Htp: made the lemmas better looking, a general note about singular chain complex

thesis/5_Homotopy.tex

@@ -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 $y = s_0 x$. 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{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{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: $d_0 y' = x + x' - x = x'$ and $d_1 y' = x + x' - x' = x$. 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 $d_0 w = x_0 + x_1 -x_1 = x_0$, 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 $w = y + z - s_0 x_1$.
+	\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)
+			&= Z_n(X) \\
-	\end{align*}
-	\begin{align*}
 		\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:
+$$ s^i(x_0, \ldots, x_n) = (x_0, \ldots, x_i + x_{i+1}, \ldots, x_n) \in \Delta^{n-1}. $$
+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:
+$$ C: \Top \tot{\text{Sing}} \sSet \tot{\Z^\ast} \sAb \tot{C} \Ch{\Ab}, $$
+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.

thesis/references.bib

@@ -48,6 +48,18 @@ $p14:	3:
 	publisher={JSTOR}
 }
 
+% theorem 4.1
+@article{eilenberg,
+	title={On the groups H ($\pi$, n), I},
+	author={Eilenberg, Samuel and Lane, Saunders Mac},
+	journal={The Annals of Mathematics},
+	volume={58},
+	number={1},
+	pages={55--106},
+	year={1953},
+	publisher={JSTOR}
+}
+
 %p303:	5.5
 @book{rotman,
 	title={An introduction to homological algebra},