In this section we discuss some formal properties of the $2$-category of categories. This will lead us to the definition of a (strict) $2$-category later.
Let us denote $\mathop{\mathrm{Ob}}\nolimits (\textit{Cat})$ the class of all categories. For every pair of categories $\mathcal{A}, \mathcal{B} \in \mathop{\mathrm{Ob}}\nolimits (\textit{Cat})$ we have the “small” category of functors $\text{Fun}(\mathcal{A}, \mathcal{B})$. Composition of transformation of functors such as
is called vertical composition. We will use the usual symbol $\circ $ for this. Next, we will define horizontal composition. In order to do this we explain a bit more of the structure at hand.
Namely for every triple of categories $\mathcal{A}$, $\mathcal{B}$, and $\mathcal{C}$ there is a composition law
coming from composition of functors. This composition law is associative, and identity functors act as units. In other words – forgetting about transformations of functors – we see that $\textit{Cat}$ forms a category. How does this structure interact with the morphisms between functors?
Well, given $t : F \to F'$ a transformation of functors $F, F' : \mathcal{A} \to \mathcal{B}$ and a functor $G : \mathcal{B} \to \mathcal{C}$ we can define a transformation of functors $G\circ F \to G \circ F'$. We will denote this transformation ${}_ Gt$. It is given by the formula $({}_ Gt)_ x = G(t_ x) : G(F(x)) \to G(F'(x))$ for all $x \in \mathcal{A}$. In this way composition with $G$ becomes a functor
To see this you just have to check that ${}_ G(\text{id}_ F) = \text{id}_{G \circ F}$ and that ${}_ G(t_1 \circ t_2) = {}_ Gt_1 \circ {}_ Gt_2$. Of course we also have that ${}_{\text{id}_\mathcal {A}}t = t$.
Similarly, given $s : G \to G'$ a transformation of functors $G, G' : \mathcal{B} \to \mathcal{C}$ and $F : \mathcal{A} \to \mathcal{B}$ a functor we can define $s_ F$ to be the transformation of functors $G\circ F \to G' \circ F$ given by $(s_ F)_ x = s_{F(x)} : G(F(x)) \to G'(F(x))$ for all $x \in \mathcal{A}$. In this way composition with $F$ becomes a functor
To see this you just have to check that $(\text{id}_ G)_ F = \text{id}_{G\circ F}$ and that $(s_1 \circ s_2)_ F = s_{1, F} \circ s_{2, F}$. Of course we also have that $s_{\text{id}_\mathcal {B}} = s$.
These constructions satisfy the additional properties
whenever these make sense. Finally, given functors $F, F' : \mathcal{A} \to \mathcal{B}$, and $G, G' : \mathcal{B} \to \mathcal{C}$ and transformations $t : F \to F'$, and $s : G \to G'$ the following diagram is commutative
in other words ${}_{G'}t \circ s_ F = s_{F'}\circ {}_ Gt$. To prove this we just consider what happens on any object $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$:
which is commutative because $s$ is a transformation of functors. This compatibility relation allows us to define horizontal composition.
Definition 4.28.1. Given a diagram as in the left hand side of: we define the horizontal composition $s \star t$ to be the transformation of functors ${}_{G'}t \circ s_ F = s_{F'}\circ {}_ Gt$.
\[ \xymatrix{ \mathcal{A} \rtwocell ^ F_{F'}{t} & \mathcal{B} \rtwocell ^ G_{G'}{s} & \mathcal{C} } \text{ gives } \xymatrix{ \mathcal{A} \rrtwocell ^{G \circ F} _{G' \circ F'}{\ \ s \star t} & & \mathcal{C} } \]
Now we see that we may recover our previously constructed transformations ${}_ Gt$ and $s_ F$ as $ {}_ Gt = \text{id}_ G \star t $ and $ s_ F = s \star \text{id}_ F $. Furthermore, all of the rules we found above are consequences of the properties stated in the lemma that follows.
Lemma 4.28.2. The horizontal and vertical compositions have the following properties
$\circ $ and $\star $ are associative,
the identity transformations $\text{id}_ F$ are units for $\circ $,
the identity transformations of the identity functors $\text{id}_{\text{id}_\mathcal {A}}$ are units for $\star $ and $\circ $, and
given a diagram
we have $ (s' \circ s) \star (t' \circ t) = (s' \star t') \circ (s \star t)$.
\[ \xymatrix{ \mathcal{A} \rruppertwocell ^ F{t} \ar[rr]_(.3){F'} \rrlowertwocell _{F''}{t'} & & \mathcal{B} \rruppertwocell ^ G{s} \ar[rr]_(.3){G'} \rrlowertwocell _{G''}{s'} & & \mathcal{C} } \]
Proof. The last statement turns using our previous notation into the following equation
According to our result above applied to the middle composition we may rewrite the left hand side as $ s'_{F''} \circ s_{F''} \circ {}_ Gt' \circ {}_ Gt $ which is easily shown to be equal to the right hand side. $\square$
Another way of formulating condition (4) of the lemma is that composition of functors and horizontal composition of transformation of functors gives rise to a functor
whose source is the product category, see Definition 4.2.20.
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