c_2}$. One can calculate that ${\displaystyle ab=(b_1+c_1)(b_2+c_2)=b_1{b}_2+b_1{c}_2+c_1{b}_2+c_1{c}_2}$, which indicates that ${\displaystyle b_1{b}_2\in {\mathcal{A}}^{-1}}$. From the fact that ${\displaystyle \mathcal{T},\mathcal{S}}$ are homomorphisms, one can obtain the following relation: ${\displaystyle \mathcal{S}\circ \mathcal{T}(b_1{c}_2+c_1{b}_2+c_1{c}_2)=\mathcal{S}\circ \mathcal{T}\left(b_1{c}_2\right)=\mathcal{ST}\left(b_1\right)·\mathcal{S}\left(\mathcal{T}\left(c_2\right)\right)=0}$, which implies that ${\displaystyle ab}$ is an ${\displaystyle \mathcal{S}\circ \mathcal{T}-}$Weyl element. ${\displaystyle \square }$

Proposition 7.

Suppose ${\displaystyle \mathcal{A}, \mathcal{B}, \mathcal{D}}$ are Banach algebras and ${\displaystyle \mathcal{B}\subset \mathcal{A}}$, ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}, \mathcal{S}:\mathcal{B}\to \mathcal{D}, \mathcal{S}\circ \mathcal{T}:\mathcal{A}\to \mathcal{D}}$ are homomorphisms and

$$ker\mathcal{T}\subset comm\left({\mathcal{A}}^{-1}\right)\bigcap comm\left({\mathcal{B}}^{-1}\right),$$

$$ker\mathcal{S}\subset comm\left({\mathcal{A}}^{-1}\right)\bigcap comm\left({\mathcal{B}}^{-1}\right)\bigcap ker(\mathcal{S}\circ \mathcal{T}).$$

If ${\displaystyle a\in \mathcal{A}}$ is a ${\displaystyle \mathcal{T}-}$Browder element, ${\displaystyle b\in \mathcal{B}}$ is an ${\displaystyle \mathcal{S}-}$Browder element, then ${\displaystyle ab\in \mathcal{B}}$ is an ${\displaystyle \mathcal{S}\circ \mathcal{T}-}$Browder element.

Proof. 

Since ${\displaystyle a\in \mathcal{A}}$ is ${\displaystyle \mathcal{T}-}$Browder element, then there exist ${\displaystyle b_1\in {\mathcal{A}}^{-1}, c_1\in ker\mathcal{T}}$ with ${\displaystyle b_1{c}_1=c_1{b}_1}$ such that ${\displaystyle a=b_1+c_1}$. Since ${\displaystyle b\in \mathcal{B}}$ is ${\displaystyle \mathcal{S}-}$Browder element, then there exist ${\displaystyle b_2\in {\mathcal{B}}^{-1}, c_2\in ker\mathcal{S}}$ with ${\displaystyle b_2{c}_2=c_2{b}_2}$ such that ${\displaystyle a=b_2+c_2}$. It follows that ${\displaystyle ab=(b_1+c_1)(b_2+c_2)=b_1{b}_2+b_1{c}_2+c_1{b}_2+c_1{c}_2}$, which induces that ${\displaystyle ab}$ is an ${\displaystyle \mathcal{S}\circ \mathcal{T}-}$Weyl element. One can check that ${\displaystyle b_1{b}_2}$ commutes with ${\displaystyle b_1{c}_2+c_1{b}_2+c_1{c}_2}$. And hence, ${\displaystyle ab}$ is an ${\displaystyle \mathcal{S}\circ \mathcal{T}-}$Browder element. ${\displaystyle \square }$

Proposition 8.

Suppose ${\displaystyle \mathcal{A}, \mathcal{B}, \mathcal{D}}$ are Banach algebras and ${\displaystyle \mathcal{B}\subset \mathcal{A}}$, ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}, \mathcal{S}:\mathcal{B}\to \mathcal{D}, \mathcal{S}\circ \mathcal{T}:\mathcal{A}\to \mathcal{D}}$ are homomorphisms, ${\displaystyle \mathcal{T}}$ satisfies the Riesz property with spectrum preserving, ${\displaystyle \mathcal{S}}$ satisfies the Riesz property, and ${\displaystyle ker\mathcal{T}\subset comm\left({\mathcal{A}}^{-1}\right)\bigcap comm\left({\mathcal{B}}^{-1}\right)}$, ${\displaystyle ker\mathcal{S}\subset comm}$${\displaystyle \left({\mathcal{A}}^{-1}\right)\bigcap comm\left({\mathcal{B}}^{-1}\right)\bigcap ker(\mathcal{S}\circ \mathcal{T})}$. If ${\displaystyle a\in \mathcal{A}}$ is a ${\displaystyle \mathcal{T}-}$Ruston element, ${\displaystyle b\in \mathcal{B}}$ is an ${\displaystyle \mathcal{S}-}$Ruston element, then ${\displaystyle ab\in \mathcal{B}}$ is an ${\displaystyle \mathcal{S}\circ \mathcal{T}-}$Ruston element.

Proof. 

It suffices to prove that in this case, the Ruston element is equivalent to the Browder element. Firstly, we prove that ${\displaystyle \mathcal{S}\circ \mathcal{T}}$ satisfies the Riesz property if ${\displaystyle \mathcal{T},\mathcal{S}}$ satisfy the Riesz property and the range of ${\displaystyle \mathcal{T}}$ is closed. Suppose ${\displaystyle a\in \mathcal{A}}$ and ${\displaystyle \mathcal{S}\circ \mathcal{T}\left(a\right)=0}$, then ${\displaystyle \mathcal{T}\left(a\right)\in ker\mathcal{S}}$. It follows that ${\displaystyle \sigma \left(\mathcal{T}\right(a\left)\right)}$ is either finite or a sequence converging to 0 since ${\displaystyle \mathcal{S}}$ satisfies the Riesz property. Thus, ${\displaystyle \sigma \left(a\right)}$ is either finite or a sequence converging to 0. Hence, ${\displaystyle \mathcal{S}\circ \mathcal{T}}$ satisfies the Riesz property. Consequently, the Browder element is a Ruston element from ([16], Remark 2.8), which indicates that ${\displaystyle ab}$ is an ${\displaystyle \mathcal{S}\circ \mathcal{T}-}$Ruston element. ${\displaystyle \square }$

Corollary 10.

Suppose ${\displaystyle \mathcal{A},\mathcal{B},\mathcal{D}}$ are Banach algebras and ${\displaystyle \mathcal{B}\subset \mathcal{A}}$, ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B},\mathcal{S}:\mathcal{B}\to \mathcal{D},\mathcal{S}\circ \mathcal{T}:\mathcal{A}\to \mathcal{D}}$ are homomorphisms, ${\displaystyle \mathcal{T}}$ and ${\displaystyle \mathcal{S}}$ are bounded below (or one to one). If ${\displaystyle a\in \mathcal{A}}$ is a ${\displaystyle \mathcal{T}-}$Weyl element, ${\displaystyle b\in \mathcal{B}}$ is an ${\displaystyle \mathcal{S}-}$Weyl element, then ${\displaystyle ab\in \mathcal{B}}$ is an ${\displaystyle \mathcal{S}\circ \mathcal{T}-}$Weyl element.

Proof. 

Since ${\displaystyle \mathcal{T}}$ and ${\displaystyle \mathcal{S}}$ are bounded below, then we obtain that ${\displaystyle ker\left(\mathcal{T}\right)=ker\left(\mathcal{S}\right)=\left\{0\right\}}$, which implies that ${\displaystyle ker\mathcal{S}\subset ker(\mathcal{S}\circ \mathcal{T})}$. It follows that ${\displaystyle ab}$ is an ${\displaystyle \mathcal{S}\circ \mathcal{T}-}$Weyl element by means of Proposition 7. ${\displaystyle \square }$

Corollary 11.

Suppose ${\displaystyle \mathcal{A},\mathcal{B},\mathcal{D}}$ are Banach algebras and ${\displaystyle \mathcal{B}\subset \mathcal{A}}$, ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B},\mathcal{S}:\mathcal{B}\to \mathcal{D},\mathcal{S}\circ \mathcal{T}:\mathcal{A}\to \mathcal{D}}$ are homomorphisms, ${\displaystyle \mathcal{T}}$ and ${\displaystyle \mathcal{S}}$ are bounded below (or one to one). If ${\displaystyle a\in \mathcal{A}}$ is a ${\displaystyle \mathcal{T}-}$Browder element, ${\displaystyle b\in \mathcal{B}}$ is an ${\displaystyle \mathcal{S}-}$Browder element, then ${\displaystyle ab\in \mathcal{B}}$ is an ${\displaystyle \mathcal{S}\circ \mathcal{T}-}$Browder element.

Proof. 

Since ${\displaystyle \mathcal{T}}$ and ${\displaystyle \mathcal{S}}$ are bounded below, then one can see that ${\displaystyle ker\mathcal{T}=\left\{0\right\}\subset comm\left({\mathcal{A}}^{-1}\right)}$${\displaystyle \bigcap comm\left({\mathcal{B}}^{-1}\right)}$, ${\displaystyle ker\mathcal{S}=\left\{0\right\}\subset comm\left({\mathcal{A}}^{-1}\right)\bigcap comm\left({\mathcal{B}}^{-1}\right)\bigcap ker(\mathcal{S}\circ \mathcal{T})}$ hold. According to Proposition 8, it follows that ${\displaystyle ab}$ is an ${\displaystyle \mathcal{S}\circ \mathcal{T}-}$Browder element. ${\displaystyle \square }$

Proposition 9.

Suppose ${\displaystyle \mathcal{A},\mathcal{B},\mathcal{D}}$ are Banach algebras and ${\displaystyle \mathcal{B}\subset \mathcal{A}}$, ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B},\mathcal{S}:\mathcal{B}\to \mathcal{D},\mathcal{S}\circ \mathcal{T}:\mathcal{A}\to \mathcal{D}}$ are homomorphisms, ${\displaystyle \mathcal{T}}$ satisfies the Riesz property with spectrum preserving, ${\displaystyle \mathcal{S}}$ satisfies the Riesz property, and ${\displaystyle ker\mathcal{S}\subset ker(\mathcal{S}\circ \mathcal{T})}$. If ${\displaystyle a\in \mathcal{A}}$ is a ${\displaystyle \mathcal{T}-}$almost Ruston element, ${\displaystyle b\in \mathcal{B}}$ is an ${\displaystyle \mathcal{S}-}$almost Ruston element, then ${\displaystyle ab\in \mathcal{B}}$ is an ${\displaystyle \mathcal{S}\circ \mathcal{T}-}$almost Ruston element.

Proof. 

It is not difficult to check that ${\displaystyle \mathcal{S}\circ \mathcal{T}}$ satisfies the Riesz property; hence, the Weyl element must be an almost Ruston element from ([16], Remark 2.8). Therefore, ${\displaystyle ab}$ is an ${\displaystyle \mathcal{S}\circ \mathcal{T}-}$almost Ruston element. ${\displaystyle \square }$

At last, we consider for any homomorphism ${\displaystyle S}$, when ${\displaystyle a\in \mathcal{A}}$ is an ${\displaystyle \mathcal{S}-}$ Fredholm (Weyl, Browder, Ruston, almost Ruston) element if ${\displaystyle a\in \mathcal{A}}$ is ${\displaystyle \mathcal{T}-}$Fredholm (Weyl, Browder, Ruston, almost Ruston) element, which indicates that Fredholm theory does not have dependence on homomorphisms. The proof of the following proposition is omitted here; one can check it by oneself.

Proposition 10.

1. Let ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ be a homomorphism. Suppose ${\displaystyle \mathcal{T}}$ is one to one, and ${\displaystyle \sigma (a,\mathcal{A})}$ is totally disconnected. Then, if ${\displaystyle a}$ is a ${\displaystyle \mathcal{T}-}$Fredholm element, then ${\displaystyle a}$ is an ${\displaystyle \mathcal{S}-}$Fredholm element for any homomorphism ${\displaystyle \mathcal{S}:\mathcal{A}\to \mathcal{D}}$.

2. Let ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ be a homomorphism with the Riesz property. Suppose the kernel of any homomorphism has the same spectral idempotent set. Then, if ${\displaystyle a}$ is a ${\displaystyle \mathcal{T}-}$Browder element, then ${\displaystyle a}$ is an ${\displaystyle \mathcal{S}-}$Browder element for any homomorphism ${\displaystyle \mathcal{S}}$ on ${\displaystyle \mathcal{A}}$.

3. Let ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ be a homomorphism with the Riesz property. Suppose the kernel of any homomorphism has the same spectral idempotent set. Then, if ${\displaystyle a}$ is a ${\displaystyle \mathcal{T}-}$Weyl element, then ${\displaystyle a}$ is an ${\displaystyle \mathcal{S}-}$Weyl element for any homomorphism ${\displaystyle \mathcal{S}}$ on ${\displaystyle \mathcal{A}}$.

4. Let ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ be a homomorphism with the Riesz property. Suppose the kernel of any homomorphism has the same spectral idempotent set. Then, if ${\displaystyle a}$ is ${\displaystyle a\mathcal{T}-}$Ruston element, then ${\displaystyle a}$ is an ${\displaystyle \mathcal{S}-}$Ruston element for any homomorphism ${\displaystyle \mathcal{S}}$ on ${\displaystyle \mathcal{A}}$.

5. Let ${\displaystyle \mathcal{T}:\mathcal{A}\to \mathcal{B}}$ be a homomorphism with the Riesz property. Suppose the kernel of any homomorphism has the same spectral idempotent set. Then, if ${\displaystyle a}$ is a ${\displaystyle \mathcal{T}-}$almost Ruston element, then ${\displaystyle a}$ is an ${\displaystyle \mathcal{S}-}$almost Ruston element for any homomorphism ${\displaystyle \mathcal{S}}$ on ${\displaystyle \mathcal{A}}$.

Proof. 

1. Since ${\displaystyle \mathcal{T}}$ is one to one and ${\displaystyle \sigma (a,\mathcal{A})}$ is totally disconnected, from Theorem 1, one can obtain that ${\displaystyle a}$ is an ${\displaystyle \mathcal{S}-}$Fredholm element for any homomorphism ${\displaystyle \mathcal{S}}$.

2 and 3. According to ([17], Theorem 5.4), it follows that ${\displaystyle a}$ is ${\displaystyle \mathcal{S}-}$Browder for any homomorphism ${\displaystyle \mathcal{S}}$. Similarly, one can obtain the corresponding result about ${\displaystyle \mathcal{S}-}$Weyl.

4 and 5. From Theorem 2, one can obtain the results. ${\displaystyle \square }$