• Consider the category $Set^\partial$ of sets with partial functions as morphisms, and the category $Set^*$, the category of pointed sets and and basepoint-preserving functions as morphisms.
• The functor $F: Set^\partial \rightarrow Set^*$sends sets to set-with-basepoint. To be very precise about basepoint considerations, since this is where the non-inversion will lie, let us say that for a set $X$, we add a basepoint $\{X \}$. So, the functor $F$ sends a set $X$ to the set a = {X} in (X U {a}, a), which expanded out is $(X \cup \{ \{ X \} \}, \{ X \})$. The functor $F$ sends a partial function $f: A \rightharpoonup B$ to based function by defining $F(f): F(A) \rightarrow F(B)$which sends undefined values $a$ to $\{B\} \in F(B)$, and is forced by definition sends the basepoint $\{ A \} \in F(A)$ to $\{ B \} \in F(B)$.
• The "inverse" functor $G: Set^* \rightarrow Set^\partial$ forgets the basepoints, and sends a function $h: (A, a) \rightarrow (B, b)$to the partial function $G(h): A \rightarrow B$ by not mapping those elements in the domain which were mapped to $b$ by $h$.
• Going from a partial function $f$ to a bottomed function $F(f)$ and then back again to partial function $G(F(f))$ forgets no information.
• On the other hand, going from a basepointed set $(A, \bot)$ for some arbitrary basepoint $\bot$will return a set with a different basepoint, $(A/ \bot \cup \{ \{ A/\bot \} \}, \{ A/\bot \})$. Note that the sets $(A, \bot)$ and $A/\bot \cup \{ \{ A/\bot\} \}$ are isomorphic, but not equal!

G(F((A, a)))= let x = remove A a in G(x)= let x = remove A a;      botnew = set([x]) in insert x botnew-- | adds the set (A - a) as an element of (A - a)= insert (remove A a) (remove A a)

Thus, we should come up with a weaker notion of equality : Adjoints!