# Isomorphism Theorems in the Primary Categories of Krasner Hypermodules

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## Abstract

**:**

## 1. Introduction

## 2. Preliminaries

**Definition**

**1.**

**Definition**

**2.**

**Definition**

**3.**

- 1.
- $(M,+)$ is a semihypergroup (associativity);
- 2.
- $(M,+)$ is commutative (commutativity);
- 3.
- there is a scalar identity ${0}_{M}$ (existence of scalar identity);
- 4.
- for every $x\in M$, there is a unique element denoted by $-x$ called the inverse of x such that ${0}_{M}\in x+(-x)$, which for simplicity, we write as ${0}_{H}\in x-x$ (existence of inverse);
- 5.
- for all $x,y,z\in M$, it holds that $x\in y+z\u27f9y\in x-z$ (reversibility).

**Definition**

**4.**

**Lemma**

**1.**

**Proposition**

**1.**

**Proposition**

**2.**

**Definition**

**5.**

- 1.
- a morphism $f:B\mapsto C$ is said to be a mono if for every $g,h:A\mapsto B$ the following implication holds:$$f\circ g=f\circ h\u27f9g=h.$$
- 2.
- a morphism $f:A\mapsto B$ is said to be an epi if for every $g,h:B\mapsto C$ the following implication holds:$$g\circ f=h\circ f\u27f9g=h.$$
- 3.
- A morphism $f:A\mapsto B$ of a category $\mathcal{C}$ is called an iso in $\mathcal{C}$ if there exists some $g:B\mapsto A$ (in $\mathcal{C}$) such that $f\circ g=i{d}_{B}$ and $g\circ f=i{d}_{A}$. In that case, g is denoted by ${f}^{-1}$.

## 3. Krasner Hypermodules

**Definition**

**6.**

- 1.
- $(R,+)$ is a canonical hypergroup;
- 2.
- $(R,\xb7)$ is a semigroup including ${0}_{R}$ as a bilaterally-absorbing element, that is ${0}_{R}\xb7x=x\xb7{0}_{R}={0}_{R}$ for all $x\in R$;
- 3.
- $(y+z)\xb7x=(y\xb7x)+(z\xb7x)$ and $x\xb7(y+z)=x\xb7y+x\xb7z$ for all $x,y,z\in R$.

**Definition**

**7.**

**Definition**

**8.**

**Definition**

**9.**

- 1.
- ${r}_{1}\ast ({a}_{1}+{a}_{2})={r}_{1}\ast {a}_{1}+{r}_{1}\ast {a}_{2}$;
- 2.
- $({r}_{1}+{r}_{2})\ast {a}_{1}={r}_{1}\ast {a}_{1}+{r}_{2}\ast {a}_{1}$;
- 3.
- $({r}_{1}\xb7{r}_{2})\ast {a}_{1}={r}_{1}\ast ({r}_{2}\ast {a}_{1})$;
- 4.
- ${0}_{R}\ast {a}_{1}={0}_{A}$.

**Remark**

**1.**

- (i)
- If A is a left Krasner hypermodule over a Krasner hyperring R, then we say that A is a left Krasner R-hypermodule. Similarly, the right Krasner R-hypermodule is defined by the map $\ast :A\times R\mapsto A$ satisfying the similar properties mentioned in Definition 9 with affection on the right.
- (ii)
- If R is a hyperring with the unit element ${1}_{R}$ and A is a Krasner R-hypermodule satisfying ${1}_{R}\ast a=a$ (resp. $a\ast {1}_{R}=a$) for all $a\in A$, then A is said to be a unitary left (resp. right) Krasner R-hypermodule.
- (iii)
- From now on, for convenience, every hyperring R is assumed to have the unit element ${1}_{R}$ and by “an R-hypermodule A”, we mean a unitary left Krasner R-hypermodule, unless otherwise stated.

**Definition**

**10.**

**Remark**

**2.**

- 1.
- Every hyperring R is an R-hypermodule, and every $I\u22b4R$ is an R-subhypermodule of R.
- 2.
- Let R be a hyperring. Every canonical hypergroup A can be considered as an R-hypermodule with the trivial external multiplication $r\ast a={0}_{A}$ for every $r\in R$ and $a\in A$.
- 3.
- Let A be an R-hypermodule and $\varnothing \ne B\subseteq A$. For every $I\u22b4R$,$$IB=\{a\in \sum _{i=1}^{m}{r}_{i}\ast {n}_{i}\mid {r}_{i}\in I,\phantom{\rule{3.33333pt}{0ex}}{n}_{i}\in B,\phantom{\rule{3.33333pt}{0ex}}m\in {\mathbb{Z}}^{+}\}$$
- 4.
- Let ${\left\{{A}_{i}\right\}}_{i\in I}$ be a family of R-subhypermodules of A. Then, ${\cap}_{i\in I}{A}_{i}\le A$.

**Definition**

**11.**

- 1.
- $f(x+y)\subseteq f\left(x\right)+f\left(y\right)$;
- 2.
- $f(r\ast x)=r\ast f\left(x\right)$

- $Ker\left(f\right)$ is an R-subhypermodule of A.
- Clearly, $Im\left(f\right)$ may not be an R-subhypermodule of B.
- For every morphism f in ${}_{{R}_{s}}h\mathbf{mod}$, $Im\left(f\right)$ is always an R-subhypermodule of the codomain of f.

**Proposition**

**3.**

**Proof.**

## 4. Main Results

**Theorem**

**1.**

**Proof.**

**Remark**

**3.**

- (i)
- An iso in the category ${}_{R}h\mathit{m}\mathit{o}\mathit{d}$ (or an R-isomorphism) is surjective and injective, i.e., bijective. For this, let $f:A\mapsto B$ be an iso in ${}_{R}h\mathit{m}\mathit{o}\mathit{d}$. Therefore, $f\circ {f}^{-1}=i{d}_{B}$ and ${f}^{-1}\circ f=i{d}_{A}$. Clearly, $f\circ {f}^{-1}=i{d}_{B}$ implies f is surjective. Furthermore, if $f\left(x\right)=f\left(y\right)$, then ${f}^{-1}\circ f=i{d}_{A}$ implies that $x=y$.
- (ii)
- $f:A\mapsto B$ is an R-isomorphism in ${}_{{R}_{s}}h\mathit{m}\mathit{o}\mathit{d}$ if and only if it is bijective. To show this fact, suppose $f:A\mapsto B$ is bijective. Then, ${f}^{-1}:B\mapsto A$ is also an R-homomorphism. Indeed, for every ${y}_{1},{y}_{2}\in B$, there are (unique) ${x}_{1},{x}_{2}\in A$ such that $f\left({x}_{i}\right)={y}_{i}$ for $i\in \{1,2\}$, and since $f({x}_{1}+{x}_{2})=f\left({x}_{1}\right)+f\left({x}_{2}\right)={y}_{1}+{y}_{2}$, we obtain ${f}^{-1}({y}_{1}+{y}_{2})={x}_{1}+{x}_{2}={f}^{-1}\left({y}_{1}\right)+{f}^{-1}\left({y}_{2}\right)$. Finally, $f(r\ast x)=r\ast f\left(x\right)$ implies ${f}^{-1}(r\ast y)=r\ast {f}^{-1}\left(y\right)$, and so, ${f}^{-1}$ is an R-homomorphism. Therefore, $f\circ {f}^{-1}=i{d}_{B}$ and ${f}^{-1}\circ f=i{d}_{A}$, and f is an R-isomorphism. The converse fact follows from (i) since every R-isomorphism in ${}_{{R}_{s}}h\mathit{m}\mathit{o}\mathit{d}$ is an R-isomorphism in ${}_{R}h\mathit{m}\mathit{o}\mathit{d}$.

**Notation**

**1.**

**Theorem**

**2.**

**Proof.**

**Theorem**

**3.**

**Proof.**

**Theorem**

**4.**

**Proof.**

**Proposition**

**4.**

**Proof.**

- (i)
- (⇐) It is clear.(⇒) Let $f\in ho{m}_{R}(A,B)$ (resp., $f\in ho{m}_{R}^{s}(A,B)$) be an R-monomorphism and $f\left(x\right)=f\left(y\right)$. Then, $f(r\ast x)=f(r\ast y)$ for an arbitrary $r\in R$. Now, define $g,h\in ho{m}_{R}^{s}(R,A)$ with $g\left(r\right)=r\ast x$ and $h\left(r\right)=r\ast y$ for each $r\in R$. Clearly, $f\circ g=f\circ h$. Since f is monic, we have $g=h$. Therefore, $f\left(1\right)=g\left(1\right)$ implies $x=y$.
- (ii)
- (⇐) Let $Ker\left(f\right)=\left\{{0}_{A}\right\}$ and $f\left(x\right)=f\left(y\right)$. Therefore, ${0}_{B}\in f\left(x\right)-f\left(y\right)=f(x-y)$. Thus, there is $z\in x-y$ such that $f\left(z\right)={0}_{A}$. By assumption, $z={0}_{A}$. Now, ${0}_{A}\in x-y$ implies $x=y$.(⇒) Let $z\in Ker\left(f\right)$ and f is injective. Therefore, $f\left(z\right)={0}_{B}$. On the other hand $f\left({0}_{A}\right)={0}_{B}$. Thus, $z={0}_{A}$ by injectivity. □

**Corollary**

**1.**

**Proposition**

**5.**

**Proof.**

- (i)
- Let $f\in ho{m}_{R}(A,B)$ be surjective, and let $g,h\in ho{m}_{R}(B,C)$. If $g\circ f=h\circ f$, then for all $a\in A$, we have $g\left(f\right(\left(a\right))=h(f\left(a\right))$. Now, let $b\in B$. Clearly, there is $x\in A$ such that $f\left(x\right)=b$. Thus, $g\left(b\right)=g\left(f\right(x\left)\right)=h\left(f\right(x\left)\right)=h\left(b\right)$. Hence, f is an R-epimorphism.
- (ii)
- (⇐) It is clear from (i).(⇒) Let $f\in ho{m}_{R}^{s}(A,B)$ be an R-epimorphism and $b\in B$. Suppose f is not surjective. Then, $f\left(A\right)\ne B$. Define $g,h:B\mapsto \frac{B}{f\left(A\right)}$ with $g\left(b\right)={0}_{\frac{B}{A}}$ and $h\left(b\right)=b+f\left(A\right)$. Then, clearly, $g\circ f=h\circ f$, but $g\ne h$. This contradiction shows that f is surjective. □

**Proposition**

**6.**

**Proof.**

**Proposition**

**7.**

**Proof.**

## 5. Conclusions

## Author Contributions

## Funding

## Conflicts of Interest

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Shojaei, H.; Fasino, D.
Isomorphism Theorems in the Primary Categories of Krasner Hypermodules. *Symmetry* **2019**, *11*, 687.
https://doi.org/10.3390/sym11050687

**AMA Style**

Shojaei H, Fasino D.
Isomorphism Theorems in the Primary Categories of Krasner Hypermodules. *Symmetry*. 2019; 11(5):687.
https://doi.org/10.3390/sym11050687

**Chicago/Turabian Style**

Shojaei, Hossein, and Dario Fasino.
2019. "Isomorphism Theorems in the Primary Categories of Krasner Hypermodules" *Symmetry* 11, no. 5: 687.
https://doi.org/10.3390/sym11050687