Injective Equivalence I'm trying to prove that these two statements are equivalent.
I've already proven that $f$ injective implies that
$$f^{-1} \left(f(B)\right) = B$$
but I need to show that
$$f^{-1} \left(f(B)\right) = B \Leftrightarrow f\left(\bigcap A_t\right)=\bigcap f\left(A_t\right).
$$
Any advice would be greatly appreciated! 
 A: The intuition is that injections map to different elements, so if $f$ is injective and $\{A_t\}$ are pairwise independent, the right-hand side will hold.
Assume $f$ is injective and let $x \in \cap A_t$, then $x \in A_t \forall t$, so $f(x) \in f(A_t) \forall t$ and thus $f(\cap A_t) \subseteq \cap f(A_t)$.
Now assume $y \in \cap f(A_t)$ so $\exists x_t \in A_t$ such that $f(x)=y$, but since $f$ is injective, $\{x_t\}_t$ really contains one element.
Now think how to prove the implication the other way with a similar intuition.
A: Note: this isn't necessarily the fastest approach, but it is often instructional to try and do these sorts of proofs without getting too bogged down in the elements, and instead use the Boolean algebra structure on the sets to make the argument as much as possible.  That is the way I have structured this advice:

It should be pretty clear that for any $f$ the following always hold:


*

*$$f\left[\bigcap A_{t}\right] \subseteq \bigcap f\left[A_{t}\right]$$

*$$B\subseteq f^{-1}[f[B]].$$ 


So you just need to show that:
$$\Big[f\left[\bigcap A_{t}\right] \supseteq \bigcap f\left[A_{t}\right]\Big]\Leftrightarrow \Big[ f^{-1}\left[f\left[B\right]\right]\subseteq B\Big].$$

For the $\Rightarrow$ direction, assume that the left hand side of the theorem holds and consider $C=\big(f^{-1}[f[B]]\setminus B\big)=\left\{x \, \left\vert (x\notin B) \& \big(\exists y\in B \, \left(f(x)=f(y)\right)\big)\right.\right\}$. Since $f^{-1}[f[B]]=B\cup C$, the result will follow if you can show that $C$ is empty. 
The assumption about $f$ tells us that $f[B]\cap f[C]=\emptyset$ (why?).  But we also know that $f[C]\subseteq f[B]$ (why?).  So, $f[C]$ must be the empty set, and you're almost there.

For the $\Leftarrow$ direction, assume that the right hand side of the theorem holds and take a family of sets $\{A_{t}\vert t\in T\}$. For each $t\in T$, we know that $f^{-1}[f[A_{t}]]\subseteq A_{t}$, and so 
$$ \bigcap_{t\in T} \left[f^{-1}[f[A_{t}]]\right]\subseteq\bigcap_{t\in T} A_{t}.$$  Applying $f$ to both sides, the result will follow if we can show that 
$$\bigcap_{t\in T}\left[f[A_{t}]\right]\subseteq f\left[\bigcap_{t\in T} \left[f^{-1}[f[A_{t}]]\right]\right]$$
(why?).
But since it is always the case that 
$$f^{-1}\left[\bigcap C\right]=\bigcap \left[ f^{-1}[C]\right],$$
this follows from (2) above (why?).
