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\title{Quantum Gravity Seminar\\Winter 2007\\
Checking that Cat is a 2-Category}
\author{Alex Hoffnung}

\begin{document}

\maketitle

\noindent Check that Cat is a (strict) 2-category.

\section*{What's in a 2-category?}
\noindent A 2-category has objects, morphisms, and 2-morphisms.  
In Cat, these will be given as follows:
\begin{itemize}
\item \textit{Objects.}  Categories will be the bjects.
\item \textit{Morphisms.} Functors between categories will be the morphisms.
\item \textit{2-Morphisms.} Natural transformations will be the 2-morphisms.
\end{itemize}

\section*{Is Cat even a category?}
\noindent  We check the basic requirements of a category:
\begin{itemize}
\item \textit{Identity morphisms.} For any category $C$ in Cat, we get an
identity functor $1_{C}\colon C\rightarrow C$.  This is the functor which
does nothing to the objects and morphisms of $C$.
\item \textit{Composition.} For any pair of functors $F\colon C\rightarrow
D$ and $G\colon D\rightarrow E$, we define
$FG\colon C\rightarrow E$ to be the composite
of $F$ and $G$.  We just need to make sure that $FG$ is a functor:
that is, it preserves identities and composition.
\begin{itemize}
\item \textit{Identities.} For any object $x$ in $C$,
\[FG(1_{x}) ~=~ G(1_{F(x)}) ~=~ 1_{FG(x)}.\]
\item \textit{Composition.} For any pair of morphisms $f\colon x\rightarrow
y$ and $g\colon y\rightarrow z$ in $C$,
\[FG(fg) ~=~ G(F(f)F(g)) ~=~ G(F(f))G(F(g)) ~=~ FG(f)FG(g).\]
\end{itemize}
Now we need the left and right unit laws to hold, along with the associative
law.
\begin{itemize}
\item \textit{Left and right unit laws.} For any functor
$F\colon C\rightarrow D$, it is clear that $1_{C}$F = F = F$1_{D}$.
\item \textit{Associative law.} For any triple of functors $F\colon
A\rightarrow B$, $G\colon B\rightarrow C$, $H\colon C\rightarrow D$,
$F(GH) = (FG)H$.  This follows from the associativity of the 
functions on objects and morphisms which comprise our functors.
\end{itemize}
\end{itemize}

\noindent So Cat is a category!  A good start.

\section*{Defining the compositions.}
\noindent Now we need to pay attention to the 2-morphisms.  In this
case, the natural transformations.  We can compose our 2-morphisms in
two ways, vertically and horizontally.  We want to show that these
composites are natural transformations and that the
interchange law holds between these two compositions.

\begin{itemize}
\item \textit{Vertical composition.} Given functors $F,G,H\colon
C\rightarrow D$ and natural transformations $\alpha\colon
F\Longrightarrow G$ and $\beta\colon G\Longrightarrow H$, we want to
define $\alpha\beta\colon F\Longrightarrow H$.

\begin{center}
\xymatrix{
  & A\ruppertwocell^f{\alpha}
    \rlowertwocell_h{\beta}
    \ar[r]^g
 & B\\
}
\end{center}

Since $\alpha$ and $\beta$ are natural transformations, we have for
every x in C, functions
\[\alpha_{x}\colon F(x)\rightarrow G(x)~and~\beta_{x}\colon G(x)\rightarrow H(x)\]
such that for all morphisms $f\colon x\rightarrow y$ in C, the following diagrams commute:
\begin{center}
\xymatrix{ & F(x)\ar[r]^{F(f)}\ar[d]_{\alpha_{x}} &
  F(y)\ar[d]^{\alpha_{y}} & & & G(x)\ar[r]^{G(f)}\ar[d]_{\beta_{x}} &
  G(y)\ar[d]^{\beta_{y}} &\\ & G(x)\ar[r]^{G(f)} & G(y) & & &
  H(x)\ar[r]^{H(f)} & H(y) & }
\end{center}

Given that the diagrams above commute and defining the composite
$(\alpha\beta)_{x}\colon F(x)\rightarrow H(x)$ for any $x$ in $C$ as
\begin{center}
\xymatrix{
  & F(x)\ar[r]^{\alpha_{x}} & G(x)\ar[r]^{\beta_{x}} & H(x) &\\
  }
\end{center}
it is clear that the following diagram commutes: 

\begin{center}
\xymatrix{
  & F(x)\ar[r]^{F(f)}\ar[d]_{\alpha_{x}} & F(y)\ar[d]^{\alpha_{y}} &\\
  & G(x)\ar[r]^{G(f)}\ar[d]_{\beta_{x}} & G(y)\ar[d]^{\beta_{y}} &\\
  & H(x)\ar[r]^{H(f)} & H(y) &   }
\end{center}
Thus, the vertical composite is natural.

\item \textit{Horizontal composition.} Given functors
$F,F^{\prime}\colon C\rightarrow D$, $G,G^{\prime}\colon D\rightarrow
E$, and natural transformations $\alpha\colon F\Longrightarrow
F^{\prime}$ and $\beta\colon G\Longrightarrow G^{\prime}$, we want to
define $\alpha\circ\beta\colon F\circ G\Longrightarrow F^{\prime}\circ
G^{\prime}$.  Given any object $x$ in $C$, we have
\[\alpha_{x}\colon F(x)\rightarrow F^{\prime}(x).\]
Under the functors $G$ and $G^{\prime}$, this gets mapped to:
\begin{center}
\xymatrix{
  & G(F(x))\ar[r]^{G(\alpha_{x})} & G(F^{\prime}(x))&\\
  }
\end{center}
and
\begin{center}
\xymatrix{ & G^{\prime}(F(x))\ar[r]^{G^{\prime}(\alpha_{x})} &
  G^{\prime}(F^{\prime}(x)).&\\ }
\end{center}
By the naturality of $\beta$, we obtain another commuting diagram which
allows us to make a well-defined choice for our composite:
\begin{center}
\xymatrix{ & G(F(x))\ar[r]^{G(\alpha_{x})}\ar[d]_{\beta_{F(x)}} &
  G(F^{\prime}(x))\ar[d]^{\beta_{F^{\prime}(x)}}&\\ &
  G^{\prime}(F(x))\ar[r]^{G^{\prime}(\alpha_{x})} &
  G^{\prime}(F^{\prime}(x)).&\\ }
\end{center}
Now we just need to check that our horizontal composition is natural.
So we need the following diagram to commute:

\begin{center}
\xymatrix{ & G(F(x))\ar[r]^{G(F(f))}\ar[d]_{\alpha\circ\beta_{x}} &
  G(F(y))\ar[d]^{\alpha\circ\beta_{y}}&\\ &
  G^{\prime}(F^{\prime}(x))\ar[r]^{G^{\prime}(F^{\prime}(f))} &
  G^{\prime}(F^{\prime}(y)).&\\ }
\end{center}

We can extend this diagram, using the definition of our horizontal
composite, to the following diagram:

\begin{center}
\xymatrix{ & G(F(x))\ar[r]^{G(F(f))}\ar[d]_{G(\alpha_{x})} &
  G(F(y))\ar[d]^{G(\alpha_{y})} &\\ &
  G(F^{\prime}(x))\ar[r]^{G(F^{\prime}(f))}\ar[d]_{\beta_{F^{\prime}(x)}}
  & G(F^{\prime}(y))\ar[d]^{\beta_{F^{\prime}(y)}}&\\ &
  G^{\prime}(F^{\prime}(x))\ar[r]^{G^{\prime}(F^{\prime}(f))} &
  G^{\prime}(F^{\prime}(y)).&\\ }
\end{center}
The upper square commutes since we have
\[F(f)\alpha_{y}~=~\alpha_{x}F^{\prime}(F)\]
by the naturality of $\alpha$, and we can then apply the functor G to
this equation.  The lower square commutes by the naturality of
$\beta$.  Thus the horizontal composite is natural.
\end{itemize}

\subsection*{The interchange law.}
\noindent Given objects $C, D, E$ in Cat and morphisms $F$, $F^{\prime},
F^{\prime\prime}\colon C\longrightarrow D$ and $G$, $G^{\prime},
G^{\prime\prime}\colon D\longrightarrow E$, if we have 2-morphisms
$\alpha\colon F\Longrightarrow F^{\prime}$, $\alpha^{\prime}\colon
F^{\prime}\Longrightarrow F^{\prime\prime}$, $\beta\colon
G\Longrightarrow G^{\prime}$, and $\beta^{\prime}\colon
G^{\prime}\Longrightarrow G^{\prime\prime}$, then we want the following
composition diagram to be unambiguous:
\begin{center}
\xymatrix{
  & C\ruppertwocell^F{\alpha}
    \rlowertwocell_F^{\prime\prime}{\alpha^{\prime}}
    \ar[r]^F^{\prime}
  & D \ruppertwocell^G{\beta}
    \rlowertwocell_G^{\prime\prime}{\beta^{\prime}}
    \ar[r]^G^{\prime}
  & E\\
}
\end{center}
By the naturality of vertical composition we get the following
commutative rectangle:
\begin{center}
\xymatrix{ & Fx\ar[r]^{Ff}\ar[d]_{\alpha_{x}} & Fy\ar[d]^{\alpha_{y}}
  &\\ & F^{\prime}x\ar[r]^{F^{\prime}f}\ar[d]_{\alpha^{\prime}_{x}} &
  F^{\prime}y\ar[d]^{\alpha^{\prime}_{y}} &\\ &
  F^{\prime\prime}x\ar[r]^{F^{\prime\prime}f} & F^{\prime\prime}y & }
\end{center}
If we apply our 3 functors from $D$ to $E$ to this rectangle, we get a big
commuting cube, which tells us that the interchange law
\[\alpha\alpha^{\prime}\circ\beta\beta^{\prime}~=~(\alpha\circ\beta)(\alpha^{\prime}\circ\beta^{\prime})\]
holds.  So we inspect this cube and see that it commutes by the
naturality of vertical composition, horizontal composition and
functoriality of $G$, $G^{\prime}$ and $G^{\prime\prime}$.
\begin{center}
\xymatrix{ & & &
  G^{\prime\prime}Fx\ar[r]^{G^{\prime\prime}Ff}\ar[d]_{G^{\prime\prime}\alpha_{x}}
  & G^{\prime\prime}Fy\ar[d]^{G^{\prime\prime}\alpha_{y}} &\\ & & &
  G^{\prime\prime}F^{\prime}x\ar[r]^{G^{\prime\prime}F^{\prime}f}\ar[d]_{G^{\prime\prime}\alpha^{\prime}_{x}}
  &
  G^{\prime\prime}F^{\prime}y\ar[d]^{G^{\prime\prime}\alpha^{\prime}_{y}}&\\
  & & &
  G^{\prime\prime}F^{\prime\prime}x\ar[r]^{G^{\prime\prime}F^{\prime\prime}f}
  & G^{\prime\prime}F^{\prime\prime}y.&\\ & &
  G^{\prime}Fx\ar[uuur]^{\beta^{\prime}_{Fx}}\ar[r]^{G^{\prime}Ff}\ar[d]_{G^{\prime}\alpha_{x}}
  &
  G^{\prime}Fy\ar[uuur]^{\beta^{\prime}_{Fy}}\ar[d]^{G^{\prime}\alpha_{y}}
  &\\ & &
  G^{\prime}F^{\prime}x\ar[uuur]^{\beta^{\prime}_{F^{\prime}x}}\ar[r]^{G^{\prime}F^{\prime}f}\ar[d]_{G^{\prime}\alpha^{\prime}_{x}}
  &
  G^{\prime}F^{\prime}y\ar[uuur]^{\beta^{\prime}_{F^{\prime}y}}\ar[d]^{G^{\prime}\alpha^{\prime}_{y}}&\\
  & &
  G^{\prime}F^{\prime\prime}x\ar[uuur]^{\beta^{\prime}_{F^{\prime\prime}x}}\ar[r]^{G^{\prime}F^{\prime\prime}f}
  &
  G^{\prime}F^{\prime\prime}y\ar[uuur]^{\beta^{\prime}_{F^{\prime\prime}y}}.&\\
  & GFx\ar[uuur]^{\beta_{Fx}}\ar[r]^{GFf}\ar[d]_{G\alpha_{x}} &
  GFy\ar[uuur]^{\beta_{Fy}}\ar[d]^{G\alpha_{y}} &\\ &
  GF^{\prime}x\ar[uuur]^{\beta_{F^{\prime}x}}\ar[r]^{GF^{\prime}f}\ar[d]_{G\alpha^{\prime}_{x}}
  &
  GF^{\prime}y\ar[uuur]^{\beta_{F^{\prime}y}}\ar[d]^{G\alpha^{\prime}_{y}}&\\
  &
  GF^{\prime\prime}x\ar[uuur]^{\beta_{F^{\prime\prime}x}}\ar[r]^{GF^{\prime\prime}f}
  & GF^{\prime\prime}y\ar[uuur]^{\beta_{F^{\prime\prime}y}}.&\\ }
\end{center}
After staring at this for a while we can believe that the
interchange law holds.
\section*{Is Cat a 2-category?}
\subsection*{The category $\Hom(C,D)$}
\noindent The first thing we need to show is that given two objects $C,
D$ in Cat, we get a category $\Hom(C,D)$.  The objects of $\Hom(C,D)$ are
functors and the morphisms are natural transformations.  Luckily, we
already know how to compose our morphisms.  For each triple $F,G,H$ of
objects in $\Hom(C,D)$ we have a function
\[\circ\colon \Hom(F,G)\times \Hom(G,H) \Longrightarrow \Hom(F,H)\]
which is just our vertical composition of natural transformations that
we defined above.  Also, for any object $F$ in $\Hom(C,D)$, we have an identity
natural transformation
\[\mathbf{1}_{F}\colon F\Longrightarrow F.\]
Given any natural transformation $\alpha\colon F\Longrightarrow G$, we
have
\[\mathbf{1}_{F}\alpha~=~\alpha~=~\alpha\mathbf{1}_{G}.\]
This is clear since for any object x in C, the natural transformations
$\mathbf{1}_{F}$ and$\alpha$ give us morphisms
\[\mathbf{1}_{F_{x}}\colon Fx\Longrightarrow Fx~~and~~\alpha_x\colon Fx\Longrightarrow Gx.\]
We know that morphisms compose and $\mathbf{1}_{F_{x}}$, the component
of our identity natural transformation will behave as an identity.
\\\\ We need to show that our morphisms, the natural transformations,
are associative.  Given any four objects $F, G, H, I$ in $\Hom(C,D)$ and
natural transformations $\alpha\colon F\Longrightarrow G, \beta\colon
G\Longrightarrow H, \gamma\colon H\Longrightarrow I$, we know that
\[(\alpha\beta)\gamma~=~\alpha(\beta\gamma),\]
since the components of natural transformations are morphisms, which
are associative.  
\\ So, we see that $\Hom(C,D)$ is indeed a category.
\subsection*{The composition functor}
\noindent Given any three objects $C, D, E$ in Cat, we define a
composition functor
\[\circ\colon \Hom(C,D)\times \Hom(D,E) \Longrightarrow \Hom(C,E),\]
by
\[(F,G)~\longmapsto FG\]
- regular functor composition - and
\[(\alpha,\beta)~\longmapsto \alpha\circ\beta\]
- horizontal composition.  While our composition notation is becoming
heavily overloaded, we will only really concern ourselves with the
vertical and horizontal compostion symbols defined in the second
section of this note, where vertical composition is juxtaposition and
horizontal composition is $\circ$.\\\\ 
Now we want to check that this `composition functor'
is actually a functor.  Given any object $(F,G)$ in
$\Hom(C,D)\times\Hom(D,E)$, we have
\[\circ(\mathbf{1}_{F},\mathbf{1}_{G})~=~\mathbf{1}_{F}\circ\mathbf{1}_{G},\]
the identity on $FG$.\\ We also need to check that the functor
respects composition.  But this is just the interchange law proved
earlier!  So we have a functor.
\subsection*{Almost there!}
Since Cat is a strict category, our associator natural ismorphisms
will all be just the identity natural isomorphism since the
associative law for functors is an equation as we saw earlier.  A
similar statement is true for the left and right unit natural
isomorphisms.\\\\ So, Cat is a 2-category.
\end{document} 
