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\documentclass{article} 
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	\begin{document}   
 
	\begin{center}   
	{\bf Equivalence of Categories \\}   
	{\small John\ C.\ Baez, April 6, 2004\\}   
	\vspace{0.3cm}   
        \end{center}

\noindent
1.  Prove that a functor $F \maps C \to D$ is an equivalence iff it is
essentially surjective, full and faithful.

\vskip 1em 
{\it Hint: let me remind you of all the necessary definitions.}

\begin{defn} \et A {\bf category} $C$ consists of:
\begin{itemize}
\item a collection $\Ob(C)$ of {\bf objects}.
\item for any pair of objects $x,y$, a set $\Hom(x,y)$ of {\bf morphisms}
from $x$ to $y$.  (If $f \in \Hom(x,y)$ we write $f \maps x \to y$.)
\end{itemize}
equipped with:
\begin{itemize}
\item for any object $x$, an {\bf identity morphism} $1_x \maps x \to x$.
\item for any pair of morphisms $f \maps x \to y$ and $g \maps y \to z$,
a morphism $fg \maps x \to z$ called the {\bf composite} of $f$ and $g$.
\end{itemize}
such that:
\begin{itemize}
\item for any morphism $f \maps x \to y$, the {\bf left and right unit laws}
hold: $1_x f = f = f 1_y$.
\item for any triple of morphisms $f \maps w \to x$, $g \maps x \to y$,
$h \maps y \to z$, the {\bf associative law} holds: $(fg)h = f(gh)$.
\end{itemize}
\end{defn}

\noindent
We usually write $x \in C$ as an abbreviation for $x \in \Ob(C)$.
An {\bf isomorphism} is a morphism $f \maps x \to y$ with an {\bf inverse},
i.e.\ a morphism $g \maps y \to x$ such that $fg = 1_x$ and $gf = 1_y$.

\begin{defn} \et Given categories $C,D$, a {\bf functor} $F \maps C \to D$
consists of:
\begin{itemize}
\item a function $F \maps \Ob(C) \to \Ob(D)$.
\item for any pair of objects $x,y \in \Ob(C)$, a function
$F \maps \Hom(x,y) \to \Hom(F(x),F(y))$.  
\end{itemize}
such that:
\begin{itemize}
\item $F$ {\bf preserves identities}: for any object $x \in C$, $F(1_x) = 1_{F(x)}$.
\item $F$ {\bf preserves composition}: for any pair of morphisms $f \maps x \to y$,
$g \maps y \to z$ in $C$, $F(fg) = F(f) F(g)$.  
\end{itemize}
\end{defn}

\noindent
It's not hard to define identity functors and composition of functors,
and to check the left and right unit law and associative law for these.

\begin{defn} \et 
Given functors $F,G \maps C \to D$, a {\bf natural transformation}
$\alpha \maps F \To G$ consists of:
\begin{itemize}
\item a function $\alpha$ mapping each object $x \in C$ to a morphism
$\alpha_x \maps F(x) \to G(x)$  
\end{itemize}
such that:
\begin{itemize}
\item
for any morphism $f \maps x \to y$ in $C$, this diagram commutes:
\[
\xymatrix{
 F(x) 
   \ar[r]^{F(f)}
   \ar[d]_{\alpha_x}
&  F(y)  
  \ar[d]^{\alpha_y}
\\
 G(x)
   \ar[r]_{G(f)}
&  G(y)
}
\]
\end{itemize}
\end{defn}

\noindent
With a little thought you can figure out how to compose natural transformations
$\alpha \maps F \to G$ and $\beta \maps G \To H$ and get a natural 
transformation $\alpha \beta \maps F \To H$.  We can also define identity 
natural transformations.
Again, it's not hard to check the left and right unit law and associativity
for these.

\begin{defn} \et Given functors $F,G \maps C \to D$, a {\bf natural
isomorphism} $\alpha \maps F \To G$ is a natural transformation that has
an {\bf inverse}, i.e.\ a natural transformation $\beta \maps G \To F$
such that $\alpha \beta = 1_F$ and $\beta \alpha = 1_G$.
\end{defn}

\noindent
It's not hard to see that a natural transformation $\alpha \maps F \To G$
is a natural isomorphism iff for every object $x \in C$, the 
morphism $\alpha_x$ is invertible.

\begin{defn} \et A functor $F \maps C \to D$ is an {\bf equivalence} if it 
has a {\bf weak inverse}, that is, a functor $G \maps D \to C$ such that 
there exist
natural isomorphisms $\alpha \maps FG \To 1_C$, $\beta \maps GF \To 1_D$.
\end{defn}

\begin{defn} \et A functor $F \maps C \to D$ is {\bf essentially surjective} if
for every object $x \in D$ there is an object $\tilde{x} \in C$ 
such that $F(\tilde{x}) \iso x$.
\end{defn}

\begin{defn} \et A functor $F \maps C \to D$ is {\bf full} if for every pair of
objects $x,y \in C$, the function $F \maps \hom(x,y) \to \hom(F(x),F(y))$
is onto. 
\end{defn}

\begin{defn} \et 
A functor $F \maps C \to D$ is {\bf faithful} if for every pair of
objects $x,y \in C$, the function $F \maps \hom(x,y) \to \hom(F(x),F(y))$
is one-to-one.
\end{defn}

{\it
I will be glad to give you further hints if you need them.
The fun part is constructing the weak inverse of a functor 
$F$ using the fact that it's essentially surjective, full and faithful.
This is a categorified version of constructing the inverse of
a function using the fact that it's surjective and injective.  
}



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