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Consider the class $ob$ consisting of two distinct objects $x$ and $y$, and the class $mor$ of morphisms consisting of the ordered pairs $(x,x)$, $(y,y)$ and $(x,y)$. I would like to show that these classes constitute a category. I am stuck on what to do for a composition operator.

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The question makes very little sense. What do you mean by "the class mor of morphisms consisting of the ordered pairs (x,x), (y,y), (x,y)"? Are these ordered pairs the arrows themselves, or are you using them to mean the homsets? Are you saying that there are only 3 arrows in the category? – Braindead Nov 4 '12 at 18:38
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The composition operation is part of the definition of a category, so it doesn't make sense to ask whether these classes constitute a category; that will depend on how you define composition.

So I'll take the question to mean: How can we define composition for these morphisms such that the objects, the morphisms and the composition together will constitute a category?

We need identities, and (assuming that your notation is meant to indicate the source and target of the morphisms) there is only one candidate for each identity, so the compositional behaviour of $(x,x)$ and $(y,y)$ is determined by that. The only composition that doesn't include one of those would be $(x,y)\circ(x,y)$, and that doesn't occur because the source and target don't match.

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So, I would have to redefine the problem, explicitly defining the composition operator? It can't somehow be constructed as if these were sets? – Dan Christensen Oct 26 '12 at 20:13
@Dan: I'm not sure what you mean by "as if these were sets", but yes, it can't be constructed, i.e. there can be different categories with the same objects and morphisms. For this to happen there need to be at least two morphisms between the same source and target; otherwise the sources and targets already uniquely determine composition. Perhaps the simplest example is objects $x,y,z$, morphisms $xx,yy,zz,xy,yz,xz_1,xz_2$, where you can freely choose whether $yz\circ xy$ should be $xz_1$ or $xz_2$. – joriki Oct 26 '12 at 20:19

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