I am trying to solve exercise 4 in the first chapter of Barr Wells "Category Theory for Computing Science" see here on page 8.
I recap here the text of the exercise.
Let $\mathcal{P}(C)$ the power set of a set $C$ and let $Rel(A,B)=\mathcal{P}(A\times B)$. Let $\phi: Rel(A,B) \to Hom(A,\mathcal{P}(B))$ be defined as follows: $$ \phi(r)(a) = \{b \in B : (a,b)\in r\}\,. $$ The first question is to show that $\phi$ is a bijection.
I am having some trouble to understand how this would be possible.
If I am interpreting the definition of $\phi$ correctly, $\phi$ takes a relation $r \in \mathcal{P}(A\times B)$ and provides a function $f:A\to \mathcal{P}(B):a\mapsto \bar{b}$. We require that that $\phi(r)(a) = \{\bar{b} \in B | (a,\bar{b}) \in r\}$, so, it gives me the impression that we are restricting to certain type of functions in $Hom(A,\mathcal{P}(B))$.
However, $Hom(A,\mathcal{P}(B))$ has functions such as $f:A\to \mathcal{P}(B):a\mapsto \emptyset$. So $\phi$ cannot be surjective.
It feels like one should have defined $\phi(r)(a) = \{\bar{b} \in \mathcal{P}(B) : (a,\bar{b}) \in r\}$.
Comments/thoughts? Am I missing something?
Hint: this is the inverse of $\phi$: \begin{align} &\operatorname{Hom}(A,\mathcal{P}(B))\to\operatorname{Rel}(A,B)& &f\mapsto\bigcup_{a\in A}(\{a\}\times f(a)) \end{align} In particular, note that $\phi(\varnothing)(a)=\varnothing$ for every $a\in A$.