I'm just beginning to learn category theory. So far, the basic examples (like Set) are making sense. But I'm having a little trouble getting my head around the fundamentals.

Suppose I try to define a category with objects A, B, and C. There's an arrow (f) from A to B, an arrow (g) from B to C, and two arrows (h1 and h2) from A to C (in addition to the identity arrows).

Now it must obey the composition axiom:

For any two arrows f : A → B, g : B → C, where src(g) = tar (f), there exists an arrow g ◦ f : A → C, ‘g following f’, which we call the composite of f with g.

It seems we have two choices for g ◦ f. Since we're not trying to give the category any meaning, it shouldn't matter which we pick. But does the choice matter in identifying the category? Does picking h1 give a "different" category than picking h2? If so, shouldn't it be part of the category's definition? If not, then is it meaningless to ask "which one is the composite?" Perhaps another way of asking this is: does composition have to mean any (one) thing?

My question probably doesn't make a lot of sense, but hopefully there are enough clues in here for someone to correct my confusion.

Edit: Perhaps restating the axiom in these different ways clarifies my confusion.

a) For any two arrows f : A → B, g : B → C, where src(g) = tar (f), there exists at least one arrow g ◦ f : A → C, ‘g following f’, which we could call the composite of f with g.

b) For any two arrows f : A → B, g : B → C, where src(g) = tar (f), there exists an arrow g ◦ f : A → C, ‘g following f’, that is the composite of f with g.

  • 2
    $\begingroup$ Yes, you have two choices. The resulting categories are "different" (in the sense that you've picked two different composition laws on the same sets of objects and morphisms), but isomorphic (via the isomorphism that switches $h_1$ and $h_2$). This is analogous to how you can write down different but isomorphic group structures on the same set. $\endgroup$ May 31, 2016 at 0:16
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    $\begingroup$ Oh, I misunderstood your question; thanks for the edit. The correct interpretation is b): composition is part of the data of a category (the most important part), and in particular $g \circ f$ is the name of a unique morphism. $\endgroup$ May 31, 2016 at 0:22
  • $\begingroup$ Thanks @QiaochuYuan! Both of your comments seem to answer my question (though the fact that you think you misunderstood the first time means that I'm probably missing something else subtle...). The book I'm reading says "The objects and arrows of a category are very often called the category’s data." Perhaps it is simply obvious from context that the composition decisions are, too. $\endgroup$
    – A_P
    May 31, 2016 at 1:06
  • $\begingroup$ Some authors use "data" to refer to things like the objects and morphisms and "structure" to refer to things like the composition. I use "data" to refer to both (that is, to the objects, and the morphisms, and the composition). (The point of this distinction is mostly to distinguish between data, or data + structures, from "properties," which are axioms that the data, or data + structure, satisfy.) $\endgroup$ May 31, 2016 at 4:12

1 Answer 1


To define a category, you have to specify what composition is in that category. It's like the multiplication operation in a group: to define a group, it's not enough to just say you have a set and it is possible to multiply elements of the set; you have to actually say what you mean by "multiply" as part of the definition of the group.

So you might define one operation $\circ_1$ for which $g\circ_1 f=h_1$, and a different operation $\circ_2$ for which $g\circ_2 f=h_2$, and these define two different categories with the same sets of objects and morphisms. (As Qiaochu commented though, these two categories are isomorphic, by just switching $h_1$ and $h_2$.)

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    $\begingroup$ +1. To drive the point home: even if there's only one possible group structure on the set you care about (up to isomorphism), you still have to specify it! If you tell me "$G=\{0, 1\}$," you haven't told me a group: you have to tell me exactly how multiplication works in $G$ (even though, of course, the result won't matter up to isomorphism). $\endgroup$ May 31, 2016 at 2:01

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