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I found this definition in http://arxiv.org/abs/hep-th/0304074 (pp.13) for 2-categories: It is a collection $\mathcal{C}_0$ of objects, $\mathcal{C}_1$ of morphisms, and $\mathcal{C}_2$ of 2-morphisms such that:

  1. $(\mathcal{C}_{0},\mathcal{C}_{1},s^{(1)},t^{(1)},id^{(1)},\bullet) $ is a small category

  2. $ (\mathcal{C}_{1},\mathcal{C}_{2},s^{(2)},t^{(2)},id^{(2)},\circ) $ is a small category

  3. $ t^{(1)}(s^{(2)}(f))=t^{(1)}(t^{(2)}(f)) $ and $ s^{(1)}(s^{(2)}(f))=s^{(1)}(t^{(2)}(f)) $ meaning that 2-morphisms are bi-gones
  4. A map $\bullet:\mathcal{C}_{2}\times_{\mathcal{C}_{0}}\mathcal{C}_{2}\rightarrow\mathcal{C}_{2} $ of horizonal composition between 2-morphisms (the vertical one is already defined by item 2.)

Further axioms which the author says are in fact the requirement for the maps $s^{(2)}$, $t^{(2)}$ and $id^{(2)}$ to induce functors from the category $ (\mathcal{C}_{0},\mathcal{C}_{2},s^{(1)}\circ s^{(2)},t^{(1)}\circ t^{(2)},id^{(1)}\circ id^{(2)},\bullet) $ to the category $(\mathcal{C}_{0},\mathcal{C}_{1},s^{(1)},t^{(1)},\bullet) $, these axioms are:

  1. $ s^{(2)}(f_{2} \bullet f_{1})=s^{(2)}(f_{2})\bullet s^{(2)}(f_{1}) $
  2. $ t^{(2)}(f_{2}\bullet f_{1})=t^{(2)}(f_{2})\bullet t^{(2)}(f_{1}) $
  3. $ id^{(2)}(id^{(1)}(s^{(1)}(s^{(2)}(f))))\bullet f=f=f\bullet id^{(2)}(id^{(1)}(t^{(1)}(t^{(2)}(f)))) $
  4. $ (f_{1}\bullet f_{2})\bullet f_{3}=f_{1}\bullet(f_{2}\bullet f_{3}) $
  5. $ id^{(2)}(g_{1})\bullet id^{(2)}(g_{2})=id^{(2)}(g_{1}\bullet g_{2}) $
  6. (f_1 \circ f'_1) \bullet (f_2 \circ f'_2) = (f_1 \bullet f_2) \circ (f'_1 \bullet f'_2)

As I see it, items 1., 2. and 5. are suffisiant to ensure that $s^{(2)}$, $t^{(2)}$ and $id^{(2)}$ induce the required functors, so my questions are:

  1. What do the remaining axioms establish?
  2. Why do we construct a category in this way, i.e: requiring $(\mathcal{C}_{0},\mathcal{C}_{1},s^{(1)},t^{(1)},id^{(1)},\bullet) $ and $ (\mathcal{C}_{1},\mathcal{C}_{2},s^{(2)},t^{(2)},id^{(2)},\circ) $ to be small categories and requiring certain functors from $ (\mathcal{C}_{0},\mathcal{C}_{2},s^{(1)}\circ s^{(2)},t^{(1)}\circ t^{(2)},id^{(1)}\circ id^{(2)},\bullet)$ to $(\mathcal{C}_{0},\mathcal{C}_{1},s^{(1)},t^{(1)},\bullet) $
  3. Is there another way to construct a 2-category?

Note that $s^{(i)}$, $t^{(i)}$ and $id^{(i)}$ are the source, target and identity maps for the given category.

  • 0
    The problem is that I have to follow the flow of ideas that the author is suggesting to grasp the final idea.2012-03-15

1 Answers 1

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First of all I guess you meant something

6) (f_1 \circ f'_1) \bullet (f_2 \circ f'_2) = (f_1 \bullet f_2) \circ (f'_1 \bullet f'_2)

This condition implies that the maps $\bullet \colon \mathcal{C}_2 \times_{\mathcal C_1} \mathcal C_2 \to \mathcal C_2$ and $\bullet \colon \mathcal C_1 \times_{\mathcal C_0} \mathcal C_1 \to \mathcal C_1$ give as a functor of kind

$(\mathcal C_1,\mathcal C_2,s^{(2)},t^{(2)},\circ^2,id^{(2)}) \times (\mathcal C_1,\mathcal C_2,s^{(2)},t^{(2)},\circ^2,id^{(2)}) \to (\mathcal C_1,\mathcal C_2,s^{(2)},t^{(2)},\circ^2,id^{(2)})$

This property cannot be deduced by the axioms 1,2 and 5 which simply gives to you information about source and target of horizontal composites.

Similarly the axiom 4 says that horizontal composition is associative, which cannot be deduced from the axioms 1,2 and 5.

The 3 axiom says that given f \colon f_1 \to f'_1 in $\mathcal C_2$ such that $f_1 \colon a \to b$ and f'_1 \colon a \to b in $\mathcal C_1$ the equality

$id^{2}( id^{1}(a)) \bullet f = f = f \bullet id^{2}( id^1 (b))$ so gives you information about the horizontal identities.

Note: the fact that $f_1$ and $f_2$ have the same sources and target follow from the condition $3$ in the begging of the question.

To address the last two questions: generally we use more compact definition, like the one said in the article you have linked above

a $2$-category is an enriched category in $\mathbf{Cat}$

the definition you gave is just an expansion of the definition above.

There are many other definition of $2$-category, more exactly there is one definition for every definition of $\infty$-category: if you interested in seeing such definition I suggest you to read the nlab where you can find more reference and link about.

In particular this definition of $2$-category can be modify to get a notion of weak-$2$-category, a.k.a. as bicategory.

Hope this help.

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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/2796/discussion-between-ineff-and-ubugnu)2012-03-16