(Apologies to Pirandello for the title.)

I have a couple of quite analogous situations, each featuring candidates for the objects, products, and coproducts of a possible category; missing still are the morphisms1 (including, of course, the identities) that would yield the corresponding bona fide categories.

I will refer to these hypothetical categories as $\mathbf{C}_1$ and $\mathbf{C}_2$.

  1. $\mathrm{ob}(\mathbf{C}_1) = \mathbb{N_0}$, the set of non-negative integers, and the product and coproduct are the usual multiplication and addition in $\mathbb{N}_0$.

  2. $\mathrm{ob}(\mathbf{C}_2) = \mathrm{ob}(\mathbf{Set})$, and the product and coproduct are the usual set-theoretic intersection and (not-necessarily-disjoint) union.

Is it possible to define morphisms for either of these cases that would render it a category? Also, if there's a standard name (as a category) for any of these cases please let me know.

Regarding the first case, I find it suggestive that $\#$ (cardinality2) maps $\mathrm{ob}(\mathbf{FinSet}) \to \mathbb{N}_0$ in such a way that $$\#(A\;\Pi\;B) = \#A \times \#B,$$ and $$\#(A \amalg B) = \#A + \#B.$$ Therefore, if there is indeed a category $\mathbf{C}_1$, it looks like $\#$ would be a good candidate for a functor $\mathbf{FinSet}\to \mathbf{C}_1$. In fact, a suitable extension of $\#$ to the morphisms of $\mathbf{FinSet}$ would provide a pretty good idea of what the morphisms of $\mathbf{C}_1$ would have to look like.

Addendum (in response to Qiaochu Yuan's answer). For example, I can see how the object called3 $\mathbf{18} = \{0,\dots, 17\} \in \mathrm{ob}(\mathbf{C}_1)$ could serve as the "common domain" of two surjections (i.e. "candidate projections") $\mathbf{18} \rightarrowtail \mathbf{3}$ and $\mathbf{18} \rightarrowtail \mathbf{6}$ (where $\mathbf{3} = \{0, \dots, 2\}, \mathbf{6} = \{0, \dots, 5\} \in \mathrm{ob}(\mathbf{C}_1)$), but it's not clear to me how one specifies these surjections in detail so that they constitute a categorical product $\mathbf{3} \; \Pi \; \mathbf{6}$.

(BTW, I think part of my difficulty here is that it's hard for me to know when the phrase "up to isomorphism" is being left out, or what exactly this tacit isomorphism is.)

Addendum 2 OK, I see now: one way to define $\mathbf{18} \rightarrowtail \mathbf{3}$ and $\mathbf{18} \rightarrowtail \mathbf{6}$, respectively, could be $n \mapsto \lfloor n/6\rfloor$ and $n \mapsto n\;\mathrm{mod}\;6$. I had not expected an asymmetric definition, but I guess there's nothing wrong with it.

1I realize that, when it comes to categories, having the objects, products, and coproducts, but not the morphisms, is a bit like having a table set with plates, silverware, napkins, etc., but no food: it may all be suggestive enough to whet one's appetite, but it's still a very long way from the real thing.

2Originally I had used the notation $\mathrm{card}(\dots)$ for cardinality, instead of $\#$, but found that the font was too similar to the regular font, and it would get lost within the surrounding text.

3Originally I had defined $\mathbf{18}$ as $\{1, \dots, 18\}$, but the current definition works better with the projections givein in Addendum 2.

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    $\begingroup$ Actually, for a poset regarded as a category (as the second example turns out to be), having either the products or the coproducts is equivalent to having the morphisms. $\endgroup$ Sep 19, 2015 at 17:55
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    $\begingroup$ Regarding your addendum: you have to pick projections from $18$ to $3$ and to $6$, as there are many possible choices. For instance, send each group of six numbers in order to one of $1,2,3$, and just take the remainder modulo $6$ for the projection to $6$. More generally, pick any bijection of $18$ with the cartesian product of $3$ and $6$-this tells you what the projections will have to be. $\endgroup$ Sep 19, 2015 at 20:50
  • $\begingroup$ @KevinCarlson: Thanks. Just now I was having thoughts in that general direction. $\endgroup$
    – kjo
    Sep 19, 2015 at 21:33

1 Answer 1

  1. Take the category of finite sets and functions. This category is commonly called $\text{FinSet}$. (Strictly speaking, to satisfy your request I should take a skeleton of this category.)

  2. Take the category of sets and inclusions (actually a large poset). I don't know if this category has a name; arguably it's usually not the category you want to work in, and you instead want to work in the category of sets and functions, which is commonly called $\text{Set}$. As far as I'm concerned, there's no such thing as the intersection or union of sets.

  • $\begingroup$ Thanks for your answer! It arrived while I was writing some additional thoughts on the first case (though originally I incorrectly wrote $\mathbf{Set}$ instead of $\mathbf{FinSet}$). Those remarks already suggest that the non-negative integers ($\mathbb{N}_0$) can be viewed as the quotient of $\mathrm{ob}(\mathbf{FinSet})$ relative to $\mathrm{card}$. What I still don't have is a clear idea of how exactly how one extends this "quotienting" $\mathrm{ob}(\mathbf{FinSet}) \to \mathbb{N}_0$ to its morphisms (and thereby define the morphisms between elements of the quotient, $\mathbb{N}_0$). $\endgroup$
    – kjo
    Sep 19, 2015 at 19:11
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    $\begingroup$ @kjo: AFAIK the construction of skeletons cannot be made functorial, and in particular it is not done by quotienting the objects by isomorphism. Instead it's done by picking one object from each isomorphism class. For example, in $\text{FinSet}$ you can pick $\{ 1, 2, ... n \}$ as the representative of the $n$-element set, and then morphisms are given by functions $\{ 1, 2, ... n \} \to \{ 1, 2, ... m \}$. $\endgroup$ Sep 19, 2015 at 19:31
  • $\begingroup$ Would the fact that the construction of skeletons cannot be made functorial imply that in fact no way to define morphisms to get a category from the first case in my question? (See also the addendum I made to my question a few minutes ago.) $\endgroup$
    – kjo
    Sep 19, 2015 at 19:56
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    $\begingroup$ @kjo: the difficulty is not defining the category (I do so above), it's defining the functor from the original category to a skeleton of it. $\endgroup$ Sep 19, 2015 at 21:40
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    $\begingroup$ @kjo: to construct a skeleton, like I said, you just have to pick a representative of each isomorphism class, then take the full subcategory on these representatives. This comes with a functor into, rather than out of, the original category. $\endgroup$ Sep 19, 2015 at 22:59

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