Pullbacks in Category of Sets and Partial Functions

Do pullbacks exist in the category of sets and partial functions?

Are the 'the same' as they are in Sets? That is, given two partial functions $f : A \to C$ and $g : B \to C$, is the pullback given by $\{ (a,b) \in A\times B ~|~ f(a)=g(b) \}$?

If not, what is a simple description of the pullback?

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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 $\times$ in based sets translates back through this equivalence to $\mathcal{C}$ and 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 $B$, or nothing.

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As a remark, the category at hand is actually complete, because the category of pointed sets is isomorphic to the comma (slice) category $(1\downarrow\mathbf{Set})$ (where 1 is a one object set). Now, in general the projection $(x\downarrow C)\to C$ of the comma category creates limits (Ex. 5.1.1 in Mac Lane), and since $\mathbf{Set}$ is complete, so is $(1\downarrow\mathbf{Set})$. – user2734 Apr 21 '10 at 12:35
Thanks. You've given me the product, but I'm still having a little trouble working out what the pullback looks like. – supercooldave Apr 21 '10 at 12:38
Oh, and I have a question: Isn't the category of sets and partial functions actually isomorphic (not just equivalent) to the category of pointed sets? – user2734 Apr 21 '10 at 12:39
@ unknown (google):No because:"the base-point is not unique". To be precise there is only one empty set but many based one point sets. – Thomas Kragh Apr 21 '10 at 12:45
@Thomas Kragh: Ah, of course. Thank you. – user2734 Apr 21 '10 at 13:05