Exercise 2.15

Question

In the situation of Exercise \href{exc:2.14}{$14$}, show that every element of $M$ can be written in the form $\mu_i(x_i)$ for some $i \in I$ and some $x_i \in M_i$.
  \par Show also that if $\mu_i(x_i) = 0$ then there exists $j \ge i$ such that $\mu_{ij}(x_i) = 0$ in $M_j$.

Amit Awasthi · Accepted Answer

\begin{proof}
  Any element in $M$ is of the form $\sum_{i \in F} x_i + D$ for some finite subset $F \subset I$. Since $I$ is directed and $F$ is finite, there exists some $j \in I$ such that $i \le j$ for all $i \in F$. By definition of $D$, $\sum_{i \in F} x_i + D = \sum_{i \in F} \mu_{ij}(x_i) + D$. Each $\mu_{ij}(x_i) \in M_j$, hence we can write
  \begin{equation*}
    \sum_{i \in F} \mu_{ij}(x_i) + D = \mu_{j}\left( \sum_{i \in F} \mu_{ij}(x_i)\right).
  \end{equation*}
  \par Now suppose $\mu_i(x_i) = 0$. This implies $x_i \in D \subset C$, and so $x_i - x_i' + \mu_{ij}(x_i') = 0$ for some $x_i' \in M_i$ and $j \ge i$. Since this equality holds in $C$, which is a direct sum, we see that the coordinate in index $i$ must equal zero, and so $x_i = x_{i}'$. Then, $\mu_{ij}(x_i') = \mu_{ij}(x_i)$ is the index $j$ coodinate, which must also equal zero, hence $\mu_{ij}(x_i) = 0$.
\end{proof}

Exercise 2.15

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Exercise 2.15

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