Relationship between $f\left(\bigcap\limits_{i \in I} A_i\right)$ and $\bigcap\limits_{i \in I} f(A_i)$. I am trying to investigate and ultimately prove the relationship between $f\left(\bigcap\limits_{i \in I} A_i\right)$ and $\bigcap\limits_{i \in I} f(A_i)$. The first thing I was able to find is that in general, regardless of the specification of $f$, we have $f\left(\bigcap\limits_{i \in I} A_i\right) \subset \bigcap\limits_{i \in I} f(A_i)$.
Proof. Let $y \in f\left(\bigcap\limits_{i \in I} A_i\right)$. Then for each $i \in I$ and corresponding $A_i \in \{A_i\}_{i \in I}$, there exists $a_i \in A_i$ such that $f(a_i) = y$. So for each $i \in I$, $y \in f(A_i)$, so $y \in \bigcap\limits_{i \in I} f(A_i)$.
Now, if $f$ is injective, then there exists a unique such $a_i$, call it $x$, so that $x \in A_i$ for all $i$, i.e., $x \in \bigcap\limits_{i \in I} A_i$, so if $y = f(x)$, $y \in f\left(\bigcap\limits_{i \in I} A_i\right)$. That gives the opposite inclusion provided that $f$ is injective, but I am not able to prove -- and am not sure if it's possible -- that equality holds if and only if $f$ is injective. In other words, why must this $a_i$ be unique? I could have $f(x) = y$ for some $x \in \bigcap\limits_{i \in I} A_i$ and $f(t) = y$ for some $t \neq x$. This depends on the domain of the function, of course, since this does have to hold for all $y$.
So I'm having trouble proving the reverse direction, mostly because I don't fully believe that injectivity is a necessary condition, but surely a sufficient condition.
 A: And now, if $f$ is not injective, take $x$ and $y$ such that $x\ne y$ and that $f(x)=f(y)=z$, take $I=\{1,2\}$, take $A_1=\{x\}$, and take $A_2=\{y\}$. Then$$f(A_1\cap A_2)=f(\emptyset)=\emptyset\quad\text{but}\quad f(A_1)\cap f(A_2)=\{z\}\ne\emptyset.$$So, whenever $f$ is not injective, the statement is false.
A: If $f : X \to Y$ is injective, then WLOG assume $X \subseteq Y$ and $f$ is the inclusion map. Then for all $A \subseteq X$, $f(A) = A$. Then clearly $f(\bigcap\limits_{i \in I} A_i) = \bigcap\limits_{i \in I} A_i = \bigcap\limits_{i \in I} f(A_i)$.
Note that the above requires $I$ inhabited.
Conversely, suppose that the image map $f : P(X) \to P(Y)$ preserves (inhabited) intersections. Then for all $x$ and $y$ such that $f(x) = f(y)$, we see that $f(\{x\} \cap \{y\}) = f(\{x\}) \cap f(\{y\}) = \{f(x)\}$. So it must be the case that $x = y$.
A: What is true is that if $f(\bigcap_{i\in I}A_i)=\bigcap_{i\in I}f(A_i)$ for all family of subsets $(A_i)_{i\in I}$, then $f$ must be injective. Indeed, $f(\{x\}\cap\{y\})=\{f(x)\}\cap\{f(y)\}$ for any $x,y$ implies that $f$ must be injective.
A: We need a bit more framing to your question.  As noted by the other two answers, if you are allowed to choose your family of sets $A_i$ or it is supposed to work for all $A_i$,  then being injective is both necessary and sufficient.
However, if you are working with a specific family of $A_i$,  you just need that the set of inputs for which injectivity fails belongs to each member $A_i$.  In other words,  your necessary and sufficient criteria will be that "
$$\{x|(\exists y)(\exists z)y\neq z\land x=f(y)=f(z)\}\subseteq A_i $$
for all $i\in I$.
