Corrigendum an Addendum: “On the axiomatic theory of spectrum II” J.J. Koliha, M. Mbekhta, V. M¨ uller*, Pak Wai Poon The main purpose of this paper is to correct the proof of Theorem 15 of [4], concerned with the stability of the class of quasi-Fredholm operators under finite rank perturbations, and to answer some open questions raised there. Recall some notations and terminology from [4]. For closed subspaces M,L of a Banach space X we write M e ⊂L ( M is essentially contained in L ) if there exists a finite-dimensional subspace F ⊂ X such that M ⊂ L + F . Equivalently, dim M/(M ∩ L) = dim(M + L)/L < ∞. Similarly we write M e =L if M e ⊂L and L e ⊂M . For a (bounded linear) operator T ∈L(X ) write R ∞ (T )= T ∞ n=0 R(T n ) and N ∞ (T )=  ∞ n=0 N (T n ). An operator T ∈L(X ) is called semi-regular (essentially semi-regular) if R(T ) is closed and N (T ) ⊂ R ∞ (T ) ( N (T ) e ⊂R ∞ (T ), respectively ) . Further, T is called quasi-Fredholm if there exists d ≥ 0 such that R(T d+1 ) is closed and R(T )+ N (T d )= R(T )+ N ∞ (T ) (equivalently, N (T ) ∩ R(T d )= N (T ) ∩ R ∞ (T )). The proof of Theorem 15 of [4] relies on the following statement (where d is an integer whose existence is postulated in the definition of quasi-Fredholm operators): if T is quasi-Fredholm and F of rank 1 then N (T ) ∩ R(T d ) ⊂ R ∞ (T + F ). This, however, need not be satisfied. Counterexample. Let H be the Hilbert space with an orthonormal basis {e 1 ,e 2 ,...}. Define T,F ∈L(H ) by Te 1 =0,Te n = e n-1 (n ≥ 2), Fe 2 = -e 1 , Fe n =0 (n = 2). Then T is quasi-Fredholm (with d = 0) and is surjective, F has rank 1, and T + F is given by (T + F )e 1 =(T + F )e 2 =0, (T + F )e n = e n-1 (n ≥ 3). It follows that R ∞ (T + F )= R(T + F ) is equal to the linear span of {e 2 ,e 3 ,...}, and N (T ) to the one-dimensional space spanned by e 1 . Thus N (T ) ⊂ R ∞ (T + F ). We proceed now to give a correct proof of Theorem 15 of [4]. Theorem. Let T ∈L(X ) be a quasi-Fredholm operator and let F ∈L(X ) be a finite-rank operator. Then T + F is also quasi-Fredholm. Proof. Clearly it is sufficient to consider only the case of dim R(F ) = 1. Thus there exist z ∈ X and ϕ ∈ X * such that Fx = ϕ(x)z (x ∈ X ). Mathematics Subject Classification: 47A10, 47A53. Keywords and phrases: quasi-Fredholm operators, ascent, descent. * Supported by the grant No. 201/96/0411 of GA ˇ CR. 1