C: made the proof with the sums nicer

thesis/4_Constructions.tex

@@ -4,26 +4,21 @@
 Comparing chain complexes and simplicial abelian groups, we see a similar structure. Both objects consists of a sequence of abelian groups, with maps in between. At first sight simplicial abelian groups have more structure, because there are maps in both directions. It is not clear how to make degeneracy maps given a chain complex, in fact it is already unclear how to define more maps (the face maps) out of one (the boundary one). Constructing a chain complex from a simplicial abelian group on the other hand seems doable.
 
 \subsection{Unnormalized chain complex}
-Given a simplicial abelian group $A$, we have a family of abelian groups $A_n$. We define a group homomorphism $\del_{n-1} : A_n \to A_{n-1}$ as:
+Given a simplicial abelian group $A$, we have a family of abelian groups $A_n$. For every $n>0$ we define a group homomorphism $\del_n : A_n \to A_{n-1}$:
-$$\del_{n-1} = d_0 - d_1 + \ldots + (-1)^n d_n \text{ for every } n > 0.$$
+$$\del_n = d_0 - d_1 + \ldots + (-1)^n d_n.$$
 \begin{lemma}
 	Using $A_n$ as the family of abelian groups and the maps $\del_n$ as boundary maps gives a chain complex.
 \end{lemma}
 \begin{proof}
-	We already have a collection of abelian groups together with maps, so the only thing to proof is $\del_n \circ \del_{n+1} = 0$. This can be done with a calculation:
+	We already have a collection of abelian groups together with maps, so the only thing to proof is $\del_n \circ \del_{n+1} = 0$. This can be done with a calculation.
-
 	\begin{align*}
-		\del_n \circ \del_{n+1} &= \textstyle (\sum_{i=0}^{n+1} (-1)^i d_i) \circ (\sum_{j=0}^{n+2} (-1)^j d_j) \\
+		\del_{n-1} \circ \del_n &= \sum_{i=0}^{n-1}  \sum_{j=0}^{n} (-1)^{i+j} d_i \circ d_j \\
-			&= \textstyle \sum_{i=0}^{n+1}  \sum_{j=0}^{n+2} (-1)^{i+j} (d_i \circ d_j) \\
+			&\eqn{1} \sum_{i=0}^{n-1} \sum_{j=0}^{i} (-1)^{i+j} d_i \circ d_j + \sum_{i=0}^{n-1} \sum_{j=i+1}^{n} (-1)^{i+j} d_i \circ d_j \\
-			&= \textstyle \sum_{i=0}^{n+1} (\sum_{j=0}^{i} (-1)^{i+j} (d_i \circ d_j) + \sum_{j=i+1}^{n+2} (-1)^{i+j} (d_i \circ d_j)) \\
+			&\eqn{2} \sum_{i=0}^{n-1} \sum_{j=0}^{i} (-1)^{i+j} d_i \circ d_j + \sum_{i=0}^{n-1} \sum_{j=i+1}^{n} (-1)^{i+j} d_{j-1} \circ d_i \\
-			&= \textstyle \sum_{i=0}^{n+1} (\sum_{j=0}^{i} (-1)^{i+j} (d_i \circ d_j) + \sum_{j=i+1}^{n+2} (-1)^{i+j} (d_{j-1} \circ d_i)) \\
+			&\eqn{3} \sum_{i=0}^{n-1} \sum_{j=0}^{i} (-1)^{i+j} d_i \circ d_j - \sum_{i=0}^{n-1} \sum_{j=i}^{n-1} (-1)^{i+j} d_j \circ d_i \\
-			&= \textstyle \sum_{i=0}^{n+1}  \sum_{j=0}^{i} (-1)^{i+j} (d_i \circ d_j) - \sum_{i=0}^{n+1} \sum_{j=i}^{n+1} (-1)^{i+j} (d_j \circ d_i) \\
+			&= \sum_{i=0}^{n-1} \sum_{j=0}^{i} (-1)^{i+j} d_i \circ d_j - \sum_{i=0}^{n-1} \sum_{j=0}^{i} (-1)^{i+j} d_i \circ d_j = 0
-			&= \textstyle \sum_{i=0}^{n+1}  \sum_{j=0}^{i} (-1)^{i+j} (d_i \circ d_j) - \sum_{i=0}^{n+1} \sum_{j=i}^{n+1} (-1)^{i+j} (d_j \circ d_i) \\
-			&= \textstyle \sum_{i=0}^{n+1}  \sum_{j=0}^{i} (-1)^{i+j} (d_i \circ d_j) - \sum_{j=0}^{n+1} \sum_{i=j}^{n+1} (-1)^{i+j} (d_i \circ d_j) \\
-			&= 0.
 	\end{align*}
-
+	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.
-	We split the inner sum in two halves and we use the simplicial equations on the second sum. Then we do a shift of indices and change the roles of $i$ and $j$ in the second sum, so that the sums have an equal range and cancel out. So indeed this is a chain complex.
 \end{proof}
 
 This construction gives a functor $C : \sAb \to \Ch{\Ab}$\todo{C: prove this? Is it an adjunction?}. And in fact we already used it in the construction of the singular chaincomplex, where we defined the boundary maps as $\del(\sigma) = \sigma \circ d_0 - \sigma \circ d_1 + \ldots + (-1)^{n+1} \sigma \circ d_{n+1}$ (on generators). The terms $\sigma \circ d_i$ are the maps given by the $\mathbf{Hom}$-functor from $\Top$ to $\Set$, in fact this $\mathbf{Hom}$-functor can be used to get a functor $Sing : \Top \to \sSet$, applying the free abelain group pointwise give a functor $\Z^\ast : \sSet \to \sAb$, and finally using the functor $C$ gives the singular chain complex.

thesis/preamble.tex

@@ -48,6 +48,9 @@
 \newcommand{\del}{\partial}
 \newcommand{\nsubgrp}{\trianglelefteq}
 
+\newcommand{\eqn}[1]{\stackrel{(#1)}{=}}
+\newcommand{\refeqn}[1]{{\small $(#1)$}}
+
 \newcommand{\todo}[1]{
 	\addcontentsline{tdo}{todo}{\protect{#1}}
 	$\ast$ \marginpar{\tiny $\ast$ #1}