TITLE: What is the "correct" category of multisets
QUESTION [18 upvotes]: During seminar the other day, when speaking about subobject classifiers, I asked if the subobject classifier for the category of multisets would have integer truth values, corresponding to the number of times and element is in the set. We attempted to show this, but quickly realized that we were not even sure of the "correct" category for multisets. 
To clarify, when I say correct I want my category to 

Have objects identified by multisets
Have maps between the multisets be on the level of elements in the multiset, and forget the order of those elements, e.g. there is only one map {111223}->{55}
The subobject classifier will behave as I had hoped, with {1} having truth value 3 in {111}

My question is;

Can you construct a category satisfying these properties?

Thanks in advance!
EDIT: First, sorry about not checking nLab, I forget about that site far too often. Second, I should say that I have a little bit of motivation for my property two. So let me clarify what I meant in property two. Given a multiset, it can be thought of as a pair $S\times\mathbb{N}$ for a set $S$. Now, when considering morphisms between multisets I want the maps $f,g:\lbrace 1122\rbrace\rightarrow\lbrace34\rbrace$ such that $f$ sends
$\begin{eqnarray*}
1&\mapsto& 3,\\
1&\mapsto& 4,\\
2&\mapsto& 3,\\
2&\mapsto& 4\\
\end{eqnarray*}$
and $g$ sends
$\begin{eqnarray*}
1&\mapsto& 4,\\
1&\mapsto& 3,\\
2&\mapsto& 4,\\
2&\mapsto& 3\\
\end{eqnarray*}$
to be the same morphism. But if $h$ sends 
$\begin{eqnarray*}
1&\mapsto& 4,\\
1&\mapsto& 4,\\
2&\mapsto& 4,\\
2&\mapsto& 3\\
\end{eqnarray*}$
then $h$ is not the same as $g$ or $f$. Further I would like it such that $\lbrace 112\rbrace$ is not a subobject of $\lbrace 12\rbrace$ but it is a subobject of $\lbrace 11122\rbrace$.
Hopefully this will clear it up.

REPLY [2 votes]: I suspect that the notion of "multiset" is actually ambiguous. For instance, different notions will be appropriate when talking about bosons vs. fermions vs. distinguishable (but intrinsically identical) particles. Ultimately, I suspect that if you really want there to be a reasonable category of multisets, you should assume that the multiple copies of a given element are distinguishable (though identical). So I'd be led to one of the following two candidates for the category of multisets:

The category $Set^\to$ of maps of sets. An object consists of two sets $A,B$ and a function $f: A \to B$; a morphism is a commuting square of maps.

The category $Set^{epi}$ of surjective maps of sets. This is the full subcategory of $Set^\to$ where $f: A \to B$ is required to be surjective.


To be clear, if $A \to B$ is an object of one of these categories, we are thinking of it as a multiset whose elements are the elements $b \in B$, with multiplicity given by the size of the fiber $f^{-1}(b) \subseteq A$. The passage from (1) to (2) is motivated by the idea that if $b \in B$ occurs with multiplicity 0, then we shouldn't think of $b$ as occurring in $B$ at all. The category $Set^{epi}$ of case (2) is equivalent to the category of equivalence relations suggested by Owen Biesel.
In case (1), we are working with the category of presheaves on the arrow category (a Grothendieck topos) and we can calculate the subobject classifier $\Omega$ in the standard way. It turns out that $\Omega$ is the multiset $\{110\}$, with the universal subobject $1 \to \Omega$ picking out one of the copies of $1$.
In case (2), we are dealing with a full subcategory $Set^{epi}$ of case the category $Set^\to$ from case (1), but I'm pretty sure this category of multisets doesn't have pullbacks. Nevertheless, any pullback square in $Set^\to$ whose objects happen to lie in $Set^{epi}$ will still be a pullback in $Set^{epi}$, so there is a non-negligible supply of pullbacks, and it's conceivable there might be a subobject classifier. But $\Omega$ from case (1) is not a subobject classifier -- it classifies subobjects in $Set^\to$, and there are almost always more of these than subobjects in $Set^{epi}$. So I don't think this category has a subobject classifier, though I wouldn't claim to have a proof.