add la5
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\documentclass{../../../lecture}
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\usepackage{enumerate}
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\begin{document}
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\begin{aufgabe}
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Es sei $K$ Körper, $M$ eine Menge und $m_0 \in M$ ein fest gewähltes
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Element. In \\$V = \text{Abb}(M, K)$ betrachten wir die Teilmengen
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$U = \{f \in V \mid f(m_0) = 0\} $ und
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\\$W = \{f \in V \mid \forall x, y \in M \colon f(x) = f(y)\} $
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Zunächst: $K$ ist K-Vektorraum mit $(K, +, 0)$. Damit wird
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$V = \text{Abb}(M, K)$ zum Vektorraum.
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\begin{enumerate}[a)]
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\item Beh.: $U \subset V$ ist Untervektorraum.
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\begin{proof}
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Seien $f_1, f_2 \in U$, $a \in K$ beliebig. Zu zeigen:
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$(f_1 + f_2)(m_0) = 0 $ und $(a f_1)(m_0) = 0$.
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\[
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(f_1 + f_2)(m_0) = f_1(m_0) + f_2(m_0) = 0 + 0 = 0
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.\] $\implies (f_1 + f_2) \in U$.
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\[
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(a f_1)(m_0) = a f_1(m_0) = a \cdot 0 = 0
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.\] $\implies (a f_1) \in U$.
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\end{proof}
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Beh.: $W \subset V$ ist Untervektorraum
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\begin{proof}
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Seien $f_1, f_2 \in W$, $a \in K$ und $x, y \in M$ beliebig.
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Zu zeigen:
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$(f_1 + f_2)(x) = (f_1 + f_2)(y)$
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und $(a f_1)(x) = (a f_1)(y)$.
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\begin{align*}
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(f_1 + f_2)(x) = f_1(x) + f_2(x) = f_1(y) + f_2(y)
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= (f_1 + f_2)(y)
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.\end{align*}
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$\implies (f_1 + f_2) \in W$.
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\[
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(a f_1)(x) = a f_1(x) = a f_1(y) = (a f_1)(y)
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.\] $\implies (a f_1) \in W$.
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\end{proof}
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\item Beh.: $U \cap W = \{0\} $
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\begin{proof}
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Sei $f \in U \cap W$ beliebig:
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\begin{align*}
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&\forall m \in M \colon f(m) = f(m_0) \land f(m_0) = 0 \\
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\implies &\forall m \in M \colon f(m) = 0 \\
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\implies &f = 0
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.\end{align*}
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\end{proof}
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\item Beh.: $V = U + W$
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\begin{proof}
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Sei $f \in V$ beliebig.
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Zu zeigen: $\exists u \in U, \exists w \in W \colon f = u + w$
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Dann wähle $u \in U$, s.d.
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\[
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u(m) = \begin{cases}
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f(m) - f(m_0) & m \neq m_0 \\
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0 & m = m_0
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\end{cases}
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.\] und $w \in W$, s.d.
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\[
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w(m) = f(m_0) \text{ } \forall m \in M
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.\]
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Damit folgt:
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\[
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f(m) = u(m) + w(m) = \begin{cases}
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f(m) - f(m_0) + f(m_0) = f(m) & m \neq m_0 \\
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0 + f(m_0) = f(m_0) & m = m_0
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\end{cases}
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.\]
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\end{proof}
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\end{enumerate}
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\end{aufgabe}
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\begin{aufgabe}
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Es sei $K$ ein Körper,
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$U = \text{Abb}\left( \{0, 1, \ldots, n\}, K\right) $ und
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$V = \text{Abb}\left( \{0, 1, \ldots, n+1\}, K\right)$.
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\begin{align*}
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\psi\colon V &\to K^{n+2} \\
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f &\mapsto \left( f(0), f(1), \ldots, f((n+1) \right) \\
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\partial\colon V &\to U \\
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f &\mapsto \left( i \mapsto (i + 1) \cdot f(i+1) \right)
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.\end{align*}
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\begin{enumerate}[a)]
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\item Beh.: $\psi$ ist linear.
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\begin{proof}
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Seien $v_1, v_2 \in V$, $a \in K$ beliebig.
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\begin{align*}
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\psi(v_1 + v_2) &= \left( (v_1+v_2)(0), (v_1 + v_2)(1), \ldots, (v_1+v_2)(n+1) \right) \\
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&= \left( v_1(0) + v_2(0), v_1(1) + v_2(1), \ldots, v_1(n+1) + v_2(n+1) \right) \\
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&= \left( v_1(0), v_1(1), \ldots, v_1(n+1) \right)
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+ \left( v_2(0), v_2(1), \ldots, v_2(n+1) \right) \\
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&= \psi(v_1) + \psi(v_2)
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.\end{align*}
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\begin{align*}
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\psi(a v_1) &= \left(a v_1(0), a v_1(1), \ldots, a v_1(n+1)\right) \\
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&= a (v_1(0), v_1(1), \ldots, v_1(n+1) \\
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&= a \psi(v_1)
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.\end{align*}
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\end{proof}
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Beh.: $\partial$ ist linear.
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\begin{proof}
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Seien $v_1, v_2 \in V$, $a \in K$ und $i \in \{0, 1, \ldots, n\} $ beliebig.
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\begin{align*}
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\partial(v_1+v_2)(i) &= (i+1) \cdot (v_1 + v_2)(i+1) \\
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&= (i+1) \cdot (v_1(i+1) + v_2(i+1)) \\
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&= (i+1)v_1(i+1) + (i+1)v_2(i+1) \\
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&= \partial(v_1)(i) + \partial(v_2)(i)
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.\end{align*}
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\begin{align*}
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\partial(a v_1)(i) &= (i + 1)(a v_1)(i+1) \\
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&= a (i+1) v_1 (i+1) \\
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&= a \partial(v_1)(i)
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.\end{align*}
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\end{proof}
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\item Beh.: $\psi$ ist Isomorphismus.
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\begin{proof}
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Zu zeigen: $\psi$ ist bijektiv.
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Seien $v_1, v_2 \in V$ mit $\psi(v_1) = \psi(v_2)$. Dann
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\begin{align*}
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&\psi(v_1) = \left( f_1(0), f_1(1), \ldots, f_1(n+1) \right)
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= \left( f_2(0), f_2(1), \ldots, f_2(n+1) \right) = \psi(v_2)\\
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\implies& f_1(k) = f_2(k) \text{ }\forall k \in \{0, \ldots, n+1\} \\
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\implies& f_1 = f_2
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.\end{align*}
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$\implies \psi$ ist injektiv.
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Sei $c = (c_0, \ldots, c_{n+1}) \in K^{n+2}$, dann ex. ein $f \in V$, s.d.
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\begin{align*}
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&f(k) = c_k \text{ } \forall k \in \{0, \ldots, n+1\} \\
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\implies &\psi(f) = c
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.\end{align*}
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$\implies \psi$ ist surjektiv.
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\end{proof}
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\item Beh.: $\partial$ surjektiv
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$\iff \text{char}K \notin \{2, \ldots, n+1\} $
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\begin{proof}
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Damit $\partial$ surjektiv ist, muss für alle $u \in U$
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ein $v \in V$ existieren, s.d. $\partial(v) = u$.
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Sei $u \in U, k \in \{0, \ldots, n\}$ beliebig, dann muss
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für $v$ gelten:
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\begin{align*}
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&\partial(v)(k) = (k + 1) \cdot v(k+1) = u(k)
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.\end{align*}
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Dies ist genau dann wohldefiniert, wenn $k+1 \neq 0$, denn genau
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dann ex. ein Inverses zu $k+1$ und damit:
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\begin{align*}
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&v(k+1) = (k+1)^{-1} \cdot u(k)
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.\end{align*}
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Bleibt zu zeigen: char$K \not\in \{2, \ldots, n+1\} \iff
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k+1 \neq 0$ $\forall k \in \{0, \ldots, n\} $.
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\begin{align*}
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&k + 1 \neq 0 \\
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\stackrel{k > 0}{\iff} & k + 1 \neq \text{char}K \\
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\stackrel{0 \le k \le n}\iff & \text{char}K = 0 \lor \text{char}K > n + 1 \\
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\iff & \text{char}K \not\in \{2, \ldots, n+1\}
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.\end{align*}
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\end{proof}
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\item Beh.: $\psi(\text{ker }\partial) =
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\left\{ (c, \underbrace{0, \ldots, 0}_{n+1\text{-mal}}) \mid c \in K\right\} $
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\begin{proof}
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Zunächst: $\text{ker }\partial$.
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Damit $r \in V$ im Kern von $\partial$ liegt, muss gelten:
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$\partial(r)(k) = 0$ $\forall k \in \{0, \ldots, n\}$
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\begin{align*}
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&\partial(r)(k) = (k+1) \cdot r(k+1) \\
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\stackrel{k+1 \neq 0}{\implies} &r(k+1) = 0
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.\end{align*}
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Damit: $r(k) = 0$ $\forall k \in \{1, \ldots, n+1\} $.
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\begin{align*}
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\psi(r) &= \left( r(0), r(1), \ldots, r(n+1) \right) \\
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&= (c, \underbrace{0, \ldots, 0}_{n+1\text{-mal}}) \text{ } \forall c \in K
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.\end{align*}
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Das heißt:
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\[
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\psi(\text{ker }\partial) = \left\{ (c, \underbrace{0, \ldots, 0}_{n+1\text{-mal}}) \mid c \in K\right\}
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.\]
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\end{proof}
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\end{enumerate}
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\end{aufgabe}
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\end{document}
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