# 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 '12 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 $dom(f)\cap dom(g)=\varnothing$? – Fosco Loregian Feb 7 '12 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. – Fosco Loregian Feb 7 '12 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 '12 at 23:30
I don't understand the definition of the morphisms in pSet. Does $D$ belong to the data? – Martin Brandenburg Feb 8 '12 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.
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 '12 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). – Fosco Loregian Feb 8 '12 at 9:47