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In beginning to learn category theory, I've tried to come up with my own examples of categories.

Could we have a category Int, where the objects are integers, and arrows, all mappings between integers?

If you can't (which I suspect might be the case), what can't be an object/arrow?

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    $\begingroup$ What do you mean by a mapping between integers? $\endgroup$ – lulu Aug 10 '17 at 19:57
  • $\begingroup$ @lulu : well if you see an integee as a finite ordinal (which is the most common definition if I'm not mistaken), "mappings between integers" makes perfetc sense $\endgroup$ – Maxime Ramzi Aug 10 '17 at 20:38
  • $\begingroup$ @lulu; that's actually part of my problem. For Set, arrows are functions with the objects as (co)domains, meaning that, for $f : A \rightarrow B$ and $g : B \rightarrow A$, $f g$ is not necessarily equivalent to $1_A$. However, for a category of integers, you can't have an integer as a (co)domain for a function (can you?). I guess I'm asking, fundamentally, how do objects and arrows interact – what is their relationship? $\endgroup$ – digitalis_ Aug 10 '17 at 20:55
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    $\begingroup$ Sure you can, you just need to define what you mean. Order defines arrows in a nice way (as the solution from @jgon points out). $\endgroup$ – lulu Aug 10 '17 at 20:59
  • $\begingroup$ @Max That would make sense for natural numbers, but not for negative integers. $\endgroup$ – Arnaud D. Aug 11 '17 at 7:57
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Sure you can. You just have to decide what the mappings between the integers are, and what composition is. For example, one can define the poset category on $\Bbb{Z}$ whose objects are the integers, and has a unique arrow $n\to m$ whenever $n\le m$. You can check that there is a composition defined by transitivity of $\le$ that makes this setup into a category.

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    $\begingroup$ To elaborate, the relation should be a preorder, i.e. reflexive and transitive. $\endgroup$ – Chris Culter Aug 10 '17 at 20:02
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    $\begingroup$ Or, you can take divisibility as the base relation. Then isomorphism in this category will be $x\cong \pm x$, products in this category will be greatest common divisor, and coproducts the lowest common multiple. ..... Or, you can take all the functions as your arrows between sets of form $\{0,\dots,n-1\}$. $\endgroup$ – Berci Aug 10 '17 at 21:17
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    $\begingroup$ Oh I meant the usual less than or equal to relation, but yeah other appropriate relations work too $\endgroup$ – jgon Aug 10 '17 at 21:32
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    $\begingroup$ Thanks! Btw, just to clarify, what is that arrow $n → m$? (My guess would be that it's not a function – something like the pair $(n,m)$.) $\endgroup$ – digitalis_ Aug 13 '17 at 13:38
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    $\begingroup$ @digitalis_ Arrows don't have to be anything more than formal arrows. Ok, I know that maybe doesn't clarify anything yet. What I mean is that as long as we have a set $\operatorname{Hom}(A,B)$ for each pair of objects $A,B$ in our category, and a composition law that takes $\operatorname{Hom}(B,C)\times \operatorname{Hom}(A,B)$ to $\operatorname{Hom}(A,C)$, the elements of the hom-sets can be anything we want. They don't have to have any other meaning beyond that imposed by the category structure. $\endgroup$ – jgon Aug 15 '17 at 3:18

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