\documentclass[a4paper,10pt]{article}

\input xy
\xyoption{all}
\xyoption{ps}
\xyoption{2cell}
\UseAllTwocells


%opening
\title{Quantum Gravity Seminar\\Homework 1\\Winter 2007}
\author{Alex Hoffnung}

\begin{document}

\maketitle

\noindent Check that Cat is a 2-category.

\section*{What's in a 2-category?}
\noindent A 2-category has the following:
\begin{itemize}
\item \textit{Objects}  In Cat, the collection of all categories will serve as objects.
\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 $1_{C}: C\rightarrow C$.  This is the functor which does nothing to the objects and morphisms of C.
\item \textit{Composition} For any pair of functors $F: C\rightarrow D$ and $G: D\rightarrow E$, $FG: C\rightarrow E$ is the composite functor of F and G.  We just need to make sure FG 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 $f:x\rightarrow y$ and $g:y\rightarrow z$ 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 along with the associative law to hold.
\begin{itemize}
\item \textit{Left and right unit laws}  For any functor $F:C\rightarrow D$, it is clear that $1_{C}$F = F = F$1_{D}$.
\item \textit{Associative law}  For any triple of functors $F:A\rightarrow B$, $G:B\rightarrow C$, $H:C\rightarrow D$, F(GH) = (FG)H.  This follows from the associativity of the object and morphism functions which compose the 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 compositions result in natural transformations and that the interchange law holds between these two compositions.

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

\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
\[\alpha_{x}: F(x)\rightarrow G(x)~and~\beta_{x}: G(x)\rightarrow H(x)\]
such that for all morphisms $f:x\rightarrow y$ 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 composition $(\alpha\beta)_{x}:F(x)\rightarrow H(x)$ 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 composition is natural.


\item \textit{Horizontal composition} Given functors $F,F^{\prime}:C\rightarrow D$,  $G,G^{\prime}:D\rightarrow E$, and natural transformations $\alpha:F\Longrightarrow F^{\prime}$ and $\beta:G\Longrightarrow G^{\prime}$, we want to define $\alpha\circ\beta:F\circ G\Longrightarrow F^{\prime}\circ G^{\prime}$.  Given any object x in C, we have
\[\alpha_{x}:F(x)\rightarrow F^{\prime}(x).\]
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 get another commuting diagram which allows us to make a well-defined choice for our composition.
\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 composition 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 we get
\[F(f)\alpha_{y}~=~\alpha_{x}F^{\prime}(F)\]
by the naturality of $\alpha$ and we just apply the functor G to this equation.  The lower square commutes by the naturality of $\beta$.  Thus the horizontal composition is natural.
\end{itemize}

\subsection*{The interchange law}
\noindent Given objects C, D, E in Cat and morphisms F, $F^{\prime}, F^{\prime\prime}:C\longrightarrow D$ and G, $G^{\prime}, G^{\prime\prime}:D\longrightarrow E$, if we have 2-morphisms $\alpha:F\Longrightarrow F^{\prime}$, $\alpha^{\prime}:F^{\prime}\Longrightarrow F^{\prime\prime}$, $\beta:G\Longrightarrow G^{\prime}$, and $\beta^{\prime}:G^{\prime}\Longrightarrow G^{\prime\prime}$, the 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}
And 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
\[\circ:Hom(F,G)\times Hom(G,H) \Longrightarrow Hom(F,H)\]
which is just our vertical composition of natural transformations that we defined above.  Also, for object F in Hom(C,D), we get an identity natural transformation
\[\mathbf{1}_{F}:F\Longrightarrow F.\]
Given any natural transformation $\alpha:F\Longrightarrow G$, 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
\[\mathbf{1}_{F_{x}}:Fx\Longrightarrow Fx~~and~~\alpha_x:Fx\Longrightarrow Gx.\]
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 $\alpha:F\Longrightarrow G, \beta:G\Longrightarrow H, \gamma:H\Longrightarrow I$, 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
\[\circ:Hom(C,D)\times Hom(D,E) \Longrightarrow Hom(C,E),\]
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 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}