Very minor fixes, up the version

ca.tex

@@ -409,7 +409,7 @@
 Ji{\v r}\'i Lebl\\[3ex]} 
 \today
 \\
-(version 1.3)
+(version 1.4)
 \end{minipage}}
 
 %\addtolength{\textwidth}{\centeroffset}
@@ -427,7 +427,7 @@
 \bigskip
 
 \noindent
-Copyright \copyright 2019--2022 Ji{\v r}\'i Lebl
+Copyright \copyright 2019--2023 Ji{\v r}\'i Lebl
 
 %PRINT
 % not for the coil version
@@ -841,7 +841,7 @@ \subsection{The geometry and topology of the plane}
 This version ought to be called
 Cauchy inequality, but lamentably that name could refer to a different
 inequality, one that we will call the Cauchy estimates.}, note: $\Re z
-\bar{w}$ is the real dot product),\index{Cauchy--Schwarz}
+\bar{w}$ is the $\R^2$ dot product),\index{Cauchy--Schwarz}
 \item
 $\sabs{z+w} \leq \sabs{z} + \sabs{w}$ \quad (Triangle inequality).%
 \index{triangle inequality!complex numbers}
@@ -16106,7 +16106,7 @@ \subsection{In the plane}
 with no limit points in $\C$ and $\{ m_k \}$ is a sequence of
 natural numbers.
 Then there exists an entire holomorphic $f \colon \C \to \C$ that
-has zeros exactly at $c_k$, with orders given by $m_k$.
+has zeros exactly at $c_k$, with orders given by~$m_k$.
 
 More precisely, suppose $\{ a_n \}$ is the sequence of nonzero $\{ c_k \}$ with points
 repeated according to the multiplicities $\{ m_k \}$ and $m$
@@ -16279,12 +16279,12 @@ \subsection{Factorization of sine}
 \pi z \prod_{n=1}^\infty 
 \left(1-\frac{z^2}{n^2}\right) .
 \end{equation*}
-That's a rather nice factorization.  Of course we still do not know if $f(z)$ is
+That's a rather nice factorization.  Of course, we still do not know if $f(z)$ is
 $\sin(\pi z)$.
 All we know is that the two have the same zeros,
 and the derivative at 0 is $\pi$ as it should be.
 As $f$ captures the zeros of $\sin(\pi z)$, we write
-(as in the factorization theorem),
+(as in the factorization theorem)
 \begin{equation*}
 \sin(\pi z)
 = 
@@ -16293,7 +16293,7 @@ \subsection{Factorization of sine}
 \pi z
 e^{g(z)}
 \prod_{n=1}^\infty 
-\left(1-\frac{z^2}{n^2}\right)
+\left(1-\frac{z^2}{n^2}\right) ,
 \end{equation*}
 for some entire function $g$.  We need to show that $g \equiv 0$.  By the
 computation of the derivative above, $g(0) = 0$.
@@ -16465,7 +16465,7 @@ \subsection{The product theorem in any open set}
 with no limit points in $U$, and $\{ m_k \}$ is a sequence of
 natural numbers.
 Then there exists a holomorphic $f \colon U \to \C$ with
-zeros exactly at $c_k$, with orders given by $m_k$.
+zeros exactly at $c_k$, with orders given by~$m_k$.
 \end{thm}
 
 \begin{proof}

changes-draft.html

@@ -3,7 +3,7 @@
 <li>In the description of the typical application of Hurwitz,
   be a bit more precise (besides fixing an erratum).
 <li>In the proof of Proposition B.3.13, use j,k for the same thing
-	as in the definition B.3.12 and Proposition B.3.11
+	as in the Definition B.3.12 and Proposition B.3.11
 	for consistency.  This required changing k to h',
 	which is more consistent with the naming of A' anyway.
 	Also use p for the fixed point rather than x for