diff --git a/sections/complete-reducibility.tex b/sections/complete-reducibility.tex
@@ -837,7 +837,7 @@ establish\dots
\end{tikzcd}
\end{center}
induces a long exact sequence of the form
- \begin{center}
+ \begin{equation}\label{eq:standard-h1-ext-seq}
\begin{tikzcd}
\cdots \rar[dashed] &
H^1(\mathfrak{g}, N) \rar &
@@ -845,19 +845,14 @@ establish\dots
H^1(\mathfrak{g}, \sfrac{M}{N}) \rar[dashed] &
\cdots
\end{tikzcd}
- \end{center}
+ \end{equation}
- Since \(0 < \dim N, \dim \sfrac{M}{N} < \dim M\) it follows
- \(H^1(\mathfrak{g}, N) = H^1(\mathfrak{g}, \sfrac{M}{N}) = 0\). The exactness
- of
- \begin{center}
- \begin{tikzcd}
- 0 \rar & H^1(\mathfrak{g}, M) \rar & 0
- \end{tikzcd}
- \end{center}
- then implies \(H^1(\mathfrak{g}, M) = 0\). Hence by induction in \(\dim M\)
- we find \(H^1(\mathfrak{g}, M) = 0\) for all finite-dimensional \(M\). We are
- done.
+ Since \(\dim N < \dim M\), it follows \(H^1(\mathfrak{g}, N) = 0\). In
+ addition, since \(\dim N > 0\), we find \(\dim \mfrac{M}{N} < \dim M\) and
+ thus \(H^1(\mathfrak{g}, \sfrac{M}{N}) = 0\). The exactness of
+ (\ref{eq:standard-h1-ext-seq}) then implies \(H^1(\mathfrak{g}, M) = 0\).
+ Hence by induction in \(\dim M\) we find \(H^1(\mathfrak{g}, M) = 0\) for all
+ finite-dimensional \(M\). We are done.
\end{proof}
We are now finally ready to prove\dots