update la5

ws2019/la/uebungen/la5.tex

@@ -2,6 +2,7 @@
 
 \usepackage{enumerate}
 \usepackage{array}
+\usepackage{mathtools}
 
 \title{Übungsblatt Nr. 5}
 \author{Christian Merten, Mert Biyikli}
@@ -202,10 +203,10 @@ $V = \text{Abb}\left( \{0, 1, \ldots, n+1\}, K\right)$.
             Bleibt zu zeigen: char$K \not\in \{2, \ldots, n+1\} \iff
             k+1 \neq 0$ $\forall k \in \{0, \ldots, n\} $.
             \begin{align*}
-                &k + 1 \neq 0 \\
-                \stackrel{k \ge  0}{\iff} & k + 1 \neq \text{char}K \\
-                \stackrel{1 \le k + 1 \le n + 1}\iff & \text{char}K = 0 \lor \text{char}K > n + 1 \\
-                \iff & \text{char}K \not\in \{2, \ldots, n+1\}
+                &k + 1 \neq 0 \\[-2mm]
+                \stackrel{\mathclap{\strut k \ge  0}}{\qquad \iff \qquad} &k + 1 \neq \text{char}K \\[-2mm]
+                \stackrel{\mathclap{\strut 1 \le k + 1 \le n + 1}}{\qquad \iff \qquad} &\text{char}K = 0 \lor \text{char}K > n + 1 \\[1mm]
+                \qquad \iff \qquad & \text{char}K \not\in \{2, \ldots, n+1\}
             .\end{align*}
         \end{proof}
     \item Bestimmen Sie $\psi(\text{ker }K) \subset K^{n+2}$.
@@ -239,7 +240,7 @@ $V = \text{Abb}\left( \{0, 1, \ldots, n+1\}, K\right)$.
                     Damit folgt:
                     \[
                         \psi(\text{ker }\partial) =
-                        \{(a, \underbrace{0, \ldots, 0}_{n+1\text{-mal}})  \mid c \in K\} 
+                        \{(a, \underbrace{0, \ldots, 0}_{n+1\text{-mal}})  \mid a \in K\} 
                     .\] 
                 \item $\text{char }K \in \{2, \ldots, n+1\} $. Dann gilt für $k = \text{char }K-1$:
                     \[
@@ -321,12 +322,12 @@ $V = \text{Abb}\left( \{0, 1, \ldots, n+1\}, K\right)$.
                 \begin{align*}
                     (*(f_1 + f_2))(\varphi) &= ((f_1 + f_2)^{*})(\varphi)
                     = \varphi \circ (f_1 + f_2)
-                    = \varphi \circ f_1 + \varphi \circ f_2
+                    \stackrel{\varphi \text{ linear}}{=} \varphi \circ f_1 + \varphi \circ f_2
                     = (*(f_1))(\varphi) + (*(f_2))(\varphi) \\
                     (*(a f_1))(\varphi)
                     &= ((a f_1)^{*})(\varphi)
                     = \varphi \circ (a f_1)
-                    = a (\varphi \circ f_1)
+                    \stackrel{\varphi \text{ linear}}{=} a \cdot (\varphi \circ f_1)
                     = a\cdot (*(f_1))(\varphi)
                 .\end{align*}
             \end{proof}