Truncated value and ramification counts

At a finite target $a$ and a point $b\in D_r$ with $f(b)=a$, [F3] writes $f-a=(z-b{)}^{m_b}h$ with $h(b)\mathrm{\ne }0$ and $m_b\ge 1$; the point contributes $1$ to $\bar{n}$ and $m_b-1$ to $n_1$, hence $m_b$ to their sum, matching its contribution to $n$. At $a=\mathrm{\infty }$, [F4] gives a pole of order $m_p$ contributing $1$ and $m_p-1$; summing the finitely many points of $D_r$ gives $n(r,a;f)=\bar{n}(r,a;f)+n_1(r,a;f)$.

The derivative satisfies $f^{\mathrm{\prime }}\equiv ̸0$: if $f^{\mathrm{\prime }}\equiv 0$, then $f$ is holomorphic with zero derivative on $\mathrm{\Omega }=\mathbb{C}\setminus P$, where $P$ is the pole set, and $\mathrm{\Omega }$ is a domain because [F5] makes $P$ closed and discrete, so $P\mathrm{\ne }\mathbb{C}$ and any two points of $\mathrm{\Omega }$ are joined by a polygonal path that meets $P$ in only finitely many points and can be detoured around them. By [F6], $f$ is constant, say $f=c$, on $\mathrm{\Omega }$. Near a pole $p\in P$ it would then follow from [F4] that $\mathrm{\mid }f\mathrm{\mid }\to \mathrm{\infty }$, contradicting $f=c$ on a punctured neighbourhood of $p$; so $P=\mathrm{\varnothing }$ and $f\equiv c$ on $\mathbb{C}$, contradicting nonconstancy.

Since $0\le \bar{n}(r,a;f)\le n(r,a;f)$ and $0\le n_1(r,a;f)\le n(r,a;f)$ pointwise by step 1.1, and $n(r,a;f)$ is finite on every bounded disc, both $\bar{n}$ and $n_1$ are finite there.

The zeros of $f^{\mathrm{\prime }}$ are locally finite and do not accumulate at poles. At a pole $p$ of order $m$, [F4] gives a local representation $f=(z-p{)}^{-m}g$ with $g$ holomorphic and $g(p)\mathrm{\ne }0$, so $f^{\mathrm{\prime }}=(z-p{)}^{-m-1}(-mg+(z-p)g^{\mathrm{\prime }})$ has a pole of order $m+1$ there and no zero in a small punctured neighbourhood. Away from the poles $f^{\mathrm{\prime }}$ is holomorphic and, by step 1.2, not identically zero on the domain $\mathbb{C}\setminus P$; hence its zeros are isolated (Zeros of a nonzero holomorphic function are isolated). A set of isolated points with no accumulation point in $\mathbb{C}$ has only finitely many members in each bounded closed disc: otherwise a sequence of distinct zeros in the disc would converge, by compactness, to a limit that is an accumulation point.

Since $n=\bar{n}+n_1$ pointwise by step 1.1, including at the centre $z=0$, and since $n_1(t,a;f)\le n(t,a;f)$ is finite for every $t>0$ by [F2], subtracting the centre terms and integrating against $dt\mathrm{/}t$ gives $N(r,a;f)=\bar{N}(r,a;f)+N_1(r,a;f)$ for every $r>0$; the definition of $N_1(r,f)$ uses the same centre regularization as the displayed formulas.

The ramification points of the sphere map are exactly the zeros of $f^{\mathrm{\prime }}$ together with the poles of order at least two. At a non-pole point $b$ where [F3] gives $f-f(b)=(z-b{)}^{m}h$ with $h(b)\mathrm{\ne }0$ and $m\ge 1$, the product rule gives $f^{\mathrm{\prime }}=(z-b{)}^{m-1}(mh+(z-b)h^{\mathrm{\prime }})$ with $mh(b)\mathrm{\ne }0$, so $f^{\mathrm{\prime }}$ has a zero of order exactly $m-1$ at $b$; hence $m\ge 2$ exactly when $f^{\mathrm{\prime }}(b)=0$, and then the ramification weight $m-1$ equals the zero order of $f^{\mathrm{\prime }}$. At a pole of order $m$, the representation of step 2.2 shows that the local degree is $m$ and the weight $m-1$, while $f^{\mathrm{\prime }}$ has no zero there. Therefore $n_1(t,f)=n(t,0;f^{\mathrm{\prime }})+{\sum }_{\mathrm{\mid }p\mathrm{\mid }\le t}(m_p-1)$ for every $t>0$, and this is finite by [F2] and step 2.2.

Steps 1.1 and 2.3 give the pointwise and integrated identities, and steps 2.1, 2.2 and 3.1 show that every count introduced above is finite on each bounded disc and that the ramification points form a locally finite divisor, so the definition is well posed.