Very minor nits

ca.tex

@@ -16338,7 +16338,7 @@ \subsection{Factorization of sine}
 inside the square brackets may differ by a constant.
 Since the far left-hand side and the far right-hand side do not have any
 logarithms in them they are clearly well-defined.
-The equality hods with $g'=0$, which is 
+The equality holds with $g'=0$, which is 
 a nice exercise in applying the residue theorem, and we
 leave it to the reader.
 
@@ -16546,14 +16546,14 @@ \subsection{The product theorem in any open set}
 \end{cor}
 
 \begin{proof}
-By the product theorem there exists a holomorphic function $h$ that has
+By the product theorem, there exists a holomorphic function $h$ that has
 zeros exactly where $f$ has poles, and of the same order.  The function $fh$
 therefore has removable singularities at all the poles of $f$.  In other
 words, there is a holomorphic $g$ such that $g=fh$.
 \end{proof}
 
 \begin{remark}
-When we introduced the corollary, we mentioned domain, but then we proved it
+When we introduced the corollary, we mentioned \myquote{domain,} but then we proved it
 for an open set.  The issue is a bit of algebra.  If $U$ is not connected,
 then the set of holomorphic functions is not an integral domain, it has zero
 divisors, and a field of fractions is only defined for integral domains.