From adbb5f5d4aec42d458d38bff517ad6dbccce8cf1 Mon Sep 17 00:00:00 2001
From: Joshua Moerman <lakseru@gmail.com>
Date: Tue, 28 May 2013 16:19:18 +0200
Subject: [PATCH] Htp: basics of eq. rel.

---
 thesis/5_Homotopy.tex | 26 ++++++++++++++++++--------
 1 file changed, 18 insertions(+), 8 deletions(-)

diff --git a/thesis/5_Homotopy.tex b/thesis/5_Homotopy.tex
index 0d5b397..4dc2239 100644
--- a/thesis/5_Homotopy.tex
+++ b/thesis/5_Homotopy.tex
@@ -9,21 +9,31 @@ When dealing with homotopy in a topological space $X$ we always need a base-poin
 
 \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.:
-	$$ Z_n(X) = \{ x \in X_n | d_i(x) = \ast \text{ for all } i < n \}. $$
+	$$ Z_n(X) = \{ x \in X_n | d_i(x) = \ast \text{ for all } i \leq n \}. $$
 	For two $n$-simplices $x, x' \in Z_n(X)$, we define $x \sim x'$ if there exists $y \in X_{n+1}$ such that:
 	\begin{align}
-		d_{n+1}(y) &= x' \\
-		d_n(y) &= x \\
-		d_i(y) &= \ast \text{ for all } i < n.
+		d_0(y) &= x \\
+		d_1(y) &= x' \\
+		d_i(y) &= \ast \text{ for all } i > 1.
 	\end{align}
 \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.
 
-\todo{Htp: Insert prove $\sim$ is eq. rel. for $\sAb$.}
-
 \todo{Htp: Discuss/picturize Kan-condition?}
 
+\begin{lemma}
+	The relation $\sim$ as defined above is an equivalence relation on $Z_n(X)$.
+\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{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 $x_0 \sim x_1$ and $x_1 \sim x_2$. Let $x, z \in X_{n+1}$ such that ... Define $w = y + z - s_0 x_1$.
+\end{proof}
+
 \begin{definition}
 	Given a simplicial abelian group $X$, we define the $n$-th homotopy group as:
 	$$ \pi_n(X) = Z_n(X) / \sim. $$
@@ -39,13 +49,13 @@ Note that this is an abelian group, because $Z_n(X)$ is a subgroup of $X_n$, and
 	By writing out the definitions of the $n$-cycles and $n$-boundaries of the normalized chain complex, we see:
 	\begin{align*}
 		\ker(\del) &= \{ x \in N(X)_n \I \del(x) = 0 \} \\
-			&= \{ x \in X_n \I d_i(x) = 0 \text{ forall } i < n \text{ and } d_n(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*}
 		\im(\del) &= \{ \del(y) \I y \in N(X)_{n+1} \} \\
-			&= \{ d_{n+1} \I y \in X_{n+1}, d_i(y) = 0 \text{ for all } i \leq n \} \\
+			&= \{ 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 \}
 	\end{align*}
 	So we see that $\pi_n(X) = Z_n(X) / \sim = \ker(\del) / \im(\del) = H_n(N(X))$.