Let $R,S$ two rings, and let $f:R\rightarrow S$ be a homomorphism.
Prove thar if $f$ is an epimorphism, then for any subset $X\subseteq R$ we have that $f(X)$ is the ideal generated by itself, that is, $f(X)=(f(X))$.
$\subseteq$) $f(X)\subseteq (f(X))$ is clear.
$\supseteq$) $(f(X))=\left\lbrace y_1 s_1+...+y_n s_n \vert y_i\in f(X), s_i\in S, i=1,...,n \right\rbrace $. Therefore an element $u\in (f(X))$, has the form $u=y_1s_1+...+y_ns_n$ where $y_i\in f(X)\subseteq S, s_i\in S$ for $i=1,...,n$. Since $f$ is surjective, $f(R)=S$ and since $S$ is a ring and $f(X)\subseteq f(R)=S$, we have that $u\in S$.
Also, there exists $x_i\in X, r_i \in R$ such that $f(x_i)=y_i$ and $f(r_i)=s_i$ respectively for all $i$. Hence $u=f(x_1)f(r_1)+...+f(x_n)f(r_n)=f(x_1r_1)+...+f(x_nr_n)\in R$ since $R$ is a ring.
I have tried to prove $u\in f(X)$, but I don't see how to do it because the product of each pair $y_is_i$ or $x_ir_i$ is in the whole ring and is not in the subset.
Also, I have other doubt. Is it necessary to consider $f$ an epimorphism?
Hope this helps you.
The statement is a little bit strange. Because the equality $f(X)=(f(X))$ might have no sense if the subset $X$ has no an algebraic structure. Considere for instance the identity function from a ring onto itself, then $id(\lbrace x\rbrace)=\lbrace x\rbrace$ and if $x$ is not the zero element, is not always true that $\lbrace x\rbrace=(x)$.
But, if you look at the ideals, then $f(X)$ is an ideal since $f$ is onto if $X$ is an ideal. In these circumstances your statement is true.
For the other contention, $(f(X))$ is the smallest (in the sense of inclusion) ideal which contains $f(X)$ because $$(f(X))=\bigcap \lbrace I\subseteq R~\mid~ I<R~\wedge f(X)\subseteq I\rbrace$$
Where $I<R$ means that $I$ is an ideal of $R$. Thus $(f(X))\subseteq f(X)$ as a consequence of $f(X)$ to be ideal and $f(X)\subseteq f(X)$.
If the function is not surjective then $f(X)=(f(X))$ might have no sense.