I am trying to find counterexample for the following claim
Let $y=f(x)$ be a function defined in a set $M$. Let $\mathcal{M}$ be a system of subsets of $M$, and let $f(\mathcal{M})$ denote the system of all images $f(A)$ of sets $A\in\mathcal{M}$. If $\mathcal{M}$ is an algebra, so is $f(\mathcal{M})$.
The relevant definitions are:
(i) The nonempty collection $\mathcal{S}$ of sets is a ring if it closed under $\Delta$ and $\cap$.
(ii) A ring $\mathcal{A}$ is called algebra if it has an element $E$ such that $E\cap A=A$ for all $A\in \mathcal{A}$
I can show that $f(\mathcal{M})$ is a ring. To prove that $f(\mathcal{M})$ is an algebra I need to show (ii) holds. I thought that since $\mathcal{M}$ is algebra it has an element $E$ that satisfies (ii) and I tried to prove that $f(E)$ satisfies (ii) for $f(\mathcal{M})$. However in general, we have $f(A)=f(E\cap A)\neq f(E)\cap f(A)$. So I am looking for a counterexample. Any help is appreciated.
I will use property of sets that say $A\subset B \iff A\cap B = A$. We know that $E\cap A=A$ for all $A\in\mathcal{M}$, so $A\subset E$ and $f(A)\subset f(E)$. Hence $f(A)\cap f(E)=f(A)$.