lie-algebras-and-their-representations

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