christian/uni
1
0
Fork 0
Code Issues Pull Requests Releases Wiki Activity

update la5

Browse Source
This commit is contained in:
christian
2019-11-20 20:29:11 +01:00
parent 902c7802fa
commit 482802ee2a
2 changed files with 5 additions and 4 deletions
Download Patch File Download Diff File Expand all files Collapse all files
ws2019/la/uebungen/la5.pdf
BIN
View File
Binary file not shown.
ws2019/la/uebungen/la5.tex
+5 -4
View File
@@ -231,7 +231,7 @@ $V = \text{Abb}\left( \{0, 1, \ldots, n+1\}, K\right)$.
Aus (c) folgt: $k+1 \neq 0 \iff \text{char K} \not\in \{2, \ldots, n+1\} $. Aus (c) folgt: $k+1 \neq 0 \iff \text{char K} \not\in \{2, \ldots, n+1\} $.
\begin{enumerate}[(i)] \begin{enumerate}[(i)]
\item $\text{char }K \not\in \{2, \ldots, n+1\} $. Dann ist $k + 1 \neq 0$, d.h. \item $\text{char }K \not\in \{2, \ldots, n+1\} $: Dann ist $k + 1 \neq 0$, d.h.
\begin{align*} \begin{align*}
&f(k+1) = 0 \text{ } \forall k \in \{0, 1, \ldots, n\} \\ &f(k+1) = 0 \text{ } \forall k \in \{0, 1, \ldots, n\} \\
\implies &f(k) = 0 \text{ } \forall k \in \{1, \ldots, n+1\} \\ \implies &f(k) = 0 \text{ } \forall k \in \{1, \ldots, n+1\} \\
@@ -242,7 +242,7 @@ $V = \text{Abb}\left( \{0, 1, \ldots, n+1\}, K\right)$.
\psi(\text{ker }\partial) = \psi(\text{ker }\partial) =
\{(a, \underbrace{0, \ldots, 0}_{n+1\text{-mal}}) \mid a \in K\} \{(a, \underbrace{0, \ldots, 0}_{n+1\text{-mal}}) \mid a \in K\}
.\] .\]
\item $\text{char }K \in \{2, \ldots, n+1\} $. Dann gilt für $k = \text{char }K-1$: \item $\text{char }K \in \{2, \ldots, n+1\} $: Dann gilt für $k = \text{char }K-1$:
\[ \[
k + 1 = \text{char } K - 1 + 1 = \text{char } K = 0_K k + 1 = \text{char } K - 1 + 1 = \text{char } K = 0_K
.\] .\]
@@ -346,9 +346,10 @@ $V = \text{Abb}\left( \{0, 1, \ldots, n+1\}, K\right)$.
.\end{align*} .\end{align*}
Wegen $f$ surjektiv gilt: $V = f(U)$ und damit: Wegen $f$ surjektiv gilt: $V = f(U)$ und damit:
\begin{align*} \begin{align*}
\forall v \in V\colon \varphi_1(v) = \varphi_2(v) & \forall v \in V\colon \varphi_1(v) = \varphi_2(v)
.\end{align*} .\end{align*}
$\implies \varphi_1 = \varphi_2$ $\implies \varphi_1 = \varphi_2$ \\
$\implies f^{*}$ injektiv
\end{proof} \end{proof}
\end{enumerate} \end{enumerate}
\end{aufgabe} \end{aufgabe}