$f$ is surjective iff $f[f^{-1}[B]]=B$ for each $B$ Question: 
Let $S$ and $T$ be sets and let $f:S\to T$. Show that $f$ is a surjection from $S$ to $T$ iff for each subset $B$ of $T$, $f[f^{-1}[B]]=B$.

So it's a surjection if for each element $t$ in $T$, there's a corresponding value $s$ in $S$ so that $f(s) = t$?
So we need to show that
1.) $f$ is a surjection from $S$ to $T$ then for each subset $B$ of $T$ $f[f^{-1}[B]] = B$.
2.) If for each subset $B$ of $T$, $f[f^{-1}[B]] = B$ then $f$ is a surjection from $S$ to $T$.
to show the biconditional?

1.) So if $f$ is a surjection, then every value in $T$ has a value in $S$ that the function maps to it. Since the inverse of a function $f:S\to T$ would be $f^{-1}: T\to S$, sending $S$ back into $f$ should then result in the original set being sent through the inverse function. 
2.) If we send $B$ through the inverse, and then put the result of that back in $f$, and we get $B$ again, that means that every value in $B$ has a value in $S$ that is mapped via $f$ to $B$. And if $B$ is a subset of $T$, then that means every value in $T$ has a value in $S$ that maps to it.
I don't think this needs to be a rigorous proof (usually those say prove and not show), so I don't know if this is enough if it's correct.
 A: First things first, I think a clarification of notation is in order. If $f:S\to T$ is a function, then $f^{-1}$ does not necessarily denote the inverse of $f$, for $f$ simply may not be invertible. Thus in general, when we are given only that $f$ is a function, we make a "notational abuse", so to speak, and consider two maps between the power sets of the domain and the codomain, viz., $f:\mathcal{P}(S)\to \mathcal{P}(T), A\mapsto f[A]:=\{f(x)|x\in A\}$ and $f^{-1}:\mathcal{P}(T)\to \mathcal{P}(S), B\mapsto f^{-1}[B]:=\{x|f(x)\in B\}$. Normally these functions are denoted by $f(\cdot)$ and $f^{-1}(\cdot)$, but in your notation you use $f[\cdot]$ and $f^{-1}[\cdot]$.

Other than that, you know the definition of surjectivity and a biconditional correctly.
In mathematics, "show" and "prove" are used synonymously (so is "demonstrate" etc.). In fact, even when you are giving an example you are also required to show that the example you give works the way you claim it to work.

Hint: $\forall B\subseteq T: f[f^{-1}[B]]\subseteq B$ for any $f$, regardless of it being surjective or not.
