3 added 43 characters in body

Pullbacks exists but are not what you describe.

The category $\mathcal{C}$ of sets and partial functions is equivalent to the category of based sets and based functions, by sending the set $A$ to $A$ disjoint union a base point $*$ and sending $f$ to the obvious based function which sends everything on which $f$ was not defined to the base-point.
The pullback in based sets are well-known and for example the product $\times$ translating in based sets translates back through this equivalence to the category of sets $\mathcal{C}$ and partial functions becomes:
$A \times_\mathcal{C} B \approx (A \times_{\textrm{set}} B) \sqcup_{\textrm{set}} A \sqcup_{\textrm{set}} B$
From the purely sets and partial function point of view this is also explainable: . Indeed, any morphism from $Z$ to this product is given by a choice for each point in $Z$ of either: a point in $A$ and a point in $B$, or a point in $A$, or a point $B$.B$, or nothing. 2 added 212 characters in body; added 23 characters in body Pullbacks exists but are not what you describe. The answer is as follows: The category$\mathcal{C}$of sets and partial functions is equivalent to the category of based sets and based functions, by sending the set$A$to$A$disjoint union a base point$*$and sending$f$to the obvious based function which sends everything on which$f$was not defined to the base-point. The pullback in based sets are well-known and for example the product$\times$but translating back through this equivalence to the category of sets and partial functions you essentially getbecomes:$A \times_\mathcal{C} B \approx (A \times_{\textrm{set}} B) \sqcup_{\textrm{set}} A \sqcup_{\textrm{set}} B$From the purely sets and partial function point of view this is also explainable: any morphism from$Z$to this product is given by either a point in$A$and a point in$B$, or a point in$A$, or a point$B$. 1 Pullbacks exists but are not what you describe. The answer is as follows: The category$\mathcal{C}$of sets and partial functions is equivalent to the category of based sets and based functions, by sending the set$A$to$A$disjoint union a base point$*$and sending$f$to the obvious based function which sends everything on which$f$was not defined to the base-point. The pullback in based sets are$\times$but translating back through this equivalence to the category of sets and partial functions you essentially get:$A \times_\mathcal{C} B \approx (A \times_{\textrm{set}} B) \sqcup_{\textrm{set}} A \sqcup_{\textrm{set}} B\$