Fix small types
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@ -225,7 +225,7 @@ $\pi = (2 3 1), f(X_1, X_2, X_3) = X_1-X_2+X_1X_3 \implies \pi f(X_1, X_2, X_3)
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\pi(i)<\pi(j)}}
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(X_{\pi(j)}-X_{\pi(i)})\Bigr)
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\Bigl(\prod_{\substack{i<j \\
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\pi(j)<\pi(i)}}(X_{\pi(i)}-X_{\pi(j)})\Bigr) \\
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\pi(j)<\pi(i)}}(X_{\pi(j)}-X_{\pi(i)})\Bigr) \\
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& = (-1)^{\abs{\{(i, j)\in[n]\times[n]:i<j\land\pi(i)>\pi(j)\}}}
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\prod_{i<j}(X_j-X_i) \\
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& = s(\pi)f(X_1, \dots, X_n) \text{ mit } \\
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@ -404,7 +404,7 @@ $\varphi$ alternierend und $a_i = a_j$ für $i\neq j \implies \varphi(a_1, \dots
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\]
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z.Z.: $\varphi$ alternierend. Seien $b_1, \dots, b_n$ linear abhängig.\\
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O.B.d.A. $b_1=\mu_2b_2+\cdots+\mu_nb_n$. Dann gilt
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\[\varphi(b_1, \dots, b_n) = \sum_{j=2}^{n}\mu j \varphi(b_j, b_2, \dots, b_n)\]
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\[\varphi(b_1, \dots, b_n) = \sum_{j=2}^{n}\mu_j \varphi(b_j, b_2, \dots, b_n)\]
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Es genügt also zu zeigen, dass
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$\varphi(b_1, \dots, b_n) = 0$ falls $b_1 = b_i, i\in\{2, \dots, n\}$.
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Dann gilt aber $\lambda_{1j}=\lambda_{ij} \forall j$.
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@ -4075,7 +4075,7 @@ Wir haben eine echte Verallgemeinerung.
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Sei $\alpha \in \Hom(V, W)$.
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\begin{itemize}
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\item $\alpha$ injektiv $\implies \alpha^\dagger = (\alpha^* \circ \alpha)^{{}^{-1}} \circ \alpha^*$
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\item $\alpha$ surjektiv $\alpha^\dagger = \alpha^* \circ (\alpha \circ \alpha^*)^{{}^{-1}}$
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\item $\alpha$ surjektiv $\implies \alpha^\dagger = \alpha^* \circ (\alpha \circ \alpha^*)^{{}^{-1}}$
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\end{itemize}
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\end{satz}
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\begin{proof}
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