The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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 $\mathop{\mathrm{Ob}}\nolimits (\textit{Cat})$ 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

\[ \xymatrix{ \mathcal{A} \rruppertwocell ^{F''}{t'} \ar[rr]_(.3){F'} \rrlowertwocell _ F{t} & & \mathcal{B} } \text{ composes to } \xymatrix{ \mathcal{A} \rrtwocell ^{F''}_ F{\ \ t \circ t'} & & \mathcal{B} } \]

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

\[ \circ : \mathop{\mathrm{Ob}}\nolimits (\text{Fun}(\mathcal{B}, \mathcal{C})) \times \mathop{\mathrm{Ob}}\nolimits (\text{Fun}(\mathcal{A}, \mathcal{B})) \longrightarrow \mathop{\mathrm{Ob}}\nolimits (\text{Fun}(\mathcal{A}, \mathcal{C})) \]

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 $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$:

\[ \xymatrix{ G(F(x)) \ar[r]^{G(t_ x)} \ar[d]_{s_{F(x)}} & G(F'(x)) \ar[d]^{s_{F'(x)}} \\ G'(F(x)) \ar[r]_{G'(t_ x)} & G'(F'(x)) } \]

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.


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 003D. Beware of the difference between the letter 'O' and the digit '0'.