# How to define a 2-category

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.

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This seems like a very strange way to do things – so strange that I can't even say immediately whether it's correct. 2-categories have been known since the 60s, so you can find plenty of (more reasonable!) definitions out there on the internet or in books. – Zhen Lin Mar 15 '12 at 18:10
The problem is that I have to follow the flow of ideas that the author is suggesting to grasp the final idea. – ubugnu Mar 15 '12 at 18:56

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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I have a problem with your explanation for the axiom 3, the 2-morphism you defined is not a bi-gone! Apart from that, what I have understood from your answer is that the 6 cited axioms are not only the requirements that $s^{(2)}$, $t^{(2)}$ and $id^{(2)}$ to be functors between the given categories, am I right? – ubugnu Mar 15 '12 at 19:01
@ubugnu yes you're right: more in detail given a triple of objects $a,b,c \in \mathcal C_0$ you have that the map $\bullet \colon \mathcal C_1(b,c) \times \mathcal C_1(a,b) \to \mathcal C_1(a,c)$ (where $\mathcal C_1(x,y) = \{ f \in \mathcal C_2 | s^1(f)=x, t^1(f)=y\}$) is a functor. – Giorgio Mossa Mar 15 '12 at 19:36
@ubugnu: about the axiom $3$, yeah it was a typo, thanks to have pointed out. – Giorgio Mossa Mar 15 '12 at 19:43
thank you for all these clarifications, so if I understand well, to define properly a 2-category (or even a higher one) we need all inside it to be functors, associative compositions (when possible)? – ubugnu Mar 15 '12 at 20:40
The point is that $n+1$-categories are (weak-)enriched $n$-category: they are objects defined in term of (weak-)$n$-categories, $n$-functors, $n$-natural transformations and so on. All these datas are required to satisfy some axioms. – Giorgio Mossa Mar 15 '12 at 22:45