@ -14,7 +14,7 @@ We will briefly define categories an functors to fix the notation. We will not p
\end{itemize}
\end{itemize}
\end{definition}
\end{definition}
Instead of writing $f \in\Hom{\cat{C}}{A}{B}$ we write $f: A \to B$, as many categories have functions as maps. There is a category $\Set$ of sets with functions, a category $\Ab$ of abelian groups with group homomorphisms, a category $\Top$ of topological spaces and continuous maps, and many more.
Instead of writing $f \in\Hom{\cat{C}}{A}{B}$ we write $f: A \to B$, as many categories have functions as maps. For brevity we sometimes write $gf$ instead of $g \circ f$. There is a category $\Set$ of sets with functions, a category $\Ab$ of abelian groups with group homomorphisms, a category $\Top$ of topological spaces and continuous maps, and many more.
\begin{definition}
\begin{definition}
A \emph{functor}$F$ from a category $\cat{C}$ and to a category $\cat{D}$ consists of a function $F_0$ from the objects of $\cat{C}$ to the objects of $\cat{D}$ and a function $F_1$ from maps in $\cat{C}$ to maps in $\cat{D}$, such that
A \emph{functor}$F$ from a category $\cat{C}$ and to a category $\cat{D}$ consists of a function $F_0$ from the objects of $\cat{C}$ to the objects of $\cat{D}$ and a function $F_1$ from maps in $\cat{C}$ to maps in $\cat{D}$, such that
In this calculation we did the following. We split the inner sum in two halves \refeqn{1} and we use the simplicial equations on the second sum \refeqn{2}. Then we do a shift of indices \refeqn{3}. By interchanging the roles of $i$ and $j$ in the second sum, we have two equal sums which cancel out. So indeed this is a chain complex.
In this calculation we did the following. We split the inner sum in two halves \refeqn{1} and we use the simplicial equations on the second sum \refeqn{2}. Then we do a shift of indices \refeqn{3}. By interchanging the roles of $i$ and $j$ in the second sum, we have two equal sums which cancel out. So indeed this is a chain complex.
\end{proof}
\end{proof}
Thus, associated to a simplicial abelian group $A$ we obtain a chain complex $M(A)$ with $M(A)_n = A_n$ and the boundary operator as above. This construction defines a functor
Thus, associated to a simplicial abelian group $A$ we obtain a chain complex $M(A)$ with $M(A)_n = A_n$ and the boundary operator as above. Following the book \cite{goerss} we will call the chain complex $M(X)$ the \emph{Moore complex} or \emph{unnormalized chain complex} of $X$. This construction defines a functor
$$ M: \sAb\to\Ch{\Ab}$$
$$ M: \sAb\to\Ch{\Ab}$$
by assigning $M(f)_n = f_n$ for a natural transformation $f: A \to B$. It follows from a nice calculation that $M(f)$ is indeed a chain map:
by assigning $M(f)_n = f_n$ for a natural transformation $f: A \to B$. It follows from a nice calculation that $M(f)$ is indeed a chain map:
Many functors to $\sAb$ can be shown to have this description.\footnote{And also many functors to $\sSet$ are of this form if we leave out all additivity requirements.} In our case we could have defined our functor $K$ as
Many functors to $\sAb$ can be shown to have this description.\footnote{And also many functors to $\sSet$ are of this form if we leave out all additivity requirements.} In our case we could have defined our functor $K$ as
$$ K'(C)=\Hom{\Ch{\Ab}}{N\Z[\Delta[-]]}{C}. $$
$$ K'(C)=\Hom{\Ch{\Ab}}{N\Z[\Delta[-]]}{C}. $$
We will not show that this functor $K'$ is isomorphic to our functor $K$ defined earlier, however we will indicate that it makes sense by writing out explicit calculations for $K'(C)_0$ and $K'(C)_1$.
We will not show that this functor $K'$ is isomorphic to our functor $K$ defined earlier, however we will indicate that it makes sense by writing out explicit calculations for $K'(C)_0$ and $K'(C)_1$. First we see that
\foreach\i/\j in {2/2, 3/1, 4/0}\draw[->] (m-1-\i) -- node {$f_\j$} (m-2-\i);
\foreach\i/\j in {2/2, 3/1, 4/0}\draw[->] (m-1-\i) -- node {$f_\j$} (m-2-\i);
\end{tikzpicture}
\end{tikzpicture}
\Bigg\}\iso C_0 = K(C)_0, $$
\Bigg\}\iso C_0 = K(C)_0, $$
because for $f_1, f_2, \ldots$ there is now choice at all, and for $f_0: \Z\to C_0$ we only have to choose an image for $1\in\Z$. In the next dimension we see:
because for $f_1, f_2, \ldots$ there is now choice at all, and for $f_0: \Z\to C_0$ we only have to choose an image for $1\in\Z$. In the next dimension we see
\foreach\i/\j in {2/2, 3/1, 4/0}\draw[->] (m-1-\i) -- node {$f_\j$} (m-2-\i);
\foreach\i/\j in {2/2, 3/1, 4/0}\draw[->] (m-1-\i) -- node {$f_\j$} (m-2-\i);
\end{tikzpicture}
\end{tikzpicture}
\Bigg\}\iso C_1 \oplus C_0 = K(C)_1, $$
\Bigg\}\iso C_1 \oplus C_0 = K(C)_1, $$
because again we can choose $f_1$ anyway we want, which gives us $C_1$. But then we are forced to choose $f_0(x, x)=\del(f_1(x))$ for all $x \in\Z$, so we are left with choosing an element $c \in C_0$ for defining $f(1,-1)= c$. Adding this gives $C_1\oplus C_0$.
because again we can choose $f_1$ anyway we want, which gives us $C_1$. But then we are forced to choose $f_0(x, x)=\del(f_1(x))$ for all $x \in\Z$, so we are left with choosing an element $c \in C_0$ for defining $f(1,-1)= c$. Adding this gives $C_1\oplus C_0$.
Joshua Moerman
12 years ago