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.