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Viewing a 2-category as a category enriched in $\mathsf{Cat}$, I can see from where comes the vertical composition: morphisms of a 2-category are objects of $\mathsf{Cat}$ and 2-morphisms of this 2-category are morphisms of $\mathsf{Cat}$, so they can be (vertically) composed if they're composable. vertical composition

What I don't see, is from where the horizontal composition comes, in the diagram enter image description here

we can get the composite $g_1 g_2$ and $g'_1 g'_2$, these will be in turn also objects of $\mathsf{Cat}$ and may eventually have no morphism between them! The horizontal composition $\alpha_1 \circ \alpha_2$ make no sens in this case.

So my question: can we always define the horizontal composition in a 2-category for all 2-morphisms?

EDIT To make my question more clear, is it suffisant to say that each "hom-collection" must be a category to define the horizontal composition of 2-morphisms?

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Regarding the edit: no, it is not sufficient, if by "hom collection" you mean each local hom-object $\hom(a, b)$. Maybe your difficulty is notational? There is the vertical composition $\circ$ in each hom-category, and there is the horizontal composition which could be denoted $\bullet$. Given $\alpha_1, \alpha_2$ as in your post, the pair $(\alpha_1, \alpha_2)$ is a morphism in $\hom(b,c)\times\hom(a,c)$. Then define $\alpha_1 \bullet\alpha_2$ to be the value $\text{comp}(a, b, c)(\alpha_1, \alpha_2)$ for the composition functor $\text{comp}(a, b, c): \hom(b, c)\times\hom(a, b)\to \hom(a,c)$. – user43208 Sep 21 '13 at 22:14
up vote 3 down vote accepted

It comes about by applying the composition functor (composition being an arrow in $\text{Cat}$),

$$\hom(b, c) \times \hom(a, b) \to \hom(a, c)$$

to a pair of morphisms, one in the hom-category $\hom(b, c)$ (where we have a morphism $\alpha_1$ between objects $g_1$, $g_1'$) and the other in the hom-category $\hom(a, b)$ (where we have a morphism $\alpha_2$ between objects $g_2$, $g_2'$).

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The point is that in my mind this is the same composition used vertically, and it still need the composability condition, where am i wrong? – ubugnu Sep 21 '13 at 17:36
@ubugnu The vertical composition takes place within a single hom-category; e.g., if $g$, $g'$, $g''$ are objects in $\hom(a, b)$ and $\alpha': g' \to g''$ and $\alpha: g \to g'$ are morphisms in $\hom(a, b)$; here the composition is across a 1-morphism $g'$. The horizontal composition as I described it above involves several hom-categories, and notice that is composition across a 0-cell $b$. I'm not quite sure what you mean by "the composability condition"; two 2-cells are horizontally composable if they have a 0-cell $b$ between them. – user43208 Sep 21 '13 at 18:28
Okey, I understand the difference but not yet the origin of the $\hom(b, c) \times \hom(a, b) \to \hom(a, c)$ functor. – ubugnu Sep 21 '13 at 18:35
Such composition functors are part of the data of a category enriched in the category $\text{Cat}$. See, and apply this to the particular situation when the monoidal category is $\text{Cat}$ with the monoidal product given by cartesian product. – user43208 Sep 21 '13 at 19:21
As both answers (yours and the one from @Giorgio Mossa) have helped me a lot to correct my conception of enriched categories, I'll mark as "Accepted" that of the one who confirm that this is correct: the composition of 1-morphisms in a 2-category is given by the object function of the (being part of the definition) functors $\circ\colon \mathbf C(Y,Z) \times \mathbf C(X,Y) \to \mathbf C(X,Z)$ (far from me to make it a competition, it's just to pick one :-)) – ubugnu Sep 23 '13 at 17:53

As you says $2$-categories are just categories enriched in $\mathbf{Cat}$.

This means that a 2-category is given by the following data:

  • a set $\mathbf C$ of the objects of the $2$-category;
  • for each pair $X,Y \in \mathbf C$ a category $\mathbf C(X,Y)$ (i.e. an object in $\mathbf {Cat}$);
  • for each triple $X,Y,Z \in \mathbf C$ a morphism in $\mathbf {Cat}$ (i.e. a functor) $$\circ \colon \mathbf C(Y,Z) \times \mathbf C(X,Y) \to \mathbf C(X,Z)$$
  • for every object $X \in \mathbf C$ a functor $I \to \mathbf C(X,X)$, where $I$ is the category with one object and so it correspond to an object of the category $\mathbf C(X,X)$.

these data are subjected to the usual axioms of assiociativity and unit.

So the morphisms of the $2$-category $\mathbf C$ are not object in $\mathbf {Cat}$, but are the objects of the categories $\mathbf C(X,Y)$, where $X$ and $Y$ ranges over $\mathbf C$. Similarly $2$-morphisms are not morphisms of $\mathbf {Cat}$ (i.e. functors) but they are morphisms in the categories $\mathbf C(X,Y)$.

Vertical composition for $2$-cells is the operation obtained by the compositions of the categories $\mathbf C(X,Y)$. More explicitly for every pair of objects $X$ and $Y$ in $\mathbf C$, a triple $f,g,h \in \mathbf C(X,Y)$ and two $2$-cells $\alpha \colon f \Rightarrow g$ and $\beta \colon g \Rightarrow h$, the vertical composite $\beta \circ \alpha \colon f \Rightarrow h$ is the composite of $\alpha$ and $\beta$ in the category $\mathbf C(X,Y)$.

Horizontal composition (as stated by user43208 above) is given by the arrow function of the functors

$$\circ\colon \mathbf C(Y,Z) \times \mathbf C(X,Y) \to \mathbf C(X,Z)$$

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Are the functors $\circ\colon \mathbf C(Y,Z) \times \mathbf C(X,Y) \to \mathbf C(X,Z)$ canonically defined like the composition of morphisms inside a single $\mathbf C(X,Y)$? – ubugnu Sep 21 '13 at 19:10
@ubugnu They are part of the data of an enriched category, in the same way that the multiplication operation $G \times G \to G$ is part of the data of a given group $G$. – user43208 Sep 21 '13 at 19:24
@ubugnu is exactly as user43208 says above. $2$-categories can be thought as a family of categories indexed by pair of objects that acts between them through the horizontal composition. – Giorgio Mossa Sep 21 '13 at 19:33
I know I abuse your generosity, things were mixed in my head! I'll say one last sentence just answer by "true" or "false" and let me clear it all alone in my head: "horizontal composition of 2-morphisms is the arrow function of the tensorial product of the monoidal category $\mathsf{Cat}$" – ubugnu Sep 21 '13 at 19:52
False (I could say more, but since you asked...) – Giorgio Mossa Sep 21 '13 at 19:58

This picture might help:

$$\begin{matrix} \star &\xleftarrow{g_1} & \star & \xleftarrow{g_2} & \star \\ ||& \mathbf{1}_{g_1} & || & \alpha_2 & || \\ \star &\xleftarrow{g_1} & \star & \xleftarrow{g_2'} & \star \\ ||& \alpha_1 & || & \mathbf{1}_{f_2} & || \\ \star &\xleftarrow{g_1'} & \star & \xleftarrow{g_2'} & \star \end{matrix}$$

So it's enough to understand horizontal composition between a 1-morphism and a 2-morphism: e.g. $g_1 \circ \alpha_2$.

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