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:
$(\mathcal{C}_{0},\mathcal{C}_{1},s^{(1)},t^{(1)},id^{(1)},\bullet) $ is a small category
$ (\mathcal{C}_{1},\mathcal{C}_{2},s^{(2)},t^{(2)},id^{(2)},\circ) $ is a small category
- $ 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
- 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:
- $ s^{(2)}(f_{2} \bullet f_{1})=s^{(2)}(f_{2})\bullet s^{(2)}(f_{1}) $
- $ t^{(2)}(f_{2}\bullet f_{1})=t^{(2)}(f_{2})\bullet t^{(2)}(f_{1}) $
- $ id^{(2)}(id^{(1)}(s^{(1)}(s^{(2)}(f))))\bullet f=f=f\bullet id^{(2)}(id^{(1)}(t^{(1)}(t^{(2)}(f)))) $
- $ (f_{1}\bullet f_{2})\bullet f_{3}=f_{1}\bullet(f_{2}\bullet f_{3}) $
- $ id^{(2)}(g_{1})\bullet id^{(2)}(g_{2})=id^{(2)}(g_{1}\bullet g_{2}) $
- $ (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:
- What do the remaining axioms establish?
- 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) $
- 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.
First of all I guess you meant something
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
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.