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A covering-surface formulation of the Ahlfors second fundamental theorem is as follows. For a regular finite bordered Riemann surface $X$ mapped holomorphically to a fixed metric bordered Riemann surface $Y$, let $S=A_f(X)/A(Y)$ be the average sheet number and $L$ the pulled-back length of the relative boundary, namely the boundary mapped into the interior of $Y$. With the modern sign convention for the Euler characteristic,where $h$ depends on $Y$ and its metric, not on $X$ or the map. The associated area comparison is $|S_D-S|\le h_{D}L$ for each fixed regular target region $D$. These formulations and the sign convention are explained in www.math.purdue.edu/~eremenko/dvi/ahl.pdf .

The island version of the Ahlfors second fundamental theorem for a disk source, which is the useful form here, says that for $q\ge 3$ fixed Jordan domains $D_j$ with disjoint closures on the Riemann sphere,Here $n_j$ counts simply connected islands of a meromorphic function over $D_j$, once each without weighting by mapping degree, andAn island of a meromorphic function is a relatively compact inverse-image component mapped properly onto its target domain; components reaching the source boundary are not counted. The simply connected island estimate is recorded, for spherical disks, in Lemma 2 of pdfs.semanticscholar.org/d247/562607e237af7c8fc6e683b77be6b94829fa.pdf . Regular target regions give the same estimate with a region-dependent constant. General Jordan domains can be enclosed in slightly larger regular Jordan domains still having disjoint closures. A simple island over an enlarged domain restricts to one over the original domain, so it suffices to prove the conclusion for regular domains.

Suppose a nonconstant meromorphic function on the plane has no simple islands over five such domains. On every counted simply connected island its proper map has integer degree at least two. The area formula therefore givesTogether with area comparison and the Ahlfors second fundamental theorem, this yieldsThe area comparison can also be seen directly. The two-form $[1_D/A(D)-1/(4\pi )]dA$ on the target sphere has zero integral and a bounded one-form primitive. One construction solves its Poisson equation: the gradient of the spherical Green function has an integrable $1/\mathrm{distance}$ singularity, so convolution against the bounded density gives a bounded primitive. Pulling back and applying Stokes theorem, or smooth approximation at the domain boundary, bounds the integral by a constant times $L(R)$.

We now derive the needed length-area exhaustion of the complex plane. The Cauchy-Schwarz inequality on a circle givesIf $L(R)/S(R)\ge \epsilon >0$ for all sufficiently large $R$, then $S(R)>0$ andIntegration makes $1/S(R)$ negative, which is impossible. Hence there are radii $R_k\to \infty$ with $L(R_k)/S(R_k)\to 0$. This contradicts $S/2\le CL$ and proves the Ahlfors five islands theorem: one of the five target domains has a bounded simply connected inverse-image component mapped biholomorphically onto it.

Four domains do not suffice. Take the Weierstrass elliptic function $\wp$ for the square lattice $\mathbb{Z}+i\mathbb{Z}$. The equationshows the three distinct finite branch values $e_1,e_2,e_3$; the fourth is infinity, because all poles are double. Equivalently, the degree-two map from the elliptic torus to the Riemann sphere has these four branch values, with all their preimages of local degree two. Choose four small disjoint Jordan domains, each containing one of these four totally ramified Weierstrass values. A degree-one island of a meromorphic function over any one would contain a preimage of its central value of local degree one, contradicting total ramification. Thus $\wp$ provides four domains with no simple island.

If $\varphi$ is identically zero there are no exceptions. Otherwise choose $x_0$ with $\varphi (x_0)>0$ and put $I_{+}=\{x\ge x_0:{\varphi }^{\prime }(x)>\varphi (x{)}^2\}$. Monotonicity implies $\varphi (x)>0$ thereafter. On this set, $1<{\varphi }^{\prime }/{\varphi }^2$, whenceAdding the finite initial interval $[0,x_0]$ provesThis derivative growth outside a finite-measure set controls exceptional radii in Nevanlinna estimates. For example let $T_s$ be the smooth spherical characteristic and $A(r)=r{T}_s^{\prime }(r)$ its increasing area-counting function. Apply the lemma to $T_s+1$ and $A+1$: outside the union of two finite-measure sets,Their logarithms are consequently $O(\log r+\log (T_s+1))$. Such derivative/area bounds in estimates from the Poisson-Jensen formula give the Nevanlinna logarithmic derivative lemma, $m(r,f^{\prime }/f)=O({\log }^{+}{T}_f+\log r)$ outside a finite-length set. The second main theorem uses that estimate as its error term. The elementary growth lemma controls that error and its exceptional set; it is not itself a substitute for the analytic estimates.

We use the truncated second main theorem stated in the preceding solution, together with the fact that a transcendental meromorphic function has $T_f(r)/\log r\to \infty$. Here is a justification of the latter fact. If there are infinitely many poles, fixing any arbitrarily large finite pole count gives $N(r;\infty )\ge A\log r-O_A(1)$. If there are only finitely many poles, multiply by a polynomial $p$ removing them to obtain a transcendental entire function $h=pf$. Its Taylor series has nonzero coefficients of arbitrarily high degrees $n$. Cauchy's coefficient estimate gives $\log M_h(r/2)\ge n\log r-O_n(1)$; the subharmonic function Poisson estimate gives $\log M_h(r/2)\le 3m(r,h)$. Since $T_h\le T_f+O(\log r)$, letting $n$ be arbitrarily large proves the claimed ratio limit. Thus the second-theorem error is $o(T_f)$ along large nonexceptional radii.

Let $N_{\mathrm{simple}}(r;a)$ count just the simple $a$-points. A multiple $a$-point of degree $m\ge 2$ contributes one to the truncated count and at least two to the full count; a simple point contributes one to each. Hence, with the same positive logarithmic weights for large radii,If each of the five values had only finitely many simple preimages, each simple counting function would be $O(\log r)$. Summing and using the first main theorem would give ${\sum }_{j=1}^5\overline{N}(r;a_j)\le (5/2)T_f+O(\log r)$. But the second theorem requires $3T_f\le \sum \overline{N}+o(T_f)$, a contradiction. This proves the stronger result five values force infinitely many simple preimages:In particular more than one simple zero is necessary, but more than one target with simple zeros is not necessary.

For the last distinction, take a nonsingular Weierstrass elliptic function scaled so that ${\wp }^{\prime 2}=4\wp (\wp -1)(\wp +1)$; it has branch values $-1,0,1,\infty$. These are the four totally ramified Weierstrass values: finite branch preimages have local degree two, and all poles are double. SetThe first four targets are the images of those branch values under the Möbius transformation. Their preimages remain double; at a lattice pole, for example, $f(z)\sim (z-z_{\ast }{)}^2$. The last target corresponds to $\wp =3$, where ${\wp }^{\prime 2}=96\ne 0$, so every such preimage is simple. The degree-two torus map attains that regular value, and periodicity produces infinitely many preimages in the plane. This nonconstant elliptic function is transcendental because it has infinitely many poles. Exactly one of the five selected finite targets therefore has simple preimages.