# On the classification of Schreier extensions of monoids with non-abelian kernel

Nelson Martins-Ferreira , Andrea Montoli , Alex Patchkoria and Manuela Sobral
From the journal Forum Mathematicum

## Abstract

We show that any regular (right) Schreier extension of a monoid M by a monoid A induces an abstract kernel Φ:MEnd(A)Inn(A). If an abstract kernel factors through SEnd(A)Inn(A), where SEnd(A) is the monoid of surjective endomorphisms of A, then we associate to it an obstruction, which is an element of the third cohomology group of M with coefficients in the abelian group U(Z(A)) of invertible elements of the center Z(A) of A, on which M acts via Φ. An abstract kernel Φ:MSEnd(A)Inn(A) (resp. Φ:MAut(A)Inn(A)) is induced by a regular weakly homogeneous (resp. homogeneous) Schreier extension of M by A if and only if its obstruction is zero. We also show that the set of isomorphism classes of regular weakly homogeneous (resp. homogeneous) Schreier extensions inducing a given abstract kernel Φ:MSEnd(A)Inn(A) (resp. Φ:MAut(A)Inn(A)), when it is not empty, is in bijection with the second cohomology group of M with coefficients in U(Z(A)).

## 1 Introduction

The classification of extensions is a classical problem in group theory. It is well known that extensions with abelian kernel inducing the same action are classified by the 2-dimensional cohomology group. The case of non-abelian kernels was studied by Schreier [25, 26]: to any group extension 0ABG1, he associated a group homomorphism Φ:GAut(A)Inn(A), called abstract kernel of the extension, and he determined conditions on such a homomorphism Φ in order to get the existence of extensions having it as abstract kernel. (The notation for group extensions is borrowed from Mac Lane’s book [16], and it is justified by the fact that we will use the multiplicative notation for the group G and the additive one for the other groups.) Later, Eilenberg and Mac Lane [10] gave an interpretation of such results in terms of cohomology: to an abstract kernel Φ can be associated an element Obs(Φ), called obstruction of the abstract kernel, of the third cohomology group H3(G,Z(A)), where Z(A) is the center of A and the left G-module structure on Z(A) is induced by Φ. Then Φ is induced by an extension if and only if Obs(Φ) is the zero element of H3(G,Z(A)). Moreover, if there is an extension inducing Φ, then the set of isomorphism classes of the extensions inducing it is in bijection with the second cohomology group H2(G,Z(A)). See, for example, [10, 16] for a detailed account of this result.

The same kind of result was then extended to other algebraic structures, such as associative algebras [12] and Lie algebras [13] over a field, rings [15], categories of interest [19], categorical groups [11, 7]. A categorical approach to this problem was initiated by Bourn in [1] and then generalized in [6, 2, 9, 8] to the context of semi-abelian [14] action accessible [3] categories.

The situation for monoid extensions is more complicated. Schreier extensions of monoids, a direct generalization of group extensions, were introduced by Rédei [24]. In [27], the Schreier extensions of a monoid M by an M-module A were classified by H2(M,A), the classical second cohomology group of M with coefficients in the M-module A. Then, in [20, 22], the Schreier extensions of a monoid M by an M-semimodule A (i.e. a commutative monoid on which M acts) have been classified by means of the second cohomology monoid H2(M,A), of a cohomology theory of monoids with coefficients in semimodules [21, 22] which generalizes the classical Eilenberg–Mac Lane cohomology of monoids. The problem of classifying Schreier extensions of monoids whose kernels are (not necessarily abelian) groups was studied in [27]. There the abstract kernel is involved in the definition of the extension because the author of [27] was not able to induce an abstract kernel, i.e. a monoid homomorphism Φ:MEnd(A)Inn(A), from a given Schreier extension 0ABM1.

In the present paper, we show how to induce an abstract kernel from a regular (see Definition 3.7) Schreier extension of monoids, a particular case of which is a Schreier extension of a monoid whose kernel is a group. More specifically, in Section 3, we associate to any regular Schreier extension of a monoid M by a monoid A a monoid homomorphism Φ:MEnd(A)Inn(A), and in Section 4, we show that there is a canonical representative of such a monoid extension, called the crossed product extension. In Section 5, we show that if the abstract kernel Φ takes values in SEnd(A)Inn(A), where SEnd(A) is the monoid of surjective endomorphisms of A, then it is possible to associate to Φ an element Obs(Φ) of the third cohomology group H3(M,U(Z(A))), where U(Z(A)) is the abelian group of invertible elements of the center Z(A) of A, and the action of M on U(Z(A)) is induced by Φ. Moreover, we show that an abstract kernel Φ is induced by an extension if and only if Obs(Φ) is the zero element of the third cohomology group. Finally, in Section 6, we show that the set Ext(M,A,Φ) of isomorphism classes of regular weakly homogeneous (resp. homogeneous) Schreier extensions of M by A (see Definition 3.11) which induce the same abstract kernel Φ:MSEnd(A)Inn(A) (resp. Φ:MAut(A)Inn(A)), when it is not empty, is in bijection with the second cohomology group H2(M,U(Z(A))) of M with coefficients in the M-module U(Z(A)). This is done, as for the classical case of extensions of groups, by showing that there is a simply transitive action of the abelian group H2(M,U(Z(A))) on the set Ext(M,A,Φ). Hence our approach is very similar to the classical one for groups, yielding a new, additional interpretation of the classical Eilenberg–Mac Lane cohomology in terms of monoid extensions.

## 2 Preliminaries

In this section, we recall some notions we need in the rest of the paper, and we fix some notations.

Given a monoid M, we will denote by Z(M) the center of M, namely

Z(M)={zMzm=mzfor allmM},

and by U(M) the group of invertible elements of M.

## Definition 2.1.

Given a monoid M and a subgroup H (i.e. a subgroup H of the group U(M)), we say that H is

1. right normal if, for all mM, mHHm, where mH={mhhH} and Hm={hmhH},

2. left normal if, for all mM, HmmH,

3. normal if it is both right and left normal, i.e., mH=Hm.

Note that H is right normal in M if and only if H is left normal in Mop.

If H is a subgroup of a monoid M, the relation on M defined by

m1m2m1=hm2for somehH

is an equivalence relation on M, called the right coset relation. The equivalence class of an element m is cl(m)=Hm. We will denote by MH the quotient set. Similarly, we can define the left coset relation.

## Proposition 2.2.

If H is right normal in M, then the operation Hm1Hm2=Hm1m2 is well defined, and (MH,,H) is a monoid.

## Proof.

If Hm1=Hm1 and Hm2=Hm2, then there exist h1,h2H such that m1=h1m1 and m2=h2m2. Hence m1m2=h1m1h2m2. Since H is right normal, there exists h3H such that m1h2=h3m1, and so

m1m2=h1m1h2m2=h1h3m1m2,

which proves that Hm1m2=Hm1m2. ∎

The same happens for the left coset relation, when H is left normal.

## Example 2.3.

If A is a monoid, End(A) is the monoid of endomorphisms of A (w.r.t. the usual composition of functions, (gf)(a)=g(f(a))), and Inn(A) is the subgroup of inner automorphisms induced by the invertible elements of A, then Inn(A) is right normal, but not left normal, in End(A). Indeed, if φEnd(A), μgInn(A), then

(φμg)(a)=φ(μg(a))=φ(gag-1)=φ(g)φ(a)φ(g)-1=μφ(g)(φ(a))=(μφ(g)φ)(a),

hence φμg=μφ(g)φ, which shows that Inn(A) is right normal in End(A). But it is not left normal, in general. A concrete counterexample is the following. If A is the symmetric group S3, consider the endomorphism f of S3 defined by

f(id)=f((123))=f((132))=id,f((12))=f((13))=f((23))=(12).

Then, for every element sS3, fμs=f, and so fInn(A)={f}, but the endomorphism μ(13)f is different from f; indeed,

μ(13)f((12))=μ(13)f((13))=μ(13)f((23))=(13)(12)(13)-1=(23),

and so Inn(A)f is not contained in fInn(A).

## Proposition 2.4.

If G is a group, then Inn(G) is normal in the monoid Epi(G) of epimorphisms of G.

## Proof.

As we observed before, Inn(G) is right normal in End(G), and so it is right normal in Epi(G), too. Let us prove that it is also left normal. If φEpi(G) and gG, let gG be such that φ(g)=g (since G is a group, φ is surjective). Then, for all xG, we have

(μgφ)(x)=gφ(x)g-1=φ(g)φ(x)φ(g)-1=φ(gxg-1)=(φμg)(x),

hence φInn(G)=Inn(G)φ. ∎

## Definition 3.1 ([24]).

Let

(3.1)E:0A𝜅B𝜎M1

be a sequence of monoids and monoid homomorphisms such that σ is a surjection, κ is an injection and κ(A)={bB|σ(b)=1} (i.e. κ is the kernel of σ). E is a (right) Schreier extension of M by A (some authors would say “A by M” ) if, for every xM, there exists an element uxσ-1(x) such that, for every bσ-1(x), there exists a unique aA such that b=κ(a)+ux. The elements ux, for xM, will be called the representatives of E. We will always choose u1=0 (we use the multiplicative notation for M and the additive one for the other monoids involved).

Note that if (3.1) is a Schreier extension, then σ is the cokernel of κ. Indeed, suppose that f:BC is a monoid homomorphism such that fκ(a)=0 for all aA. Define a map g:MC by putting g(x)=f(b), bσ-1(x). If σ(b1)=x=σ(b2), then b1=κ(a1)+ux and b2=κ(a2)+ux, whence f(b1)=f(ux)=f(b2). Hence g is well defined. Clearly, g is a monoid homomorphism and gσ=f. The uniqueness of such a homomorphism g is also clear.

## Example 3.2.

Let be the commutative monoid of natural numbers, with the usual sum, and let Cm(t) denote the multiplicative cyclic group of order m with generator t. The sequence 0m¯𝑝Cm(t)1, where m¯(1)=m and p(1)=t, is a Schreier extension of Cm(t) by , with representatives given by 0,1,,m-1.

From now on, we will treat κ just as an inclusion.

## Proposition 3.3.

Let E be a Schreier extension as in (3.1), with representatives ux, xM. An element bσ-1(x) is another representative of x for E if and only if b=g+ux for some gU(A).

## Proof.

Since ux is a representative, there exists a unique aA such that b=a+ux. Moreover, if b is a representative for E, then there is a unique aA such that ux=a+b. Hence we get b=a+a+b. By the uniqueness in the Schreier condition, we get a+a=0. Similarly, from the equality ux=a+a+ux, we get a+a=0, and so a is invertible. Conversely, if b=g+ux with gU(A), then, for every bσ-1(x), there exists a unique aA such that

b=a+ux=a-g+b.

Moreover, if a1+b=a2+b, then a1+g+ux=a2+g+ux; then the uniqueness in the Schreier condition implies a1+g=a2+g, and hence a1=a2 because g is invertible. ∎

## Lemma 3.4.

Let E be a Schreier extension as in (3.1), with representatives ux, xM. For aA, let a be the unique element in A such that ux+a=a+ux. If aU(A), then aU(A), too.

## Proof.

There exists a unique a′′A such that ux+(-a)=a′′+ux. From the equality ux+a=a+ux, we obtain ux=a+ux-a=a+a′′+ux, and the uniqueness in the Schreier condition implies a+a′′=0. Similarly, from the equality ux+(-a)=a′′+ux, we get ux=a′′+ux+a=a′′+a+ux, from where we obtain a′′+a=0. ∎

## Proposition 3.5.

Let E be a Schreier extension as in (3.1), and let ux,uy,vx,vy be representatives, for x,yM. If ux+uy is a representative, then so is vx+vy.

## Proof.

Thanks to Proposition 3.3, we know that there exist g1,g2U(A) such that vx=g1+ux, vy=g2+uy. Moreover, there exists a unique hA such that ux+g2=h+ux, and such an h is invertible thanks to the previous lemma. Then we have vx+vy=g1+ux+g2+uy=g1+h+ux+uy, with g1+hU(A). Then the thesis follows from Proposition 3.3. ∎

Let E be a Schreier extension as in (3.1), with representatives ux, xM. We already observed that, for all aA, there is a unique element aA such that ux+a=a+ux. This defines a map φ(x):AA sending a to a.

## Proposition 3.6.

The following statements hold:

1. for every xM, we have that φ(x)End(A);

2. if vx is another representative, and ψ(x):AA is the induced endomorphism of A, then ψ(x)=μgφ(x) with gU(A).

## Proof.

(a) From the obvious equality ux+0=0+ux, we get φ(x)(0)=0. Moreover, on one hand

ux+a1+a2=φ(x)(a1+a2)+ux,

while, on the other hand,

ux+a1+a2=φ(x)(a1)+ux+a2=φ(x)(a1)+φ(x)(a2)+ux.

By the uniqueness, we get that φ(x)(a1+a2)=φ(x)(a1)+φ(x)(a2).

(b) From Proposition 3.3, we know that vx=g+ux with gU(A). Moreover, for all aA,

vx+a=ψ(x)(a)+vx.

Therefore,

vx+a=g+ux+a=g+φ(x)(a)+ux=g+φ(x)(a)-g+vx.

This means that ψ(x)(a)=g+φ(x)(a)-g=(μgφ(x))(a) for all aA.∎

The previous proposition implies that, for a Schreier extension E as in (3.1), there is an induced well-defined map

(3.2)Φ:MEnd(A)Inn(A),

given by Φ(x)=cl(φ(x)), such that Φ(1)=cl(idA) (see Proposition 2.2 and Example 2.3). In order to have that Φ is a monoid homomorphism, we need an additional assumption.

## Definition 3.7.

Let E be a Schreier extension as in (3.1). We say that E is a regular Schreier extension if whenever ux and uy are representatives for E, then so is ux+uy (such extensions are called normal Schreier extensions in [22]).

## Proposition 3.8.

If E is a regular Schreier extension, then the map (3.2) is a monoid homomorphism.

## Proof.

Let x,yM, and let ux,uy and uxy be representatives. We have the corresponding

φ(x),φ(y),φ(xy)End(A)

with

ux+a=φ(x)(a)+ux,uy+a=φ(y)(a)+uy,uxy+a=φ(xy)(a)+uxy

for all aA. Since E is regular, ux+uy is a representative, hence ux+uy=g+uxy for some gU(A). On one hand, we have

ux+uy+a=g+uxy+a=g+φ(xy)(a)+uxy,

while, on the other hand,

ux+uy+a=ux+φ(y)(a)+uy=φ(x)(φ(y)(a))+ux+uy=φ(x)(φ(y)(a))+g+uxy.

Therefore,

g+φ(xy)(a)=φ(x)(φ(y)(a))+g,

whence

φ(x)(φ(y)(a))=g+φ(xy)(a)-g.

This means that φ(x)φ(y)=μgφ(xy), i.e. Φ(x)Φ(y)=Φ(xy). ∎

## Definition 3.9.

Given a regular Schreier extension E as in (3.1), the induced monoid homomorphism (3.2) is called the abstract kernel induced by the extension E. More generally, we will call abstract kernel any such homomorphism, even when it is not induced by an extension.

The following proposition gives examples of regular Schreier extensions.

## Proposition 3.10.

Let E be a Schreier extension as in (3.1) such that A is a group (such extensions are called special Schreier extensions in [4, 5, 17, 18]). Then every element of B is a representative, and therefore E is regular.

## Proof.

Let xM, and let ux be a representative. For every bσ-1(x), there exists (a unique) aA such that b=a+ux. Being A a group, a is invertible. Then it follows from Proposition 3.3 that b is a representative. Thus, every element of B is a representative, and hence E is regular. ∎

Note that the extension of Example 3.2 serves as an example of Schreier extension which is not regular.

## Definition 3.11.

A Schreier extension E as in (3.1), with representatives ux, xM, is

1. weakly homogeneous if, for all bσ-1(x), there exists aA such that b=ux+a,

2. homogeneous if, for all bσ-1(x), there is a unique such a.

Note that, thanks to Proposition 3.3, this definition does not depend on the choice of representatives. (Indeed, for any representative vx, we have ux=g+vx, gU(A). If (a) holds, then

b=-g+g+b=-g+ux+a=vx+a.

If (b) holds, then we have

vx+a1=vx+a2g+vx+a1=g+vx+a2ux+a1=ux+a2a1=a2.)

The following proposition is a generalization of [5, Proposition 3.8], where only split extensions were considered.

## Proposition 3.12.

Let E be a Schreier extension as in (3.1), with representatives ux, xM. Let φ(x):AA be the induced endomorphism of A relative to the element xM. Then

1. E is weakly homogeneous if and only if φ(x) is surjective for all xM;

2. E is homogeneous if and only if φ(x)Aut(A) for all xM.

## Proof.

(a) Suppose that E is weakly homogeneous, and consider xM. Given aA, there exists aA such that a+ux=ux+a, from which we obtain that φ(x)(a)=a, and so φ(x) is surjective. Conversely, suppose that φ(x) is surjective. Given bσ-1(x), there exists a unique aA such that b=a+ux (because E is Schreier). The surjectivity of φ(x) implies the existence of aA such that φ(x)(a)=a. Hence we have ux+a=φ(x)(a)+ux=a+ux=b.

(b) Suppose that E is homogeneous. We already know that, for all xM, φ(x) is surjective. Suppose that φ(x)(a1)=φ(x)(a2). Then ux+a1=φ(x)(a1)+ux=φ(x)(a2)+ux=ux+a2, and the uniqueness in the definition of a homogeneous Schreier extension implies that a1=a2, and so φ(x) is injective. Conversely, suppose that φ(x)Aut(A). We already know that E is weakly homogeneous. If ux+a1=ux+a2, then φ(x)(a1)+ux=φ(x)(a2)+ux. Being E Schreier, this implies that φ(x)(a1)=φ(x)(a2), and the injectivity of φ(x) gives us that a1=a2. ∎

From now on, SEnd(A) denotes the monoid of surjective endomorphisms of a monoid A.

The previous proposition shows that a regular weakly homogeneous Schreier extension E as in (3.1) induces a monoid homomorphism Φ:MSEnd(A)Inn(A), while a regular homogeneous Schreier extension induces a monoid homomorphism Φ:MAut(A)Inn(A).

The following result is a generalization of [5, Proposition 3.4].

## Proposition 3.13.

If E:0A𝜅B𝜎M1 is a regular Schreier extension and M is a group, then E is homogeneous.

## Proof.

Given representatives ux, xM, with u1=0, consider the induced endomorphisms φ(x):AA. If φ(x)(a1)=φ(x)(a2), then ux+a1=ux+a2, whence ux-1+ux+a1=ux-1+ux+a2. Since E is regular, ux-1+ux is a representative of 1; hence it is an invertible element of A (by Proposition 3.3). This implies that a1=a2, and thus φ(x) is injective. Moreover, since E is regular, φ(x)φ(x-1)=μgφ(xx-1)=μgφ(1)=μg for some gU(A) (see Proposition 3.8). Being μg an automorphism, we deduce that φ(x) is surjective. Then the thesis follows from Proposition 3.12. ∎

## Example 3.14.

Consider the sequence A𝜅A×gC2(t)𝜎C2(t), where A is any monoid, C2(t) is the cyclic group of order 2 with generator t, g is a fixed element of U(Z(A)), C2(t) acts on the monoid A in a way that tg=g, and A×gC2(t) is the cartesian product A×C2(t) with the monoid operation defined by

(a1,1)+(a2,1)=(a1+a2,1),(a1,1)+(a2,t)=(a1+a2,t),
(a1,t)+(a2,1)=(a1+ta2,t),(a1,t)+(a2,t)=(a1+ta2+g,1).

It is straightforward to check that this operation is associative and that (0,1) is the neutral element. The morphism σ is just the canonical projection, while κ(a)=(a,1). This sequence is a regular homogeneous Schreier extension. In order to show that it is Schreier, it suffices to choose the representatives u1=(0,1) and ut=(0,t). Thanks to Proposition 3.5, regularity is proved just by observing that the element

ut+ut=(0,t)+(0,t)=(g,1)=(g,1)+(0,1)=(g,1)+u1

is a representative since (g,1) is invertible (see Proposition 3.3). The previous proposition implies that the extension is homogeneous.

Several other examples of Schreier and homogeneous Schreier extensions may be found in [4, 5].

## 4 The crossed product extension

Let E:0A𝜅B𝜎M1 be a regular Schreier extension, with representatives ux, xM. Being E regular, we know that, for all x,yM, the element ux+uy is a representative. Thanks to Proposition 3.3, we get that there exists a unique element f(x,y)U(A) such that ux+uy=f(x,y)+uxy. This defines a map

f:M×MU(A)such thatf(x,1)=f(1,y)=0

for all x,yM (because we are assuming that u1=0). Then we have, on one hand,

ux+uy+uz=f(x,y)+uxy+uz=f(x,y)+f(xy,z)+uxyz,

and, on the other hand,

ux+uy+uz=ux+f(y,z)+uyz=φ(x)(f(y,z))+ux+uyz=φ(x)(f(y,z))+f(x,yz)+uxyz,

where φ:MEnd(A) is the map defined by ux+a=φ(x)(a)+ux (as we explained in the previous section). Whence

(4.1)φ(x)(f(y,z))+f(x,yz)=f(x,y)+f(xy,z)for allx,y,zM.

Furthermore, for every x,yM and every aA, we have, on one hand,

ux+uy+a=ux+φ(y)(a)+uy=φ(x)φ(y)(a)+ux+uy=φ(x)φ(y)(a)+f(x,y)+uxy,

and, on the other hand,

ux+uy+a=f(x,y)+uxy+a=f(x,y)+φ(xy)(a)+uxy,

whence

φ(x)φ(y)(a)+f(x,y)=f(x,y)+φ(xy)(a).

Being f(x,y) invertible, the last equality implies that

φ(x)φ(y)(a)=f(x,y)+φ(xy)(a)-f(x,y).

Thus φ(x)φ(y)=μf(x,y)φ(xy) for all x,yM.

## Proposition 4.1.

Let monoids M,A and maps φ:MEnd(A), f:M×MU(A) such that, for all x,y,zM,

φ(1)=idA,f(x,1)=f(1,y)=0,φ(x)φ(y)=μf(x,y)φ(xy),φ(x)(f(y,z))+f(x,yz)=f(x,y)+f(xy,z),

be given. Then the set [A,φ,f,M] of all pairs (a,x)A×M with the operation defined by

(a1,x)+(a2,y)=(a1+φ(x)(a2)+f(x,y),xy)

is a monoid, and the sequence

A𝑖[A,φ,f,M]𝑝M,i(a)=(a,1),p(a,x)=x,

is a regular Schreier extension of M by A, called the crossed product extension, such that the induced monoid homomorphism Φ:MEnd(A)Inn(A) sends xM to the equivalence class of φ(x). Furthermore, a pair (a,x) is a representative if and only if aU(A).

## Proof.

It is straightforward to show that the operation is associative and that (0,1) is its neutral element. The maps i and p are clearly monoid homomorphisms, p is surjective and i is injective, and the image of i is the kernel of p. Let us show that we get a regular Schreier extension. For any xM, we consider the element ux=(0,x). These elements are representatives: indeed, every element (a,x)A×M can be written as (a,x)=(a,1)+(0,x), and such writing is unique because the equality (a1,1)+(0,x)=(a2,1)+(0,x) implies (a1,x)=(a2,x), and hence a1=a2. So the extension is Schreier. Proposition 3.3, together with the equality (a,x)=(a,1)+(0,x), implies that (a,x) is a representative if and only if aU(A). It remains to show that the extension is regular. For all x,yM, we have

ux+uy=(0,x)+(0,y)=(0+φ(x)(0)+f(x,y),xy)=(f(x,y),xy)=(f(x,y),1)+(0,xy)=(f(x,y),1)+uxy,

and then, since f(x,y)U(A), ux+uy is a representative by Proposition 3.3. Hence, thanks to Proposition 3.5, the extension is regular. Furthermore, for all aA, we have

ux+i(a)=(0,x)+(a,1)=(φ(x)(a),x)=(φ(x)(a),1)+(0,x)=i(φ(x)(a))+ux,

which means that Φ sends xM to the equivalence class of φ(x). ∎

## Remark 4.2.

If, in the previous proposition, we have that φ:MSEnd(A), then the crossed product extension is weakly homogeneous. Indeed, every element (a,x)A×M can be written as (a,x)=(0,x)+(a,1), where aA is such that φ(x)(a)=a (such an element exists since φ(x) is surjective).

If we have that φ:MAut(A), then the crossed product extension is homogeneous. Indeed, if

(0,x)+(a1,1)=(0,x)+(a2,1),

then

(φ(x)(a1),x)=(φ(x)(a2),x)φ(x)(a1)=φ(x)(a2)a1=a2.

(See Definition 3.11 and the note after it.)

We recall from [23] the following version of the Short Five Lemma for monoid extensions.

## Proposition 4.3 ([23, Proposition 4.5]).

Consider the following commutative diagram of monoid homomorphisms:

where the two rows are Schreier extensions and the homomorphism β sends representatives to representatives. Then,

1. if α and γ are injective, then β also is;

2. if α and γ are surjective, then β also is;

3. if α and γ are isomorphisms, then β is an isomorphism, too.

This fact allows us to prove the following.

## Proposition 4.4.

Given an abstract kernel Φ:MEnd(A)Inn(A), where A and M are monoids, fix an endomorphism φ(x)Φ(x) for every element xM (with φ(1)=idA). Then every regular Schreier extension E as in (3.1) which induces the abstract kernel Φ is isomorphic to the crossed product extension A𝑖[A,φ,f,M]𝑝M.

## Proof.

We take representatives vx, xM, of E, with v1=0. Then, for all aA and all xM, we get that vx+a=ψ(x)(a)+vx for some ψ(x)Φ(x). Then, for each xM, φ(x)=μg(x)ψ(x) for some g(x)U(A), i.e., for all aA,

φ(x)(a)=g(x)+ψ(x)(a)-g(x).

We define new representatives by putting ux=g(x)+vx for xM. Choosing g(1)=0, we get u1=0. Since E is regular, for all x,yM, ux+uy is a representative, hence ux+uy=f(x,y)+uxy with f(x,y)U(A). Furthermore, for all aA and all xM,

ux+a=g(x)+vx+a=g(x)+ψ(x)(a)+vx=g(x)+ψ(x)(a)-g(x)+ux,

that is, ux+a=φ(x)(a)+ux. Then the maps φ:MEnd(A) and f:M×MU(A) satisfy the conditions of Proposition 4.1 (see the considerations before this proposition) and therefore we have the crossed product extension [A,φ,f,M]. Consider now the diagram

where the map β is defined by β(b)=(a,x), where σ(b)=x and a is the unique element of A such that b=a+ux. Then β is a monoid homomorphism: clearly, β(0)=β(0+u1)=(0,1), and moreover,

β(a1+ux+a2+uy)=β(a1+φ(x)(a2)+ux+uy)=β(a1+φ(x)(a2)+f(x,y)+uxy)=(a1+φ(x)(a2)+f(x,y),xy)=(a1,x)+(a2,y)=β(a1+ux)+β(a2+uy).

Furthermore,

βκ(a)=β(a+u1)=(a,1)=i(a)andpβ(a+ux)=p(a,x)=x=σ(a+ux);

hence the diagram is commutative. Finally, β(ux)=(0,x), and if wx is another representative of E, then wx=g+ux, gU(A), whence β(wx)=i(g)+(0,x), and so the representatives are preserved by β (see Proposition 3.3). Thanks to Proposition 4.3, β is an isomorphism. ∎

## 5 The obstruction of an abstract kernel

The aim of this section is to show that, to any abstract kernel Φ:MSEnd(A)Inn(A) (resp. Φ:MAut(A)Inn(A)), it is possible to associate an element of the third Eilenberg–Mac Lane cohomology group of M with coefficients in the M-module U(Z(A)), called the obstruction of Φ. Moreover, we will show that the abstract kernel Φ:MSEnd(A)Inn(A) (resp. Φ:MAut(A)Inn(A)) is induced by a regular weakly homogeneous (resp. homogeneous) Schreier extension if and only if its obstruction is the zero element of the cohomology group. In order to do this, we first describe how to get from Φ a structure of M-module on U(Z(A)).

## Proposition 5.1.

Given an abstract kernel Φ:MSEnd(A)Inn(A), where A and M are monoids, the center Z(A) of A is an M-semimodule w.r.t. the action defined by xc=φ(x)(c) for xM, cZ(A) and φ(x)Φ(x).

## Proof.

We first show that xcZ(A) for all xM, cZ(A) and φ(x)Φ(x). Consider an element aA. Being φ(x) surjective, there exists aA such that φ(x)(a)=a. Then

a+φ(x)(c)=φ(x)(a)+φ(x)(c)=φ(x)(a+c)=φ(x)(c+a)=φ(x)(c)+φ(x)(a)=φ(x)(c)+a.

Now, it remains to show that the definition above does not depend on the choice of the representative φ(x) of the class Φ(x) in the quotient SEnd(A)Inn(A). To do that, consider another representative ψ(x)SEnd(A). Then there is an element gU(A) such that ψ(x)=μgφ(x). So we get

ψ(x)(c)=μgφ(x)(c)=g+φ(x)(c)-g=φ(x)(c)+g-g=φ(x)(c),

where we are using that φ(x)(c)Z(A). This concludes the proof. ∎

## Corollary 5.2.

Given an abstract kernel Φ:MSEnd(A)Inn(A), where A and M are monoids, the group U(Z(A)) of A is an M-module w.r.t. the action defined by xg=φ(x)(g) for xM, gU(Z(A)) and φ(x)Φ(x).

## Proof.

It is immediate to observe that if gU(Z(A)), then xg is also invertible, with inverse x(-g), so the action of M on Z(A) restricts to U(Z(A)). ∎

Now we describe how to associate an obstruction to an abstract kernel. Given a monoid homomorphism Φ:MSEnd(A)Inn(A), we choose a representative φ(x)Φ(x) for any xM, with φ(1)=idA. We have that

φ(x)φ(y)=μf(x,y)φ(xy)

for some f(x,y)U(A), with f(x,1)=f(1,y)=0. Now, given x,y,zM, we have, on one hand,

φ(x)φ(y)φ(z)=φ(x)μf(y,z)φ(yz)=μφ(x)(f(y,z))φ(x)φ(yz)=μφ(x)(f(y,z))μf(x,yz)φ(xyz)=μφ(x)(f(y,z))+f(x,yz)φ(xyz),

and, on the other hand,

φ(x)φ(y)φ(z)=μf(x,y)φ(xy)φ(z)=μf(x,y)μf(xy,z)φ(xyz)=μf(x,y)+f(xy,z)φ(xyz).

Comparing the two expressions, and using the fact that φ(xyz) is surjective, we get the equality

μφ(x)(f(y,z))+f(x,yz)=μf(x,y)+f(xy,z),

namely μφ(x)(f(y,z))+f(x,yz)-(f(x,y)+f(xy,z))=idA, which tells us that

φ(x)(f(y,z))+f(x,yz)-(f(x,y)+f(xy,z))U(Z(A)).

This means that there exists a unique element k(x,y,z)U(Z(A)) such that

φ(x)(f(y,z))+f(x,yz)=k(x,y,z)+f(x,y)+f(xy,z).

Clearly, k(x,y,1)=k(x,1,z)=k(1,y,z)=0.

## Definition 5.3.

The function k:M×M×MU(Z(A)) we get this way is the obstruction of the abstract kernel Φ.

## Proposition 5.4.

An obstruction k of an abstract kernel Φ as above is a 3-cocycle of the cohomology of M with coefficients in the M-module U(Z(A)).

## Proof.

Given elements x,y,z,tM, we compute the expression φ(x)(φ(y)(f(z,t))+f(y,zt))+f(x,yzt) in two different ways. On one hand, we have

φ(x)(φ(y)(f(z,t))+f(y,zt))+f(x,yzt)=φ(x)(k(y,z,t)+f(y,z)+f(yz,t))+f(x,yzt)=xk(y,z,t)+φ(x)(f(y,z))+φ(x)(f(yz,t))+f(x,yzt)=xk(y,z,t)+k(x,y,z)+f(x,y)+f(xy,z)-f(x,yz)+k(x,yz,t)+f(x,yz)+f(xyz,t)-f(x,yzt)+f(x,yzt)=xk(y,z,t)+k(x,y,z)+k(x,yz,t)+f(x,y)+f(xy,z)+f(xyz,t),

where the last equality holds since k takes values in the center of A. On the other hand, we have

φ(x)(φ(y)(f(z,t))+f(y,zt))+f(x,yzt)=φ(x)φ(y)(f(z,t))+φ(x)(f(y,zt))+f(x,yzt).

Since φ(x)φ(y)=μf(x,y)φ(xy), this is equal to

f(x,y)+φ(xy)(f(z,t))-f(x,y)+φ(x)(f(y,zt))+f(x,yzt)=f(x,y)+k(xy,z,t)+f(xy,z)+f(xyz,t)-f(xy,zt)-f(x,y)+k(x,y,zt)+f(x,y)+f(xy,zt)-f(x,yzt)+f(x,yzt)=k(xy,z,t)+k(x,y,zt)+f(x,y)+f(xy,z)+f(xyz,t),

where, once again, the last equality holds since k takes values in the center of A. Comparing the two expressions, and using the fact that f takes values in U(A), we obtain the equality

xk(y,z,t)+k(x,yz,t)+k(x,y,z)=k(xy,z,t)+k(x,y,zt).

Since k(x,y,1)=k(x,1,z)=k(1,y,z)=0, we have that k is a 3-cocycle. ∎

In the construction of the obstruction of an abstract kernel Φ, we used the fact that, given x,yM, there exists an element f(x,y)U(A) such that φ(x)φ(y)=μf(x,y)φ(xy). Such an element is not unique. However, if we replace it with an f(x,y) with the same properties, the cohomology class of the corresponding 3-cocycle k is the same.

## Proposition 5.5.

Consider an abstract kernel Φ:MSEnd(A)Inn(A), with chosen representatives φ(x)Φ(x) for any xM, with φ(1)=idA. If, for any x,yM, we have

φ(x)φ(y)=μf(x,y)φ(xy)=μf(x,y)φ(xy)

with f(x,1)=0=f(1,y) and f(x,1)=0=f(1,y), then the 3-cocycles k and k constructed using f and f are cohomologous.

## Proof.

From the equality μf(x,y)φ(xy)=μf(x,y)φ(xy), we get μf(x,y)=μf(x,y) because φ(xy) is surjective. This means that μf(x,y)-f(x,y)=idA. Hence

h(x,y)=f(x,y)-f(x,y)U(Z(A)),

so we get a map h:M×MU(Z(A)) such that h(x,1)=0=h(1,y). From the equality

f(x,y)=h(x,y)+f(x,y),

valid for all x,yM, and from the definition of the cocycles k and k, we get

k(x,y,z)=φ(x)(f(y,z))+f(x,yz)-f(xy,z)-f(x,y)=φ(x)(h(y,z)+f(y,z))+h(x,yz)+f(x,yz)-[h(xy,z)+f(xy,z)]-[h(x,y)+f(x,y)]=φ(x)(f(y,z))+f(x,yz)-f(xy,z)-f(x,y)+xh(y,z)-h(xy,z)+h(x,yz)-h(x,y)=k(x,y,z)-δ2h(x,y,z).

Thus k-k=δ2h. ∎

Conversely, starting with cohomologous cocycles:

## Proposition 5.6.

Consider an abstract kernel Φ:MSEnd(A)Inn(A), with chosen representatives φ(x)Φ(x) for any xM, with φ(1)=idA. Let f:M×MU(A) be a map with φ(x)φ(y)=μf(x,y)φ(xy) and f(x,1)=0=f(1,y) for any x,yM, and let k:M×M×MU(Z(A)) be the 3-cocycle induced by f. If k′′ is a 3-cocycle which is cohomologous to k, then there exists a map f′′:M×MU(A), with f′′(x,1)=0=f′′(1,y), such that φ(x)φ(y)=μf′′(x,y)φ(xy) and the 3-cocycle induced by f′′ is precisely k′′.

## Proof.

By assumption, there exists h:M×MU(Z(A)), with h(x,1)=0=h(1,y), such that k-k′′=δ2h. We define f′′:M×MU(A) by putting f′′(x,y)=h(x,y)+f(x,y). Clearly, f′′(x,1)=0=f′′(1,y). Moreover, μf′′(x,y)=μh(x,y)μf(x,y)=μf(x,y) since h(x,y)U(Z(A)). Therefore, φ(x)φ(y)=μf′′(x,y)φ(xy). Furthermore, for any x,y,zM, we have

φ(x)(f′′(y,z))+f′′(x,yz)-f′′(xy,z)-f′′(x,y)=φ(x)(h(y,z)+f(y,z))+h(x,yz)+f(x,yz)-[h(xy,z)+f(xy,z)]-[h(x,y)+f(x,y)]=φ(x)(f(y,z))+f(x,yz)-f(xy,z)-f(x,y)+xh(y,z)-h(xy,z)+h(x,yz)-h(x,y)=k(x,y,z)-δ2h(x,y,z)=k′′(x,y,z).

It remains to check what happens if, given an abstract kernel Φ:MSEnd(A)Inn(A), we consider two different representatives φ(x) and φ(x) of Φ(x).

## Proposition 5.7.

Consider an abstract kernel Φ:MSEnd(A)Inn(A), with chosen representatives φ(x)Φ(x) for any xM, with φ(1)=idA. Let f:M×MU(A) be a map with φ(x)φ(y)=μf(x,y)φ(xy) and f(x,1)=0=f(1,y) for any x,yM, and let k:M×M×MU(Z(A)) be the 3-cocycle induced by f. If one chooses other representatives φ(x)Φ(x), again with φ(1)=idA, there exists a map f:M×MU(A), with f(x,1)=0=f(1,y), such that φ(x)φ(y)=μf(x,y)φ(xy) and its induced 3-cocycle is precisely k.

## Proof.

Since φ(x),φ(x)Φ(x) for all xM, they differ by an inner automorphism of A. In other terms, there is a map g:MU(A), with g(1)=0, such that φ(x)=μg(x)φ(x). Then, for x,yM, we get

φ(x)φ(y)=μg(x)φ(x)μg(y)φ(y)=μg(x)μφ(x)(g(y))φ(x)φ(y)=μg(x)μφ(x)(g(y))μf(x,y)φ(xy)=μg(x)μφ(x)(g(y))μf(x,y)μg(xy)-1φ(xy)=μg(x)μφ(x)(g(y))μf(x,y)μ-g(xy)φ(xy)=μg(x)+φ(x)(g(y))+f(x,y)-g(xy)φ(xy).

Thus, defining f(x,y)=g(x)+φ(x)(g(y))+f(x,y)-g(xy), we obtain that φ(x)φ(y)=μf(x,y)φ(xy), and obviously f(x,1)=0=f(1,y). It remains to check that the induced 3-cocycle is k. We have

φ(x)(f(y,z))+f(x,yz)-f(xy,z)-f(x,y)=φ(x)[g(y)+φ(y)(g(z))+f(y,z)-g(yz)]+g(x)+φ(x)(g(yz))+f(x,yz)-g(xyz)-[g(xy)+φ(xy)(g(z))+f(xy,z)-g(xyz)]-[g(x)+φ(x)(g(y))+f(x,y)-g(xy)]=μg(x)[φ(x)(g(y))+φ(x)φ(y)(g(z))+φ(x)(f(y,z))-φ(x)(g(yz))]+g(x)+φ(x)(g(yz))+f(x,yz)-g(xyz)+g(xyz)-f(xy,z)-φ(xy)(g(z))-g(xy)+g(xy)-f(x,y)-φ(x)(g(y))-g(x)=g(x)+φ(x)(g(y))+φ(x)φ(y)(g(z))+φ(x)(f(y,z))-φ(x)(g(yz))-g(x)+g(x)+φ(x)(g(yz))+f(x,yz)-f(xy,z)-φ(xy)(g(z))-f(x,y)-φ(x)(g(y))-g(x)=g(x)+φ(x)(g(y))+φ(x)φ(y)(g(z))+φ(x)(f(y,z))+f(x,yz)-f(xy,z)-φ(xy)(g(z))-f(x,y)-φ(x)(g(y))-g(x)=g(x)+φ(x)(g(y))+μf(x,y)φ(xy)(g(z))+k(x,y,z)+f(x,y)-φ(xy)(g(z))-f(x,y)-φ(x)(g(y))-g(x)=k(x,y,z)+g(x)+φ(x)(g(y))+f(x,y)+φ(xy)(g(z))-f(x,y)+f(x,y)-φ(xy)(g(z))-f(x,y)-φ(x)(g(y))-g(x)=k(x,y,z),

and this concludes the proof. ∎

The previous propositions give the following.

## Theorem 5.8.

Any abstract kernel Φ:MSEnd(A)Inn(A) determines in an invariant way an element Obs(Φ) of the third cohomology group H3(M,U(Z(A))) of the monoid M with coefficients in the M-module U(Z(A)). An abstract kernel Φ:MSEnd(A)Inn(A) is induced by a regular weakly homogeneous Schreier extension if and only if Obs(Φ) is the zero element of H3(M,U(Z(A))).

## Proof.

The fact that the element Obs(Φ)H3(M,U(Z(A))) is uniquely determined is a consequence of the previous propositions. If the abstract kernel Φ is induced by a regular weakly homogeneous Schreier extension, we observed at the beginning of Section 4 that there exists a map f:M×MU(A) such that f(x,1)=0=f(1,y) and

φ(x)(f(y,z))+f(x,yz)=f(x,y)+f(xy,z)for allx,y,zM.

Hence the element Obs(Φ) associated to the abstract kernel Φ induced by the extension is zero. Conversely, if the obstruction of an abstract kernel Φ:MSEnd(A)Inn(A) is zero, then the crossed product extension built in Proposition 4.1 (which is weakly homogeneous by Remark 4.2) induces Φ. ∎

In particular, for abstract kernels which factor through Aut(A)Inn(A), i.e. abstract kernels of the form Φ:MAut(A)Inn(A), we get the following.

## Theorem 5.9.

Any abstract kernel Φ:MAut(A)Inn(A) determines in an invariant way an element Obs(Φ) of the third cohomology group H3(M,U(Z(A))) of the monoid M with coefficients in the M-module U(Z(A)). An abstract kernel Φ:MAut(A)Inn(A) is induced by a regular homogeneous Schreier extension if and only if Obs(Φ) is the zero element of H3(M,U(Z(A))).

If the monoid A is a group, then SEnd(A)=Epi(A) and every Schreier extension of M by A is regular (see Proposition 3.10). Recalling that such extensions are called special Schreier in [4, 5, 17, 18], it is worth mentioning the following particular case of the previous theorems.

## Corollary 5.10.

Let M be a monoid and A a group. Any abstract kernel Φ:MEpi(A)Inn(A) (resp. Φ:MAut(A)Inn(A)) determines in an invariant way an element Obs(Φ) of the third cohomology group H3(M,Z(A)) of the monoid M with coefficients in the M-module Z(A). An abstract kernel Φ:MEpi(A)Inn(A) (resp. Φ:MAut(A)Inn(A)) is induced by a weakly homogeneous (resp. homogeneous) special Schreier extension if and only if Obs(Φ) is the zero element of H3(M,Z(A)).

We observe that the particular case described in the previous corollary could also be obtained from the results of [27].

## 6 The classification of regular weakly homogeneous and regular homogeneous Schreier extensions

In this section, we show that the set Ext(M,A,Φ) of isomorphism classes of regular weakly homogeneous (resp. homogeneous) Schreier extensions (3.1) which induce the same abstract kernel Φ:MSEnd(A)Inn(A) (resp. Φ:MAut(A)Inn(A)), when it is not empty, is in bijection with the second cohomology group H2(M,U(Z(A))) of M with coefficients in the M-module U(Z(A)). In order to do this, we show that there is a simply transitive action of the abelian group H2(M,U(Z(A))) on the set Ext(M,A,Φ).

We start by recalling that an action of a group G on a set S is simply transitive if, for all s,sS, there exists a unique gG such that gs=s. Given a simply transitive action of G on S, every element sS determines then a bijection α:GS, defined by α(g)=gs.

Suppose now that an abstract kernel Φ:MSEnd(A)Inn(A) is induced by a regular weakly homogeneous Schreier extension (3.1), i.e. that the set Ext(M,A,Φ) is not empty. For every xM, we choose a representative φ(x)Φ(x), with φ(1)=idA. We define an action of H2(M,U(Z(A))) on Ext(M,A,Φ) as follows. Given elements cl(h)H2(M,U(Z(A))) and cl(E)Ext(M,A,Φ), Proposition 4.4 tells us that E is isomorphic to a crossed product extension [A,φ,f,M], where f:M×MU(A) is a map with f(x,1)=0=f(1,y) and φ(x)φ(y)=μf(x,y)φ(xy), and such that the equality (4.1) holds. Consider the function h+f:M×MU(A) defined by (h+f)(x,y)=h(x,y)+f(x,y). Clearly,

(h+f)(x,1)=0=(h+f)(1,y),

and, since h(x,y)U(Z(A)), we also have

φ(x)φ(y)=μ(h+f)(x,y)φ(xy).

Furthermore,

φ(x)(h(y,z)+f(y,z))+h(x,yz)+f(x,yz)=φ(x)(h(y,z))+φ(x)(f(y,z))+h(x,yz)+f(x,yz)=φ(x)(h(y,z))+h(x,yz)+φ(x)(f(y,z))+f(x,yz)=h(x,y)+h(xy,z)+f(x,y)+f(xy,z)=h(x,y)+f(x,y)+h(xy,z)+f(xy,z),

where we are using that h(x,yz),h(xy,z)U(Z(A)), that h is a 2-cocycle and equality (4.1). Thus

φ(x)(h+f)(y,z)+(h+f)(x,yz)=(h+f)(x,y)+(h+f)(xy,z).

Thanks to this equality, we can build the crossed product extension [A,φ,h+f,M], which is weakly homogeneous by Remark 4.2. The action of H2(M,U(Z(A))) on Ext(M,A,Φ) we are looking for is then defined by

(6.1)cl(h)cl(E)=cl([A,φ,h+f,M]).

## Theorem 6.1.

The action (6.1) is well defined and simply transitive.

## Proof.

We first prove that the action is well defined, i.e. that it does not depend on the choice of the representatives. If cl(h)=cl(h) and cl(E)=cl(E), with E isomorphic to the crossed product extension [A,φ,f,M], then cl(h)cl(E)=cl([A,φ,h+f,M]). Since cl(E)=cl(E), there exists a commutative diagram

where ζ is an isomorphism. For every xM, we have ζ(0,x)=(r(x),x) with r(x)U(A). Indeed, ζ sends representatives to representatives, and (a,x) is, by Proposition 4.1, a representative if and only if aU(A). Moreover, (r(1),1)=ζ(0,1)=(0,1), whence r(1)=0. So we get a map r:MU(A) with r(1)=0. Furthermore, for all aA and xM, we have

ζ(a,x)=ζ((a,1)+(0,x))=ζ(a,1)+ζ(0,x)=(a,1)+(r(x),x)=(a+φ(1)r(x)+f(1,x),x)=(a+r(x),x).

Using this equality, one gets

ζ((a1,x)+(a2,y))=ζ(a1+φ(x)(a2)+f(x,y),xy)=(a1+φ(x)(a2)+f(x,y)+r(xy),xy),
ζ((a1,x)+(a2,y))=ζ(a1,x)+ζ(a2,y)=(a1+r(x),x)+(a2+r(y),y)=(a1+r(x)+φ(x)(a2)+φ(x)(r(y))+f(x,y),xy).

Comparing the two expressions, we obtain

(6.2)a1+φ(x)(a2)+f(x,y)+r(xy)=a1+r(x)+φ(x)(a2)+φ(x)(r(y))+f(x,y).

Moreover, since cl(h)=cl(h), there is a 1-cochain γ:MU(Z(A)) such that

(6.3)h(x,y)+γ(xy)=h(x,y)+xγ(y)+γ(x).

Consider now the diagram

where the map ξ is defined by ξ(a,x)=(γ(x)+a+r(x),x). Clearly, the diagram is commutative. Using the equalities (6.2) and (6.3), and the fact that h,h and γ take values in U(Z(A)), it is straightforward to check that ξ is a monoid homomorphism. Moreover, it sends representatives to representatives since ξ(a,x)=(γ(x)+a+r(x),x), and γ(x)+a+r(x)U(A) whenever aU(A) (see Proposition 4.1). Then Proposition 4.3 implies that ξ is an isomorphism. This shows that the action is well defined. It is obviously an action since

(cl(h)+cl(h))cl(E)=cl(h)(cl(h)cl(E))andcl(0)cl(E)=cl(E).

The next step of the proof consists in showing that the action is simple, namely,

cl(h1)cl(E)=cl(h2)cl(E)cl(h1)=cl(h2).

If cl(h1)cl(E)=cl(h2)cl(E), we have a commutative diagram

where η is an isomorphism. As we did for ζ in the first part of the proof, one can check that

η(a,x)=(a+b(x),x),withb:MU(A),b(1)=0.

Let us prove that b is in fact a 1-cochain with h1-h2=δ1b. If aA and xM, then there exists aA such that φ(x)(a)=a because φ(x)SEnd(A). Then we get

(a+b(x),x)=η(a,x)=η(φ(x)(a),x)=η((0,x)+(a,1))=η(0,x)+η(a,1)=(b(x),x)+(a,1)=(b(x)+φ(x)(a),x)=(b(x)+a,x),

hence a+b(x)=b(x)+a, which means that b(x)U(Z(A)) for all xM. Moreover,

η((0,x)+(0,y))=η(h1(x,y)+f(x,y),xy)=(h1(x,y)+f(x,y)+b(xy),xy),
η((0,x)+(0,y))=η(0,x)+η(0,y)=(b(x),x)+(b(y),y)=(b(x)+φ(x)(b(y))+h2(x,y)+f(x,y),xy).

Therefore,

h1(x,y)+f(x,y)+b(xy)=b(x)+φ(x)(b(y))+h2(x,y)+f(x,y),

whence

h1(x,y)-h2(x,y)=φ(x)(b(y))-b(xy)+b(x)=δ1b(x,y),

and this tells us that cl(h1)=cl(h2) and the action is simple.

It remains to prove that it is transitive, i.e. that, for all cl(E),cl(E)Ext(M,A,Φ), there exists

cl(h)H2(M,U(Z(A)))such thatcl(h)cl(E)=cl(E).

Given cl(E),cl(E)Ext(M,A,Φ), we know from Proposition 4.4 that E and E are isomorphic to crossed product extensions [A,φ,f,M] and [A,φ,f,M] respectively, where, for all x,yM, the following equalities hold:

φ(x)φ(y)=μf(x,y)φ(xy)=μf(x,y)φ(xy).

Being φ(xy) surjective, this implies that μf(x,y)-f(x,y)=idA, and so f(x,y)-f(x,y)U(Z(A)). Let us then define the function h:M×MU(Z(A)) by putting

h(x,y)=f(x,y)-f(x,y).

A straightforward calculation (using equality (4.1) and the fact that h takes values in U(Z(A))) shows that h is a 2-cocycle. Then we get

cl(h)cl(E)=cl([A,φ,h+f,M])=cl([A,φ,f,M])=cl(E),

and the action is transitive. ∎

The previous theorem gives then the desired bijection between Ext(M,A,Φ) and H2(M,U(Z(A))).

## Corollary 6.2.

For any fixed cl(E)Ext(M,A,Φ), the map from H2(M,U(Z(A))) to Ext(M,A,Φ) which sends cl(h) to cl(h)cl(E) is bijective.

If U(Z(A))=0 (in particular, if U(A)=0 or Z(A)=0), then both H2(M,U(Z(A))) and H3(M,U(Z(A))) are the trivial groups. This means that, for every abstract kernel Φ:MSEnd(A)Inn(A), Obs(Φ)=0. Hence we get the following.

## Corollary 6.3.

If U(Z(A))=0, then, for every abstract kernel Φ:MSEnd(A)Inn(A), there exists, up to isomorphism, a unique weakly homogeneous Schreier extension of M by A which induces Φ. If U(A)=0, then also Inn(A)=0, and so the abstract kernel is a monoid homomorphism Φ:MSEnd(A), i.e. an action of M on A. In this case, the unique weakly homogeneous extension is the semidirect product of M and A via the action Φ.

It is immediate to see that the results of this section are valid, in particular, for abstract kernels of the form Φ:MAut(A)Inn(A) and regular homogeneous Schreier extensions. Let us state them explicitly.

## Theorem 6.4.

Given an abstract kernel Φ:MAut(A)Inn(A), if the set Ext(M,A,Φ) of isomorphism classes of regular homogeneous Schreier extensions of M by A which induce Φ is not empty, then (6.1) is a simply transitive action of the abelian group H2(M,U(Z(A))) on Ext(M,A,Φ). This action induces a bijection between Ext(M,A,Φ) and H2(M,U(Z(A))).

## Corollary 6.5.

If the monoid A is such that U(Z(A))=0, for every abstract kernel Φ:MAut(A)Inn(A), there exists, up to isomorphism, a unique homogeneous Schreier extension of M by A which induces Φ. If U(A)=0, then also Inn(A)=0, and so the abstract kernel is a monoid homomorphism Φ:MAut(A), i.e. an action of M on A. In this case, the unique homogeneous extension is the semidirect product of M and A via the action Φ.

Finally, note that if A and M are both groups, then Theorem 6.4 turns into the classical cohomological classification of group extensions with non-abelian kernel (see, e.g., [16]).

Communicated by Frederick R. Cohen

Award Identifier / Grant number: UID/MAT/00324/2019

Award Identifier / Grant number: FR-18-10849

Funding statement: This work was partially supported by the Centre for Mathematics of the University of Coimbra – UID/MAT/00324/2019, by ESTG and CDRSP from the Polytechnical Institute of Leiria – UID/Multi/04044/2019, funded by the Portuguese Government through FCT/MCTES and co-funded by the European Regional Development Fund through the Partnership Agreement PT2020. The second author was partially supported by the Programma per Giovani Ricercatori “Rita Levi-Montalcini”, funded by the Italian government through MIUR. The third author was supported by the Shota Rustaveli National Science Foundation of Georgia (SRNSFG), grant FR-18-10849, “Stable Structures in Homological Algebra”.

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