Thesis: some more diagrams

thesis/2_ChainComplexes.tex

@@ -36,9 +36,9 @@ In order to see why we are interested in the construction of homology groups, we
 	The topology on $\Delta^n$ is the subspace topology.
 \end{definition}
 
-In particular $\Delta^0$ is simply a point, $\Delta^1$ a line and $\Delta^2$ a triangle. There are nice inclusions $\Delta^n \inject \Delta^{n+1}$ which we need later on. For any $n \in \N$ we define:
+In particular $\Delta^0$ is simply a point, $\Delta^1$ a line and $\Delta^2$ a triangle. There are nice inclusions $\Delta^n \mono \Delta^{n+1}$ which we need later on. For any $n \in \N$ we define:
 \begin{definition}
-	For $i \in \{0, \ldots, n+1\}$ the $i$-th face map $\delta^i : \Delta^n \inject \Delta^{n+1}$ is defined as:
+	For $i \in \{0, \ldots, n+1\}$ the $i$-th face map $\delta^i : \Delta^n \mono \Delta^{n+1}$ is defined as:
 	$$ \delta^i (x_0, \ldots, x_n) = (x_0, \ldots, x_{i-1}, 0, x_{i+1}, \ldots, x_n) \text{ for all } x \in \Delta^n.$$
 \end{definition}
 

thesis/3_SimplicialAbelianGroups.tex

@@ -115,20 +115,22 @@ Note that indeed $\Hom{\DELTA}{X}{[n]} \in \Set$, because the collection of morp
 		\delta^1(\delta_0\sigma_0) &= \delta_0 \sigma_0 \delta_1 = \delta_0 \\
 		\delta^1(\delta_1\sigma_0) &= \delta_0 \sigma_0 \delta_1 = \delta_1.
 	\end{align*}
-	\begin{tikzpicture}
+
+	$$ \Delta[1] =
+	\begin{tikzpicture}[baseline=-0.5ex]
 	\matrix (m) [matrix of math nodes] { 
-		\Delta[1] =  & \{x_0, x_1\} & \{\sigma^0 x_0, \id, \sigma^0 x_1\} & \{ \} & \cdots \\
+		\{x_0, x_1\} & \{\sigma^0 x_0, \id, \sigma^0 x_1\} & \{ \} & \cdots \\
 	}; 
 
-	\foreach \r in {-5, 5} \draw [raise line=\r, <-] (m-1-2) -> (m-1-3);
-	\foreach \r in {0} \draw [raise line=\r, ->] (m-1-2) -> (m-1-3);
+	\foreach \r in {-5, 5} \draw [raise line=\r, <-] (m-1-1) -> (m-1-2);
+	\foreach \r in {0} \draw [raise line=\r, ->] (m-1-1) -> (m-1-2);
 
-	\foreach \r in {-10, 0, 10} \draw [raise line=\r, <-] (m-1-3) -> (m-1-4);
-	\foreach \r in {-5, 5} \draw [raise line=\r, ->] (m-1-3) -> (m-1-4);
+	\foreach \r in {-10, 0, 10} \draw [raise line=\r, <-] (m-1-2) -> (m-1-3);
+	\foreach \r in {-5, 5} \draw [raise line=\r, ->] (m-1-2) -> (m-1-3);
 
-	\foreach \r in {-15, -5, 5, 15} \draw [raise line=\r, <-] (m-1-4) -> (m-1-5);
-	\foreach \r in {-10, 0, 10} \draw [raise line=\r, ->] (m-1-4) -> (m-1-5);
-	\end{tikzpicture}
+	\foreach \r in {-15, -5, 5, 15} \draw [raise line=\r, <-] (m-1-3) -> (m-1-4);
+	\foreach \r in {-10, 0, 10} \draw [raise line=\r, ->] (m-1-3) -> (m-1-4);
+	\end{tikzpicture}.$$
 \end{example}
 
 As we are interested in simplicial abelian group, it would be nice to make these $n$-simplices into simplicial abelian groups. We have seen how to make an abelian group out of any set using the free abelian group. We can use this functor $\Z[-] : \Set \to \Ab$ to induce a functor $\Z^\ast[-] : \sSet \to \sAb$ as shown in the diagram~\ref{fig:diagram_Z}.

thesis/4_Constructions.tex

@@ -52,4 +52,41 @@ Now this is a functor, because it is a composition of functors. Furthermore it i
 $$ \Hom{A}{F \Z^{\ast} \Delta (-)}{-} : A \to \sAb $$
 where we are supposed to fill in the second argument first, leaving us with a simplicial abelian group.
 
-Now we know that $\Ch{\Ab}$ is an abelian group and we have actually two functors $C, N : \sAb \to \Ch{\Ab}$, so we now have functors from $\Ch{\Ab} \to \sAb$.
+Now we know that $\Ch{\Ab}$ is an abelian group and we have actually two functors $C, N : \sAb \to \Ch{\Ab}$, so we now have functors from $\Ch{\Ab} \to \sAb$. Of course we will be interested in the one using $N$. So we define the functor:
+$$ K(C) = \Hom{\Ch{\Ab}}{N\Z^\ast\Delta[-]}{C} \in \sAb. $$
+This definitions is very abstract, but luckily we can also give a more explicit definition. By writing it out for low dimensions we see:
+
+$$ K(C)_0 = \Hom{\Ch{\Ab}}{N\Z^\ast\Delta[0]}{C} = \Bigg\{
+\begin{tikzpicture}[baseline=-0.5ex]
+	\matrix (m) [matrix of math nodes, row sep=1em, column sep=1em] { 
+		\cdots  & 0 & 0 & \Z  \\
+		\cdots  & C_2 & C_1 & C_0 \\
+	}; 
+
+	\foreach \x in {1, 2}
+		\foreach \i/\j in {1/2, 2/3, 3/4} \draw[->] (m-\x-\i) -- (m-\x-\j);
+	
+	\foreach \i/\j in {2/2, 3/1, 4/0} \draw[->] (m-1-\i) -- node {$f_\j$} (m-2-\i);
+\end{tikzpicture}
+\Bigg\} \iso C_0, $$
+because for $f_1, f_2, \ldots$ there is now choice at all, and for $f_0 : \Z \to C_0$ we only have to choose an image for $1 \in \Z$. In the next dimension we see:
+
+$$ K(C)_1 = \Hom{\Ch{\Ab}}{N\Z^\ast\Delta[1]}{C} = \Bigg\{
+\begin{tikzpicture}[baseline=-0.5ex]
+	\matrix (m) [matrix of math nodes, row sep=1em, column sep=1em] { 
+		\cdots  & 0 & \Z & \Z^2  \\
+		\cdots  & C_2 & C_1 & C_0 \\
+	}; 
+
+	\foreach \x in {1, 2}
+		\foreach \i/\j in {1/2, 2/3, 3/4} \draw[->] (m-\x-\i) -- (m-\x-\j);
+	
+	\foreach \i/\j in {2/2, 3/1, 4/0} \draw[->] (m-1-\i) -- node {$f_\j$} (m-2-\i);
+\end{tikzpicture}
+\Bigg\} \iso C_1 \oplus C_0, $$
+
+because again we can choose $f_1$ anyway we want, which gives us $C_1$. But then we are forced to choose $f_0(x, x) = \del(f_1(x))$ for all $x \in \Z$, so we are left with choosing an element $c \in C_0$ for defining $f(1,-1) = c$. Adding this gives $C_1 \oplus C_0$. This pattern can be continued and gives the following result:
+
+\begin{proposition}
+	For any chain complex $C$ we have $K(C)_n \iso \bigoplus_{[n] \epi [p]} C_p$.
+\end{proposition}

thesis/preamble.tex

@@ -40,7 +40,8 @@
 \newcommand{\eps}{\varepsilon}
 \newcommand{\I}{\,\mid\,}
 \newcommand{\then}{\Rightarrow}
-\newcommand{\inject}{\hookrightarrow}
+\newcommand{\mono}{\hookrightarrow}
+\newcommand{\epi}{\twoheadrightarrow}
 \newcommand{\del}{\partial}
 \newcommand{\nsubgrp}{\trianglelefteq}