A map relating pre-image and power set Let $S$ and $T$ be two sets. If $R$ is a relation on $S \times T$, then $\forall t\ \in T$ we have the pre-image:
$$ R^{[t]}= \{s \in S \mid sRt\} \subseteq S$$ 
(a) We wish to prove that the relation $\{(t, R^{[t]})\mid t \in T\}$ is a map from T to the power set $\mathcal{P}(S)$ of $S$. 
So what I think we have to prove is that whenever $a, b\in T$ and $x\in  \mathcal{P}(S)$, such that $(a,x)$ and $(b,x)$  are in the relation, we can show $a=b$, this proves that the relation is a map. Is this correct and what would be the first steps?
(b) As a second question we are asked the reverse. What if we have a map: $f: T\rightarrow \mathcal{P}(S)$ then $R_f= \{ (s,t) \in f(t)\}$ is a relation on $S\times T$, where $R_f[t]=f.$
This one just gave me a big headache so far.
 A: For your first question: To show that a relation $R \subseteq S \times T$ is a map (say it represents some function $f: S \to T$), you have to prove that for every $s \in S$, there exists exactly one $t \in T$ (which would then be $f(s)$) for which $s\,R\,t$. Hence, you would have to show that there exists such a $t$, and that for two such $t_1, t_2 \in T$ such that $s\,R\,t_1$ and $s\,R\,t_2$ we must have $t_1 = t_2$.
The first point is given directly from the definition of our relation $F := \{(t, R^{[t]})\mid t \in T\} \in T \times \mathcal P(S)$, since $R^{[t]} \subset S$ is such a $p \in \mathcal P(s)$ such that $s\,F\,p$, since by definition $(t, s) \in F$. The second point also follows pretty quickly from the definition, since if $t\,F\,s' \iff (t, s') \in F$ for some $s' \in \mathcal P(S)$ we must by definition have that $(t, s')$ is of the form $(t', R^{[t']})$ for some $t' \in T$. But then we need to have $t' = t$ and $s' = R^{[t']}$, which is then just $R^{[t]}$, which is our $p$, so that this $p$ is unique and $F$ is indeed a map.
For some intuition on what this actually means, this map $F$ associates to every $t \in T$ its preimage under $R$. The second question then asks whether this also goes in the reverse direction, that is, whether for a given map $f: T \to \mathcal P(S)$, there is a relation $R_f \subseteq S \times T$ such that $f$ is the map giving preimages under this relation $R_f$ as described above. If we fix some $t \in T$, we would then need that $f(t)$ be equal to $R_f^{[t]}$. Since we are looking for a relation $R_f$ satisfying this, this means that for any $t \in T$ we must have $R_f^{[t]} = f(t)$. Given $s \in S$ then, this means that $s$ lies in $R_f^{[t]}$ if and only if $s$ lies in $f(t)$, so that $s\,R_f\,t \iff s \in f(t)$. Hence we can write our relation as
$$R_f = \{(s, t) \in S \times T \mid s \in f(t)\}.$$
So for every $t \in T$ everything that relates to $t$ and thus lies in the preimage of $t$ is exactly everything in $f(t)$.
I hope this clears up some of your confusion. The definition for $R_f$ that you gave in your question does not seem to work, since $f$ as a relation would be a subset of $T \times \mathcal P(S)$ and $f(t)$ for some $t \in T$ would be a subset of $S$ and not a set of pairs in $S \times T$, so there might be some mistake there, which is maybe where the confusion came from.
