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\newcommand{\maps}{\colon}
\newcommand{\tensor}{\otimes}
\newcommand{\C}{\textit{\textbf{C}}}
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\title{Quantum Gravity Seminar\\Homework 2}
\author{Alex Hoffnung}
\date{October 20, 2006}
\begin{document}
\maketitle
\noindent Given a category $\C$ with finite products,
show it can be turned into a monoidal category.
\section*{Universal property of products}
\noindent Given objects $A_{1},A_{2},\ldots,A_{n}$ in \textbf{\textit{C}},
for some n, there exists an object A, called the product, with morphisms
$\pi_{i}\maps A \rightarrow A_{i}$, for each i, such that for every
object X in \textbf{\textit{C}} with morphisms $\phi_{i} \maps A
\rightarrow A_{i}$, for each i, there exists a unique morphism
$\phi \maps X \rightarrow A$ such that
\begin{center}
\xymatrix{
& X\ar@<1ex>[dr]_{\phi_{i}}\ar[r]^{\phi} & A \ar[d]^{\pi_{i}} \\
& & A_{i} }
\end{center}
commutes. i.e., $\pi_{i}\circ\phi$=$\phi_{i}$, for all i.
\\\\
\subsection*{Isomorphism of products}
It will be useful to first show that any two products of a given collection of objects are isomorphic.\\
Consider two products D and $D^{\prime}$ of the objects A and B, each with projection maps to A and B.
\begin{center}
\xymatrix{
& & D\ar@{-->}[d]^{\exists !\phi}\ar[dl]\ar[dr] &\\
& A & D^{\prime}\ar@{-->}[d]^{\exists !\phi^{\prime}}\ar[l]\ar[r] & B &\\
& & D\ar[ul]\ar[ur] }
\end{center}
\noindent By the universal property of products there exist unique maps
$\phi\maps D \rightarrow D^{\prime}$ and $\phi^{\prime}\maps D^{\prime}
\rightarrow D$ such that each triangle commutes. There also exists a
unique map from $D$ to $D$ which makes the square commute. It is easily
checked that the composition $\phi^{\prime}\circ\phi$ makes the square
commute. Since the identity satisfies this property we have
$\phi^{\prime}\circ\phi = 1_{D}$. Thus, $D$ and $D^{\prime}$ are
isomorphic. This proof generalizes
easily to products of any finite number of objects.
\subsection*{The product of no objects is terminal}
Consider the product of no objects, which we call $1$.
We would like to show that this is a terminal object in our category, i.e.,
for every object $X$ in $\C$
there is exactly one morphism to $1$.
\begin{center}
\xymatrix{
& X\ar@{-->}[d]^{!} &\\
& 1 }
\end{center}
For any object the diagram above vacuously satisfies the hypotheses of
the universal property of products, so there exists a unique map into
1.\\\
\subsection*{The tensor functor}
We define the map $\tensor \maps \C \times \C \rightarrow \C$ both on
objects and morphisms. Given a pair of objects $A$ and $B$ and assuming
the axiom of choice, the tensor map chooses a product of $A$ and $B$ with
projection maps to $A$ and $B$. (In what follows, the arrows which are
these projections will be not be labelled since it is clear what they
are.) Since we have already shown that all products of $A$ and $B$ are
isomorphic, we will call the chosen product $A \tensor B$ and the map of
objects is well-defined up to isomorphism. Given a pair of morphisms
$f\maps A\rightarrow A^{\prime}$ and $g\maps B\rightarrow
B^{\prime}$, consider the diagram
\begin{center}
\xymatrix{
& & A\tensor B\ar@{-->}[ddd]^{\exists !f\tensor g}\ar[dl]\ar[dr] &\\
& A\ar[d]^{f} & & B\ar[d]^{g} \\
& A^{\prime} & &B^{\prime} \\
& & A^{\prime}\tensor B^{\prime}\ar[ul]\ar[ur] & }
\end{center}
and we define $\tensor (f,g)$ to be the unique map $f \tensor g\maps A \tensor B \rightarrow A^{\prime} \tensor B^{\prime}$ in the diagram.
\\\\
\subsubsection*{The identity}
Similarly, if we consider the identity maps $1_{A}\maps A\rightarrow A$
and $1_{B} \maps B \rightarrow B$, the following diagram gives the unique morphism $1_{A}\tensor 1_{B}$ from A$\tensor$B to A$\tensor$B such that the diagram commutes.
\begin{center}
\xymatrix{
& & A\tensor B\ar@{-->}[ddd]^{\exists !}\ar[dl]\ar[dr] &\\
& A\ar[d]^{1_{A}} & & B\ar[d]^{1_{B}}\\
& A & & B \\
& & A\tensor B\ar[ul]\ar[ur] & }
\end{center}
Since the identity map $1_{A\tensor B}$ satisfies this diagram, we have $1_{A}\tensor 1_{B}$ = $1_{A\tensor B}$.
\\
\subsubsection*{Composition}
Given morphisms $f\maps A\rightarrow B$, $g\maps B\rightarrow C$,
$f^{\prime} \maps A^{\prime}\rightarrow B^{\prime}$,
$g^{\prime} \maps B^{\prime}\rightarrow C^{\prime}$.
We want to show $\tensor$ respects composition of morphisms.
\begin{center}
\xymatrix{
& & A\ar[r]^{f} & B\ar[r]^{g} & C &\\
& & & B\tensor B^{\prime}\ar[drr]^{g\tensor g^{\prime}}\ar[u]\ar@{-}[d] & & \\
& A\tensor A^{\prime}\ar@{-->}[rrrr]^{\exists ! (f\circ g)\tensor (f^{\prime}\circ g^{\prime})}\ar[uur]\ar[urr]^{f\tensor f^{\prime}}\ar[dr] & & \ar[d] & & C\tensor C^{\prime}\ar[uul]\ar[dl] \\
& & A^{\prime}\ar[r]^{f^{\prime}} & B^{\prime}\ar[r]^{g^{\prime}} & C^{\prime} & }
\end{center}
The definition of tensors of morphisms given above tells us that each small square of this diagram commutes giving $f\tensor f^{\prime}$ and $g\tensor g^{\prime}$. The full diagram itself defines $(g\circ f)\tensor (g^{\prime}\circ f^{\prime})$ by the same tensor definition. Since all the small squares commute, $g \tensor g^{\prime}\circ f\tensor f^{\prime}$ satsifies the commutivity of the larger square. Thus, by uniqueness, we must have $(g\circ f)\tensor (g^{\prime}\circ f^{\prime})$ = $(g\tensor g^{\prime})\circ (f\tensor f^{\prime})$.
\\
\section*{The unitors}
Since $1\tensor A$ and A are both products of 1 and $A$,
we have an isomorphism $l_{A} \maps 1\tensor A\rightarrow A$
for every object A in our category given by the diagram
\begin{center}
\xymatrix{
& & 1 &\\
& 1\tensor A\ar@{-->}[r]^{\exists ! l_{A}}\ar[ur]^{!}\ar[dr] & A\ar[u]^{!_{A}}\ar[d]_{1_{A}} &\\
& & A & }
\end{center}
\noindent Similarly, we have an isomorphism $r_{A}\maps A\tensor 1\rightarrow A$. Note that the above diagram says that $l_{A}$ must be the very same projection chosen by the functor from $A\tensor 1$ to A. The same is true for $r_{A}$.
\subsection*{Naturality}
Given the functors $F\maps \C \rightarrow \C$
defined by $A \longmapsto 1\tensor A$, $f \longmapsto 1\tensor f$
and the identity functor, we now show that $l_{A}$ is natural. That is,
given objects $A$ and $B$ in $C$ and a morphism $f$ in $C$ from $A$ to $B$,
we want the following diagram to commute: \\
\begin{center}
\xymatrix{
& F(A) = 1\tensor A\ar[r]^{F(f)=1\tensor f}\ar[d]_{l_{A}} & 1\tensor B = F(B)\ar[d]^{l_{B}} &\\
& Id(A) = A\ar[r]^{Id(f)=f} & B = Id(B) }
\end{center}
But extending this diagram slightly we have the diagram which defined the tensor functor on morphisms
\begin{center}
\xymatrix{
& 1\ar[r]^{!} & 1 &\\
& 1\tensor A\ar[r]^{1\tensor f}\ar[d]_{l_{A}}\ar[u] & 1\tensor B\ar[d]^{l_{B}}\ar[u] &\\
& A\ar[r]^{f} & B }
\end{center}
which commutes. The bottom square is the same diagram as our naturality square above and the squares commute as desired. A similar argument gives us that $r_{A}$ is natural.
\section*{The associators}
We now define ismorphisms of products called the associators. Given objects A, B and C in \textbf{\textit{C}}, we consider the product diagram
\begin{center}
\xymatrix{
& & & (A\tensor B)\tensor C\ar@{-->}[dddd]^{\exists !\alpha_{A,B,C}}\ar[dl]\ar[dr] &\\
& & A\tensor B\ar[dl]\ar[d] & & C &\\
& A & B & & B\tensor C\ar[ll]\ar[u] &\\
& & & & & &\\
& & & A\tensor(B\tensor C)\ar[uull]\ar[uul]\ar[uur] & & }
\end{center}
and call the unique map $\alpha_{A,B\tensor C}$ the associator.
As with the other morphisms given from product diagrams, this map is an
isomorphism.
\subsection*{Naturality}
As with the unitors, we need to show the associator is a natural
isomorphism. So given functors $F$ and $G$ which take objects $A$,
$B$, and $C$ to $(A\tensor B)\tensor C$ and $A\tensor(B\tensor C)$
respectively, and morphisms $f\maps A\rightarrow A^{\prime}$,
$g\maps B\rightarrow B^{\prime}$, and $h\maps C\rightarrow C^{\prime}$
to $(f\tensor g)\tensor h$ and $f\tensor(g\tensor h)$ respectively, we
must show the following diagram commutes.
\begin{center}
\xymatrix{
& (A\tensor B)\tensor C\ar[r]^{(f\tensor g)\tensor h}\ar[d]^{\alpha_{A,B,C}} & (A^{\prime}\tensor B^{\prime})\tensor C^{\prime}\ar[d]^{\alpha_{A^{\prime},B^{\prime},C^{\prime}}} &\\
& A\tensor(B\tensor C)\ar[r]^{f\tensor(g\tensor h)} & A^{\prime}\tensor(B^{\prime}\tensor C^{\prime}) & }
\end{center}
Each of the morphisms in this square are defined as the unique morphism that makes some diagram commute. These diagrams use the universal property of products with projections onto either A, B, and C or $A^{\prime}$, $B^{\prime}$, and $C^{\prime}$. Since we have morphisms between each object and its primed partner, respectively, it can be shown that this diagram is made up of several smaller commuting diagrams, therefore it commutes. The explicit calculations are left out since they are cumbersome and not very instructive.
\section*{The unit laws}
We want the following diagram to hold.
\begin{center}
\xymatrix{
& (A\tensor1)\tensor B\ar[rr]^{\alpha_{A\tensor1,B}}\ar[dr]_{r_{A}\tensor1_{B}} & & A\tensor(1\tensor B)\ar[dl]^{1_{A}\tensor l_{B}} &\\
& & A\tensor B & }
\end{center}
The appropriate definition for $\alpha_{A,1,B}$ comes from the following diagram.
\begin{center}
\xymatrix{
& & & (A\tensor 1)\tensor B\ar@{-->}[dddd]^{\exists !\alpha_{A,1,B}}\ar[dl]\ar[dr] &\\
& & A\tensor 1\ar[dl]\ar[d] & & B &\\
& A & 1 & & 1\tensor B\ar[ll]\ar[u] &\\
& & & & & &\\
& & & A\tensor(1\tensor B)\ar[uull]\ar[uul]\ar[uur] & & }
\end{center}
So, we have
\begin{center}
\xymatrix{
& & & (A\tensor 1)\tensor B\ar@{-->}[dddd]^{\exists !\alpha_{A,1,B}}\ar@{-->}[ddl]^{r_{A}\tensor1_{B}}\ar[dl]\ar[dr] &\\
& & A\tensor 1\ar[dl] & & B &\\
& A & A\tensor B\ar[l]\ar[urr] & & 1\tensor B\ar[u] &\\
& & & & & &\\
& & & A\tensor(1\tensor B)\ar[uull]\ar@{-->}[uul]_{1_{A}\tensor l_{B}}\ar[uur] & & }
\end{center}
$r_{A}\tensor1_{B}$ and $1_{A}\tensor l_{B}$, respectively, make the top and bottom of the diagram commute since these are the diagrams in the respective product diagram definitions. Since the associator also makes this diagram commute, we have that all smaller diagrams here commute. $(1_{A}\tensor l_{B})\circ\alpha_{A,1,B}$ and $r_{A}\tensor1_{B}$ are both maps from $(A\tensor1)\tensor B$ to $A\tensor B$ and they both make the following portion of the above diagram commute.
\begin{center}
\xymatrix{
& & & (A\tensor 1)\tensor B\ar@{-->}[ddl]\ar[dl]\ar[dr] &\\
& & A\tensor 1\ar[dl] & & B &\\
& A & A\tensor B\ar[l]\ar[urr] & & & }
\end{center}
So, the two morphisms are equal and the desired diagram commutes.
\section*{The pentagon equation}
Finally, given objects A, B, C, and D, the associators must satisfy the following diagram.
\begin{center}
\xymatrix{
& & ((A\tensor B)\tensor C)\tensor D\ar[dl]_{\alpha_{A\tensor B,C,D}}\ar[dr]^{\alpha_{A,B,C}\tensor1_{D}} & &\\
& (A\tensor B)\tensor(C\tensor D)\ar[d]^{\alpha_{A,B,C\tensor D}}& & (A\tensor(B\tensor C))\tensor D\ar[d]^{\alpha_{A,B\tensor C,D}} & \\
& A\tensor(B\tensor(C\tensor D)) & & A\tensor((B\tensor C)\tensor D)\ar[ll]^{1_{A}\tensor\alpha_{B,C,D}} & }
\end{center}
Each of the objects in the pentagon is a product of the four objects
$A,B,C,$ and $D$ with the appropriate projection maps. All of the
morphisms in the pentagon equation are associators or tensor products
of associators and identity maps. Therefore, they all satisfy
commutative product diagrams with projections onto each of the four
objects. Drawing each of these product diagrams gives a tangle of
morphisms and triangles which make up the pentagon. Since each
diagram comprising the pentagon commutes, it is a simple (yet messy)
exercise to see that the pentagon actually commutes.\\
\end{document}