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C: functoriality/additivty/lemma

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Joshua Moerman 12 years ago
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  1. 1
      thesis/3_SimplicialAbelianGroups.tex
  2. 21
      thesis/4_Constructions.tex

1
thesis/3_SimplicialAbelianGroups.tex

@ -102,6 +102,7 @@ Note that because of the third equation, the degeracy maps $s_i$ are injective.
An element $x \in X_{n+1}$ is \emph{degenerate} if it lies in the image of $s_i : X_n \to X_{n+1}$ for some $i$. An element is called \emph{non-degenerate} if this is not the case. An element $x \in X_{n+1}$ is \emph{degenerate} if it lies in the image of $s_i : X_n \to X_{n+1}$ for some $i$. An element is called \emph{non-degenerate} if this is not the case.
\end{definition} \end{definition}
\begin{lemma} \begin{lemma}
\label{le:non-degenerate}
We can write any $x \in X_n$ uniquely as $x = \beta^\ast y$ for some surjective map $\beta : [n] \epi [m]$ and $y \in X_m$ non-degenerate. We can write any $x \in X_n$ uniquely as $x = \beta^\ast y$ for some surjective map $\beta : [n] \epi [m]$ and $y \in X_m$ non-degenerate.
\end{lemma} \end{lemma}
\begin{proof} \begin{proof}

21
thesis/4_Constructions.tex

@ -51,7 +51,20 @@ $$ \del = d_0|_{N(A)_n}. $$
\end{proof} \end{proof}
We will call this chain complex $N(A)$ the \emph{normalized chain complex} of $A$. We will call this chain complex $N(A)$ the \emph{normalized chain complex} of $A$.
\todo{C: functoriality} \begin{lemma}
The above construction gives a functor $N: \sAb \to \Ch{\Ab}$. Furthermore $N$ is additive.
\end{lemma}
\begin{proof}
Given a map $f: A \to B$ of simplicial abelian groups, we consider the restrictions:
$$ f_n |_{N(A)_n} : N(A)_n \to B_n. $$
Because $f_n$ commutes with the face maps we get:
$$ d_i(f_n(x)) = f_{n-1}(d_i(x)) = 0, $$
for $i>0$ and $x \in N(A)_n$. So the restriction also restircts the codomain, i.e. $f_n |_{N(A)_n} : N(A)_n \to N(B)_n$ is well-defined. Furthermore it commutes with the boundary operator, since $f$ itself commutes with all face maps. This gives functoriality $N(f): N(A) \to N(B)$.
Let $f, g: A \to B$ be two maps, then
$$ N(f+g) = (f+g)|_{N(A)} = f|_{N(A)} + g|_{N(A)} = N(f) + N(g). $$
By recalling that in both categories addition of maps was defined pointwise, we have additivity of $N$.
\end{proof}
\begin{example} \begin{example}
We will look at the normalized chain complex of $\Z[\Delta[0]]$. Recall that it looked like: We will look at the normalized chain complex of $\Z[\Delta[0]]$. Recall that it looked like:
@ -113,10 +126,10 @@ We can extend the above lemmas to a more general statement.
\end{proof} \end{proof}
\begin{lemma} \begin{lemma}
\label{le:decomp4} \label{le:decomp4}
For $\beta \neq \gamma$ we have $\beta^\ast(N(X))_p \cap \gamma^\ast(N(X))_q = 0$. Let $\beta : [n] \epi [m]$ and $\gamma : [n] \epi [m']$ be two maps such that $\beta \neq \gamma$. Then we have $\beta^\ast(N(X))_p \cap \gamma^\ast(N(X))_q = 0$.
\end{lemma} \end{lemma}
\begin{proof} \begin{proof}
Follows from $x = \beta^\ast y$ uniquely for a non-degenerate $y$. \todo{C: proof this in chapter about sAb} Note that $N(X)_i$ only contains non-degenerate $i$-simplices (and $0$). For $x \in \beta^\ast(N(X))_p \cap \gamma^\ast(N(X))_q$ we have $x = \beta^\ast y = \gamma^\ast y'$, where $y$ and $y'$ are non-degenerate. By lemma~\ref{le:non-degenerate} we know that every $n$-simplex is \emph{uniquely} determined by a non-degenerate simplex and a surjective map. For $x \neq 0$ this gives a contradiction.
\end{proof} \end{proof}
Again the former lemma of these two lemmas proofs the existence of a decomposition and the latter proofs the uniqueness. So combining this gives: Again the former lemma of these two lemmas proofs the existence of a decomposition and the latter proofs the uniqueness. So combining this gives:
@ -132,7 +145,7 @@ And by considering $X_n$ as a whole we get:
$X_n = \bigoplus_{[n] \epi [p]} N(X)_p$. $X_n = \bigoplus_{[n] \epi [p]} N(X)_p$.
\end{corollary} \end{corollary}
Using corollary~\ref{cor:decomp} we can proof a nice categorical fact about $N$, which we will use later on. \todo{C: $N$ is add.} Using corollary~\ref{cor:decomp} we can proof a nice categorical fact about $N$, which we will use later on.
\begin{lemma} \begin{lemma}
The functor $N$ is fully faithful, i.e.: The functor $N$ is fully faithful, i.e.:
$$ N : \Hom{\sAb}{A}{B} \iso \Hom{\Ch{\Ab}}{N(A)}{N(B)}. $$ $$ N : \Hom{\sAb}{A}{B} \iso \Hom{\Ch{\Ab}}{N(A)}{N(B)}. $$