How to interpret $1 \to 0$ in ${\bf Set}^\mathrm{op}$, and ${\bf Set}^\mathrm{op}$  itself? How to interpret the morphism $1 \to 0$  in ${\bf Set}^\mathrm{op}$, dual to $\bf Set$, with the standard meanings of the initial and terminal objects? Since the objects have the same interpretation in the dual, can  ${\bf Set}^\mathrm{op}$ be interpreted (and if so uniquely?) as the category of sets and partial functions?
 A: It is a remarkable fact that $\textbf{Set}^\textrm{op}$ is actually a completely concrete category: it is naturally equivalent to the category of complete atomic boolean algebras via the contravariant power set functor. Thus, an object $X$ in $\textbf{Set}^\textrm{op}$ secretly stands for its powerset $P X$, and a morphism $X \to Y$ in $\textbf{Set}^\textrm{op}$ is then a homomorphism of complete boolean algebras $P X \to P Y$. (More precisely, if $f : Y \to X$ is a map in $\textbf{Set}$, then the corresponding homomorphism $P f : P X \to P Y$ is the one that sends a subset $U \subseteq X$ to its preimage $f^{-1} U \subseteq Y$.)
A: If $f\colon A\to B$ in $\mathbf{Set}$, then the morphism in the opposite category $f^{op}\colon B\to A$ can be thought of as a multivalued partial function.  That is, it is only defined on $\operatorname{im} f$ and the image of $f^{op}(b)$ is the set $f^{-1}(b)$.
If we try and recover exactly which conditions we need to place on our multivalued partial functions so that every one corresponds uniquely to a honest function of sets, we find that $\mathbf{Set}^{op}$ is equivalent to the category whose objects are sets and whose morphisms are surjective multivalued partial functions with the property that $f(a)$ and $f(b)$ are disjoint sets for $a\neq b$.  If $f\colon B\to A$ is a morphism as defined above, the corresponding function $g\colon A\to B$ of sets is given by $g(a)=b$ where $b$ is the unique element such that $a\in f(b)$.
This is doubtfully a useful interpretation...
