@ -5,8 +5,7 @@ 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?}
\subsection{Homotopy groups}
\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.:
$$ Z_n(X)=\{ x \in X_n | d_i(x)=\ast\text{ for all } i \leq n \}. $$
@ -16,6 +15,7 @@ When dealing with homotopy in a topological space $X$ we always need a base-poin
d_1(y) &= x' \\
d_i(y) &= \ast\text{ for all } i > 1.
\end{align}
We will call $y$ the \emph{homotopy} and notate $y: x \sim x'$.
\end{definition}
Of course we would like $\sim$ to be an equivalence relation, however this is not true for all simplicial sets. For example there is in general no reason for symmetry, existence of a $1$-simplex $y$ from $x$ to $x'$ does not give us a $1$-simplex $y'$ from $x'$ to $x$. One can give an precise condition on when it is a equivalence relation, the so called Kan-condition. In our case of abelien groups, however, we can prove this directly.
@ -23,15 +23,16 @@ Of course we would like $\sim$ to be an equivalence relation, however this is no
\todo{Htp: Discuss/picturize Kan-condition?}
\begin{lemma}
The relation $\sim$ as defined above is an equivalence relation on $Z_n(X)$.
The relation $\sim$ as defined above is an equivalence relation on $Z_n(X)$. Furthermore it is compatible with addition.
\end{lemma}
\begin{proof}
\todo{Htp: Make this a bit nicer}
\emph{Reflexivity}. Let $x \in Z_n(X)$, define $y = s_0 x$. Now calculate $d_0 y = d_1 y = x$, because of the simplicial equations. And $d_i y =0$ for all $i > 1$, because $x \in Z_n(X)$.
\emph{Reflexivity}. Let $x \in Z_n(X)$, define $y = s_0 x$. By considering the simplicial identities $d_0 s_0=\id$ and $d_1 s_0=\id$, it follows that $d_0 y = d_1 y = x$. Furthermore $d_i y = d_i s_0 x = s_0 d_{i-1} x =0$ for all $i > 1$, because $x \in Z_n(X)$.
\emph{Symmetry}. Let $x, x' \in Z_n(X)$ with $y: x \sim x'$. Define $y' = s_0 x + s_0 x' - y$, then by using linearity: $d_0 y' = x + x' - x = x'$ and $d_1 y' = x + x' - x' = x$. For $i>1$ we again get $d_i y' =0$, because $x \in Z_n(X)$.
\emph{Symmetry}. Let $x, x' \in Z_n(X)$ with $x \sim x'$. Let $y \in X_{n+1}$ such that $d_0 y = x$, $d_1 y = x'$ and $d_i y =0$ for all $i > 1$. Define $y' = s_0 x + s_0 x' - y$, then by using linearity:$d_0y' = x + x' - x = x'$ and $d_1y' = x+ x' - x' = x$. Again we get $d_i y' =0$, because $x \in Z_n(X)$.
\emph{Transitivity}. Let $x_0, x_1, x_2\in Z_n(X)$ with $y: x_0\sim x_1$ and $z: x_1\sim x_2$. Define $w = y + z - s_0 x_1$. By linearity we have$d_0w = x_0+ x_1-x_1= x_0$, similarly $d_1 w = x_2$. Again for $i>1$ we have $d_i w =0$.
\emph{Transitivity}. Let $x_0, x_1, x_2\in Z_n(X)$ with $x_0\sim x_1$ and $x_1\sim x_2$. Let $x, z \in X_{n+1}$ such that ... Define $w = y + z - s_0x_1$.
\emph{Addition}. Let $y: x_0\sim x_1$ and $z: x_2\sim x_3$. Then by linearity $y + g: x_0+ x_2\sim x_1+ x_3$ and $-y: -x_0\sim-x_1$.
\end{proof}
\begin{definition}
@ -51,9 +52,7 @@ Note that this is an abelian group, because $Z_n(X)$ is a subgroup of $X_n$, and
\ker(\del) &= \{ x \in N(X)_n \I\del(x) = 0 \}\\
&= \{ x \in X_n \I d_i(x) = 0 \text{ forall } i > 0 \text{ and } d_0(x) = 0 \}\\
&= \{ x \in X_n \I d_i(x) = 0 \text{ forall } i \leq n \}\\
&= Z_n(X)
\end{align*}
\begin{align*}
&= Z_n(X) \\
\im(\del) &= \{\del(y) \I y \in N(X)_{n+1}\}\\
&= \{ d_0 y \I y \in X_{n+1}, d_i(y) = 0 \text{ for all } i > 0 \}\\
&= \{ x \in N(X)_n \I x \sim 0 \}
@ -65,8 +64,22 @@ Note that this is an abelian group, because $Z_n(X)$ is a subgroup of $X_n$, and
For a chain complex $C$ we have $H_n(C)\iso\pi_n(K(C))$
\end{corollary}
\begin{proof}
By the established equivalence we have:
By the established equivalence we have for any chain complex $C$:
$$\pi_n(K(C))\iso H_n(N(K(C)))\iso H_n(C). $$
\end{proof}
\subsection{Topology}
In section~\ref{sec:Constructions} we saw that we can construct a functor $G: \cat{C}\to\sSet$ if we are provided a functor the other way around. If we can define a functor $F: \DELTA\to\Top$, then for any space $X$ we have a simplicial set $\Hom{\Top}{F-}{X}: \DELTA^{op}\to\Set$. In section~\ref{sec:Chain Complexes}, we already defined the \emph{topological $n$-simplex}$\Delta^n$ and face maps $\delta^i : \Delta^n \mono\Delta^{n+1}$. We can similarly define degeneracy maps $s^i: \Delta^n \to\Delta^{n-1}$ as:
The reader is invited to check the cosimplicial identities himself and conclude that we now have a functor $F: \DELTA\to\Top$, and hence we have a functor $S: \Top\to\sSet$ given by:
$$\text{Sing}(X)_n =\Hom{\Top}{\Delta^n}{X}. $$
Recall construction of the singular chain complex in section~\ref{sec:Chain Complexes}:
$$ C_n(X)=\Z[\Hom{\cat{Top}}{\Delta^n}{X}]. $$
Where the boundary map was given as an alternating sum. Looking more closely we see that this construction decomposes as:
where the last functor is the \emph{unnormalized chain complex}. All the categories involved have a notion of homotopy. In topological spaces this is the known notion where $f, g:X \to Y$ are homotopic if there exists a homotopy $H:I \times X \to Y$ with the appropriate properties. In simplicial sets (or simplicial abelian groups) we only saw the notion of homotopy groups, but there exists a more general notion of homotopy, as discussed in the overview of Friedman \cite{friedman}. And finally in chain complexes we saw homology groups, but this category also has a more general notion of chain homotopy, which can be found in any book on homological algebra such as in the book of Weibel \cite{weibel}.
It is known that for any simplicial abelian group both the normalized and unnormalized chain complex have the same homology groups. More precisely for any simplicial abelian group $X$ we have:
$$ H_n(N(X))\iso H_n(C(X))\quad\text{for all } n \in\N. $$
This is for example proven in \cite[Theorem 4.1]{eilenberg}. So this assures that the homology groups of the singular chain complex of a space are really the homotopy groups of the simplicial abelian group which is in the background.
Joshua Moerman
12 years ago