In other words a chain complex is the following diagram.
$$\cdots\to C_4\to C_3\to C_2\to C_1\to C_0$$
Of course we can make this more general by taking for example $R$-modules instead of abelian groups. We will later see which kind of algebraic objects make sense to use in this definition. The boundary operators give rise to certain subgroups, because all groups are abelian, subgroups are normal subgroups.
Of course we can make this more general by taking for example $R$-modules instead of abelian groups. We will later see which kind of algebraic objects make sense to use in this definition\todo{Ch: Will I discuss ab. cat. ?}. The boundary operators give rise to certain subgroups, because all groups are abelian, subgroups are normal subgroups.
\begin{definition}
Given a chain complex $C$ we define the following subgroups:
@ -5,6 +5,8 @@ We've already seen homology in chain complexes. We can of course now translate t
When dealing with homotopy in a topological space $X$ we always need a base-point $\ast\in X$. This is also the case for homotopy in simplicial sets. We will notate the chosen base-point of a simplicial set $X$ with $\ast\in X_0$. Note that it is a $0$-simplex, but in fact the base-point is present in all sets $X_n$, because we can consider its degenerate simplices $s_0(\ldots(s_0(\ast))\ldots)\in X_n$, we will also denote these elements as $\ast$. Of course in our situation we are concerned about simplicial abelien groups, where there is an obvious choice for the base-point, namely $0$.
\todo{Htp: Do I want to define homotopy between maps?}
\begin{definition}
Given a simplicial set $X$ with base-point $\ast$, we define $Z_n(X)$ to be the set of $n$-simplices with the base-point as boundary, i.e.:
$$ Z_n(X)=\{ x \in X_n | d_i(x)=\ast\text{ for all } i < n \}. $$
Joshua Moerman
12 years ago