\documentclass[14pt]{beamer}

% beamer definieert 'definition' al, maar dan engels :(
% fix van:
% http://tex.stackexchange.com/questions/38392/how-to-rename-theorem-or-lemma-in-beamer-to-another-language
\usepackage[dutch]{babel}
\uselanguage{dutch}
\languagepath{dutch}
\deftranslation[to=dutch]{Definition}{Definitie}

\usepackage{array}

\input{../thesis/preamble}

\title{Dold-Kan correspondentie
	\huge $$ \Ch{\cat{Ab}} \simeq \cat{sAb} $$}
\author{Joshua Moerman}
\institute[Radboud Universiteit Nijmegen]{Begeleid door Moritz Groth}
\date{}

\begin{document}


\begin{frame}
	\titlepage
\end{frame}


\begin{frame}
	\frametitle{Wat is $\Ch{\cat{Ab}}$?}
	\begin{definition}
	Een \emph{ketencomplex} $C$ bestaat uit abelse groepen met groepshomomorfisme:
	$$ \cdots \to C_4 \tot{\del_3} C_3 \tot{\del_2} C_2 \tot{\del_1} C_1 \tot{\del_0} C_0 $$

	zodat $\del_n \circ \del_{n+1} = 0$ voor alle $n \in \N$.
	\end{definition}
\end{frame}


\begin{frame}
	\frametitle{Voorbeeld}
	\centering \vspace{-0.5cm}
	Bekijk $\Delta^n \tot{f} X$,\, dwz.\, \raisebox{-.2\height}{\includegraphics{simplex_in_X}}
	\bigskip
	\bigskip

	\includegraphics<1>{singular_chaincomplex1}
	\includegraphics<2>{singular_chaincomplex2}
	\includegraphics<3>{singular_chaincomplex3}
\end{frame}


\begin{frame}
	\frametitle{Interessant?}
	Gegeven een ketencomplex $C$: \\
	$ \cdots \tot{\del_2} C_2 \tot{\del_1} C_1 \tot{\del_0} C_0 $ met $\del_n \circ \del_{n+1} = 0$
	\bigskip\bigskip
	

	Dan geldt $im(\del_{n+1}) \trianglelefteq ker(\del_n)$

	Definieer: $H_n(C) = ker(\del_{n-1}) / im(\del_n)$

	met $ker(\del_{-1}) = C_0$
\end{frame}


\begin{frame}
	\frametitle{Voorbeeld}
	\raisebox{-.2\height}{\includegraphics[width=0.7\textwidth]{singular_chaincomplex_small}}, $ H_1 = \frac{ker(\del_0)}{im(\del_1)} $?
	\bigskip

	\begin{tabular}{m{0.3\textwidth} m{0.7\textwidth}}
		\includegraphics<1>{singular_homology1}
		\includegraphics<2->{singular_homology2}
		&
		$\sigma_1 - \sigma_2 + \sigma_3 \in ker (\del_0) $ \newline
		\visible<2->{$\del_1(\tau) = \sigma_1 - \sigma_2 + \sigma_3 $ \newline
		Dus $ \sigma_1 + \sigma_2 - \sigma_3 \in im (\del_1) $ \newline
		Dus $ 0 = [\sigma_1 - \sigma_2 + \sigma_3] \in H_1 $}
	\end{tabular}
	\bigskip

	\visible<3->{
	\begin{tabular}{m{0.3\textwidth} m{0.7\textwidth}}
		\includegraphics{singular_homology3}
		&
		$ \sigma_1 - \sigma_2 + \sigma_3 \in ker (\del_0) $ \newline
		Maar $ \sigma_1 - \sigma_2 + \sigma_3 \not \in im (\del_1) $ \newline
		Dus $ 0 \neq [\sigma_1 - \sigma_2 + \sigma_3] \in H_1 $
	\end{tabular}
	}

\end{frame}


\begin{frame}
	\frametitle{Dold-Kan Correspondentie}
	\begin{center}
	{\Large $ \Ch{\cat{Ab}} \simeq \cat{sAb} $}

	verder:
	{\Large $$ H_n(N(X)) \iso \pi_n(X) $$}
	waarbij $N : \cat{sAb} \tot{\simeq} \Ch{\cat{Ab}}$.
	\end{center}
\end{frame}


\begin{frame}
	\begin{center}
	\Huge Vragen?
	\end{center}
\end{frame}


\end{document}
