# Category on integers with the usual product and coproduct?

I've just been introduced to category theory. I understand the basic definitions, and I'm trying to get some intuition on how categories tick. I'm wondering: is there a category $C$ such that:

• Its objects are either the integers, the positive integers, or the nonnegative integers (I'm fine with any one of these)
• Its category-theoretic product is the same as the usual product
• Its coproduct is the same as the usual addition?

Intuitively, I think the answer is yes, but I can't come up with a construction.

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You may be interested in the concept of categorification ncatlab.org/nlab/show/vertical+categorification. In my opinion this is one of the most fundamental principals of modern mathematics. You ask for a categorification of the semiring or rig (ncatlab.org/nlab/show/rig) of natural numbers. The answer is the 2-rig of finite sets (ncatlab.org/nlab/show/2-rig). – Martin Brandenburg Aug 26 '13 at 12:44

For the non-negative integers, take the category of finite sets. (The exponential here is also the same as the usual exponential.)

For the integers, the answer is no. More generally, the category-theoretic coproduct never has nontrivial inverses. See this math.SE answer.

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For the category of finite sets, in order to translate it to the non-negative integers, do we label a set with its size? If so, doesn't that category have more than one "1", more than one "2", etc...? If I naively identify every set with the same size to be the same object, I don't think this still works. (And I have to add a lot more morphisms.) But I could be wrong. – Lopsy Aug 26 '13 at 1:02
Oh - when it was presented to me, I heard that the category of groups doesn't identify isomorphic objects with each other; like, the group {0,1,2,3} under addition mod 4 and the group {1,i,-1,-i} under multiplication are different objects, even though they're isomorphic. Is this a normal thing to do in category theory? If so, I like your finite sets construction. – Lopsy Aug 26 '13 at 1:04
I suppose technically we want the skeleton of $\sf Set$, but having extra isomorphic copies of objects is not such a big deal. – anon Aug 26 '13 at 1:07
@Lopsy: yes, we take isomorphism classes. – Qiaochu Yuan Aug 26 '13 at 1:10
@Lopsy: or, as anon says, we can take the skeleton of $\text{FinSet}$, which can be taken to have objects the non-negative integers. But as a category theorist I freely identify equivalent categories in general. – Qiaochu Yuan Aug 26 '13 at 4:01