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Tag 003D

4.27. Formal properties

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 $\mathop{\rm Ob}\nolimits(\textit{Cat})$ the class of all categories. For every pair of categories $\mathcal{A}, \mathcal{B} \in \mathop{\rm Ob}\nolimits(\textit{Cat})$ we have the ''small'' category of functors $\text{Fun}(\mathcal{A}, \mathcal{B})$. Composition of transformation of functors such as $$ \xymatrix{ \mathcal{A} \rruppertwocell^{F''}{t'} \ar[rr]_(.3){F'} \rrlowertwocell_F{t} & & \mathcal{B} } \text{ composes to } \xymatrix{ \mathcal{A} \rrtwocell^{F''}_F{~~t \circ t'} & & \mathcal{B} } $$ 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 $$ \circ : \mathop{\rm Ob}\nolimits(\text{Fun}(\mathcal{B}, \mathcal{C})) \times \mathop{\rm Ob}\nolimits(\text{Fun}(\mathcal{A}, \mathcal{B})) \longrightarrow \mathop{\rm Ob}\nolimits(\text{Fun}(\mathcal{A}, \mathcal{C})) $$ 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 $$ \text{Fun}(\mathcal{A}, \mathcal{B}) \longrightarrow \text{Fun}(\mathcal{A}, \mathcal{C}). $$ 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 $$ \text{Fun}(\mathcal{B}, \mathcal{C}) \longrightarrow \text{Fun}(\mathcal{A}, \mathcal{C}). $$ 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 $$ {}_{G_1}({}_{G_2}t) = {}_{G_1\circ G_2}t, ~(s_{F_1})_{F_2} = s_{F_1 \circ F_2}, \text{ and }{}_H(s_F) = ({}_Hs)_F $$ 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 $$ \xymatrix{ G \circ F \ar[r]^{{}_Gt} \ar[d]_{s_F} & G \circ F' \ar[d]^{s_{F'}} \\ G' \circ F \ar[r]_{{}_{G'}t} & G' \circ F' } $$ in other words ${}_{G'}t \circ s_F = s_{F'}\circ {}_Gt$. To prove this we just consider what happens on any object $x \in \mathop{\rm Ob}\nolimits(\mathcal{A})$: $$ \xymatrix{ G(F(x)) \ar[r]^{G(t_x)} \ar[d]_{s_{F(x)}} & G(F'(x)) \ar[d]^{s_{F'(x)}} \\ G'(F(x)) \ar[r]_{G'(t_x)} & G'(F'(x)) } $$ which is commutative because $s$ is a transformation of functors. This compatibility relation allows us to define horizontal composition.

Definition 4.27.1. Given a diagram as in the left hand side of: $$ \xymatrix{ \mathcal{A} \rtwocell^F_{F'}{t} & \mathcal{B} \rtwocell^G_{G'}{s} & \mathcal{C} } \text{ gives } \xymatrix{ \mathcal{A} \rrtwocell^{G \circ F} _{G' \circ F'}{~~s \star t} & & \mathcal{C} } $$ we define the horizontal composition $s \star t$ to be the transformation of functors ${}_{G'}t \circ s_F = s_{F'}\circ {}_Gt$.

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 4.27.2. The horizontal and vertical compositions have the following properties

  1. $\circ$ and $\star$ are associative,
  2. the identity transformations $\text{id}_F$ are units for $\circ$,
  3. the identity transformations of the identity functors $\text{id}_{\text{id}_\mathcal{A}}$ are units for $\star$ and $\circ$, and
  4. given a diagram $$ \xymatrix{ \mathcal{A} \rruppertwocell^F{t} \ar[rr]_(.3){F'} \rrlowertwocell_{F''}{t'} & & \mathcal{B} \rruppertwocell^G{s} \ar[rr]_(.3){G'} \rrlowertwocell_{G''}{s'} & & \mathcal{C} } $$ we have $ (s' \circ s) \star (t' \circ t) = (s' \star t') \circ (s \star t)$.

Proof. The last statement turns using our previous notation into the following equation $$ s'_{F''} \circ {}_{G'}t' \circ s_{F'} \circ {}_Gt = (s' \circ s)_{F''} \circ {}_G(t' \circ t). $$ 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 $$ (\circ, \star) : \text{Fun}(\mathcal{B}, \mathcal{C}) \times \text{Fun}(\mathcal{A}, \mathcal{B}) \longrightarrow \text{Fun}(\mathcal{A}, \mathcal{C}) $$ whose source is the product category, see Definition 4.2.20.

    The code snippet corresponding to this tag is a part of the file categories.tex and is located in lines 4392–4633 (see updates for more information).

    \section{Formal properties}
    \label{section-formal-cat-cat}
    
    \noindent
    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.
    
    \medskip\noindent
    Let us denote $\Ob(\textit{Cat})$ the class of all categories.
    For every pair of categories
    $\mathcal{A}, \mathcal{B} \in \Ob(\textit{Cat})$
    we have the ``small'' category of functors
    $\text{Fun}(\mathcal{A}, \mathcal{B})$.
    Composition of transformation of functors such as
    $$
    \xymatrix{
    \mathcal{A}
    \rruppertwocell^{F''}{t'}
    \ar[rr]_(.3){F'}
    \rrlowertwocell_F{t}
    & &
    \mathcal{B}
    }
    \text{ composes to }
    \xymatrix{
    \mathcal{A}
    \rrtwocell^{F''}_F{\ \ t \circ t'}
    & &
    \mathcal{B}
    }
    $$
    is called {\it vertical} composition. We will use the usual
    symbol $\circ$ for this. Next, we will define {\it horizontal}
    composition. In order to do this we explain a bit more
    of the structure at hand.
    
    \medskip\noindent
    Namely for every triple
    of categories $\mathcal{A}$, $\mathcal{B}$, and $\mathcal{C}$
    there is a composition law
    $$
    \circ : \Ob(\text{Fun}(\mathcal{B}, \mathcal{C}))
    \times
    \Ob(\text{Fun}(\mathcal{A}, \mathcal{B}))
    \longrightarrow
    \Ob(\text{Fun}(\mathcal{A}, \mathcal{C}))
    $$
    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?
    
    \medskip\noindent
    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
    $$
    \text{Fun}(\mathcal{A}, \mathcal{B})
    \longrightarrow
    \text{Fun}(\mathcal{A}, \mathcal{C}).
    $$
    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 ${}_{\text{id}_\mathcal{A}}t = t$.
    
    \medskip\noindent
    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
    $$
    \text{Fun}(\mathcal{B}, \mathcal{C})
    \longrightarrow
    \text{Fun}(\mathcal{A}, \mathcal{C}).
    $$
    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 $s_{\text{id}_\mathcal{B}} = s$.
    
    \medskip\noindent
    These constructions satisfy the additional properties
    $$
    {}_{G_1}({}_{G_2}t) = {}_{G_1\circ G_2}t,
    \ (s_{F_1})_{F_2} = s_{F_1 \circ F_2},
    \text{ and }{}_H(s_F) = ({}_Hs)_F
    $$
    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
    $$
    \xymatrix{
    G \circ F \ar[r]^{{}_Gt} \ar[d]_{s_F}
    &
    G \circ F' \ar[d]^{s_{F'}} \\
    G' \circ F \ar[r]_{{}_{G'}t}
    &
    G' \circ F'
    }
    $$
    in other words ${}_{G'}t \circ s_F =  s_{F'}\circ {}_Gt$.
    To prove this we just consider what happens on
    any object $x \in \Ob(\mathcal{A})$:
    $$
    \xymatrix{
    G(F(x)) \ar[r]^{G(t_x)} \ar[d]_{s_{F(x)}}
    &
    G(F'(x)) \ar[d]^{s_{F'(x)}} \\
    G'(F(x)) \ar[r]_{G'(t_x)}
    &
    G'(F'(x))
    }
    $$
    which is commutative because $s$ is a transformation
    of functors. This compatibility relation allows us
    to define horizontal composition.
    
    \begin{definition}
    \label{definition-horizontal-composition}
    Given a diagram as in the left hand side of:
    $$
    \xymatrix{
    \mathcal{A}
    \rtwocell^F_{F'}{t}
    &
    \mathcal{B}
    \rtwocell^G_{G'}{s}
    &
    \mathcal{C}
    }
    \text{ gives }
    \xymatrix{
    \mathcal{A}
    \rrtwocell^{G \circ F} _{G' \circ F'}{\ \ s \star t}
    & &
    \mathcal{C}
    }
    $$
    we define the {\it horizontal} composition $s \star t$ to be the
    transformation of functors ${}_{G'}t \circ s_F =  s_{F'}\circ {}_Gt$.
    \end{definition}
    
    \noindent
    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.
    
    \begin{lemma}
    \label{lemma-properties-2-cat-cats}
    The horizontal and vertical compositions have the following
    properties
    \begin{enumerate}
    \item $\circ$ and $\star$ are associative,
    \item the identity transformations $\text{id}_F$
    are units for $\circ$,
    \item the identity transformations of the identity functors
    $\text{id}_{\text{id}_\mathcal{A}}$
    are units for $\star$ and $\circ$, and
    \item given a diagram
    $$
    \xymatrix{
    \mathcal{A}
    \rruppertwocell^F{t}
    \ar[rr]_(.3){F'}
    \rrlowertwocell_{F''}{t'}
    & &
    \mathcal{B}
    \rruppertwocell^G{s}
    \ar[rr]_(.3){G'}
    \rrlowertwocell_{G''}{s'}
    & &
    \mathcal{C}
    }
    $$
    we have $ (s' \circ s) \star (t' \circ t) = (s' \star t') \circ (s \star t)$.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    The last statement turns using our previous notation into the following
    equation
    $$
    s'_{F''}
    \circ
    {}_{G'}t'
    \circ
    s_{F'}
    \circ
    {}_Gt
    =
    (s' \circ s)_{F''}
    \circ
    {}_G(t' \circ t).
    $$
    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.
    \end{proof}
    
    \noindent
    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
    $$
    (\circ, \star) :
    \text{Fun}(\mathcal{B}, \mathcal{C})
    \times
    \text{Fun}(\mathcal{A}, \mathcal{B})
    \longrightarrow
    \text{Fun}(\mathcal{A}, \mathcal{C})
    $$
    whose source is the product category,
    see Definition \ref{definition-product-category}.

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