where P(X, Y ) is the subspace of HomA(X, Y ) consisting of the A-homomorphisms which factorize through projective A-modules

VOL. 72 1997 NO. 1

ON LOCALLY BOUNDED CATEGORIES STABLY EQUIVALENT TO THE REPETITIVE ALGEBRAS OF

TUBULAR ALGEBRAS

BY

ZYGMUNT P O G O R Z A L Y (TORU ´N)

1. Introduction. Throughout the paper K is a fixed algebraically closed field. By an algebra we mean a finite-dimensional K-algebra, which we shall assume, without loss of generality, to be basic and connected. For an algebra A, we shall denote by mod(A) the category of finitely generated right A-modules, and by mod(A) the stable category of mod(A). Recall that the objects of mod(A) are the objects of mod(A) without projective direct summands, and for any two objects X, Y in mod(A) the space of morphisms from X to Y in mod(A) is Hom_A(X, Y ) = HomA(X, Y )/P(X, Y ), where P(X, Y ) is the subspace of Hom_A(X, Y ) consisting of the A-homomorphisms which factorize through projective A-modules. For every f ∈ HomA(X, Y ) we shall denote by f its coset modulo P(X, Y ). Two algebras A and B are said to be stably equivalent if their stable module categories mod(A) and mod(B) are equivalent.

Following [5, 11] we shall say that a module T in mod(A) is a tilting (respectively, cotilting) module if it satisfies the following conditions:

(1) Ext²_A(T, −) = 0 (respectively, Ext²_A(−, T ) = 0);

(2) Ext¹_A(T, T ) = 0;

(3) the number of nonisomorphic indecomposable summands of T equals the rank of the Grothendieck group K0(A).

Two algebras A and B are said to be tilting-cotilting equivalent if there exist a sequence of algebras A = A0, A1, . . . , Am, Am+1 = B and a sequence of modules T_Aⁱ_i, 0 ≤ i ≤ m, such that Ai+1 = EndAi(Tⁱ) and Tⁱ is either a tilting or a cotilting module.

Following Gabriel [9], a K-category R is called locally bounded if the following conditions are satisfied:

1991 Mathematics Subject Classification: Primary 16G20.

Supported by Polish Scientific Grant KBN 590/PO3/95/08.

[123]

(a) different objects are not isomorphic;

(b) the algebra R(x, x) of endomorphisms of x is local for every object x in R;

(c) P

y∈RdimKR(x, y) < ∞ and P

y∈RdimKR(y, x) < ∞ for every object x in R.

Interesting examples of locally bounded K-categories are the repetitive algebras introduced by Hughes and Waschb¨usch in [12]. For an algebra A denote by D = HomK(−, K) the standard duality on mod(A). Recall that the repetitive algebra bA of A is the selfinjective, locally finite-dimensional matrix algebra without identity defined by

A =b



























· 0

· ·

· ·

· Ai−1

Ei−1 Ai

Ei Ai+1

· ·

· ·

0 · ·



























where matrices have only finitely many nonzero entries, Ai = A, Ei =

ADAA for all integers i, all the remaining coefficients are zero, and the multiplication is induced from the canonical bimodule structure of DA and the zero morphism DA ⊗ADA → 0.

One of the interesting problems concerning repetitive algebras is a clas- sification of locally bounded K-categories which are stably equivalent to a given repetitive algebra. The problem was studied by several authors (see [1, 2, 14, 20, 21]). Wakamatsu proved in [21] that if A is tilting-cotilting equivalent to B then bA is stably equivalent to bB. Peng and Xiao proved in [14] that if H is a hereditary algebra and Λ is a locally bounded K-category which is stably equivalent to bH, then there is an algebra B tilting-cotilting equivalent to H such that bB ∼= Λ. We shall prove the following theorem on locally bounded K-categories stably equivalent to the repetitive algebras of tubular algebras in the sense of Ringel [18].

Theorem. Let A be a tubular algebra. A locally bounded K-category Λ is stably equivalent to bA if and only if Λ is isomorphic to the repetitive algebra bB of a tubular algebra B which is tilting-cotilting equivalent to A.

Our proof of the above result rests heavily on the main results obtained in [15, 16] for trivial extension algebras. In the case when Λ is a repetitive algebra the above theorem has been proved in [2].

We shall use freely results about Auslander–Reiten sequences which can be found in [3].

2. Preliminaries

2.1. Following Ringel [18], the canonical tubular algebras of type (2, 2, 2, 2) are defined by the quiver

•

•

• •

•

•

α2

JJJJ^β2JJJJJJ$$

XXXXXXXXXX++

α2

tttttttttt::

β2

gggggggggg33

γ2

XXXXXXXXXX++

δ2

JJJJJJJJJJ$$

γ2

gggggggggg33

δ2

tttttttttt::

with the relations α1α2+β1β2+γ1γ2= 0 and α1α2+kβ1β2+δ1δ2= 0, where k is some fixed element from K \ {0, 1}. The canonical tubular algebras of type (p, q, r) = (3, 3, 3), (2, 4, 4) or (2, 3, 6) are given by the quiver

• • •

• • • • •

• • •

α2 //

αp

@@@@

α1 @

~~~~~>>

β1 //

γ1

@@@@

@

β2 // ^βq //

γ2 // ~~~~^γr~>>

with α1α2. . . αp+ β1β2. . . βq+ γ1γ2. . . γr = 0.

2.2. For the repetitive algebra bA the identity morphisms Ai → A_i−1, Ei→ E_i−1 induce an automorphism νA of bA which is called the Nakayama automorphism. Moreover, the orbit space bA/(νA) has the structure of a finite-dimensional K-algebra which is isomorphic to the trivial extension T (A) of A by its minimal injective cogenerator bimodule ADAA. This is the algebra whose additive structure coincides with that of the group A ⊕ DA, and whose multiplication is defined by the formula (a, f )(b, g) = (ab, ag+f b) for a, b ∈ A, f, g ∈ ADAA. Thus bA is a Galois cover in the sense of [9] of the selfinjective algebra T (A) with the infinite cyclic group (νA) generated by νA.

2.3. A locally bounded K-category R is said to be locally support-finite [6] if for every indecomposable projective R-module P , the set of isomor- phism classes of indecomposable projective R-modules P⁰such that there ex- ists an indecomposable finite-dimensional R-module M with HomR(P, M ) 6=

0 6= HomR(P⁰, M ) is finite. Of particular interest is the fact that the repet- itive algebra bA of a tubular algebra A is locally support-finite (see [13]). A locally bounded K-category is said to be triangular if its ordinary quiver has no oriented cycles.

2.4. Following Gabriel (see [9]), for a locally bounded K-category R and a torsion-free group G of K-automorphisms of R acting freely on the objects of R, R/G is the quotient category whose objects are the G-orbits of the objects of R. Moreover, there is a covering functor F : R → R/G which maps any object x of R to its G-orbit G · x. F induces the push- down functor Fλ : mod(R) → mod(R/G), which preserves indecompos- ables and Auslander–Reiten sequences, maps projective R-modules to pro- jective R/G-modules and preserves projective resolutions. Furthermore, if R is locally support-finite then Fλ is dense and induces a bijection between the set (ind(R)/∼=)/G of the G-orbits of the isomorphism classes of finite- dimensional indecomposable R-modules and the set ind(R/G)/∼= of the iso- morphism classes of finite-dimensional indecomposable R/G-modules [6].

2.5. Let ΩR : mod(R) → mod(R) be Heller’s loop-space functor for a selfinjective locally bounded K-category R. Then ΩRτ_R⁻¹ΩR(S) is simple for every simple R-module S, where τ_R⁻¹ stands for the Auslander–Reiten translate TrD on mod(R). Thus we obtain a permutation of the isomor- phism classes of the simple R-modules. This permutation induces a K- automorphism νR of R in an obvious way. We denote by (νR) the infinite cyclic group of K-automorphisms of R generated by νR.

3. Preparatory results

3.1. Throughout this section we shall assume that R1 and R2 are self- injective locally bounded K-categories which are locally support-finite and have no indecomposable projective modules of length 2. Moreover, there is a fixed equivalence functor Φ : mod(R1) → mod(R2).

3.2. Proposition. If M is an indecomposable nonprojective finite-di- mensional R1-module then Φ(τR1(M )) ∼= τR2(Φ(M )) and Φ(ΩR1(M )) ∼= ΩR2(Φ(M )).

P r o o f. A direct adaptation of the arguments from the proofs of Propo- sition 2.4 and Theorem 4.4 of [4].

3.3. Lemma. If τR⁻¹1(M ) 6∼= Ω_R⁻²₁(M ) for every indecomposable nonpro- jective finite-dimensional R1-module M then (νR2) acts freely on the objects of R2.

P r o o f. We have to show that ΩR2τ_R⁻¹₂ΩR2(S) 6∼= S for every simple R2- module S. Suppose to the contrary that there exists a simple R2-module S

with ΩR2τ_R⁻¹₂ΩR2(S) ∼= S. Then there exists a nonprojective indecompos- able finite-dimensional R1-module M such that Φ(M ) ∼= S, and we infer by Proposition 3.2 that ΩR1τ_R⁻¹₁ΩR1(M ) ∼= M , which contradicts our assump- tion, because this isomorphism implies τ_R⁻¹₁(M ) ∼= Ω_R⁻²₁(M ).

3.4. Lemma. Let F1 : mod(R1) → mod(R1) and F2 : mod(R2) → mod(R2) be exact equivalences satisfying the following conditions:

(a) If F_i^s: mod(Ri) → mod(Ri), i = 1, 2, are defined by F_i^s(X) = Fi(X) for X ∈ mod(Ri), F_i^s(f ) = Fi(f ) for f : X → Y in mod(Ri), then F_i^s are well-defined functors which are equivalences.

(b) For every object X ∈ mod(R1), F₁^s(X) ∼= Φ⁻¹F₂^sΦ(X), where Φ⁻¹ is a fixed quasi-inverse of Φ.

Then F₁^s and Φ⁻¹F₂^sΦ are isomorphic functors.

P r o o f. In the first step of the proof we show that for every short exact sequence

0 → U → X^w → V → 0^p

in mod(R1) with all terms without projective direct summands there are w⁰ : Φ⁻¹F₂^sΦ(U ) → Φ⁻¹F₂^sΦ(X) and p⁰ : Φ⁻¹F₂^sΦ(X) → Φ⁻¹F₂^sΦ(V ) such that the following sequences are exact in mod(R1):

0 → F₁^s(U )^F−→ F^1(w) ₁^s(X)^F−→ F^1(p) ₁^s(V ) → 0, 0 → Φ⁻¹F₂^sΦ(U )^w

0

→ Φ⁻¹F₂^sΦ(X) ^p

0

→ Φ⁻¹F₂^sΦ(V ) → 0,

where w⁰ = Φ⁻¹F₂^sΦ(w) and p⁰ = Φ⁻¹F₂^sΦ(p). The exactness of the first sequence is obvious by the definition of F₁^s, because F1is exact.

In order to show the exactness of the second, we first show that w⁰ is a monomorphism, where w⁰ is any representative of the coset Φ⁻¹F₂^sΦ(w).

Suppose to the contrary that w⁰ is not a monomorphism. Then w⁰= w⁰₂w⁰₁ with w⁰₁ : Φ⁻¹F₂^sΦ(U ) → im(w⁰) an epimorphism and w⁰₂ : im(w⁰) → Φ⁻¹F₂^sΦ(X) a monomorphism. Since w is a monomorphism, we infer by [17;

Lecture 3] that w 6= 0. Thus w⁰ = w⁰₂w⁰₁ 6= 0 and there are W ∈ mod(R₁) and w1 : U → W , w2 : W → X such that Φ⁻¹F₂^sΦ(wi) = w⁰_i, i = 1, 2, because Φ⁻¹F₂^sΦ is an equivalence. Since w⁰₁ is a proper epimorphism, we have the following inequality for lengths: l(im(w⁰)) < l(Φ⁻¹F₂^sΦ(U )). But F1 is an additive exact equivalence, hence F1 preserves the lengths of R1- modules. Therefore F₁^spreserves the lengths of R1-modules without projec- tive direct summands and so does Φ⁻¹F₂^sΦ, because F₁^s(M ) ∼= Φ⁻¹F₂^sΦ(M ) for any M ∈ mod(R1) by the assumption of our lemma. Consequently, l(W ) = l(im(w⁰)) < l(U ). But w − w2w1 factorizes through a projective R1-module, say P . Thus there are q1 : U → P and q2: P → X such that w − w2w1 = q2q1. Since w is a monomorphism, there is q⁰₁ : X → P such

that q1 = q⁰₁w. Then w − w2w1 = q2q1 = q2q₁⁰w and w − q2q⁰₁w = w2w1. Hence (idX− q₂q₁⁰)w = w2w1. But (idX− q₂q⁰₁)w is a monomorphism, be- cause idX − q₂q⁰₁ is an isomorphism. Therefore we obtain a contradiction, because the monomorphism (idX− q₂q₁⁰)w factorizes through the module W of length smaller than U . Consequently, w⁰ is a monomorphism.

Dually one proves that p⁰ is an epimorphism, where p⁰ is any represen- tative of the coset Φ⁻¹F₂^sΦ(p).

Since Φ⁻¹F₂^sΦ preserves the lengths of R1-modules without projective direct summands, showing that p⁰w⁰= 0 is sufficient to show that the consid- ered sequence is exact. Since pw = 0, we have pw = 0. Thus p⁰w⁰= 0. Hence there are a projective R1-module P and morphisms q1 : Φ⁻¹F^sΦ(U ) → P and q2 : P → Φ⁻¹F₂^sΦ(V ) such that p⁰w⁰ = q2q1. Since w⁰ is a monomor- phism and p⁰is an epimorphism, there are morphisms q₂⁰ : P → Φ⁻¹F₂^sΦ(X) and q₁⁰ : Φ⁻¹F₂^sΦ(X) → P such that p⁰w⁰= q2q1= p⁰q₂⁰q⁰₁w⁰. Then putting w⁰⁰= (idX− q₂⁰q⁰₁)w⁰ we obtain p⁰w⁰⁰= 0 and w⁰⁰= w⁰.

In the second step of the proof we show that there is an isomorphism f : F₁^s → Φ⁻¹F₂^sΦ given by a family (f (X))X∈mod(R1) of isomorphisms in mod(R1) such that for every morphism u : X → Y in mod(R1) the diagram

F₁^s(X) ^{f (X)}−→ Φ⁻¹F₂^sΦ(X)

F₁^s(u)↓ ↓^{Φ−1F2sΦ(u)} F₁^s(Y ) ^{f (Y )}−→ Φ⁻¹F₂^sΦ(Y )

commutes. We construct a family (f (X))X∈mod(R1) such that for every X ∈ mod(R1) there is an isomorphism fX in mod(R1) with fX= f (X) and such that for every short exact sequence

0 → U → X^w → V → 0^p in mod(R1) the diagram with exact rows

0 → F₁^s(U ) ^F1→^(w) F₁^s(X) ^F→^1(p) F₁^s(V ) → 0

↓^fU ↓^fX ↓^fV

0 → Φ⁻¹F₂^sΦ(U ) ^w

0

→ Φ⁻¹F₂^sΦ(X) ^p

0

→ Φ⁻¹F₂^sΦ(V ) → 0 commutes, where w⁰, p⁰ are as in the first step of the proof. This condition is called the commutativity condition for fX.

Our construction will run inductively on the length of X in mod(R1). If l(X) = 1 then X is a simple R1-module. Fix an isomorphism fX = f (X) : F₁^s(X) → Φ⁻¹F₂^sΦ(X). Let u : X → X be a nonzero morphism. Since X is simple, u is an automorphism. Thus Φ⁻¹F₂^sΦ(u) = v, where v is an

automorphism. But u is multiplication by ku∈ K^∗= K \ {0}. Since F₁^s(idX) = id_F^s

1(X) and Φ⁻¹F₂^sΦ(idX) = id_Φ⁻¹_F^s

2Φ(X), it follows that for u = idX· k_u we have

F₁^s(u) = id_F^s

1(X)· k_u and Φ⁻¹F₂^sΦ(idX· k_u) = id_Φ⁻¹_F^s

2Φ(X)· k_u. Thus for any f (X) we have f (X)F₁^s(u) = Φ⁻¹F₂^sΦ(u)f (X).

Now consider two isomorphic simple modules X, Y such that X 6= Y . For every isomorphism class [X] of a simple R1-module X fix a representative, say X. For every Y isomorphic to X fix an isomorphism uY : X → Y . Then fix an isomorphism fX : F₁^s(X) → Φ⁻¹F₂^sΦ(X), and for every Y ∈ [X] define fY : F₁^s(Y ) → Φ⁻¹F₂^sΦ(Y ) by the formula

fY = f (Y ) = Φ⁻¹F₂^sΦ(uY)f (X)F₁^s(u⁻¹_Y ),

where fY is an arbitrary fixed representative of the coset f (Y ). If u : Z → Y is an isomorphism with Y, Z ∈ [X] then for Z = X we have u = uY · ku for some ku∈ K^∗. Thus F₁^s(u) = F₁^s(uY) · ku and Φ⁻¹F₂^sΦ(u) = Φ⁻¹F₂^sΦ(uY) · ku. Therefore f (Y ) = Φ⁻¹F₂^sΦ(uY)f (X)F₁^s(u⁻¹_Y ), which implies that f (Y ) = (Φ⁻¹F₂^sΦ(uY) · ku)f (X)(F₁^s(u⁻¹_Y ) · k_u⁻¹) = Φ⁻¹F₂^sΦ(u)f (X)F₁^s(u⁻¹_Y ).

Thus f (Y )F₁^s(u) = Φ⁻¹F₂^sΦ(u)f (X).

Now consider the case Y = X. Then u = u⁻¹_Z · k⁻¹_u for some ku∈ K^∗. Thus F₁^s(u) = F₁^s(u⁻¹_Z ) · k⁻¹_u and Φ⁻¹F₂^sΦ(u) = Φ⁻¹F₂^sΦ(u⁻¹_Z ) · k_u⁻¹. There- fore f (Z) = Φ⁻¹F₂^sΦ(uZ)f (X)F₁^s(u⁻¹_Z ), which implies

f (Z)⁻¹= F₁^s(uZ)f (X)⁻¹Φ⁻¹F₂^sΦ(u⁻¹_Z )

= (F₁^s(uZ) · ku)f (X)⁻¹(Φ⁻¹F₂^sΦ(u⁻¹_Z ) · k⁻¹_u )

= F₁^s(u)⁻¹f (X)⁻¹Φ⁻¹F₂^sΦ(u).

Then

f (Z) = (Φ⁻¹F₂^sΦ(u))⁻¹f (X)F₁^s(u) and

Φ⁻¹F₂^sΦ(u)f (Z) = f (X)Φ⁻¹F₂^sΦ(u).

Finally, consider the case Z 6= X 6= Y . Then uY · k_u= uZ for some ku∈ K^∗. Moreover, we infer by the above considerations that f (Z)F₁^s(uZ) = Φ⁻¹F₂^sΦ(uZ)f (X) and f (Y )F₁^s(uuZ) = Φ⁻¹F₂^sΦ(uuZ)f (X). But F₁^s(uuZ)

= F₁^s(u)F₁^s(uZ) and Φ⁻¹F₂^sΦ(uuZ) = Φ⁻¹F₂^sΦ(u)Φ⁻¹F₂^sΦ(uZ). Then we get

f (Y )F₁^s(u)f (Z)⁻¹f (Z)F₁^s(uZ) = Φ⁻¹F₂^sΦ(u)Φ⁻¹F₂^sΦ(uZ)f (X) and f (Y )F₁^s(u)f (Z)⁻¹= Φ⁻¹F₂^sΦ(u). Consequently,

f (Y )F₁^s(u) = Φ⁻¹F₂^sΦ(u)f (Z),

and for simple R1-modules X the family (f (X)) is constructed.

Assume now that a family (f (X)) is constructed for every X ∈ mod(R1) with l(X) ≤ n. Consider Y ∈ mod(R1) with l(Y ) = n + 1. Let S be a simple submodule of Y . For the nonsplittable short exact sequence

0 → S→ Y^w → Y /S → 0,^p

where w is the inclusion monomorphism and p is the canonical epimorphism, we have the short exact sequences

0 → F₁^s(S)^F−→ F^1(w) ₁^s(Y )^F−→ F^1(p) ₁^s(Y /S) → 0, 0 → Φ⁻¹F₂^sΦ(S) ^w

0

→ Φ⁻¹F₂^sΦ(Y ) ^p

0

→ Φ⁻¹F₂^sΦ(Y /S) → 0

as in the first step of our proof. Let fS be an isomorphism such that fS = f (S). Let f_{Y /S} be an isomorphism such that f_{Y /S} = f (Y /S). Let P be the projective cover of F₁^s(Y /S). Then there is an epimorphism π : P → F₁^s(Y /S). Furthermore, fY /Sπ : P → Φ⁻¹F₂^sΦ(Y /S) is an epimorphism too, because fY /S is an isomorphism. Thus there are mor- phisms π1 : P → F₁^s(Y ) and π2 : P → Φ⁻¹F₂^sΦ(Y ) such that F1(p)π1 = π and p⁰π2 = fY /Sπ. The morphisms π1, π2 are epimorphisms, because top(F₁^s(Y )) ∼= top(F₁^s(Y /S)) and top(Φ⁻¹F₂^sΦ(Y )) ∼= top(Φ⁻¹F₂^sΦ(Y /S)).

Moreover, there is a submodule L of P such that there is an epimorphism κ : L → F₁^s(S) and F1(w)κ = π1|_L. Observe that p⁰π2(t) = 0 for every t ∈ L, because p⁰π2(t) = fY /Sπ(t) = fY /SF1(p)π1(t) = fY /SF1(p)F1(w)κ(t) = 0.

Thus im(π2|_L) ⊂ im(w⁰). Then π2|_L= w⁰fSκ · k for some k ∈ K^∗. Changing w⁰ if necessary, we may assume that π2|_L= w⁰fSκ, because if p⁰w⁰ = 0 then p⁰w⁰· k⁻¹ = 0.

We define an isomorphism fY : F₁^s(Y ) → Φ⁻¹F₂^sΦ(Y ) in the following way. For y ∈ F₁^s(Y ) we can find t ∈ P such that π1(t) = y. Then we put fY(y) = π2(t). Since ker(π1) ⊂ L and ker(π2) ⊂ L, we have ker(π1) = ker(π2) = ker(κ) because π2|_L= w⁰fSκ and π1|_L= F1(w)κ. Therefore fY

is a well-defined R1-homomorphism. Since ker(π1) = ker(π2), fY is an isomorphism. It is easy to see that the diagram

0 → F₁^s(S) ^F−→^1(w) F₁^s(Y ) ^F−→^1(p) F₁^s(Y /S) → 0

↓^fS ↓^fY ↓^{fY /S}

0 → Φ⁻¹F₂^sΦ(S) ^w

0

→ Φ⁻¹F₂^sΦ(Y ) ^p

0

→ Φ⁻¹F₂^sΦ(Y /S) → 0 commutes.

Suppose now that we have a short exact sequence 0 → U→ Y^a → V → 0.^b If im(w) is contained in im(a) then there are R1-morphisms i : S → U and r : Y /S → V such that the diagram

0 → S →^w Y →^p Y /S → 0

↓ⁱ || ↓^r

0 → U →^a Y →^b V → 0

commutes. Moreover, we deduce from the first step of the proof that there are short exact sequences

0 → F₁^s(U )^F−→ F^1(a) ₁^s(Y )^F−→ F^1(b) ₁^s(V ) → 0, 0 → Φ⁻¹F₂^sΦ(U ) ^a

0

→ Φ⁻¹F₂^sΦ(Y ) ^b

0

→ Φ⁻¹F₂^sΦ(V ) → 0.

By the inductive assumption for some r⁰ : Φ⁻¹F₂^sΦ(Y /S) → Φ⁻¹F₂^sΦ(V ) such that r⁰ = Φ⁻¹F₂^sΦ(r) we have r⁰fY /S = fVF1(r). Then r⁰fY /SF1(p) = fVF1(r)F1(p). Since F1(r)F1(p) = F1(b), we have fVF1(b) = r⁰fY /SF1(p) = r⁰p⁰fY, because it was shown above that f_{Y /S}F1(p) = p⁰fY. Observe that b⁰ can be chosen in such a way that r⁰p⁰ = b⁰. Indeed, since b = rp, we have b⁰ = Φ⁻¹F₂^sΦ(b) = Φ⁻¹F₂^sΦ(rp) = r⁰p⁰. Suppose that b⁰− r⁰p⁰ 6= 0.

Then b⁰ − r⁰p⁰ factorizes through a projective R1-module Q. Since b⁰ is an epimorphism by the first step of our proof and b⁰ − r⁰p⁰ = q2q1 with q1: Φ⁻¹F₂^sΦ(Y ) → Q, q2: Q → Φ⁻¹F₂^sΦ(V ), there is q⁰₂: Q → Φ⁻¹F₂^sΦ(Y ) such that q2q1 = b⁰q⁰₂q1. Therefore r⁰p⁰ = b⁰ − b⁰q⁰₂q1. Thus put b⁰⁰ = b⁰(idΦ⁻¹F₂^sΦ(Y )− q⁰₂q1). Then b⁰⁰= b⁰and b⁰⁰is an epimorphism. Moreover, if we put a⁰⁰= (idΦ⁻¹F₂^sΦ(Y )−q₂⁰q1)⁻¹then a⁰⁰= a⁰and a⁰⁰is a monomorphism with b⁰⁰a⁰⁰= 0. Since b⁰⁰= r⁰p⁰, we get fVF1(b) = b⁰⁰fY.

We deduce from the last commutative diagram by the snake lemma that there is a commutative diagram with exact rows

0 → S →ⁱ U →^c U/S → 0

|| ↓^a ↓^v

0 → S →^w Y →^p Y /S → 0.

By the inductive assumption v⁰fU/S = fY /SF1(v) for some v⁰. Thus v⁰f_U/SF1(c) = f_{Y /S}F1(v)F1(c).

Therefore v⁰fU/SF1(c) = fY /SF1(p)F1(a) and fY /SF1(p)F1(a) = p⁰fYF1(a), since we proved that fY /SF1(p) = p⁰fY. Now observe that for a suitable c⁰ we have f_U/SF1(c) = c⁰fU by the inductive assumption. But we may assume that v⁰c⁰ = p⁰a⁰⁰. Indeed, suppose to the contrary that p⁰a⁰⁰− v⁰c⁰ 6= 0 but p⁰a⁰⁰− v⁰c⁰ = 0. Thus this difference factorizes through a projective R1- module, say Q1. Then there are z1 : Φ⁻¹F₂^sΦ(U ) → Q1 and z2 : Q1 → Φ⁻¹F₂^sΦ(Y /S) such that p⁰a⁰⁰− v⁰c⁰ = z2z1. Since p⁰ is an epimorphism by the first step of our proof, there is z₂⁰ : Q1 → Φ⁻¹F₂^sΦ(Y ) such that p⁰z₂⁰ = z2. Then replacing a⁰⁰ by a⁰₁ = a⁰⁰− z₂⁰z1 we obtain p⁰a⁰₁ = v⁰c⁰.

Moreover, observe that a⁰₁is well-defined, because it is a monomorphism by the first step of the proof and b⁰⁰a⁰₁= r⁰p⁰a⁰₁= r⁰v⁰c⁰= 0 since r⁰v⁰= 0.

Hence we may assume that p⁰a⁰⁰−v⁰c⁰ = 0. Therefore we obtain v⁰c⁰fU = p⁰a⁰⁰fU. Furthermore,

p⁰a⁰⁰fU = v⁰c⁰fU = v⁰f_U/SF1(c) = f_{Y /S}F1(v)F1(c)

= f_{Y /S}F1(p)F1(a) = p⁰fYF1(a).

Thus p⁰(a⁰⁰fU−f_YF1(a)) = 0. Then d = (a⁰⁰fU−f_YF1(a)) : U → Φ⁻¹F₂^sΦ(Y ) and im(d) ⊂ ker(p⁰) = im(w⁰). Thus dF1(i) = 0, because dF1(i) = a⁰⁰fUF1(i)

− f_YF1(a)F1(i) = a⁰⁰i⁰fS − f_YF1(w). But a⁰⁰i⁰ = w⁰. Indeed, if a⁰⁰i⁰− w⁰ 6= 0 then it is a monomorphism by simplicity of Φ⁻¹F₂^sΦ(S). On the other hand, we know that a⁰⁰i⁰− w⁰ = 0. Therefore we find that a monomorphism factorizes through a projective module, which is impossible by [17; Lecture 3]. Then a⁰⁰i⁰fS− fYF1(w) = w⁰fS− fYF1(w) = 0.

Now we can consider the decompositions of K-spaces F₁^s(Y ) = im(F1(w))

⊕ Y⁰and Φ⁻¹F₂^sΦ(Y ) = im(w⁰) ⊕ Y⁰⁰. Since fY is an R1-isomorphism, fY is a K-linear isomorphism. Since w⁰fS = fYF1(w) and p⁰fY = fY /SF1(p), fY

restricted to Y⁰is a K-linear isomorphism of Y⁰to Y⁰⁰. But if z ∈ im(F1(a))

∩ Y⁰ then fY(z) ∈ Y⁰⁰. Furthermore, we can consider the decomposition of the K-space F₁^s(U ) = im(F1(w)) ⊕ U⁰. Then by the inductive assumption for the decomposition Φ⁻¹F₂^sΦ(U ) = im(i⁰) ⊕ U⁰⁰ the restriction of fU to U⁰ is a K-linear isomorphism between U⁰ and U⁰⁰. Since a⁰⁰i⁰ = w⁰, we get a⁰⁰fU(z) ∈ Y⁰⁰, where z ∈ im(F1(w)) ∩ Y⁰. Thus im(a⁰⁰fU− f_YF1(a)) ⊂ Y⁰⁰, and so a⁰⁰fU − f_YF1(a) = 0.

Now consider the case when im(a) does not contain im(w). First assume that U is simple. Then we have the following commutative diagram with exact rows and columns:

0 0

↓ ↓

U = U

↓^a ↓^a1

0 → S →^w Y →^p Y /S → 0

|| ↓^b ↓^b1

0 → S ^w→¹ V →^p1 V /S → 0

↓ ↓

0 0

By the inductive assumption,

a⁰₁fU = fY /SF1(a1) = fY /SF1(p)F1(a) = p⁰fYF1(a),

where a⁰₁ = Φ⁻¹F₂^sΦ(a1) satisfies the required condition by the inductive assumption. We may assume that p⁰₁b⁰ = b⁰₁p⁰, where a⁰, b⁰ are so chosen