update la5
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@@ -231,7 +231,7 @@ $V = \text{Abb}\left( \{0, 1, \ldots, n+1\}, K\right)$.
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Aus (c) folgt: $k+1 \neq 0 \iff \text{char K} \not\in \{2, \ldots, n+1\} $.
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\begin{enumerate}[(i)]
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\item $\text{char }K \not\in \{2, \ldots, n+1\} $. Dann ist $k + 1 \neq 0$, d.h.
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\item $\text{char }K \not\in \{2, \ldots, n+1\} $: Dann ist $k + 1 \neq 0$, d.h.
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\begin{align*}
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&f(k+1) = 0 \text{ } \forall k \in \{0, 1, \ldots, n\} \\
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\implies &f(k) = 0 \text{ } \forall k \in \{1, \ldots, n+1\} \\
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@@ -242,7 +242,7 @@ $V = \text{Abb}\left( \{0, 1, \ldots, n+1\}, K\right)$.
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\psi(\text{ker }\partial) =
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\{(a, \underbrace{0, \ldots, 0}_{n+1\text{-mal}}) \mid a \in K\}
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.\]
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\item $\text{char }K \in \{2, \ldots, n+1\} $. Dann gilt für $k = \text{char }K-1$:
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\item $\text{char }K \in \{2, \ldots, n+1\} $: Dann gilt für $k = \text{char }K-1$:
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\[
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k + 1 = \text{char } K - 1 + 1 = \text{char } K = 0_K
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.\]
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@@ -346,9 +346,10 @@ $V = \text{Abb}\left( \{0, 1, \ldots, n+1\}, K\right)$.
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.\end{align*}
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Wegen $f$ surjektiv gilt: $V = f(U)$ und damit:
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\begin{align*}
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\forall v \in V\colon \varphi_1(v) = \varphi_2(v)
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& \forall v \in V\colon \varphi_1(v) = \varphi_2(v)
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.\end{align*}
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$\implies \varphi_1 = \varphi_2$
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$\implies \varphi_1 = \varphi_2$ \\
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$\implies f^{*}$ injektiv
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\end{proof}
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\end{enumerate}
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\end{aufgabe}
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