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Let object of the category be integers and for objects $a$ and $b$, define: $$ \operatorname{hom}\left(a, b\right) = \left\{ f \in \mathbb{Z} | a - f = b\right\} $$ Then composition of $f \in \operatorname{hom} \left(a,b\right)$ and $g \in \operatorname{hom} \left(b,c\right)$ is addition which is associative. Also for all integers $x$, $ \exists 0 \in \operatorname{hom}\left(x,x\right)$ which acts like an identity, but then obviously all objects have the same identity. Does this prevent us from considering $\mathbb{Z}$ with subtraction as a category?

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Yes, by (my usual) definition of category one should be able to recover $a$ and $b$ from $f\in \operatorname{hom}(a,b)$. But iIf one repairs this problem for your category it becomes simply a category with countably many objects and a unique morphism $a\to b$ for any pair of objects. – Hagen von Eitzen Oct 6 '12 at 11:53
It is rather curious to have the same values (integers) both as objects and as morphisms. And it is irrelevant what morphisms are called (the only point of your definition is that all homsets are singletons), but you cannot freely call two different morphisms by the same name (for general reasons that have little to do with category theory). Some violation of this rule is accepted (for zero morphisms in an Abelian category for instance) as long as it is understood that having the same name does not imply being the same thing. – Marc van Leeuwen Oct 6 '12 at 13:28

I suggest to change to $\operatorname{hom}(a, b) = \{ f \in \mathbb{Z} | a=b+f\}$, this will facilitate generalization.

then obviously all objects have the same identity. Does this prevent us from considering Z with subtraction as a category?

It depends on what formal definition of a category you choose. If there it is required that each morphism has a unique source object and a unique target object, then we define a category in the other way: a morphism is $(a, f, b)$ such that $a=b+f$.

You can generalize this definition to any monoid $M$ and further to any action $T$ of a monoid $M$ on a set $X$. Objects are $X$, the condition for $f$ to be a morphism is $a=T(f)(b)$. It is called the transport category of $T$.

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No. One can define a category on vertex set $X$ using $\hom:X\times X\to Set$ and associative composition, without the assumption that $\hom(a,b)$ and $\hom(x,y)$ need be disjoint..

However, the category you obtain, as Hagen also wrote, contains exactly one arrow $a\to b$ for any pairs $(a,b)$, and hence every arrow is invertible (i.e., it is a grupoid).

The same construction applies on any set H: let $\hom(x,y):=\{(x,y)\}$, i.e. consider the full graph (with loops) on $H$..

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In Categories, Allegories, Freyd and Scedrov give a way of formally presenting a category that addresses this.

The idea is to present the category in terms of objects and proto-morphisms, and a ternary relation of the form "the proto-morphism $f$ can be construed as going from $A$ to $B$". So the arrows of a category so presented is the graph of this relation -- i.e. the class of all triples $(A,f,B)$ that satisfy the ternary relation.

Furthermore, composition is phrased in terms of proto-morphisms, rather than morphisms. (it is possible but rare for composition of proto-morphisms to be multi-valued -- but it has to be single valued, of course, in the context of the product of actual arrows)

Freyd, Scedrov then list a bunch of axioms that I won't list here; they're just the category axioms stated in these terms in a diagrammatic language.

The point is, $4$ and $7$ in your example are proto-morphisms; we can see that $4+7=11$, but we haven't actually talked about arrows yet. However, at our leisure we can consider $4$ as a morphism from $13$ to $17$, and then we're talking about an actual arrow.

This sort of thing is only important in an arrow-oriented formulation of categories. In an object-and-homset formulation, it doesn't matter if multiple homsets overlap; it's perfectly fine for the identity map to be the same for every object. Heck, it's fine if the entire category only has a single arrow!

An example of why this is fine is composition: it is not a (partial) operation on arrows: it is a family of functions $\circ_{xyz} : \hom(y,z) \times \hom(x,y) \to \hom(x,z)$. The math doesn't get "confused" by arrows in different homsets being equal, because by using $f \circ_{xyz} g$, we "remember" that $f$ goes from $y$ to $z$ and $g$ goes from $x$ to $y$.

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