Trace of a holomorphic differential along a nonconstant map to the sphere

For every compact $K\subseteq \widehat{\mathbb{C}}$, [F5] makes $K$ closed; continuity of $F$ from [F1] makes $F^{-1}(K)$ closed in compact $X$, hence compact by [F1]. Thus $F$ is proper in the stated sense, so [F2] supplies its degree $n\ge 1$, its finite branch-value set $R$, and its finite-sheeted covering away from $R$.

On a disk $V$ evenly covered off $R$, each inverse branch ${\phi }_j$ is holomorphic, so [F4] gives a holomorphic pullback ${\phi }_j^{\ast }\omega$ and their finite sum is holomorphic. At a point $y\in V\cap V^{\mathrm{\prime }}$, both sheet lists contain each point of $F^{-1}(y)$ exactly once. Match the branches whose values at $y$ coincide; local uniqueness of an inverse to a biholomorphism makes each matched pair equal on a neighbourhood of $y$, and finitely many branches let us shrink to one common neighbourhood. Thus the lists differ there only by a permutation, so their sums agree. Hence the local definitions give a well-defined holomorphic differential on $\widehat{\mathbb{C}}\setminus R$.

Fix $y_0\in R$. Its fibre is finite by [F2], say $F^{-1}(y_0)=\{x_1,\dots ,x_s\}$. Fix one target chart $t$ centred at $y_0$; applying the local normal form to each chart expression in this same target chart gives pairwise disjoint source coordinate neighbourhoods $U_i$ with $t\circ F=z_i^{e_i}$ and $e_i=e_{x_i}(F)$. The degree formula gives ${\sum }_i{e}_i=n$. The complement $X\setminus {\bigcup }_i{U}_i$ is compact; its image is compact by continuity and therefore closed in the Hausdorff sphere, and it omits $y_0$. Shrink the target disk $V$ about $y_0$ to miss that image and so that each local power model is defined over $V$. Then every point over $V$ lies in one of the $U_i$, so the local computations below account for every inverse branch over $V\setminus \{y_0\}$.

Let $\sigma$ be a smooth singular path simplex in the complement of $R$. By [F2] that complement is covered by degree-$n$ local biholomorphic sheets; for each of the $n$ points over its initial endpoint, [F8] gives one lift. On each subinterval lying in an evenly covered neighbourhood, the lift is the holomorphic inverse branch composed with $\sigma$, hence is smooth; a finite subdivision as in the path-lifting construction makes each lift a smooth singular path chain. Summing the $n$ lifts defines ${\mathrm{Tr}}_F(\sigma )$. Linearity defines it on finite chains.

Suppose $\gamma$ runs from regular $b$ to regular $a$. Each lift contributes its terminal point minus its initial point to the boundary by [F7]. Lifting the reverse path gives the inverse endpoint correspondence, so the terminal points are exactly the fibre over $a$, once each, and the initial points are exactly the fibre over $b$, once each. Thus $\mathrm{\partial }{\mathrm{Tr}}_F(\gamma )=F^{\ast }(a)-F^{\ast }(b)$; regularity makes every ramification weight in these two fibres equal to $1$.

In $U_i$, write $\omega =h_i(z_i)d{z}_i$ with $h_i(z_i)={\sum }_{m\ge 0}{c}_{i,m}{z}_i^m$ by [F17]. On any simply connected sector of the punctured target disk, choose a branch of $t^{1\mathrm{/}{e}_i}$; the $e_i$ inverse branches are $z_i={\zeta }^k{t}^{1\mathrm{/}{e}_i}$, where $\zeta$ is a primitive $e_i$-th root of unity, and changing the root branch only permutes them. Their contribution to the coefficient of $dt$ in the trace is ${\sum }_{k=0}^{e_i-1}{h}_i({\zeta }^k{t}^{1\mathrm{/}{e}_i})\frac{{\zeta }^k}{e_i}{t}^{1\mathrm{/}{e}_i-1}={\sum }_{q\ge 1}{c}_{i,q{e}_i-1}{t}^{q-1},$ because ${\sum }_{k=0}^{e_i-1}{\zeta }^{k(m+1)}$ is zero unless $e_i\mid (m+1)$, when it equals $e_i$. Termwise summation is valid inside the convergent Taylor radius, so this branch-independent power series is holomorphic at $t=0$. Summing over the finitely many $i$ extends the trace holomorphically over $y_0$; repeating at each point of finite $R$ proves the extension claim.

If two holomorphic differentials extend the trace, their difference is zero on the complement of finite $R$, which accumulates at every point of $R$. The identity theorem [F10] applied to local coefficient functions makes the difference zero near every point of $R$ as well, so the extension is unique.

To show that every holomorphic differential $\eta$ on the sphere is zero, write $\eta =g(z)dz$ on $\mathbb{C}$. In the infinity coordinate $w=1\mathrm{/}z$, its coefficient is $-g(1\mathrm{/}w)w^{-2}$ and is holomorphic at $w=0$ by [F4, F5]. Thus $g(z)=O(\mathrm{\mid }z{\mathrm{\mid }}^{-2})$ as $\mathrm{\mid }z\mathrm{\mid }\to \mathrm{\infty }$; it is bounded outside a disk. By [F15] it is continuous, and [F16] makes its image of a closed disk bounded, so $g$ is bounded on all of $\mathbb{C}$. Liouville's theorem [F6] makes $g$ constant; its limit at infinity is zero, so $g=0$. Applied to the extension from step 2.1, this proves clause 2.

As an independent AC-dependent check, [F14] gives genus $g=0$ for the sphere. Under the full AC hypothesis of [F12], Riemann–Roch at $D=0$ gives $h^0(K_{\widehat{\mathbb{C}}})=i(0)=g$ after [F13] trivializes ${\mathcal{O}}_X(0)$ and identifies holomorphic sections of $K$ with holomorphic differentials. Hence it also gives $\mathrm{\Omega }(\widehat{\mathbb{C}})=0$, agreeing with the direct choice-free proof in step 3.2.

On each evenly covered subinterval, [F9] identifies the integral of $\omega$ over each lifted segment with the integral of the corresponding inverse-branch pullback along the base segment. Summing over the $n$ starting points gives the integral of ${\sum }_j{\phi }_j^{\ast }\omega =F_{\ast }\omega$ there; adding the finitely many subintervals and simplices yields ${\int }_{{\mathrm{Tr}}_F(\gamma )}\omega ={\int }_{\gamma }{F}_{\ast }\omega$. Step 3.2 makes the right side zero, proving clause 3 with multiplicities retained even when lifts trace the same geometric subset.

If $a\mathrm{\ne }b$, the stated rational function $m_{a,b}$ has one simple zero at $a$, one simple pole at $b$, and no other zero or pole, including at $\mathrm{\infty }$ by [F5]. In a local coordinate at $x$, [F3] writes $F$ as a power $z^{e_x(F)}$; composing with a simple zero or pole therefore gives order $e_x(F)$ or $-e_x(F)$ by [F11], respectively, and order zero elsewhere. Consequently $\mathrm{div}(m_{a,b}\circ F)=F^{\ast }(a)-F^{\ast }(b)$ under the definitions in the statement. If $a=b$, both sides are zero because $m_{a,a}=1$. Combining this with step 1.5 proves clause 4.