For sets $A$ and $B$, consider a partial function from $A$ to $B$ to be a pair $(S,f)$ where $S \subseteq A$ and $f$ is a normal function $S \rightarrow B$. This captures the notion that the partial function is only 'defined' on $S$. Let $\mathbf{Par}$ be the category of sets with maps being partial functions. Let $\mathbf{Set}_*$ be the category of sets with distinguished elements, with maps being functions that preserve distinguished elements.

Theorem:

$$\mathbf{Par}$$ is equivalent to $$\mathbf{Set}_*$$. *Proof*: First, we write down a functor $$T : \mathbf{Par} \rightarrow \mathbf{Set}_*$$ that 'completes' a partial function into a regular (total function) by adding a special element to the sets. In particular, for a set $$A$$ let $$T(A) = (A \cup \set{\bot_A}, \bot_A)$$. We add an element to the set and quite literally *distinguish* it because it is special. For a map $$(S, f) : A \rightarrow B$$, let$$

T(S,f) = a \mapsto \left{ \begin{array}{lr} f(s) & a \in S \ \bot_B & \text{otherwise} \end{array} \right.

$$Note that this map takes $$\bot_A$$ to $$\bot_B$$ so it does preserve distinguished elements. Define $$N : \mathbf{Set}_* \rightarrow \mathbf{Par}$$ as follows: For a pointed set $$(A, x)$$ let $$N(A, x) = A \backslash x$$. For a point-preserving map $$f : (A, x) \rightarrow (B, y)$$ let $$N(f) = (f^{-1}(B \backslash y), \hat{f})$$, where $$\hat{f}$$ is simply the restriction of $$f$$ to the set $$f^{-1}(B \backslash y)$$. This is well-defined precisely because $$f^{-1}(B \backslash y)$$ certainly does not contain $$x$$, and certainly $$f$$ does not send anything in $$f^{-1}(B \backslash y)$$ to $$y$$! We wish to define $$\eta : 1 \rightarrow NT$$. So for $$A$$ a set $$A$$, we need the component $$\eta_A : A \rightarrow NTA$$, but since $$NTA = A$$, we just have $$\eta_A : A \rightarrow A$$. We make the obvious choice then and let $$\eta_A = (A, 1_A)$$, which is the identity arrow on $$A$$ in $$\mathbf{Par}$$. Let's verify that this is actually a natural transformation. Let $$(S,f) : A \rightarrow B$$ be a map in $$\mathbf{Par}$$. We need $$\eta_B \circ (S, f) = NT(S, f) \circ \eta_A$$, but this follows immediately because $$NT(S, f) = (S, f)$$ and $$\eta_A$$ and $$\eta_B$$ are identity arrows. So we have naturality. Furthermore every $$\eta_A$$ is clearly an isomorphism so $$\eta$$ is a natural isomorphism. We will now define $$\varepsilon : TN \rightarrow 1$$. We need $$\varepsilon_{(A, x)} : TN(A, x) \rightarrow (A, x)$$ a point-preserving map. Note that $$TN(A, x)$$ is the set $$A$$ with $$x$$ replaced by $$\bot_A$$, and $$\bot_A$$ distinguished instead of $$x$$. So, $$\varepsilon_{(A,x)}$$ be the map sending $$\bot_A \mapsto x$$ and $$a \mapsto a$$ for all other elements $$a$$. Let $$f : (A,x) \rightarrow (B,y)$$ be a map in $$\mathbf{Set}_*$$. To check naturality we need$$

\varepsilon{(B, y)} \circ TN(f) = f \circ \varepsilon{(A,x)} : TN(A,x) \rightarrow (B,y)

$$Let's examine $$TN(f)$$ -- it can be described explicity as:$$

TN(f) = a \mapsto \left{ \begin{array}{lr} f(a) & f(a) \neq y \ \bot_B & \text{otherwise} \end{array} \right.

$$

Let $a \in TN(A,x)$. We will consider where $a$ is sent by the left-hand side and the right-hand side by considering cases. First suppose $a \neq \bot_A$, and $f(a) = y$. Then the right-hand side sends $a$ to $y$, and the left-hand side sends $a \mapsto \bot_B \mapsto y$. Suppose $a \neq \bot_A$ and $f(a) \neq y$. Then the right-hand side sends $a$ to $f(a)$, and so does the left. Finally, suppose $a = \bot_A$. Then the right-hand side sends $a \mapsto x \mapsto y$. The left-hand side sends $a \mapsto \bot_B \mapsto y$. So the left and right-hand side are identical in all cases, so they must represent the same map. Then, we have naturality. Furthermore, $TN(A,x)$ and $(A,x)$ are certainly equal cardinality, so they are in bijection and all components of $\varepsilon$ are isomorphisms. Thus $\varepsilon$ is a natural isomorphism.

So, we have shown an equivalence between $\mathbf{Par}$ and $\mathbf{Set}_*$.