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Presentation: one more image, and refinements. Done?

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Joshua Moerman 11 years ago
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3 changed files with 960 additions and 10 deletions
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  1. BIN
      images/singular_chaincomplex_small.pdf
  2. 951
      images/singular_chaincomplex_small.svg
  3. 19
      presentation/presentation.tex

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19
presentation/presentation.tex
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@ -40,7 +40,7 @@
\begin{frame}
\frametitle{Voorbeeld}
\centering \vspace{-0.5cm}
Bekijk $\Delta^n \tot{f} X$,\, dwz...\, \raisebox{-.2\height}{\includegraphics{simplex_in_X}}
Bekijk $\Delta^n \tot{f} X$,\, dwz.\, \raisebox{-.2\height}{\includegraphics{simplex_in_X}}
\bigskip
\bigskip
@ -52,31 +52,30 @@
\begin{frame}
\frametitle{Interessant?}
Gegeven een ketencomplex $C$:
$$ \cdots \to C_4 \tot{\del_3} C_3 \tot{\del_2} C_2 \tot{\del_1} C_1 \tot{\del_0} C_0 $$
met $\del_n \circ \del_{n+1} = 0$
\bigskip
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 $ker(\del_0) = C_0$ per conventie)
met $ker(\del_{-1}) = C_0$
\end{frame}
\begin{frame}
\frametitle{Voorbeeld}
$ \cdots \to C_1 \tot{\del_0} C_0 $, wat is $ H_1 = \frac{ker(\del_0)}{im(\del_1)} $?
\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
$\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}