Prove that if $\mathcal C⊆\mathcal P(X)$ is totally closed with respect intersection then $p_f[\cal C]$ isn't totally closed with respect intersection

Question

Given a set $X$ we say that a collection $\cal C$ in $X$ is totally closed with respect intersection if the intersection of any subcollection $\cal S$ of $\cal C$ is in $\cal C$, that is we say that $\cal C$ is totally closed with respect intersection if any $\cal S\subseteq C$ is such that $\bigcap S$ is in $\cal C$. Now if $f$ is a function from a set $X$ to a set $Y$ then we observe that the position $$ p_f(U):=f[U] $$ with $U\in\mathcal P(X)$ defines a function from $\mathcal P(X)$ to $\mathcal P(Y)$ so that I observed that if $f$ is injective then $$ p_f[\mathcal C]\equiv\big\{f[C]:C\in\mathcal C\big\} $$ is totally closed with respect intersection since if $f$ is injective then for any $\mathcal S\subseteq\mathcal P(X)$ the equality $$ f\biggl[\bigcap_{S\in\mathcal S}S\biggl]=\bigcap_{S\in\mathcal S}f[S] $$ holds; however, I was not able to prove or disprove if $p_f[\cal C]$ is totally closed with respect intersection even when $f$ is not injective so that I thought to put a specific question: could someone help me giving a counterexample, please?

Answer 1

It's not true.

Let $A=\{1, 2, 3, 4\}, B=\{1,2,3\}$.

Let $f:A\to B$ be defined by $f(1)=1, f(2)=f(3)=2, f(4)=3$.

Take the collection $\{\{1,2\}, \{3, 4\}, \emptyset\}$ in $A$.

It maps to $\{\{1, 2\}, \{2, 3\},\emptyset\}$ in $B$. $\{1,2\}\cap\{2,3\}=\{2\}$.