most recent 30 from http://mathoverflow.net 2013-05-20T07:01:30Z http://mathoverflow.net/feeds/question/22036 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/22036/pullbacks-in-category-of-sets-and-partial-functions supercooldave 2010-04-21T09:47:17Z 2010-04-21T12:41:41Z

<p>Do pullbacks exist in the category of sets and partial functions? </p> <p>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 $\{ (a,b) \in A\times B ~|~ f(a)=g(b) \}$? </p> <p>If not, what is a simple description of the pullback?</p>

http://mathoverflow.net/questions/22036/pullbacks-in-category-of-sets-and-partial-functions/22046#22046 Thomas Kragh 2010-04-21T12:14:57Z 2010-04-21T12:41:41Z

<p>Pullbacks exists but are not what you describe.</p> <p>The answer is as follows:</p> <p>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.</p> <p>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:</p> <p>$A \times_\mathcal{C} B \approx (A \times_{\textrm{set}} B) \sqcup_{\textrm{set}} A \sqcup_{\textrm{set}} B$</p> <p>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.</p>