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?


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 $\times$ 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.

  • 2
    $\begingroup$ 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 $(1\downarrow\mathbf{Set})$. $\endgroup$
    – user2734
    Apr 21 '10 at 12:35
  • $\begingroup$ Thanks. You've given me the product, but I'm still having a little trouble working out what the pullback looks like. $\endgroup$ Apr 21 '10 at 12:38
  • $\begingroup$ 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? $\endgroup$
    – user2734
    Apr 21 '10 at 12:39
  • 1
    $\begingroup$ @ 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. $\endgroup$ Apr 21 '10 at 12:45
  • $\begingroup$ @Thomas Kragh: Ah, of course. Thank you. $\endgroup$
    – user2734
    Apr 21 '10 at 13:05

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