Let $N$ be the set of natural numbers. A number $a$ divides a number $b$ (written $a|b$), if there exists a number $c$ such that $b=ca$ (in this case $a$ is said to be a factor or divisor of $b$). The greatest divisor of $a$ and $b$ (written $\left(a,b\right)=d$) is a number $d$ such that $d|a,d|b$, and for any number $c$ such that $c|a,c|b$, we have $c\le d$. A number $a$ is said to be a prime number if and only if $a>1$ and is divisible by 1 and itself. We denote the set of prime numbers by $P$. Two numbers $a$ and $b$ are said to be relatively prime if $\left(a,b\right)=1$ [1] , [2] .

Lemma 1.1 [3] : Every $n\in N,n>1$, can be written as a product of prime numbers.

Theorem 1.1 [3] : The Unique Factorization Theorem. Any natural number grater than one can be written as a product of primes in one and only one way.

i.e. for any $n\in N,n>1$, can be written exactly in one way in the form $n=p_1^{e_1}{p}_2^{e_2}\cdots p_k^{e_k}$ where $e_i\ge 0,i=1,2,\cdots ,k$, each $p_i$ is prime, and $p_i\ne p_j$. We call this representation the prime-power decomposition of $n$.

For any $a\in N$, the set of all multiple numbers of $a$ is $a\uparrow =\left\{na:n\in N\right\}$. For any $A\subseteq N$ the set of all multiple of elements of $A$ is $A\uparrow =\left\{n\in N:\exists a\in A,a|n\right\}$. We say that $A$ is an upward closed subset of $N$ if $A=A\uparrow$. The set of upward closed subsets of $N$ is denoted by $\mu =\left\{A\subseteq N:A=A\uparrow \right\}$ [4] .

Lemma 1.2 [4] : For any $A\subseteq N$, $A\uparrow ={\displaystyle {\cup }_{a\in A}a\uparrow }$.

Let $P$ be the set of prime numbers and let $L_0=\left\{1\right\}$, $L_1=\left\{a:a\in P\right\}$, $L_2=\left\{a_1{a}_2:a_1,a_2\in L_1\right\}$, $\cdots$, $L_n=\left\{a_1{a}_2\cdots a_n:a_1,a_2,\cdots ,a_n\in L_1\right\}$, $\cdots$. Then we say that $L_i$, where $i\ge 0$ are the levels of $N$. A number $a\in N$ is in the level $L_n$ if $a=p_1^{e_1}{p}_2^{e_2}\cdots p_k^{e_k}$, where $p_i\in P$, $0\le e_i\le n$, $i=1,2,\cdots ,k$, and $e_1+e_2+\cdots +e_k=n$ [5] .

Lemma 1.3 [5] : (a) $N={\displaystyle {\cup }_{i=0}^{\infty }{L}_i}$

(b) $L_i\cap L_j=\varphi$ where $i\ne j$ $\left({\displaystyle {\cap }_{i=0}^{\infty }{L}_i}=\varphi \right)$

If $f:N\to N$ is a function $A\subseteq N$, then the restriction function of $f$ on $A$ is a function $f_{|}{}_A:A\to N$ such that $f\left(a\right)=f_{|}{}_A\left(a\right)$ for any $a\in A$. A function $f:N\to N$ is injective if $a_1=a_2$ whenever $f\left(a_1\right)=f\left(a_2\right)$. It is surjective if for any $b\in N$, there exists $a\in N$ such that $f\left(a\right)=b$. A function that is both injective and surjective is called bijective. If $f:N\to N$ is bijective, then $f^{-1}:N\to N$ is also a bijective function, and it is called the inverse function of $f$. A function $f:N\to N$ is called increasing if and only if $f\left(a\right)\le f\left(b\right)$ whenever $a\le b$ [6] [7] .

Lemma 1.4 If $f:N\to N$ is a function. Then

(a) $f_{|}{}_{L_i}\cap f_{|}{}_{L_j}=\varphi$, $\forall$ $i\ne j$

(b) $f={\displaystyle {\cup }_{i=0}^{\infty }{f}_{|}{}_{L_i}}$

Proof: (a) By (Lemma 1.3 (b)), $L_i\cap L_j=\varphi$ for $i\ne j$.

Therefore, $f_{|}{}_{L_i}\left(L_i\right)\cap f_{|}{}_{L_j}\left(L_j\right)=\varphi$, which implies $f_{|}{}_{L_i}\cap f_{|}{}_{L_j}=\varphi$.

(b) Let $\left(a,f\left(a\right)\right)\in f$. Since $N={\displaystyle {\cup }_{i=0}^{\infty }{L}_i}$, we have $a\in L_i,f\left(a\right)=f_{|}{}_{L_i}\left(a\right)$ for some $i$. Therefore $\left(a,f_{|}{}_{L_i}\left(a\right)\right)\in f_{|}{}_{L_i}$, which implies $f\subseteq {\displaystyle {\cup }_{i=0}^{\infty }{f}_{|}{}_{L_i}}$.

On the other hand, since $f_{|}{}_{L_i}\subset f$ $\forall$ $i=0,1,2,\cdots$, we have ${\displaystyle {\cup }_{i=0}^{\infty }{f}_{|}{}_{L_i}}\subseteq f$. Thus, $f={\displaystyle {\cup }_{i=0}^{\infty }{f}_{|}{}_{L_i}}$. ◼

As a consequence of dividing $N$ into infinitely many disjoint levels, we can divide any function $f:N\to N$ into infinitely many disjoint functions, which can be obtained by restricting $f$ to $L_i$ for all $i=0,1,2,\cdots$. If each function $f_{|}{}_{L_i}$ sends numbers to the same level, and the image of any number in level $L_i$ under the effect of $f_{|}{}_{L_i}$ is a product of images of its prime decomposition under the function $f_{|}{}_{L_1}$, then $f_{|}{}_{L_1}$ generates the function $f$, and it is called a levely multiplicative function.

Definition 2.1: A function $f:N\to N$, is said to be levely function if and only if $f\left(a\right)\in L_i$ for all $a\in L_i,i=0,1,2,\cdots$.

i.e. $f_{|}{}_{L_i}\left(L_i\right)\subseteq L_i$, $f_{|}{}_{L_0}\left(1\right)=1$

Example 2.1: If $f:N\to N$ is defined by $f\left(n\right)=2^i,n\in L_i,i=2j$, and $f\left(n\right)=3^i,n\in L_i,i=2j+1,j=0,1,2,\cdots$, then $f$ is a levely function.

Definition 2.2: A levely function $f:N\to N$ is said to be a multiplicative function or a function that is generated by $f_{|}{}_{L_1}=\left\{\left(a,f_{|}{}_{L_1}\left(a\right)\right):a\in L_1\right\}$, if and only if $f_{|}{}_{L_i}\left(a\right)=f_{|}{}_{L_i}\left(a_1{a}_2\cdots a_i\right)=f_{|}{}_{L_1}\left(a_1\right)f_{|}{}_{L_1}\left(a_2\right)\cdots f_{|}{}_{L_1}\left(a_i\right)$ for any $f_{|}{}_{L_i},i=2,3,\cdots$, and for any $a=a_1{a}_2\cdots a_i\in L_i$, where $a_1,a_2,\cdots ,a_i\in L_1$.

i.e. $\begin{array}{l}{f}_{|}{}_{L_i}=\{\left(a_1{a}_2\cdots a_i,f_{|}{}_{L_1}\left(a_1\right)f_{|}{}_{L_1}\left(a_2\right)\cdots f_{|}{}_{L_1}\left(a_i\right)\right):a_1,a_2,\cdots ,a_i\in L_1,\\ f_{|}{}_{L_1}\left(a_1\right),f_{|}{}_{L_1}\left(a_2\right),\cdots ,f_{|}{}_{L_1}\left(a_i\right)\in f_{|}{}_{L_1}\left(L_1\right)\}\end{array}$,

$f={\displaystyle {\cup }_{i=0}^{\infty }{f}_{|}{}_{L_i}}$

When $f:N\to N$ is a levely multiplicative function, we will write $f{|}_{L_i}:L_i\to L_i$ for all $i=0,1,2,\cdots$.

And

$f_{|}{}_{L_i}\left(L_i\right)=\left\{f_{|}{}_{L_1}\left(a_1\right)f_{|}{}_{L_1}\left(a_2\right)\cdots f_{|}{}_{L_1}\left(a_i\right):f_{|}{}_{L_1}\left(a_1\right),f_{|}{}_{L_1}\left(a_2\right),\cdots ,f_{|}{}_{L_1}\left(a_i\right)\in f_{|}{}_{L_1}\left(L_1\right)\right\}$. for all $i\ge 1$.

Example 2.2: Let $f_{|}{}_{L_1}:L_1\to L_1$ be defined by $f_{|}{}_{L_1}\left(n\right)=2$ $\forall$ $n\in L_1$.

Then, $f_{|}{}_{L_1}$ generates all the functions $f_{|}{}_{L_i}$, where $i\ge 2$ as follows:

$\begin{array}{c}{f}_{|}{}_{L_i}\left(n\right)=f_{|}{}_{L_i}\left(p_1{p}_2\cdots p_i\right)\\ =f_{|}{}_{L_1}\left(p_1\right)\cdot f_{|}{}_{L_1}\left(p_2\right)\cdots f_{|}{}_{L_1}\left(p_i\right)\\ =2\cdot 2\cdots 2=2^i\end{array}$

And the levely multiplicative function $f:N\to N$ is defined by $f\left(n\right)=2^i$ where $n\in L_i$.

Theorem 2.1: Every function $f_{|}{}_{L_1}:L_1\to L_1$ generates a unique levely multiplicative function $f:N\to N$, $f={\displaystyle {\cup }_{i=0}^{\infty }{f}_{|}{}_{L_i}}$.

Proof: By (Definition 2.1) and (Lemma 1.4 (b)) $f={\displaystyle {\cup }_{i=0}^{\infty }{f}_{|}{}_{L_i}}$ is generated by $f_{|}{}_{L_1}$. To prove that $f$ is unique, let $f_1$ and $f_2$ be levely multiplicative functions generated by $f_{|}{}_{L_1}$. Then, for any $a=a_1{a}_2\cdots a_i\in L_i,i\ge 1$, we have

$\begin{array}{c}{f}_1\left(a\right)=f_{|}{}_{L_i}\left(a\right)\\ =f_{|}{}_{L_i}\left(a_1{a}_2\cdots a_i\right)\\ =f_{|}{}_{L_1}\left(a_1\right)f_{|}{}_{L_1}\left(a_2\right)\cdots f_{|}{}_{L_1}\left(a_i\right)\end{array}$

and

$\begin{array}{c}{f}_2\left(a\right)=f_{|}{}_{L_i}\left(a\right)\\ =f_{|}{}_{L_i}\left(a_1{a}_2\cdots a_i\right)\\ =f_{|}{}_{L_1}\left(a_1\right)f_{|}{}_{L_1}\left(a_2\right)\cdots f_{|}{}_{L_1}\left(a_i\right)\end{array}$

Therefore, $f_1={\displaystyle {\cup }_{i=0}^{\infty }{f}_{|}{}_{L_i}}=f_2$. Hence, $f$ is unique. ◼

Lemma 2.1: If $f:N\to N$ is a levely multiplicative function, then

(a) $f_{|}{}_{L_i}\left(a\right)f_{|}{}_{L_j}\left(b\right)=f_{|}{}_{L_{i+j}}\left(ab\right)$ $\forall$ $a\in L_i,b\in L_j$

(b) $f\left(ab\right)=f\left(a\right)f\left(b\right)$ $\forall$ $a,b\in N$

Proof: (a) Let $a=a_1{a}_2\cdots a_i\in L_i,b=b_1{b}_2\cdots b_j\in L_j$. Then,

$\begin{array}{c}{f}_{|}{}_{L_i}\left(a\right)f_{|}{}_{L_j}\left(b\right)=f_{|}{}_{L_i}\left(a_1{a}_2\cdots a_i\right)f_{|}{}_{L_j}\left(b_1{b}_2\cdots b_j\right)\\ =f_{|}{}_{L_1}\left(a_1\right)f_{|}{}_{L_1}\left(a_2\right)\cdots f_{|}{}_{L_i}\left(a_i\right)f_{|}{}_{L_1}\left(b_1\right)f_{|}{}_{L_1}\left(b_2\right)\cdots f_{|}{}_{L_1}\left(b_j\right)\\ =f_{|}{}_{L_{i+j}}\left(a_1{a}_2\cdots a_i{b}_1{b}_2\cdots b_j\right)\\ =f_{|}{}_{L_{i+j}}\left(ab\right)\end{array}$

(b) Let $a,b\in N$, then there are some $i,j$ such that $a=a_1{a}_2\cdots a_i\in L_i$, $b=b_1{b}_2\cdots b_j\in L_j$, so $ab=a_1{a}_2\cdots a_i{b}_1{b}_2\cdots b_j\in L_{i+j}$. By (a) we have

$\begin{array}{c}f\left(ab\right)=f_{|}{}_{L_{i+j}}\left(ab\right)\\ =f_{|}{}_{L_i}\left(a\right)f_{|}{}_{L_j}\left(b\right)\\ =f\left(a\right)f\left(b\right)\end{array}$ ◼

Corollary 2.1: If $f:N\to N$ is a levely multiplicative function, then:

(a) $f_{|}{}_{L_1}\left(a_1\right)f_{|}{}_{L_2}\left(a_2\right)\cdots f_{|}{}_{L_n}\left(a_n\right)=f_{|}{}_{L_{1+2+\cdots +n}}\left(a_1{a}_2\cdots a_n\right)$

(b) $f_{|}{}_{L_i}\left(a_i\right)f_{|}{}_{L_{i+1}}\left(a_{i+1}\right)\cdots f_{|}{}_{L_j}\left(a_j\right)=f_{|}{}_{L_{i+\left(i+1\right)+\cdots +j}}\left(a_i{a}_{i+1}\cdots a_j\right)$, where $i<j$, $\forall$ $a_i\in L_i,a_{i+1}\in L_{i+1},\cdots ,a_j\in L_j$

(c) $f\left(a_1{a}_2\cdots a_n\right)=f\left(a_1\right)f\left(a_2\right)\cdots f\left(a_n\right)$ $\forall$ $a_1,a_2,\cdots ,a_n\in N$

(d) $f\left(a^i\right)={\left(f\left(a\right)\right)}^i\in f\left(L_i\right)$ $\forall$ $a^i\in L_i$

Proof: (a) We will use mathematical induction.

The result is obvious when $n=1$.

By (Lemma 2.1 (a)) when $n=2$, we have $f_{|}{}_{L_1}\left(a_1\right)f_{|}{}_{L_2}\left(a_2\right)=f_{|}{}_{L_{1+2}}\left(a_1{a}_2\right)$.

Suppose that it is true for $n=k-1$, $f_{|}{}_{L_1}\left(a_1\right)f_{|}{}_{L_2}\left(a_2\right)\cdots f_{|}{}_{L_{k-1}}\left(a_{k-1}\right)=f_{|}{}_{L_{1+2+\cdots +\left(k-1\right)}}\left(a_1{a}_2\cdots a_{k-1}\right)$

We will show that this implies it is true for $n=k$

$\begin{array}{c}{f}_{|}{}_{L_1}\left(a_1\right)f_{|}{}_{L_2}\left(a_2\right)\cdots f_{|}{}_{L_k}\left(a_k\right)=\left(f_{|}{}_{L_1}\left(a_1\right)f_{|}{}_{L_2}\left(a_2\right)\cdots f_{|}{}_{L_{k-1}}\left(a_{k_1}\right)\right)\left(f_{|}{}_{L_k}\left(a_k\right)\right)\\ =f_{|}{}_{L_{1+2+\cdots +\left(k-1\right)}}\left(a_1{a}_2\cdots a_{k-1}\right)f_{|}{}_{L_k}\left(a_k\right)\\ =f_{|}{}_{L_{1+2+\cdots +k}}\left(a_1{a}_2\cdots a_k\right)\end{array}$

(b) Similar to (a) by induction. However, we will start with the number $i$ instead of 1.

If $j=i$ it is obvious.

By (Lemma 2.1 (a)), in case of $j=i+1$, we have $f_{|}{}_{L_i}\left(a_i\right)f_{|}{}_{L_{i+1}}\left(a_{i+1}\right)=f_{|}{}_{L_{i+\left(i+1\right)}}\left(a_i{a}_{i+1}\right)$.

If we suppose that $f_{|}{}_{L_i}\left(a_i\right)f_{|}{}_{L_{i+1}}\left(a_{i+1}\right)\cdots f_{|}{}_{L_{j-1}}\left(a_{j-1}\right)=f_{|}{}_{L_{i+\left(i+1\right)+\cdots +\left(j-1\right)}}\left(a_i{a}_{i+1}\cdots a_{j-1}\right)$.

then

$\begin{array}{c}{f}_{|}{}_{L_i}\left(a_i\right)f_{|}{}_{L_{i+1}}\left(a_{i+1}\right)\cdots f_{|}{}_{L_j}\left(a_j\right)=\left(f_{|}{}_{L_i}\left(a_i\right)f_{|}{}_{L_{i+1}}\left(a_{i+1}\right)\cdots f_{|}{}_{L_{j-1}}\left(a_{j-1}\right)\right)\left(f_{|}{}_{L_j}\left(a_j\right)\right)\\ =f_{|}{}_{L_{i+\left(i+1\right)+\cdots +\left(j-1\right)}}\left(a_i{a}_{i+1}\cdots a_{j-1}\right)f_{|}{}_{L_j}\left(a_j\right)\\ =f_{|}{}_{L_{i+\left(i+1\right)+\cdots +\left(j-1\right)+j}}\left(a_i{a}_{i+1}\cdots a_{j-1}{a}_j\right)\end{array}$

(c) Similar to (a)

(d) Let $a^i\in L_i$. Then, by (b) we have

$\begin{array}{c}f\left(a^i\right)=f\left(a\cdots a\right)\\ =f\left(a\right)\cdots f\left(a\right)\\ ={\left(f\left(a\right)\right)}^i\in f\left(L_i\right)\end{array}$ ◼

Theorem 2.2: If $f:N\to N$ is a levely multiplicative function, then for any $a=a_1^{e_1}{a}_2^{e_2}\cdots a_n^{e_n}\in L_n$, where $0\le e_j\le n$, $j=1,2,\cdots ,n$, $e_1+e_2+\cdots +e_n=n$, and $a_1,a_2,\cdots ,a_n\in P$ in the unique decomposition of prime powers, we have $f\left(a\right)={\left(f\left(a_1\right)\right)}^{e_1}{\left(f\left(a_2\right)\right)}^{e_2}\cdots {\left(f\left(a_n\right)\right)}^{e_n}\in f\left(L_n\right)$ in the unique decomposition of the prime powers.

Proof: Let, $a=a_1^{e_1}{a}_2^{e_2}\cdots a_n^{e_n}\in L_n$, where $0\le e_j\le n$, $j=1,2,\cdots ,n$, $e_1+e_2+\cdots +e_n=n$, and $a_1,a_2,\cdots ,a_n\in P$. Then, by (Corollary 2.1 (b), (c)), we have

$\begin{array}{c}f\left(a\right)=f\left(a_1^{e_1}{a}_2^{e_2}\cdots a_n^{e_n}\right)\\ =f\left(a_1^{e_1}\right)f\left(a_2^{e_2}\right)\cdots f\left(a_n^{e_n}\right)\\ ={\left(f\left(a_1\right)\right)}^{e_1}{\left(f\left(a_2\right)\right)}^{e_2}\cdots {\left(f\left(a_n\right)\right)}^{e_n}\in f\left(L_n\right)\end{array}$

Now, by (Theorem 1.1), since $a=a_1^{e_1}{a}_2^{e_2}\cdots a_n^{e_n}$ is unique,

$f\left(a\right)={\left(f\left(a_1\right)\right)}^{e_1}{\left(f\left(a_2\right)\right)}^{e_2}\cdots {\left(f\left(a_n\right)\right)}^{e_n}$

is also unique. ◼

Furthermore, one of the characteristics that distinguishes the levely multiplicative function $f:N\to N$, which is generated by $f_{|}{}_{L_1}$, is that if $f_{|}{}_{L_1}$ is a bijective function, then all $f_{|}{}_{L_i},i\ge 2$, and $f$ are bijective.

Theorem 2.3: If $f:N\to N$ is a levely multiplicative function and $f_{|}{}_{L_1}:L_1\to L_1$ is injective, then:

(a) $f_{|}{}_{L_i}:L_1\to L_1$ is bijective, $\forall$ $i=1,2,3,\cdots$

(b) $f$ is bijective.

Proof: (a) First, we will show that $f_{|}{}_{L_i}:L_i\to L_i$ is injective, $\forall$ $i=2,3,\cdots$

Let $a,b\in L_i,a\ne b$, where $a=a_1{a}_2\cdots a_i,b=b_1{b}_2\cdots b_i$, $a_1,a_2,\cdots ,a_i,b_1,b_2,\cdots ,b_i\in L_1$. Since, $a\ne b$, there exists $a_n\in L_1,1\le n\le i$, and $b_m\in L_i,1\le m\le i$, such that $a_n\ne b_m$. Since $f_{|}{}_{L_1}$ is injective, $f_{|}{}_{L_1}\left(a_n\right)\ne f_{|}{}_{L_1}\left(b_m\right)$. By (Theorem 2.2), we have

$\begin{array}{c}{f}_{|}{}_{L_i}\left(a\right)=f_{|}{}_{L_1}\left(a_1\right)f_{|}{}_{L_1}\left(a_2\right)\cdots f_{|}{}_{L_1}\left(a_n\right)\cdots f_{|}{}_{L_1}\left(a_i\right)\\ \ne f_{|}{}_{L_1}\left(b_1\right)f_{|}{}_{L_1}\left(b_2\right)\cdots f_{|}{}_{L_1}\left(b_m\right)\cdots f_{|}{}_{L_1}\left(b_i\right)\\ =f_{|}{}_{L_i}\left(b\right)\end{array}$

Hence, $f_{|}{}_{L_i}$ is injective for all $i=2,3,\cdots$

To prove that $f_{|}{}_{L_i}$ is surjective for all $i=2,3,\cdots$, let $b=b_1{b}_2\cdots b_i\in L_i$, so $b_1,b_2,\cdots ,b_i\in L_1$. Since $f_{|}{}_{L_1}$ is surjective, there exist $a_1,a_2,\cdots ,a_i\in L_1$ such that $f_{|}{}_{L_1}\left(a_1\right)=b_1,f_{|}{}_{L_1}\left(a_2\right)=b_2,\cdots ,f_{|}{}_{L_1}\left(a_i\right)=b_i$.

Therefore, $f_{|}{}_{L_1}\left(a_1\right)f_{|}{}_{L_1}\left(a_2\right)\cdots f_{|}{}_{L_1}\left(a_i\right)=b_1{b}_2\cdots b_i$,

and $f_{|}{}_{L_i}\left(a_1{a}_2\cdots a_i\right)=b$, where $a_1{a}_2\cdots a_i\in L_i$.

So $f_{|}{}_{L_i}$ is surjective. Hence, $f_{|}{}_{L_i}$ is bijective for all $i=1,2,\cdots$

(b) First, we show that $f$ is injective.

i) If $a,b\in N,a\ne b$, and $a\in L_i,b\in L_j,i\ne j$, then $f\left(a\right)=f_{|}{}_{L_i}\left(a\right)\in f_{|}{}_{L_i}\left(L_i\right)$, $f\left(b\right)=f_{|}{}_{L_j}\left(b\right)\in f_{|}{}_{L_j}\left(L_j\right)$, and since $f_{|}{}_{L_i}\left(L_i\right)\cap f_{|}{}_{L_j}\left(L_j\right)=\varphi$, we have $f\left(a\right)\ne f\left(b\right)$

ii) If $a,b\in N,a\ne b$, and $a,b\in L_i$, then by (a), we have

$f\left(a\right)=f_{|}{}_{L_i}\left(a\right)\ne f_{|}{}_{L_i}\left(b\right)=f\left(b\right).$

Hence, $f$ is injective.

Now, by (a) $f_{|}{}_{L_i}\left(L_i\right)=L_i,\forall i=0,1,2,\cdots$. Therefore

$\begin{array}{c}f\left(N\right)=f\left({\displaystyle \underset{i=0}{\overset{\infty }{\cup }}{L}_i}\right)\\ ={\displaystyle \underset{i=0}{\overset{\infty }{\cup }}f\left(L_i\right)}\\ ={\displaystyle \underset{i=0}{\overset{\infty }{\cup }}{f}_{|}{}_{L_i}\left(L_i\right)}\\ ={\displaystyle \underset{i=0}{\overset{\infty }{\cup }}{L}_i}=N\end{array}$

Hence, $f$ is surjective and therefore bijective. ◼◼

Corollary 2.2 If $f:N\to N$ is a levely multiplicative and a bijective function, then $f^{-1}:N\to N$ is levely multiplicative and bijective.

Let $f:N\to N$ be a levely multiplicative function. If $a|b$, then $f\left(a\right)|f\left(b\right)$ is necessary, but $f\left(a\right)|f\left(b\right)$ is not sufficient for $a|b$. For example, in (Example 2.2) $2=f\left(2\right)|f\left(5\right)=2$, but $2\cancel{|}5$.

For $f\left(a\right)|f\left(b\right)$ to be necessary and sufficient for $a|b$, $f$ must be bijective.

Theorem 3.1 Let $f:N\to N$ be a levely multiplicative function.

(a) If $a|b$, then $f\left(a\right)|f\left(b\right)$.

(b) If $f$ is bijective, then $a|b$ if and only if $f\left(a\right)|f\left(b\right)$.

Proof: (a) Let $a,b\in N,a|b$. Then there exists $c\in N$ such that $b=ca$. Therefore $f\left(b\right)=f\left(ca\right)=f\left(c\right)f\left(a\right)$. Thus, $f\left(a\right)|f\left(b\right)$.

(b) ( $\Rightarrow$) By (a).

( $\Leftarrow$) Let $a,b\in N$, $f\left(a\right)|f\left(b\right)$. Then there exists $c\in N$ such that $f\left(b\right)=cf\left(a\right)$. Since $f$ is surjective, there exist $d\in N$ such that $c=f\left(d\right)$. Therefore, $f\left(b\right)=f\left(d\right)f\left(a\right)=f\left(da\right)$. Since $f$ is injective, $b=da$. Hence, $a|b$. ◼

As a result of Theorem 3.1, the next theorem and corollary show that a bijective, levely multiplicative, and increasing function preserves the greatest common divisor for any two numbers in $N$ and the relative prime numbers.

Theorem 3.2 Let $f:N\to N$ be a levely multiplicative, bijective, and increasing function, then $\left(a,b\right)=d$ if and only if $\left(f\left(a\right),f\left(b\right)\right)=f\left(d\right)$.

Proof: ( $\Rightarrow$) Let $\left(a,b\right)=d$.Then $d|a$ and $d|b$. By (Theorem 3.1), we have $f\left(d\right)|f\left(a\right)$ and $f\left(d\right)|f\left(b\right)$. If $f\left(c\right)|f\left(a\right)$ and $f\left(c\right)|f\left(b\right)$, and we suppose that $f\left(c\right)\ge f\left(d\right)$, then by (Theorem 3.1), we have $c|a$, and $c|b$' Therefore $c\le d$. But $f$ is an increasing function, so we have a contradiction. Hence $f\left(d\right)\le f\left(c\right)$, and $\left(f\left(a\right),f\left(b\right)\right)=f\left(d\right)$.

( $\Leftarrow$) Let $\left(f\left(a\right),f\left(b\right)\right)=f\left(d\right)$. Then $f\left(d\right)|f\left(a\right)$ and $f\left(d\right)|f\left(b\right)$. By (Theorem 3.1) we have $d|a,d|b$. If we suppose $c|a$ and $c|b$, and we suppose $c\ge d$, then by (Theorem 3.1), we have $f\left(c\right)|f\left(a\right)$ and $f\left(c\right)|f\left(b\right)$. Therefore $f\left(c\right)\le f\left(d\right)$. But $f$ is an increasing function, so we have a contradiction. Hence, $d\le c$. Thus, $\left(a,b\right)=d$. ◼

Corollary 3.1: Let $f:N\to N$ be a levely multiplicative, bijective and increasing function, then $\left(a,b\right)=1$ if and only if $\left(f\left(a\right),f\left(b\right)\right)=1$.

Proof: In (Theorem 3.2) when $d=1$ ◼

Now, when $f:N\to N$ is a levely multiplicative function, in case $f$ is not bijective function, it is not necessary that $f\left(a\uparrow \right)=f\left(a\right)\uparrow$. For instance, from (Example 2.2) we have

$\begin{array}{c}f\left(2\uparrow \right)=f\left(\left\{2n:n\in N\right\}\right)\\ =\left\{2^n:n\in N\right\}\end{array}$

but

$\begin{array}{c}f\left(2\right)\uparrow =2\uparrow \\ =\left\{2n:n\in N\right\}\end{array}$

Theorem 3.3: If $f:N\to N$ is a levely multiplicative and bijective function, then:

(a) $f\left(a\uparrow \right)=f\left(a\right)\uparrow$

(b) $f\left(A\uparrow \right)=f\left(A\right)\uparrow$

Proof: (a) Let $a\uparrow =\left\{na:n\in N\right\}$.

Since $f$ is bijective, we get

$\begin{array}{c}f\left(a\uparrow \right)=\left\{f\left(na\right):n\in N\right\}\\ =\left\{f\left(n\right)f\left(a\right):n\in N\right\}\\ =f\left(a\right)\uparrow \end{array}$

(b) Let $A\subseteq N$. By (Lemma 1.2), we have $A\uparrow ={\displaystyle {\cup }_{a\in A}a\uparrow }$. By (a), we have.

$\begin{array}{c}f\left(A\uparrow \right)=f\left({\displaystyle \underset{a\in A}{\cup }a\uparrow }\right)\\ ={\displaystyle \underset{f\left(a\right)\in f\left(A\right)}{\cup }f\left(a\uparrow \right)}\\ ={\displaystyle \underset{f\left(a\right)\in f\left(A\right)}{\cup }f\left(a\right)\uparrow }\\ =f\left(A\right)\uparrow \end{array}$ ◼

Corollary 3.2: If $f:N\to N$ is a levely multiplicative and bijective function and $A\subseteq N$ is upward closed, then $f\left(A\right)=f\left(A\right)\uparrow$

Proof: Let $A\subseteq N$, $A=A\uparrow$. By (Theorem 3.3 (b)) we have

$\begin{array}{c}f\left(A\right)=f\left(A\uparrow \right)\\ =f\left(A\right)\uparrow \end{array}$

Thus, $f\left(A\right)$ is upward closed. ◼