## Does $\bf pSet$ admit products?

The question is in the title. The category $\bf pSet$ of partial functions has sets as objects and $\hom(X,Y)$ is the set of all triples $(X,Y,f)$ such that there exists $D\subseteq X$ and $f\colon D\to Y$. Composition of arrows is composition of relations.

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I don't understand the confusion. Can you not define the product of objects $A$ and $B$ (sets) as the normal Cartesian product $A\times B$ with the normal projection maps (functions are clearly partial functions), and the product for partial functions $f,g$ (defined on $U\subset A$ and $V\subset B$ respectively) to be the partial function $f\times g$ defined on $U\times V\subset A\times B$? Also, maybe curly brackets do not appear because your/the author's notation isn't very good? – William Feb 7 2012 at 23:11
Given partial functions $f\colon X\to A, g\colon X\to B$ I should be able to define a unique $u\colon X\to A\times B$ such that the right diagram commute. But what if $dom(f)\cap dom(g)=\varnothing$? – tetrapharmakon Feb 7 2012 at 23:17
The answer is that the only admissible function $\varnothing\to A\times B$ is the empty one. But now the diagram doesn't commute. – tetrapharmakon Feb 7 2012 at 23:18
If curly braces don't show up, or just general weirdness is happening, try putting backticks around the dollar signs:  – Owen Biesel Feb 7 2012 at 23:30
I don't understand the definition of the morphisms in pSet. Does $D$ belong to the data? – Martin Brandenburg Feb 8 2012 at 10:14

If I'm reading your definition correctly, this category looks equivalent to the category ${\bf Set_*}$ of pointed sets and basepoint-preserving functions (the equivalence is by removing the basepoint from each pointed set: you're left with an ordinary set and possibly partial functions). So you should be able to take the ordinary product in ${\bf Set_*}$ and then pass it through the equivalence: the product of $X$ and $Y$ in ${\bf pSet}$ should be the disjoint union of the cartesian products $X\times Y$, $X\times\{*\}$, and $Y\times\{*\}$. The partial projection to $X$ is given by projection from $X\times Y$ and $X\times\{*\}$ and undefined on $Y\times{*}$, and the partial projection to $Y$ is similar.

And indeed, this works: if $C$ has partial functions $f$ and $g$ to $X$ and $Y$ respectively, then we get a partial function to $(X\times Y)\sqcup (X\times\{*\})\sqcup (Y\times\{*\})$ given by $c\mapsto (f(c),g(c))$ if both exist, $(f(c),*)$ or $(*,g(c))$ if only one does, and undefined if neither exists.

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Can any partial function be seen as a map between "punctured" sets? – tetrapharmakon Feb 7 2012 at 23:21
Yes: just add a point to the source and target sets, then complete the partial function by sending every unassigned point (including the new source basepoint) to the target basepoint. Then when you "puncture" the sets again (nice word choice), you're left with the partial function you started with. – Owen Biesel Feb 7 2012 at 23:27
Yes, that's what Owen means when he says that $\text{pSet}$ is equivalent to the category of pointed sets. The equivalence takes a pointed set to the subset obtained by removing the distinguished point. – Qiaochu Yuan Feb 7 2012 at 23:28
Thank you Owen and Qiaochu! I suspected there was a link between the two categories ($\bf pSet$ admits a zero objects, which seems quite strange if you ignore that equivalence). – tetrapharmakon Feb 8 2012 at 9:47