What $f(A)$ means on this theorem?

Question

I trying to understand this paper: https://www.researchgate.net/publication/225302984_On_Intuitionistic_Anti-Fuzzy_Submodule_of_a_Module

and I want to prove this theorem:

If $f:M\to N$ be a surjective module homomorphism and if $A=(\mu_A,\nu_A)$ be IFSM of $R$-module M, then $f(A)$ is IFSM of $R$-module $N$.

This is several preliminaries for proving this theorem.

Definition. Let $X$ be non empty set. An intuitionistic fuzzy set (IFS) $A$ in $X$ is defined as an object of the following form $$A=\{(x, \mu_A(x),\nu_A(x))\mid x\in X\}$$ which $\mu_A(x)$ and $\nu_A(x)$ respectively is two functions defined by $$\mu_A: X\to [0,1]$$ and $$\nu_A: X\to [0,1],$$ which $$0\leq \mu_A(x)+\nu_A(x)\leq 1$$ for every $x\in X$. The functions $\mu_A(x)$ and $\nu_A(x)$ define the degree of membership and degree of non-membership respectively, for every $x\in X$.

Definition. Let $M$ be a modules over a ring $R$. An IFS $A = (\mu_A , \nu_A)$ of $M$ is called intuitionistic fuzzy submodule (IFSM) if

1. $\mu_A(0)=1$ and $\nu_A(0)=0$.
2. $\mu_A(x+y)\geq \min(\mu_A(x),\mu_A(y))$ and $\nu_A(x+y)\leq \max(\mu_A(x),\mu_A(y))$
3. $\mu_A(rx)\geq \mu_A(x)$ and $\nu_A(rx)\leq \nu_A(x)$ for all $x,y\in M$ and $r\in R$.

I want to try proving Theorem 2.14 and I don't know what $f(A)$ means.

I think $f(A)$ is not the sets of image of $A$, because $f:M\to N$ and $A=(\mu_A,\nu_A)$ not subset of $M$.

Now I guessing $$f(A)=(\mu_A(f(x)),\nu_A(f(x))),$$ for all $x\in M$.

This is my try for proving that theorem, based on Definition of $IFSM(M)$.

Let $0_M$ be identity on $M$ and $0_N$ be identity on $N$.

Given:

$f:M\to N$ be a surjective module homomorphism, so \begin{align} f(x+y)=f(x)+f(y) \text{ and } f(rx)=rf(x) \end{align} for all $x,y\in M$ and $r\in R$.

and $A=(\mu_A,\nu_A)$ be IFSM of $R$-module M, so

1. $\mu_A(0_M)=1$ and $\nu_A(0_M)=0$.
2. $\mu_A(x+y)\geq \min(\mu_A(x),\mu_A(y))$ and $\nu_A(x+y)\leq \max(\mu_A(x),\mu_A(y))$
3. $\mu_A(rx)\geq \mu_A(x)$ and $\nu_A(rx)\leq \nu_A(x)$ for all $x,y\in M$ and $r\in R$.

Now we will to prove $f(A)=(\mu_A(f(x)),\nu_A(f(x)))$ be IFSM of $R$-module N.

First, I will prove $\mu_A(0_N)=1.$

\begin{align} \mu_A(0_N)&=\mu_A(f(0_M)) \text{ since $f$ is homomorphism} \end{align} But here I can't associating with $\mu_A(0_M)=1$, so I can't conclude $\mu_A(0_N)=1.$

So, my question is:

Since I can't prove this theorem, is that true if I guessing $f(A)=(\mu_A(f(x)),\nu_A(f(x)))$? If wrong, what the definition of $f(A)$?

Any help will appreciated.

I know for this theorem the reference is taken from

D.K. Basnet, “ Topic in Intuitionistic fuzzy algebra”, Lambert Academic
Publication , 2011 , ISBN: 978-3-8443-9147-3.

But since I don't have this book, I can't understand what $f(A)$ mean.