Ramification count from the derivative divisor

(Finite target) Let $b$ be a point with $f(b)=a\in \mathbb{C}$ and local degree $m\ge 1$. By [F3], $f-a=(z-b{)}^{m}h$ with $h(b)\mathrm{\ne }0$; 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 $b$. Thus $b$ contributes $m-1$ to $n(t,0;f^{\mathrm{\prime }})$ and, when $m\ge 2$, exactly $m-1$ to $n_1(t,a;f)$; when $m=1$ both contributions vanish.

(Poles) Let $p$ be a pole of order $m\ge 1$. By [F4], $f=(z-p{)}^{-m}g$ 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 $m+1$ and no zero at $p$. Hence $p$ contributes $m$ to $n(t,\mathrm{\infty };f)$, $2m-(m+1)=m-1$ to $2n(t,\mathrm{\infty };f)-n(t,\mathrm{\infty };f^{\mathrm{\prime }})$, and, when $m\ge 2$, exactly $m-1$ to $n_1(t,\mathrm{\infty };f)$.

(Disjointness for a target set) Let $A$ be a finite set of distinct sphere targets. For $a\in A$, $n_1(t,a;f)$ is a sum of weights $m_b-1$ over points $b$ with $f(b)=a$ and $m_b\ge 2$; for distinct $a$ these point sets are disjoint, and each such $b$ is a ramification point of the sphere map with the same local degree $m_b$, so its weight appears in $n_1(t,f)$. Hence $n_1(t,a;f)\ge 0$ and ${\sum }_{a\in A}{n}_1(t,a;f)\le n_1(t,f)$ for every $t>0$.

(Weight identity) Every ramification point of the sphere map is either a non-pole point with local degree $m\ge 2$, where step 1.1 makes the weight $m-1$ equal to the order of the zero of $f^{\mathrm{\prime }}$, or a pole of order $m\ge 2$, handled by step 1.2; conversely every zero of $f^{\mathrm{\prime }}$ is a non-pole point of local degree $\nu +1\ge 2$ with weight $\nu$. Therefore, for every $t>0$, $n_1(t,f)=n(t,0;f^{\mathrm{\prime }})+2n(t,\mathrm{\infty };f)-n(t,\mathrm{\infty };f^{\mathrm{\prime }}),$ both sides being finite sums of nonnegative weights.

(Integration) The identity of step 2.1 holds at $t=0$ as well, since $t\mapsto n(t,\cdot )$ is a right-continuous step function and the centre value is included in each term. Multiplying by the centre-regularisation $n(0)\log r+{\int }_0^r(\cdot -n(0))\frac{dt}{t}$ is linear, so $N_1(r,f)=N(r,0;f^{\mathrm{\prime }})+2N(r,\mathrm{\infty };f)-N(r,\mathrm{\infty };f^{\mathrm{\prime }}),$ and all terms are finite by [F5].

(Integrate the inequality) Each ramification point $b\mathrm{\ne }0$ contributes its nonnegative weight times $\log (r\mathrm{/}\mathrm{\mid }b\mathrm{\mid })$ when $\mathrm{\mid }b\mathrm{\mid }\le r$, while a point at $0$ contributes its weight times $\log r$. For $r\ge 1$ all these coefficients are nonnegative, so the pointwise inclusion of step 1.3 yields ${\sum }_{a\in A}{N}_1(r,a;f)\le N_1(r,f)$.