Why is the set of partial maps a set?

Question

In the nLab page for partial map classifiers it is stated: two partial maps (between $A$ and $B$) are considered equal if they are related by an isomorphism of spans; in this way we obtain a set of partial maps $\text{Par}_{\mathbf{c}}(A,B)$.

Why should $\text{Par}_{\mathbf{C}}(A,B)$ be a set?

It then goes on to say that we can compose a partial map $A\leftarrow D\rightarrow B$ with a map $B\to B'$ in the obvious way (i assume $A\leftarrow D\rightarrow B'$ with $D\to B'$ the composition of $D\to B$ and $B\to B'$) and that we can compose $A\rightharpoonup B$ with a map $A'\to A$ by pulling back the mono $D\to A$ along $A'$. This way we get a functor $\text{Par}_{\mathbf{C}}(-,-):\mathbf{C}^{op}\times\mathbf{C}\to\textbf{Set}$. The paragraph finishes saying: In this way $\text{Par}_{\mathbf{C}}(-,-)$ becomes a profunctor from $\mathbf{C}$ to itself. (In fact, it is the hom-set of another category with the same objects as $\mathbf{C}$).

I guess that proving that $\text{Par}_{\mathbf{C}}(-,-)$ is a functor is not really difficult but I'm having trouble understanding the last part in brackets: does it refer to a functor from a category $\mathbf{C}_{par}$ whose objects are those of $\mathbf{C}$ and whose morphisms are partial maps?

Can someone explain the paragraphs in the nLab with some detail please?

Thanks!

Answer 1

There's a bit of slipperiness here on the n-Lab's part; the key passage is

"Two partial maps are considered equal if they are related by an isomorphism of spans."

$Par_{\bf C}(A,B)$ is not literally a set. However, it is "morally" a set: up to isomorphism, in the appropriate categories there are only set-many monomorphisms into $A$ (and in a partial map $A\leftarrow D\rightarrow B$ the left map has to be a monomorphism). So really $Par_{\bf C}(A,B)$ should be understood either as a set-sized object consisting of isomorphism classes, or as a set of representatives from those classes. The first approach is implicit in the n-Lab page, which in fact abuses equality appropriately as per the quote above. The second approach is rather evil, but is necessary if for whatever reason we are leery of set-sized collections of classes (and are working at a sufficient level of generality that no workaround is satisfactory).

This also addresses your other point: the morphisms of the new category are "partial maps up to partial map equality."

Answer 2

I post this as an extended comment, since nLab only uses the direct hom-set style of representability, which may be making it hard to see why talking about representability isn't circular. Another, more concrete way to state what a partial map classifier is is to define it for any $X$ as an object $\tilde{X}$ (what nLab calls $X_\bot$) and a monomorphism $X\rightarrowtail \tilde{X}$ such that for any partial map $A\overset{a}{\leftarrowtail} A'\overset{f}{\to} X$ there is exactly one $\xi:A\to\tilde{X}$ with $$\require{AMScd}\begin{CD}A' @>a>>A \\ @VfVV @VV\xi V \\ X @>>> \tilde{X} \end{CD}$$ a pullback. Note that I have not had to say anything about the size of the collection of partial maps. If we are in a locally small category, so that there are only a set's worth of morphisms $A\to \tilde{X}$, it's a consequence of the property of our $X\rightarrowtail\tilde{X}$ that there are, up to isomorphism, only a set's worth of partial maps.

Notice that this is the same situation one is in with respect to subobjects and subobject classifiers. The existence of a subobject classifier tells us there's an an isomorphism $\mathrm{Sub}(X)\simeq\mathcal{C}(-,\Omega)$, but there's no reason to think that $\mathrm{Sub}(X)$ is a set. But given the definition of a subobject classifier in terms of unique (pullback) fillers for $$\require{AMScd}\begin{CD} S @>m>> X \\ @VVV \\ 1 @>>\top> \Omega \end{CD}$$ the essential smallness of $\mathrm{Sub}(X)$ follows as long as our category is locally small.

Alternatively, we could always use the natural isomorphism of hom-sets in a larger universe, allowing $\mathrm{Par}_{\mathcal{C}}(A,X)$ to be a (possibly) large set, and then local smallness of $\mathcal{C}$ establishes that it is isomorphic to a small set.