Exercise 2.20

Question

Keeping the same notation as in Exercise \href{exc:2.14}{$14$}, let $N$ be
  any $N$-module.
  Then $(M_i \otimes N, \mu_{ij} \otimes 1)$ is a direct system; let
  $P = \displaystyle\lim_{\longrightarrow}(M_i \otimes N)$ be its direct limit.
  For each $i \in I$ we have a homomorphism
  $\mu_i \otimes 1\colon M_i \otimes N 	o M \otimes N$, hence by Exercise
  \href{exc:2.16}{$16$} a homomorphism $\psi\colon P 	o M \otimes N$.
  Show that $\psi$ is an isomorphism, so that
  \begin{equation*}
    \lim_{\longrightarrow} (M_i \otimes N) \cong \left( \lim_{\longrightarrow} M_i ight) \otimes N.
  \end{equation*}

Amit Awasthi · Accepted Answer

\begin{proof}
  For each $i \in I$, let $g_i\colon M_i 	imes N 	o M_i \otimes N$ be the
  canonical bilinear mapping.
  Since $g_i = g_j \circ \mu_{ij}$ for $i \le j$, by the universal property of
  direct limits (Exercise 2.16) we then have the unique homomorphism
  $g\colon M 	imes N 	o P$.
  This is bilinear, since any $am+b \in M$ is contained in some $M_i$ by
  Exercise 2.15, and so
  $g(am+b,n) = g_i(am+b,n) = ag_i(m,n) + g_i(b,n)$ by the bilinearity of $g_i$;
  the $N$ case follows similarly.
  Thus, by the universal property of tensor products we have the unique
  homomorphism $\varphi\colon M \otimes N 	o P$.
  \par We claim $\varphi \circ \psi = \psi \circ \varphi = \operatorname{id}$.
  Let $x \in P$; then, by Exercise 2.15 it is contained in some
  $M_i \otimes N$, and is of the form $(m_i \otimes n)$.
  Thus,
  \begin{equation*}
    (\varphi \circ \psi)(x) = (\varphi \circ \psi)(m_i \otimes n)
    = \varphi(m_i \otimes n) = m_i \otimes n \in P.
  \end{equation*}
  Likewise, $x \in M \otimes N$ implies $x = m_i \otimes n$ for some $m_i \in M_i$
  for some $i$ by Exercise 2.15, and so
  \begin{equation*}
    (\psi \circ \varphi)(x) = (\psi \circ \varphi)(m_i \otimes n) = m_i \otimes
    n = x.\qedhere
  \end{equation*}
\end{proof}

Exercise 2.20

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Exercise 2.20

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