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Kyungpook Mathematical Journal 2024; 64(4): 607-618

Published online December 31, 2024 https://doi.org/10.5666/KMJ.2024.64.4.607

Copyright © Kyungpook Mathematical Journal.

Uniformly S-Projective Modules and Uniformly S-Projective Uniformly S-Covers

Hwankoo Kim*, Najib Mahdou, El Houssaine Oubouhou, Xiaolei Zhang

Division of Computer Engineering, Hoseo University, Asan 31499, Republic of Korea
e-mail: hkkim@hoseo.edu
Department of Mathematics, Faculty of Science and Technology of Fez, Box 2202,
University S.M. Ben Abdellah Fez, Morocco
e-mail: mahdou@hotmail.com
Department of Mathematics, Faculty of Science and Technology of Fez, Box 2202,
University S.M. Ben Abdellah Fez, Morocco
e-mail: hossineoubouhou@gmail.com
School of Mathematics and Statistics, Shandong University of Technology, Zibo 255049, China
e-mail: zxlrghj@163.com

Received: December 31, 2023; Revised: September 17, 2024; Accepted: September 17, 2024

Recently, Zhang and Qi introduced and studied the concept of uniformly S-projective (u-S-projective) modules, where S represents a multiplicative subset of a ring. In this paper, we first derive a u-S version of the well-known result that a module is projective if and only if it is a direct summand of a free module. We then provide a characterization of u-S-projective modules in terms of projective modules, specifically when S is regular, and then extend the projective basis lemma to the u-S setting. Finally, we show that a u-S-projective u-S-cover is characterized by a u-S-superfluous submodule, analogous to the way a projective cover is characterized by a superfluous submodule.

Keywords: u-S-exact sequence, u-S-isomorphism, u-S-projective module

This paper assumes that all rings are commutative with nonzero identity and that all modules are unitary. For completeness, we begin with some definitions and notations that will be used throughout this paper. For a ring R, let U(R), reg(R), Z(R), and Nil(R) denote the sets of unit elements, regular elements, zero-divisors, and nilpotent elements of R, respectively. A nonempty subset S of R is called a multiplicative subset if 1S, 0S, and for each a,bS, we have abS.

Let S be a multiplicative subset of R and M an R-module. Define

torS(M)={xMthere exists sS such that sx=0}.

Then, torS(M) is a submodule of M, called the S-torsion submodule of M, and M is said to be S-torsion if torS(M)=M. Similarly, M is said to be S-torsion-free if torS(M)=0. If S is the multiplicative subset consisting of all non-zero-divisors of R, then for any R-module M, we write tor(M)=torS(M). Thus, S-torsion modules are called torsion modules, and S-torsion-free modules are called torsion-free modules. Note that flat modules are torsion-free according to [3, Proposition 2.5.4], so free modules and projective modules are torsion-free.

Let R be a ring and S a multiplicative subset of R. Recall from [3, Definition 1.6.10] that an R-module T is called a u-S-torsion module if there exists an element sS such that sT=0. Let M, N, and L be R-modules.

  • An R-homomorphism f:MN is called a u-S-monomorphism (resp., u-S-epimorphism) if Ker(f) (resp., Coker(f)) is a u-S-torsion module. An R-homomorphism f:MN is called a u-S-isomorphism if f is both a u-S-monomorphism and a u-S-epimorphism.

  • An R-sequence MfNgL is said to be u-S-exact if there exists an element sS such that sKer(g)Im(f) and sIm(f)Ker(g). A u-S-exact sequence 0ABC0 is called a short u-S-exact sequence.

Let R be a ring and S a multiplicative subset of R. Recall from [5] that an R-module P is said to be u-S-projective if the induced sequence

0HomR(P,A)HomR(P,B)HomR(P,C)0

is u-S-exact for any u-S-exact sequence 0ABC0. By [5, Theorem 2.9], an R-module P is u-S-projective if and only if for any short exact sequence 0AfBgC0, the induced sequence

0HomR(P,A)f*HomR(P,B)g*HomR(P,C)0

is u-S-exact, if and only if ExtR1(P,M) is u-S-torsion for any R-module M. We denote the class of all u-S-projective modules by u-S-P, and the class of all projective modules by P.

In this paper, we further investigate u-S-projective modules. More precisely, in Section 2, we begin by giving the u-S-version of the well-known result that the necessary and sufficient condition for the projectivity of a module is that it is a direct summand of a free module. To this end, we define u(niformly)-S-free modules as follows: An R-module F is said to be u-S-free if F is u-S-isomorphic to some free module. We show that an R-module P is u-S-projective if and only if it is a direct summand of a u-S-free module (Theorem 2.6). Next, we extend the projective basis lemma to the u-S-version (Theorem 2.10). Clearly, if an R-module P is u-S-isomorphic to some projective module, then P is u-S-projective. Meanwhile, Theorem 2.11 shows that the converse holds if S is regular.

We then provide a characterization of u-S-projective modules in terms of projective modules. More precisely, we show that, when S is regular, an R-module P is u-S-projective if and only if there exists sS and a projective submodule P of P such that sPPP. With these characterizations, we present the u-S-versions of several well-known results (see Corollaries 2.14, 2.15, 2.16, 2.17, and 2.18).

Finally, in Section 3, we show that a u-S-projective u-S-cover is characterized by a u-S-superfluous submodule, analogous to how a projective cover is characterized by a superfluous submodule.

We begin this section with the following proposition, which characterizes when every S-torsion exact sequence (resp., S-torsion monomorphism, S-torsion epimorphism, S-torsion isomorphism) is exact (resp., a monomorphism, an epimorphism, an isomorphism).

Proposition 2.1. Let R be a ring and S a multiplicative subset of R. Then the following conditions are equivalent:

  • Every u-S-exact sequence is exact;

  • Every u-S-monomorphism is a monomorphism;

  • Every u-S-epimorphism is an epimorphism;

  • Every u-S-isomorphism is an isomorphism;

  • Every element of S is a unit.

Proof. The implications (1)(2)&(3)(4), and (5)(1)&(2) are straightforward.

(2)(5): Let aS and consider the homomorphism f:R/Ra0. Since Ker(f)=R/Ra, it follows that aKer(f)=0. Hence, Ker(f) is a u-S-torsion R-module, and thus f is a u-S-monomorphism. By (2), f is a monomorphism, so R/Ra=Ker(f)=0, and therefore a is a unit.

(3)(5): Let aS and consider the homomorphism f:0R/Ra. Since Coker(f)=R/Ra, f is a u-S-epimorphism, and hence it is an epimorphism. Consequently, 0=(f)=R/Ra, so a is a unit.

(4)(5): Let aS. Since a(R/Ra)=0, it is easy to verify that R/Ra is u-S-isomorphic to {0}, and hence by (4), R/Ra=0. Therefore, a is a unit.

Remark 2.2.

  • Let f:MN be an R-homomorphism of R-modules M and N. If M is S-torsion-free, then f is a u-S-monomorphism if and only if f is a monomorphism.

  • An R-module M is u-S-torsion if and only if M is u-S-isomorphic to {0}.

Recall that a ring R is called a φ-von Neumann regular ring if (R,Nil(R)) is a local ring (see [2, Theorem 4.2] or [6, Theorem 4.1]).

Theorem 2.3. The following conditions are equivalent for a ring R:

  • Every u-S-monomorphism is a monomorphism for every multiplicative subset S of R;

  • Every u-S-epimorphism is an epimorphism for every multiplicative subset S of R;

  • Every u-S-isomorphism is an isomorphism for every multiplicative subset S of R;

  • R is a φ-von Neumann regular ring.

Proof. (1)&(2)(3): These are trivial.

(1)(4): Let aRNil(R). Set S:={annN} and consider the homomorphism f:R/Ra0. Since Ker(f)=R/Ra, it follows that aKer(f)=0. Hence, Ker(f) is a u-S-torsion R-module, so f is a u-S-monomorphism. Thus, by (2), f is a monomorphism. Therefore, R/Ra=Ker(f)=0, and hence a is a unit. Thus, every non-nilpotent element in R is a unit, and so (R,Nil(R)) is a local ring. Therefore, R is a φ-von Neumann regular ring.

(2)(4) and (3)(4): These are analogous to (1)(4).

(4)(1)&(2): Let S be a multiplicative subset of R. If there exists sSNil(R), then there exists a positive integer n such that 0=snS, a contradiction. Therefore, SRNil(R)=U(R), and so every element in S is a unit. Thus, every u-S-monomorphism (resp., u-S-epimorphism) is a monomorphism (resp., an epimorphism).

Since the proof of the following proposition is very similar to the classical cases, we omit the proof.

Proposition 2.4. Let R be a ring, S a multiplicative subset of R, and {Pi,Qi}i be a finite family of modules such that Pi is u-S-isomorphic to Qi for all i. Then Pi is u-S-isomorphic to Qi.

However, if Pi is u-S-isomorphic to Qi for all i, this does not necessarily mean that Pi is u-S-isomorphic to Qi, as the following example shows.

Example 2.5. Let R:=Z be the ring of integers, p a prime in Z, and S:={pnnN}. Let Mn:=Z/pn for each n1. Then Mn is u-S-isomorphic to {0}. Set N:=n=1Mn. However, it is easy to see that N is not u-S-torsion, and therefore it is not u-S-isomorphic to {0}.

An R-module F is said to be u-S-free if F is u-S-isomorphic to some free module.

Theorem 2.6. Let R be a ring and S a multiplicative subset of R. Then an R-module P is u-S-projective if and only if P is a direct summand of a u-S-free module.

In order to prove Theorem 2.6, we need the following lemmas, which are of independent interest.

Now, recall the notion of u-S-split u-S-short exact sequences from [5]. Let

ξ:0AfBgC0

be a u-S-short exact sequence. Then ξ is said to be u-S-split (with respect to s) if there exist sS and an R-homomorphism f:BA such that f(f(a))=sa for any aA, that is, ff=sIdA. Obviously, any split exact sequence is u-S-split.

Lemma 2.7. Let R be a ring and S a multiplicative subset of R. A short exact sequence

ξ:0AfBgC0

is u-S-split if and only if there exist sS, an R-homomorphism f:BA, and an R-homomorphism g:CB such that gg=sIdC and ff=sIdA.

Proof. By definition, there exist sS and an R-homomorphism f:BA such that ff=sIdA. Now, define a map g:CB as follows. For zC, choose yB such that g(y)=z, and define

g(z)=sy-f(f(y)).

If g(y)=g(y), then there exists xA such that f(x)=y-y. Thus,

sy-f(f(y))-sy-f(f(y))=s(y-y)-f(f(y-y)) =sf(x)-f(f(f(x))) =sf(x)-sf(x)=0,

so g is well-defined. It can also be verified that g is linear. Finally, if g(y)=z, we have

g(g(z))=g(sy-f(f(y)))=sz,

since gf=0 and g(y)=z. Hence, gg=sIdC.

Lemma 2.8. Let R be a ring and S a multiplicative subset of R. If a short exact sequence

ξ:0AfBgC0

is u-S-split, then B is u-S-isomorphic to AC.

Proof. By Lemma 2.7, there exist sS, an R-homomorphism f:BA, and an R-homomorphism g:CB such that gg=sIdC and ff=sIdA. Define a map

β:BAC bsf(b)-fgg(b),sg(b).

Let bKer(β). Then sf(b)-fgg(b)=0 and sg(b)=0, which implies s2f(b)=0. On the other hand, since sbKer(g)=(f), there exists aA such that sb=f(a). Hence,

s2a=sff(a)=s2f(b)=0,

and thus s3b=s2f(a)=0. Consequently, s3Ker(β)=0, so β is a u-S-monomorphism.

Now, let (a,c)AC. Then,

βf(a)+g(c)=sff(a)+sfg(c)-fggf(a)-fggg(c),sgg(c) =sff(a)+sfg(c)-sfg(c),s2c =s2a,s2c.

Hence, s2(a,c)(β), so β is a u-S-epimorphism. Therefore, β is a u-S-isomorphism.

Lemma 2.9. Let R be a ring, S a multiplicative subset of R, and {Pi} a finite family of modules. Then Pi is u-S-projective if and only if each Pi is u-S-projective.

Proof. By [3, Theorem 3.3.9 (2)], for every R-module N, we have

ExtR1Pi,NExtR1Pi,N.

On the other hand, it is easy to see that ExtR1Pi,N is u-S-torsion if and only if all ExtR1Pi,N are u-S-torsion. Hence, the conclusion follows immediately from [5, Theorem 2.9].

Proof. [Proof of Theorem 2.6]

Suppose P is u-S-projective. By [3, Theorem 1.6.7], we may pick a free module F and an epimorphism g:FP. Set P:=Ker(g). Then the sequence

0PFP0

is exact. Thus, it is u-S-split by [5, Theorem 2.10], and hence F:=PP is u-S-isomorphic to F by Lemma 2.8.

Conversely, assume that P is a direct summand of a u-S-free module, that is, there exists a free module F and a submodule P of F such that PP is u-S-isomorphic to F. Since F is free, it is projective, and hence F is u-S-projective. Then PP is u-S-projective by [5, Proposition 2.14 (3)]. Consequently, P is u-S-projective by Lemma 2.9.

Theorem 2.10. [u-S-projective basis lemma]

Let R be a ring and S a multiplicative subset of R. Then a module P is u-S-projective if and only if there exist sS, elements {xiiΓ}P, and R-homomorphisms {fiiΓ}P*:=Hom(P,R) such that:

  • if xP, then almost all fi(x)=0;

  • if xP, then sx=ifi(x)xi.

In this case, {xi,fiiΓ} is called a u-S-projective basis of P.

Proof. Suppose P is u-S-projective and let φ:FP be an epimorphism, where F is free with a basis {eiiΓ}. Then there is a homomorphism f:PF such that φf=s1P, that is, the following diagram commutes:

Set xi:=φ(ei) for each iΓ. If xP, then f(x) has a unique expression f(x)=iriei, where riR and almost all ri=0. Define fi:PR by fi(x)=ri. Then fiP* and fi(x)=0 for almost all i. Moreover, we have:

sx=φf(x)=φiriei=irixi=ifi(x)xi.

Conversely, let F be free with a basis {eiiΓ} and φ:FP be a homomorphism with φ(ei)=xi. Define h:PF by h(x)=ifi(x)ei. Thus, φh=s1P. Hence, the sequence

0Ker(φ)FP0

is u-S-split. Consequently, F is u-S-isomorphic to PKer(φ) by Lemma 2.8. Therefore, P is u-S-projective by Theorem 2.6.

We say that a multiplicative subset S of a ring R is regular if Sreg(R).

Theorem 2.11. Let R be a ring and S a regular multiplicative subset of R. Then an R-module is u-S-projective if and only if it is u-S-isomorphic to some projective module.

Proof. Let P be a u-S-projective module. Then P is a direct summand of a u-S-free module, meaning there exist a free module F and a submodule P of F such that PP is u-S-isomorphic to F. Let f:FPP be a u-S-isomorphism. Then there exists sS such that sKer(f)=0 and s(PP)(f).

Since F is free and S is a regular multiplicative subset of R, it follows that sKer(f)=0 implies Ker(f)=0. Hence, F is isomorphic to f(F), and thus f(F) is free. On the other hand, f(F) is a submodule of PP, and therefore there exist submodules M (resp., M) of P (resp., P) such that f(F)=MM. Since s(PP)(f)=MM, we conclude that sPMP. Thus, the inclusion MP is a u-S-isomorphism, and M is projective (since it is a direct summand of the free module f(F)).

The converse can be deduced immediately using [5, Proposition 2.14 (3)].

Recall from [1] that an R-module M is said to be strongly S-projective if there exists a projective submodule P of M such that sMP for some sS.

Theorem 2.12. Let R be a ring, S a regular multiplicative subset of R, and M an R-module. Then M is u-S-isomorphic to some projective module if and only if M is strongly S-projective.

Proof. Suppose that M is u-S-isomorphic to some projective module P. Let f:PM be a u-S-isomorphism. Since P is torsion-free and Sreg(R), P is S-torsion-free. Thus, f is a monomorphism. Hence, f(P)P is a projective module. Since f is a u-S-isomorphism, it is also a u-S-epimorphism, meaning that Coker(f) is u-S-torsion. Therefore, there exists sS such that sMf(P)M.

Conversely, assume that M is strongly S-projective. Then there exist sS and a projective module P such that sMPM. Now consider the following short exact sequence:

0PMM/P0.

Since s(M/P)=0, it follows that M/P is u-S-torsion. Thus, P and M/P are u-S-projective by [5, Corollary 2.11]. Therefore, M is u-S-projective by [5, Proposition 2.14 (2)].

Combining Theorems 2.6, 2.10, 2.11, and 2.12, we obtain the following corollary.

Corollary 2.13. Let R be a ring, S a regular multiplicative subset of R, and P an R-module. Then the following statements are equivalent:

  • P is u-S-projective;

  • P is a direct summand of a u-S-free module;

  • P has a u-S-projective basis;

  • P is u-S-isomorphic to some projective module;

  • There exist sS and a projective submodule P of P such that sPPP, that is, P is strongly S-projective.

Corollary 2.14. Let R be a ring, S a multiplicative subset of R, and P1 and P2 be u-S-projective R-modules. Then P1RP2 is u-S-projective.

Proof. We choose modules P1 and P2 such that F1=P1P1 and F2=P2P2 are u-S-free. Then

F1RF2=P1RP2P1RP2P1RP2P1RP2.

On the other hand, F1 (resp., F2) is u-S-isomorphic to some free R-module F1 (resp., F2). Thus, F1RF2 is u-S-isomorphic to F1RF2. Since F1RF2 is free by [3, Theorem 2.2.7], we conclude that P1RP2 is u-S-projective according to Theorem 2.6.

Corollary 2.15. Let f:RT be a ring homomorphism, S a multiplicative subset of R, and P a u-S-projective R-module. Then TRP is a u-f(S)-projective T-module. More precisely, if {xi,fiiΓ} is a u-projective basis of P, then {1xi,giiΓ} is a u-f(S)-projective basis of TRP, where

giktkyk=ktkfi(yk),tkT,ykP.

Proof. Let yP. Then sy=ifi(y)xi for almost all fi(y)=0. Hence, gi(ty)=tfi(y), where tT. Thus, for zTRP, almost all gi(z)=0. Noting that

igi(ty)1xi=itfi(y)1xi=tifi(y)xi=tsy,

it follows that {1xi,giiΓ} is a u-f(S)-projective basis of TRP.

Corollary 2.16. Let R be a ring, S a multiplicative subset of R, and P a u-S-projective R-module.

  • Let I be an ideal of R such that IS=. Then P/IP is a u-S¯-projective R/I-module, where S¯={s+IR/IsS}.

  • If T is a multiplicative subset of R, then PS is a u-T-projective RT-module, where T=s1RTsS.

Using Theorem 2.12 and [3, Theorem 2.3.14], we can establish the following corollary.

Corollary 2.17. [u-S-version of the Kaplansky theorem]

Let R be a local ring and S a regular multiplicative subset of R. Then every u-S-projective module is u-S-free.

Using Theorems 2.11, 2.12, and [3, Corollary 5.2.7], we can establish the following corollary.

Corollary 2.18. Let R be an integral domain and S a multiplicative subset of R. Then every u-S-projective ideal is S-finite.

Let R be a ring and S a multiplicative subset of R.

Definition 3.1. Let M be an R-module and A a class of R-modules.

  • A map fHomR(A,M) with AA is called an A-u-S-precover of M if HomR(A',f):HomR(A',A)HomR(A,M) is a u-S-epimorphism for any A'A.

  • An A-u-S-precover f of M is called an A-u-S-cover of M if f=fα implies that α is a u-S-isomorphism for each αEndR(A).

  • If every R-module has an A-u-S-precover, then A is called a u-S-precovering class.

  • If every R-module has an A-u-S-cover, then A is called a u-S-covering class.

Trivially, if σ:AA' is a u-S-isomorphism and f:A'M is an A-u-S-precover (resp., A-u-S-cover), then fσ:AM is also an A-u-S-precover (resp., A-u-S-cover).

Lemma 3.2. Let M be an R-module, and let f:AM and f':A'M be A-u-S-covers of M. Then A is u-S-isomorphic to A'.

Proof. Since f and f' are A-u-S-precovers of M, there exist R-homomorphisms g:A'A and h:AA' such that f'=s1fg and f=s2f'h for some s1,s2S. That is, we have the following diagram:

Setting s:=s1s2, we have f=sfhg and f'=sf'gh. Thus, sgh and shg are u-S-isomorphisms. Therefore, A is u-S-isomorphic to A'.

Proposition 3.3. Let M be an R-module. Then the A-u-S-cover of M, if it exists, is a u-S-direct summand of any A-u-S-precover of M.

Proof. Let f:AM be an A-u-S-cover of M and g:A'M be an A-u-S-precover of M. Then we have the following diagram:

Thus, f=s1gt1 and g=s2ft2 for some s1,s2S. We have f=s1s2ft2t1. Therefore, t:=s1s2t2t1:AA is a u-S-isomorphism. By [4, Proposition 1.1], there exists a u-S-isomorphism t':AA such that t't=sIdA for some sS. Hence, the R-homomorphism s1s2t1t':AA' is a u-S-monomorphism, and t2s1s2t1t'=sIdA. Therefore, s1s2t1t' is u-S-split.

Proposition 3.4. Let R be a ring, S a multiplicative subset of R, and M an R-module. Then the following statements hold.

  • An R-homomorphism f:PM is a u-S-projective u-S-precover if and only if f is a u-S-epimorphism and P is u-S-projective.

  • The class u-S-P of all u-S-projective modules is u-S-precovering.

Proof. (1) Suppose that f:PM is a u-S-epimorphism with P u-S-projective. Let P' be a u-S-projective module. Then HomR(P',f):HomR(P',P)HomR(P',M) is a u-S-epimorphism. Thus, f is a u-S-projective u-S-precover.

Now, suppose that f:PM is a u-S-projective u-S-precover. Let g:P'M be an epimorphism with P' projective. Then there is an R-homomorphism t:P'P such that g=sft for some sS. Thus, sf is an epimorphism. Consequently, f is a u-S-epimorphism.

(2) Let M be an R-module. Then there is an epimorphism π:PM with P projective. Thus, π is a u-S-projective u-S-precover by (1).

Obviously, any P-precover is a P-u-S-precover, and any P-cover is a P-u-S-cover for any multiplicative subset S of R. However, the converse is not true.

Remark 3.5. Note that the class u-S-P of all u-S-projective modules is not precovering in general. Indeed, let R:=Z and S:={pnnN} for some prime pZ. Let M:=nNZ/pn, and let PfM be a u-S-projective precover of M. Let in:Z/pnM be the natural embedding map. Then there exists an R-homomorphism tn:Z/pnP such that in=ftn. We also have an R-homomorphism g:MP such that tn=gin by the universal property of direct sums. Thus, in=ftn=fgin, which implies fg=IdM. Therefore, M is a direct summand of P.

Since M is not u-S-projective by [5, Example 2.15], P cannot be u-S-projective by [5, Proposition 2.14], leading to a contradiction.

Definition 3.6. Let M be an R-module. A submodule N of M is said to be u-S-superfluous if, whenever U is a submodule of M such that s1MN+U for some s1S, we have s2MU for some s2S.

Theorem 3.7. Let R be a ring, S a multiplicative subset of R, and M an R-module. Then every u-S-epimorphism φ:PM with P u-S-projective is a u-S-projective u-S-cover if and only if Ker(φ) is a u-S-superfluous submodule of P.

Proof. Suppose the u-S-epimorphism φ:PM is a u-S-projective u-S-cover. Let U be a submodule of P such that s1PKer(φ)+U for some s1S. Consider the composition φi:UM, where i:UP is the natural embedding map. Then s1φ(P)φ(U). Since φ is a u-S-epimorphism, φi is a u-S-epimorphism. Since P is u-S-projective, there is an R-homomorphism f:PU such that φis2f=φ for some s2S. So s2if=is2f is a u-S-isomorphism. It is easy to verify that i is a u-S-isomorphism, since i is a monomorphism. Thus, s3PU for some s3S.

Conversely, suppose

0Ker(φ)PφM0

is a u-S-short u-S-exact sequence, where Ker(φ) is a u-S-superfluous submodule of P. Consider the following commutative diagram:

Then we have (f)+Ker(φ)=P. Thus, there is an element s1S such that s1P(f). Hence, f is a u-S-epimorphism. Considering the u-S-exact sequence

0Ker(f)PfP0,

we have an R-homomorphism f':PP such that ff'=s2IdP for some s2S. We claim that f' is a u-S-epimorphism. Indeed, φf'=φff'=s2φ. Then for any xP, we have φ(f'(x)-s2x)=0. Thus, s2P(f')+Ker(φ). So s3P(f') for some s3S, and f' is a u-S-epimorphism.

Let x be an element in Ker(f). Then there is yP such that s3x=f'(y). Thus, 0=s2s3f(x)=y. Therefore, s3Ker(f)=0. Consequently, f is a u-S-isomorphism.

The authors would like to thank the reviewer for his/her comments. H. Kim was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF), funded by the Ministry of Education (2021R1I1A3047469).

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