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Glue two copies of $[0,w]\times [0,h]$ along their left, bottom and right sides. Their unglued top sides form the outer boundary of a disk; the two bottom corners are marked interior points. Let $\Gamma$ be the loops surrounding both points, and ${\Gamma }^{\ast }$ the boundary-to-boundary crosscuts separating them. Then $\lambda (\Gamma )\lambda ({\Gamma }^{\ast })=4$.

Each vertical seam joins a marked point to the outer boundary, so the Jordan curve theorem forces a loop surrounding both points to meet both seams. In the flat conformal metric, its total horizontal travel is therefore at least $2w$. Doubled horizontal segments attain this length. Each separating crosscut meets the bottom seam and starts and ends at height $h$, so its total vertical travel is at least $2h$; doubled vertical segments attain it. The total area is $2wh$.

For an arbitrary density, integrate the lengths of the doubled horizontal segments over their height. The Cauchy-Schwarz inequality yields $L_{\rho }(\Gamma {)}^{2}h\le 2w{A}_{\rho }$. Doubled vertical segments similarly give $L_{\rho }({\Gamma }^{\ast }{)}^{2}w\le 2h{A}_{\rho }$. These upper bounds equal the flat-metric ratios, proving both displayed formulas. Mapping the upper and lower halves of a disc to these rectangles gives the corresponding duality for a simply connected domain with two marked interior points.

Given a compact planar set $K$ and an open neighborhood $V$, a finite square-grid construction gives a compact polygonal neighborhood inside $V$ with finitely many complementary components. Remove thin polygonal corridors disjoint from $K$ joining holes that lie in the same component of $\mathbb{C}\setminus K$, and joining holes in its unbounded component to the exterior. Corridors can also include a prescribed point from each of finitely many specified bounded components. There remains a neighborhood whose bounded holes lie in distinct bounded components of $\mathbb{C}\setminus K$, each containing its chosen point. Standard small polygonal rounding gives a finite union of planar surfaces with boundary. The index map from its first integral homology group to one integer per bounded hole is an isomorphism. Triangulate a surrounding disc: a one-cycle bounds a two-chain whose coefficients are constant in each complementary component. Zero hole coefficients give a filling supported inside the neighborhood; the face chain of each hole supplies a dual cycle. The cycles can be sums of boundary curves in different components. This uses finite triangulation, not the general Jordan curve theorem.

On vertices $0,\dots ,7$, join cyclic indices differing by one or two modulo eight. The resulting planar graph has a square face bounded in order by $0,2,4,6$. Deleting at most three vertices cannot disconnect it: each break in the surviving cyclic chain requires two consecutive deleted vertices, and two separate breaks require four deletions. It is therefore four-connected. Disjoint paths joining $0$ to $4$ and $2$ to $6$ would cross in the complementary disk of the square face, contrary to the Jordan curve theorem. Thus four-connectivity does not guarantee two-linkage.

A Jordan domain is the bounded complementary component of a closed simple curve in the plane. The Jordan curve theorem separates the plane into this inside component and one unbounded outside component. A conformal bijection from the disc or complex upper half-plane to this simply connected domain extends to a homeomorphism of their closed conformal boundaries by the Caratheodory boundary extension theorem.

A simple triod is the union of three arcs that share one endpoint and are otherwise disjoint. The Moore triod theorem states that every pairwise disjoint family of simple triods in the plane is countable. One proof chooses data from a countable basis of open disks around each branch point; the Jordan curve theorem shows that two disjoint triods cannot determine the same finite basis data, giving an injection into a countable set of finite tuples.

Here a periodic solution must mean a nonconstant periodic orbit: constant equilibria are not excluded by the stated divergence hypothesis. Suppose such an orbit exists. Since the vector field is continuously differentiable, uniqueness for the autonomous differential equation makes its trajectory a simple closed curve $C$, bounding a region $D$ by the Jordan curve theorem. The Green theorem givesAlong the orbit, $dx=fdt$ and $dy=gdt$, so the line integral is identically zero. Reversing orientation if necessary does not change this conclusion. The area integral is strictly positive because its continuous integrand is positive everywhere and $D$ has positive area. This contradiction proves the Bendixson-Dulac criterion in this case.

For the second-order equation set $y=x^{\prime }$, giving $x^{\prime }=y$, $y^{\prime }=-y-\epsilon y^3-\sin x$. Choose the Dulac function $\rho =-1$. ThenThus there is no nonconstant periodic solution for $\epsilon >0$. The equilibria $x(t)=k\pi$ do remain solutions. As a direct check, the mechanical energy $H=y^2/2+1-\cos x$ obeys $H^{\prime }=-y^2-\epsilon y^4$, whose integral over a period could vanish only if $y\equiv 0$.

The group $G(K)$ consists of the nonvanishing continuous functions, and $E(K)$ is a subgroup because multiplication is commutative: $e^g{e}^h=e^{g+h}$. The objective is a logarithm decomposition with finitely many integer obstructions, one for each bounded complementary component. No local connectivity of $K$ is assumed.

We use two elementary finite-neighborhood facts, which make the outline independent of the general Jordan theorem. Given a neighborhood $V$ of $K$, take a finite grid neighborhood $N\subset V$ with $K\subset \mathrm{int}N$. It can be chosen to be a finite union of polygonal surfaces with boundary. Its complement has finitely many bounded holes. Holes that lie in the same component of $\mathbb{C}\setminus K$ can be joined by thin polygonal corridors disjoint from $K$, which are removed from $N$. A hole lying in the unbounded component is similarly joined to the exterior. For the remaining finitely many components, corridors may be continued to include their chosen representatives $a_i$. Small polygonal adjustments preserve $K\subset \mathrm{int}N$ and $N\subset V$. Thus every remaining bounded hole lies in a distinct bounded component of $\mathbb{C}\setminus K$ and contains its representative. One may also arrange this for any prescribed finite list of bounded components by first keeping their representative points out of the grid neighborhood. All paths used are compact and disjoint from $K$, so sufficiently thin corridors remain disjoint from $K$.

For such a finite polygonal $N$, winding indices identify the homology group $H_1(N;\mathbb{Z})$ with the free abelian group on its bounded holes. Here is a finite-chain explanation. Triangulate a large disc with $N$ as a subcomplex. Every integral one-cycle bounds a two-chain in the disc. Its filling coefficient is constant on each component of the complement of $N$, is zero on the exterior, and is the winding index there. If all bounded-hole indices vanish, the filling is supported in $N$, so the cycle already bounds in $N$. Conversely the sum of the oriented faces in one hole has its chain boundary in $N$ and has index one in that hole and zero in the others. Thus cycles ${\gamma }_j$ can be chosen with $\mathrm{wind}({\gamma }_j,a_i)={\delta }_{ij}$. These can be integral sums of boundary curves, which matters when components of $N$ are nested. This proves the needed statement using finite triangulation, not separation of an arbitrary simple closed curve. Together these are finite-hole neighborhoods of a planar compact set.

Now extend $f\in G(K)$ to a continuous $F$ near $K$, using the Tietze extension theorem on its real and imaginary parts. Since ${\min }_K|f|>0$, shrink the neighborhood until $F$ has no zeros, and choose $N$ within it as above. The winding of $F$ on the cycles ${\gamma }_j$ defines integers $k_j$. Setwhere only the finitely many represented holes are used. The winding of $F/r$ on every cycle of $N$ is zero. On each component choose a local complex logarithm at a base point and continue it along paths. Zero winding number around every loop makes the continued value independent of the path; local logarithm charts make it continuous. The continuous logarithm lifting criterion therefore gives $F/r=e^g$ on $N$, and restriction gives $f=e^{g{|}_K}r{|}_K$. Thus $G(K)=E(K)G(\mathbf{a})$, even when $K$ has infinitely many holes: each individual function uses finitely many factors.

To prove uniqueness, suppose a finite product $r=\prod (z-a_i{)}^{k_i}$ equals $e^g$ on $K$. Extend $g$ continuously to a neighborhood. There $r{e}^{-g}$ is close to $1$ after shrinking, so it has its small-branch logarithm. Hence $r$ itself has a continuous function giving a complex logarithm on that neighborhood. Choose $N$ there detecting all the finitely many $a_i$ as separate holes. Its dual cycles give $\mathrm{wind}(r\circ {\gamma }_j,0)=k_j$, while a function with a logarithm has zero winding number. Therefore every $k_j=0$. This proves both freeness of the listed rational functions and $E(K)\cap G(\mathbf{a})=\{1\}$. All subgroups commute, soThis is planar logarithm factorization. If there are no bounded components the second factor is trivial.

For the Jordan curve deduction, let $J$ be homeomorphic to the circle. Parametrize it by $\gamma :[0,1]\to J$ with just the two endpoints identified. For $f\in G(J)$, lift a logarithm of $f\circ \gamma$ along the interval. Its endpoint increment is $2\pi in$. The integer $n$ is additive, every integer is attained by a function with $f(\gamma (t))=e^{2\pi int}$, and $n=0$ precisely when the lift descends to a logarithm on $J$. Hence $G(J)/E(J)\cong \mathbb{Z}$. The factorization says this is free abelian on the bounded complementary components; comparison of ranks, or tensoring with $\mathbb{Q}$, shows that exactly one bounded component exists. There is also the unique unbounded component, so the complement has exactly two components.

We still need the common-boundary assertion. A proper closed arc has trivial exponential quotient because every nonvanishing function on an interval admits a logarithm. Factorization consequently makes the complement of such an arc connected, and since it is open it is path-connected. Fix $p\in J$ and any neighborhood $V$ of $p$, and choose a small open arc $A\subset J\cap V$ containing $p$. The remainder $J\setminus A$ is a proper closed arc. Join a point in each of the two complementary components by a path avoiding $J\setminus A$. This path must meet $J$, and every such meeting lies in $A$. The segment before its first meeting lies in the first region, and the segment after its last meeting lies in the second; continuity makes both regions meet $V$. Thus every point of $J$ lies on both boundaries. Conversely either boundary is contained in the closed set $J$. There are exactly two regions, one bounded and one unbounded, with common boundary $J$, the Jordan curve theorem. This is Jordan separation from the exponential quotient; it does not assume the theorem in the factorization proof.

Write $s=A(S)=4\pi$, $L=L(\partial U)$, and $a=\min \{A(U),A(S\setminus U)\}$. We first prove a constant-version spherical isoperimetric inequality directly, including regions with several boundary components.

If $L\ge 1$, the arithmetic bound $A(U)A(S\setminus U)\le s^2/4=4{\pi }^2$ already gives $A(U)A(S\setminus U)\le 4{\pi }^2{L}^2$. If $L=0$, a region with smooth boundary is empty or the whole connected sphere up to boundary-null changes, and its area product is zero. For $0<L<1$, write the smooth boundary as finitely many disjoint simple closed curves ${\gamma }_j$ of lengths $l_j$. Fix a point $p_j$ on each curve. The shorter boundary arc from $p_j$ to any other point has length at most $l_j/2$, so ${\gamma }_j$ is contained in the closed geodesic ball of radius $l_j/2$ about $p_j$.

The complement of that small ball is connected and lies entirely on one side of ${\gamma }_j$. The Jordan curve theorem therefore gives a disk $E_j$ bounded by ${\gamma }_j$ contained in the ball. Its area satisfiesThe indicator function of $U$ and the sum ${\sum }_j{1}_{E_j}$ have the same jumps modulo two across every boundary curve. Their difference modulo two is consequently constant throughout the sphere off the boundary. Thus either $U$ or its complement lies in ${\bigcup }_j{E}_j$, up to its area-zero boundary. This also handles nested curves and disconnected regions. It follows thatCombining the cases, one valid universal choice is $\kappa =4{\pi }^2$. Sharpness is not required.

For the inclusion estimate put $x=A(V)$, $d=A(D)>0$ and $y=A(V\cap D)$. Splitting $D$ into its portions inside and outside $V$ givesBecause $0\le A(D\setminus V)\le s-x$ and $0\le A(D\cap V)\le x$, this expression lies between $-x(s-x)$ and $x(s-x)$. Apply the preceding spherical isoperimetric inequality to $V$ and divide by $sd$:Thus the constant in the spherical inclusion area discrepancy is independent of the nonempty open set $D$.

Let $g$ and $\tilde{g}$ be the old and new smooth Riemannian metrics. Their area forms satisfyThe determinant ratio is invariant under changes of coordinates, so $\theta$ is a globally defined smooth function. The layer cake representation follows from $\theta (x)={\int }_0^{\infty }{1}_{\{\theta (x)>t\}}dt$ and Tonelli's theorem:Smoothness makes every $D_t$ open. Compactness of the sphere gives $0<m=\min \theta \le M=\max \theta <\infty$. Multiply the inclusion bound by $A(D_t)$ before integrating; for empty $D_t$ the resulting bound is trivially zero. For $t<m$, $D_t=S$ and the discrepancy is exactly zero; for $t>M$, $D_t$ is empty. ThereforeDividing by $\tilde{A}(S)>0$ provesThis constant depends only on the two metrics. If $\theta$ is constant the normalized areas agree exactly, as the bound also shows.

A linked graph is $k$-linked if it has at least $2k$ vertices and, for every choice of distinct terminals $s_1,\dots ,s_k,t_1,\dots ,t_k$, it has pairwise vertex-disjoint graph paths joining $s_i$ to $t_i$ for all $i$. The pairing is prescribed, which is stronger than the unpaired conclusion of the Menger theorem.

We prove the rooted dense minor linkage step carefully. We first prove the rooted branch-set reduction. Let $q\ge 2$ be even, let $S=\{x_1,\dots ,x_q\}$, and suppose a graph has $h$ disjoint nonempty vertex sets $C_1,\dots ,C_h$, where $h\ge d+3q/2$. Suppose each $G[C_i]$ is connected or every one of its graph components meets $S$, and each $C_i$ is adjacent to all but at most $d$ of the other $C_j$ not meeting $S$. Assume also that no rooted separation of a graph of order below $q$ avoids $d+1$ of these sets. Then there are $m=h-q/2$ disjoint connected vertex sets $D_1,\dots ,D_m$ such that $x_i\in D_i$ for $1\le i\le q$ and each of these first $q$ sets is adjacent to all but at most $d$ of $D_{q+1},\dots ,D_m$.

Here adjacency of sets means existence of an edge between them. A rooted separation of a graph is $(A,B)$ with $A\cup B=V(G)$, $S\subseteq A$, no edges between $A\setminus B$ and $B\setminus A$, and order $|A\cap B|$. It avoids $C_i$ when $A\cap C_i=\varnothing$. We prove the reduction by induction, first on the vertex count and then on the edge count; any counterexample chosen minimal in this ordering will be impossible.

We may delete edges joining two vertices of $S$. They cannot affect rooted separations of a graph, since $S$ lies wholly on the first side. Within a branch set, their deletion can only split a graph component into graph components each containing an endpoint in $S$, so the branch-set hypothesis also persists. An isolated vertex outside $S$ and all branch sets can be deleted. An isolated vertex $v\in S$ is also impossible, since $(S,V(G)\setminus \{v\})$ has order $q-1$ and avoids at least $h-q\ge d+q/2\ge d+1$ sets. Thus in a minimal counterexample $S$ is independent, and every isolated vertex outside $S$ belongs to a singleton branch set.

Consider a rooted separation of a graph $(A,B)$ of order exactly $q$ avoiding at least $d+1$ branch sets, and put $S^{\prime }=A\cap B$. Restrict to $G^{\prime }=G[B]-E(G[S^{\prime }])$ and $C_i^{\prime }=C_i\cap B$. Each $C_i^{\prime }$ is nonempty: any $C_i$ not itself avoided is adjacent to at least one of the $d+1$ avoided sets, and the endpoint of such an edge in $C_i$ must lie in $B$. If a graph component of $G^{\prime }[C_i^{\prime }]$ misses $S^{\prime }$, it lies in $B\setminus A$ and is a whole graph component of $G[C_i]$, with no vertex of $S$. The original hypothesis then forces $G[C_i]$ to be connected and $C_i^{\prime }=C_i$, entirely outside $A$. Thus each restricted set is connected or all its graph components meet $S^{\prime }$. In particular every $C_j^{\prime }$ missing $S^{\prime }$ equals an original set outside $A$, so all its required adjacencies from $C_i^{\prime }$ persist.

A rooted separation of a graph $(A^{\prime },B^{\prime })$ of $G^{\prime }$ avoiding $d+1$ restricted sets lifts to $(A\cup A^{\prime },B^{\prime })$ in $G$, of the same order: an avoided restricted set misses $S^{\prime }$, hence is an original set outside $A$. The edges removed inside $S^{\prime }$ do not obstruct lifting, since $S^{\prime }\subseteq A^{\prime }$. This proves the required separation hypothesis for $G^{\prime }$. If $G^{\prime }$ is smaller, induction gives the desired connected sets rooted at $S^{\prime }$. Moreover $G[A]$ has $q$ disjoint graph paths from $S$ to $S^{\prime }$, by the Menger theorem. Indeed a separator of smaller order between these sets would give a rooted separation of a graph of $G$ of smaller order still avoiding those $d+1$ original sets. Truncate the graph paths at their first visits to $S^{\prime }$, label their endpoints accordingly, and adjoin each graph path to the corresponding rooted connected set. They meet $B$ only in their distinct endpoints, so this gives the desired sets for $G$, a contradiction. Consequently in a minimal counterexample every such separation of order $q$ has $B=V(G)$ and $A=S$, an independent set.

Now contract any edge whose endpoints do not belong to two different branch sets. Its endpoints cannot both lie in $S$, so the image of $S$ still has size $q$. The branch-set conditions persist. If the contraction created a forbidden rooted separation of a graph of order below $q$, its lift would have order below $q$, except possibly when the contracted vertex lies in its separator. In that case the lifted order increases by one and is at most $q$. Equality would put the contracted edge inside $A$, contradicting the just-proved independence of $A$ for an order-$q$ separation avoiding $d+1$ sets. Thus the separation hypothesis also persists. Applying induction in the contracted graph and expanding the contracted vertex would produce the required disjoint connected sets. Hence every edge in a minimal counterexample joins two different branch sets.

Every vertex outside the branch sets would be isolated, since every edge joins two branch sets; such an isolated vertex could be deleted. Thus the branch sets cover the graph. Each nonsingleton branch set is therefore an independent set, and all its vertices belong to $S$, because each of its singleton graph components must meet $S$. Let $C$ be the union of the nonsingleton branch sets, and write $c=|C|\le q$. There is a matching in a graph from $C$ into $V(G)\setminus S$. Otherwise the Hall marriage theorem gives $X\subseteq C$ with neighbourhood $Y\subseteq V(G)\setminus S$ satisfying $|Y|<|X|$. Since $S$ is independent, $(S\cup Y,V(G)\setminus X)$ is a rooted separation of a graph of order $q-|X|+|Y|<q$. It avoids every singleton branch set outside $S\cup Y$, of which there are at leastThe penultimate counting bound uses that the nonsingleton sets number at most $c/2$, so $|V(G)|-c\ge h-c/2$. This contradicts the separation hypothesis.

Use that matching in a graph to make a two-vertex connected set for each terminal in $C$; terminals outside $C$ remain singleton sets. There are at leastunused vertices, all singleton original branch sets. Choose $m-q$ of them for the remaining $D_i$. Every rooted $D_i$ contains a singleton original branch set: either its terminal, or its terminal's matched neighbour. That singleton misses at most $d$ of the selected nonterminal singleton sets. This constructs exactly the desired family, contradicting minimality and proving the branch-set reduction.

Return to the given $22k$-connected graph. By the allowed assumption it has a graph minor $H$ with $2\delta (H)\ge h+4k-1$, where $h=|H|$. Put $d=h-1-\delta (H)$. ThenTake a branch-set model of $H$ and the $2k$ prescribed terminals as $S$. The branch sets are connected and each misses at most $d$ of the others. The separation hypothesis holds with $q=2k$: a separation of order below $2k$ avoiding a branch set would separate a surviving terminal from that set, contrary even to $2k$-connectivity. Since $h\ge d+3k$, the reduction applies. It produces $m=h-k$ sets with $s_i\in D_i$, $t_i\in D_{k+i}$, each terminal set missing at most $d$ of the $m-2k$ nonterminal sets. For each pair $D_i,D_{k+i}$ at leastnonterminal sets are adjacent to both. Choose different such sets for the $k$ pairs successively. The connected graph property inside the three sets and the two joining edges give a graph path from $s_i$ to $t_i$; take a simple graph path within their union. Different graph paths use disjoint sets. We have provedIn fact the argument proves the stronger transfer statement with $2k$-connectivity and the stated dense graph minor.

For the negative assertion, it suffices to give a value of $k$ for which the proposed general implication fails. Take $k=2$ and the graph on cyclic indices $0,\dots ,7$ in which two indices are adjacent when their cyclic distance is one or two. This is the four-connected planar obstruction to two-linkage. Draw the even cycle in a graph $0,2,4,6$ as the outer square, the odd cycle in a graph $1,3,5,7$ as an inner square rotated by $45$ degrees of vertices, and join each odd vertex to the two neighbouring even vertices; the annulus is triangulated without crossings. The outer face has vertices $0,2,4,6$ in this order.

Deleting at most three vertices leaves the graph connected. In cyclic order, consecutive surviving vertices are joined unless separated by a run of at least two deleted vertices. Disconnection requires two distinct such runs, hence at least four deletions. Conversely deleting the four neighbours of a vertex isolates it, so the vertex connectivity is exactly four. Disjoint graph paths joining $0$ to $4$ and $2$ to $6$ cannot exist: both would lie in the closed disk complementary to the outer face, and their endpoints alternate on its boundary. A first such graph path is a crosscut of the disk; the Jordan curve theorem puts the other two endpoints on opposite sides, forcing the second graph path to meet it. ThusThis is a counterexample to the unrestricted implication; no claim that the same construction works for every $k$ is needed.

We first derive the exterior area theorem for univalent functions directly from Green's theorem, then obtain the coefficient bound needed for the Koebe quarter theorem.

Suppose $G$ is univalent on $|\zeta |>1$ and has a Laurent seriesFor $R>1$, the image of $R{e}^{it}$, $0\le t\le 2\pi$, is a regular simple closed curve. By the Jordan curve theorem the curve has one bounded and one unbounded complementary component. The image of $|\zeta |>R$ lies in the unbounded component, since it is connected, avoids the curve by injectivity and contains points near infinity. It fills that component: any finite boundary limit of the image has preimages bounded by $G(\zeta )/\zeta \to 1$, and a convergent preimage subsequence must land on $|\zeta |=R$ or else give an interior image point. Thus no boundary of the image lies inside that complementary component. The curve is positively oriented about its bounded interior, since the exterior is on the right of $R{e}^{it}$ and a holomorphic map with nonzero derivative preserves orientation. Green's formula gives its enclosed area asFor completeness, the positive Fourier mode is $R{e}^{it}$ and the negative modes are $b_n{R}^{-n}{e}^{-int}$. On multiplying by the conjugate series and integrating, all unequal-frequency products integrate to zero by Fourier orthogonality. The equal-frequency products give $i{R}^2$ and $-in|b_n{|}^2{R}^{-2n}$, respectively; the constant term gives zero. The Laurent series and its differentiated series converge uniformly on the circle, since $R>1$, so these operations are justified. Nonnegativity of $A_R$ yields $\sum n|b_n{|}^2{R}^{-2n}\le R^2$. Letting $R\downarrow 1$, or first doing so for every finite partial sum, givesNo regularity of the map on the unit circle has been assumed.

Now let $f(z)=z+a_2{z}^2+\cdots$ be a normalized univalent function. Its only zero is at zero, so $f(z)/z$, extended there by the value one, is holomorphic and nowhere zero. On the simply connected disc it has a holomorphic square root with value one at zero. Its odd square-root transform isThis is an odd function which is holomorphic and normalized. It is also univalent: $h(z)=h(w)$ gives $f(z^2)=f(w^2)$ and hence $z=\pm w$. If $z=-w$, oddness gives $h(z)=-h(z)$, so $h(z)=0$, forcing $z=w=0$. Thus $G(\zeta )=1/h(1/\zeta )$ is univalent on the exterior disc, with expansionApplying the area bound to its first negative coefficient proves the second coefficient bound for normalized univalent functions, $|a_2|\le 2$.

Let $w_0$ be any point omitted by $f$. It is nonzero. Postcomposition with a Möbius map gives another normalized univalent functionwhose denominator does not vanish. Applying the second coefficient bound both to $f$ and to $f_{w_0}$ givesEvery omitted point therefore has modulus at least $1/4$, provingThis establishes the Koebe quarter theorem from the allowed area formula.

Sharpness is witnessed by the Koebe function $k(z)=z/(1-z{)}^2$. It has the required value and derivative at zero. The equality $k(z)=k(w)$ simplifies to $(z-w)(1-zw)=0$, and $|zw|<1$ makes it injective. Moreover,Thus $-1/4$ is omitted, so no larger universal radius is possible. In fact $4k(z)+1=((1+z)/(1-z){)}^2$ shows its image is the plane slit along $(-\infty ,-1/4]$.

Without univalence take a real $\lambda >4$ andIt is entire, with $f_{\lambda }(0)=0$ and $f_{\lambda }^{\prime }(0)=1$, but it omits $-1/\lambda$, a point of modulus less than $1/4$, since the exponential never vanishes. It is genuinely not univalent on the disc: the distinct points $\pm \pi i/\lambda$ lie there and both give exponential value $-1$. Even its derivative never vanishes, illustrating why local conformality alone does not give the quarter-disc inclusion.

A hyperbolic Riemann surface in potential theory is a noncompact Riemann surface admitting a positive Green function on a Riemann surface. With pole $p$, this is a positive function harmonic off $p$, having local form $-\log |\zeta |$ plus a harmonic function in a coordinate $\zeta (p)=0$, and obtained as the minimal such positive function by exhaustion. This is a noncircular definition for the present proof. For a simply connected surface it is equivalent to disk conformal type; for arbitrary surfaces potential-theoretic hyperbolicity and disk universal-cover type should be distinguished.

The unit disk is an example. At pole zero its function is $G(z,0)=-\log |z|$: it is positive in the punctured disk, harmonic there, has the correct logarithmic singularity, and tends to zero at the boundary. If another positive function has the same singularity and is harmonic elsewhere, subtract $G$ and apply the maximum principle for harmonic functions on disks of radius $r<1$; the boundary lower bound is $\log r$. Letting $r\uparrow 1$ shows that the other function dominates $G$. Thus $G$ is indeed the positive Green function on a Riemann surface. At a general pole it is

Here is a Green-function exhaustion proof of disk uniformization. The main analytic idea is to build conformal maps on bordered disks from their Green functions and pass to an extremal limit. The positive global Green function prevents that limit from becoming constant.

First exhaust the noncompact simply connected surface $R$ by relatively compact domains $R_n$ containing $p$, each with a smooth Jordan boundary. This topological step does not use conformal uniformization: take regular neighborhoods of finitely many coordinate disks covering successive compact sets and fill bounded complementary components. Simple connectedness and the Jordan curve theorem make the filled neighborhoods topological disks; enlarge them to be nested.

Each $R_n$ has a zero-boundary Green function on a Riemann surface $G_n(\cdot ,p)$. One analytic construction, independent of conformal uniformization, is as follows. Equip the compact bordered domain with a smooth conformal metric. Multiply $-\log |\zeta |$ by a smooth cutoff function equal to one near $p$ and zero near the boundary; call the result $s$. Distributionallywhere $k$ is smooth and supported away from $p$. Solve $\Delta v=k$ with zero boundary data. This auxiliary Dirichlet problem can be obtained by minimizing $\frac{1}{2}\int |\nabla v{|}^2+\int kv$ over functions with zero boundary trace: the Poincaré inequality gives coercivity, minimization in a Hilbert space gives a weak solution, and interior and smooth-boundary elliptic regularity give a smooth solution. Then $G_n=s+v$ is harmonic away from $p$, has the logarithmic pole, and vanishes at the boundary. It is positive by the maximum principle for harmonic functions, applied after removing a sufficiently small circle around its positive pole. This construction uses ordinary linear elliptic existence, not the desired uniformization theorem. Equivalently, the same bordered-domain Green functions can be constructed by the Perron method for the Dirichlet problem using local logarithmic barriers.

On $R_n\setminus \{p\}$ choose local harmonic conjugates $G_n^{\ast }$. Their period around the pole is $-2\pi$, and every closed loop is homologous to an integer multiple of that loop. Thereforeis a single-valued holomorphic function. It extends across $p$ with one simple zero and has no others. Since $G_n>0$, $|f_n|<1$; since $G_n=0$ on the boundary, $|f_n|\to 1$ there. Thus $f_n:R_n\to \mathbb{D}$ is proper. For any $a\in \mathbb{D}$, choose a smooth interior contour sufficiently close to the boundary that $|f_n|>|a|$ everywhere on it. The contour surrounds $p$ and all possible preimages of $a$, because $|f_n|>|a|$ throughout a thin boundary collar. By Rouché's theorem, $f_n-a$ has the same number of zeros inside as $f_n$, namely one. Hence $f_n$ is onto and one-to-one, with nonzero derivative: it is a biholomorphism from $R_n$ to the disk.

Write near $p$Multiply $f_n$ by a unit constant so that, in the fixed coordinate, $f_n^{\prime }(p)=e^{-c_n}>0$. Domain monotonicity gives $G_n\le G_{n+1}$: their difference is harmonic even at $p$ and nonnegative on $\partial R_n$. The global positive Green function on a Riemann surface $G_R$ similarly majorizes every $G_n$. Comparing the regular parts shows that $c_n$ increases and is bounded above by the finite regular part of $G_R$ at $p$. Consequently

On each fixed compact subdomain, the $f_n$ are eventually bounded holomorphic functions. Montel theorem and a diagonal subsequence therefore give a locally uniform convergence limit $F:R\to \mathbb{D}$ with $F(p)=0$ and $F^{\prime }(p)=d>0$. The maximum modulus principle keeps its values strictly inside the disk. It is injective: for fixed $q$, the functions $f_n-f_n(q)$ have no zeros away from $q$; Hurwitz's theorem implies that their nonconstant limit has no other zeros. This is the locally uniform limit of univalent functions argument.

There is also an extremality property. For any holomorphic function $g:R\to \mathbb{D}$ with $g(p)=0$, the function $g\circ f_n^{-1}$ is a disk self-map fixing zero. The Schwarz lemma gives $|g^{\prime }(p)|\le f_n^{\prime }(p)$. Passing to the limit yields $|g^{\prime }(p)|\le d=F^{\prime }(p)$.

Finally suppose $F$ omits $a\in \mathbb{D}$. Necessarily $a\ne 0$. Apply the disk Möbius transformation ${\varphi }_a(w)=(w-a)/(1-\overline{a}w)$. The nowhere-zero function ${\varphi }_a\circ F$ has a global holomorphic logarithm because $R$ is simply connected, hence a holomorphic square root $h$. It maps into the disk and $|h(p)|=\sqrt{|a|}$. Normalize it by setting $g={\varphi }_{h(p)}\circ h$, so $g(p)=0$. Direct differentiation givesThe strict inequality follows from $(1-\sqrt{|a|}{)}^2>0$. This contradicts extremality. Therefore $F$ omits no point, and we conclude

A simply connected space is path-connected and has trivial fundamental group, equivalently every based loop is null-homotopic. The Whitehead theorem says that a map of CW complexes inducing isomorphisms on all homotopy groups is a homotopy equivalence. Its homological Whitehead theorem form says that a map between simply connected CW complexes, in particular finite ones, is a homotopy equivalence if it induces isomorphisms on integral homology. The latter implication follows from the relative Hurewicz theorem and the ordinary Whitehead theorem; homotopy equivalences necessarily induce homology isomorphisms.

For the attached-cell space $X$, the cellular homology chain complex in positive degrees isDualizing gives the cellular cohomology complex $\mathbb{Z}\stackrel{0}{\to }\mathbb{Z}\stackrel{t\mapsto (2t,3t)}{\to }{\mathbb{Z}}^2$. The last map is injective and its image is primitive because $\gcd (2,3)=1$. Its cokernel is $\mathbb{Z}$, explicitly via $(a,b)\mapsto 3a-2b$. HenceThe van Kampen theorem gives ${\pi }_1(X)=⟨a\mid a^2=1,a^3=1⟩=1$. Moreover $H_2(X)=\ker (23)$ is generated by $(3,-2)$. Since $X$ is simply connected, the Hurewicz theorem gives ${\pi }_2(X)\cong H_2(X)$. Choose a map $S^2\to X$ representing that generator. It induces an isomorphism on $H_2$, and on $H_0$, while all other reduced homology groups vanish on both sides. The homological Whitehead theorem proves $\boxed{X\simeq S^2}$. This is a special case of coprime two-cell attachments to a circle.

It is not homeomorphic to the sphere. Indeed, if it were, its embedded one-skeleton would be a Jordan circle in $S^2$, with the two open cell interiors as its complementary regions. The Jordan curve theorem and its Schoenflies form identify the closures of those regions with discs. A characteristic map, homeomorphic on the interior of such a disc, has local degree $+1$ or $-1$ at an interior point; the boundary homomorphism in relative homology then makes its attaching map on the circle have degree $+1$ or $-1$. This contradicts degrees $2$ and $3$. This argument does not require choosing special covering representatives of the attaching maps.

Finally, two compact examples with the same additive cohomology areBoth have $H^0=\mathbb{Z}$, $H^1={\mathbb{Z}}^2$, $H^2=\mathbb{Z}$ and zero higher cohomology, by their cellular chain complexes. But their fundamental groups are respectively ${\mathbb{Z}}^2$ and the nonabelian free group on two generators, so they cannot be homotopy equivalent. This illustrates that cohomology groups do not determine homotopy type.

We use Zhang's argument, obtaining the critical nonpercolation conclusion from uniqueness rather than assuming it as part of the Harris-Kesten theorem. For independent bond percolation on ${\mathbb{Z}}^2$, apply part (ii) to independent site percolation on its line graph; open sites correspond to open bonds. The line graph is amenable and of finite type. Therefore an infinite bond cluster, if it exists, is almost surely unique. The same holds for the dual bond percolation model.

Assume for contradiction that $\theta (1/2)>0$. The planar dual graph is another square lattice, and its open bonds correspond to closed primal bonds, also with parameter $1/2$. Both models then have an infinite percolation cluster almost surely, by part (i). As a disc $D_R$ expands, the probability that it meets an infinite cluster tends to one, for each model. Choose $R$ avoiding all primal and dual vertices and all intersections of either graph with the four endpoints of the following arcs. Divide the boundary circle into four equal arcs $A_1,A_2,A_3,A_4$ in cyclic order. Let $L_i$ be the event of an infinite primal open ray whose last exit from $D_R$ is in $A_i$; define $L_i^{\ast }$ similarly for an infinite dual open ray. After that exit the entire ray stays outside the disc. The existence of a ray follows by finite branching in an infinite locally finite graph.

Each $L_i$ is increasing in the primal bond states, and their union has probability tending to one. Four-fold rotation symmetry makes their probabilities equal. The FKG inequality for their decreasing complements gives the square-root trick for positively associated eventsThe left side tends to zero, so $\mathbb{P}(L_i)\to 1$. Apply the same argument to the dual states to get $\mathbb{P}(L_i^{\ast })\to 1$. A union bound, requiring no independence between the models, now shows that for sufficiently large $R$ the alternating eventhas positive probability.

The alternating exterior-arm separation lemma says that the two primal rays on this event cannot be joined by a primal open path lying outside $D_R$. Indeed, such a joining path contains a simple crosscut of the exterior between the two primal exit points. Join that crosscut to a chord inside the disc. The Jordan curve theorem puts one of the alternating dual exit points on the bounded side. Its dual ray, which is restricted to the exterior of the disc and runs to infinity, would have to cross the open primal crosscut. A dual open bond cannot cross a primal open bond. This is impossible.

Let $F$ be the finite set of primal bonds whose drawn edges meet $D_R$, and let $E$ be the projection of the alternating-arm event onto the states outside $F$. Because $F$ is finite, this projection is a finite union of measurable events; $\mathbb{P}(E)>0$. Every configuration in $E$ admits an assignment on $F$ witnessing the original four arms. Now close all bonds of $F$. The infinite tails of the two primal rays beyond their first exterior vertices survive; any path joining the surviving tails would now avoid the disc. Appending their original exterior segments back to their exit points would give the forbidden exterior connection. Thus the modified configuration contains at least two infinite percolation clusters. By independence, or the finite-energy property of Bernoulli percolation, $\mathbb{P}(E\cap \{F\text{ all closed}\})=2^{-|F|}\mathbb{P}(E)>0$. This contradicts the uniqueness obtained from part (ii). Therefore

For bond percolation, let $C(0)$ be the percolation cluster of the origin. Its percolation probability isWe prove $\theta (1/2)=0$ by alternating arms at the self-dual parameter, using the uniqueness of the infinite percolation cluster proved in Question 4. Suppose instead that $\theta (1/2)>0$.

At $p=1/2$, declaring a dual edge open exactly when its crossing primal edge is closed produces independent dual bond percolation with the same parameter on ${\mathbb{Z}}^2+(1/2,1/2)$. It therefore also has an infinite percolation cluster almost surely, and at most one. The two preliminary arguments below imply that, for a sufficiently large square, the probability of two primal infinite exterior arms starting through its top and bottom sides exceeds $3/4$. Applying the same arguments to the dual lattice gives probability greater than $3/4$ for dual infinite exterior arms starting through the left and right sides. Thus all four arms occur together with positive probability, by the union bound.

Here are precise boundary conventions for the alternating exterior-arm separation lemma. Use the primal graph vertex box $[-N,N{]}^2$, the dual graph vertex box $[-N+1/2,N-1/2{]}^2$, and the common continuous square with sides at $\pm (N+1/4)$. The primal arms first cross its top and bottom; the dual arms first cross its left and right. Subsequent graph vertices of every arm lie outside this continuous square. Large $N$ gives the required probabilities for both boxes, by the same rotational symmetry and boundary-side estimate. Primal open paths and dual open paths cannot cross, since the two states of a crossing edge are complementary.

The two primal arms cannot meet in the exterior. If they did, a finite simple primal crosscut in the exterior would join the top and bottom entry points. Together with one of the two arcs of the square boundary it would enclose a bounded region containing one of the dual entry points. That dual exterior arm could not reach infinity without crossing the primal crosscut or the forbidden square interior. This is the alternating exterior-arm separation lemma. Thus the two primal arms are disjoint.

Now modify only the primal edges with both endpoints in $[-N,N{]}^2$, making them open. Their grid supplies a simple interior path joining the two primal arm roots. The unchanged exterior arms and this connector form a proper doubly infinite simple primal path. Such a path separates the plane into two sides, by the Jordan curve theorem applied to its one-point compactification in the sphere. The left and right dual arms lie on opposite sides. Preserve their tails from their first exterior dual graph vertices; none of those tails crosses a modified edge. Their original boundary attachment edges may be destroyed, but the infinite exterior tails themselves remain. They now belong to two distinct infinite dual clusters, since any dual connection between them would have to cross the doubly infinite primal path.

This modification has positive probability by the finite-energy property of Bernoulli percolation: condition on the exterior configuration and open the finitely many interior edges. The exterior projections of the four-arm event have positive probability. The resulting positive-probability event with two infinite dual clusters contradicts uniqueness of the infinite percolation cluster on the dual lattice. Therefore

Only the selected paths are drawn. The black connector lies in the finitely modified box; the two blue infinite dual tails remain outside and on opposite sides.

The proof uses self-duality, symmetry, positive association, finite-energy property of Bernoulli percolation and uniqueness; it does not assume the Harris-Kesten theorem as a black box. The boundary language means an infinite ray whose first graph vertex is on the side and all later graph vertices are outside the graph vertex box. It cannot mean literal intersection of two disjoint sets, a closed box boundary and its strict complement.

A parametrization is $X(\theta ,z)=(\cosh z\cos \theta ,\cosh z\sin \theta ,z)$. It gives a homeomorphism $S^1\times \mathbb{R}\to S$, with inverse obtained from the height and angular position. Its first fundamental form isFor a surface of revolution with radius $r(z)$, the Gaussian curvature is $-r^{\prime \prime }/[r(1+r^{\prime 2}{)}^2]$. Thus $\boxed{K=-{\mathrm{sech}}^{4}z<0}$. The waist $z=0$ is a closed geodesic: ${\Gamma }_{\theta \theta }^z=-\tanh z$ vanishes there, and $\theta$ can be parametrized at constant speed.

For the uniqueness of a closed geodesic on a negatively curved cylinder, use Gauss-Bonnet theorem with piecewise geodesic boundary:A geodesic disk with smooth boundary would require curvature integral $2\pi$. A disk bounded by a geodesic monogon would require $\pi +\alpha >0$, and a geodesic bigon would require $\alpha +\beta >0$. All are impossible when $K<0$.

Here is the topological reduction needed to apply this observation without assuming simplicity. Lift a closed geodesic to the universal covering plane of the cylinder. A self-intersection of its lift supplies an innermost monogon by the Jordan curve theorem, so the lift is embedded. A contractible closed geodesic would similarly supply a geodesic disk or monogon, so its winding number is nonzero. Its lift is consequently a proper line, invariant under a nonzero power of the deck translation. Two distinct translates of that line cannot intersect: periodicity would repeat an intersection, and innermost arcs between intersections would bound a bigon. The disjoint translates are ordered transversely, with the deck translation preserving that order. A finite cyclic permutation of this ordered family is impossible, so the lift is invariant under the primitive deck translation too. Thus the original geodesic has a simple image and is at most a repeated traversal of it.

Likewise two distinct simple essential closed geodesics cannot intersect: their periodic lifts would produce a bigon. If disjoint, the Jordan curve theorem on the cylinder says they bound a compact annulus. Its two boundaries have zero geodesic curvature and $\chi =0$, so Gauss-Bonnet gives $\int KdA=0$, again impossible. Therefore there is at most one closed geodesic image. Reparametrizations, reversed orientation and multiple traversals are understood to describe the same geometric geodesic.

On $S_{0,3}$, a simple proper arc joining two distinct specified punctures is unique up to isotopy relative to its ends. Indeed, a small regular neighborhood of the arc and its two ends has one boundary curve separating those two punctures from the third; the Jordan curve theorem gives the unique such separation, and the disc it bounds gives the isotopy between any two choices.

Choose the three joining arcs, one for each puncture pair, with disjoint interiors. They form an ideal triangle graph whose complement consists of two discs. A pure homeomorphism fixes all three endpoints and carries each arc to an isotopic arc. The simultaneous-isotopy lemma makes it fix the three arcs, and the Alexander trick on each complementary disc makes it isotopic to the identity. Therefore

Fix punctures $p,q,r$. There is one isotopy class ${\ell }_p$ of essential arc from $p$ back to $p$: by the Jordan curve theorem, it separates $q$ from $r$, and any two such arcs are isotopic relative to the punctures.

Altogether $\mathcal{A}(S_{0,3})$ has six vertices:The joining arcs $a_{pq},a_{pr},a_{qr}$ span a triangle. For each puncture $p$, the three verticesspan another triangle. These are all the maximal simplices: ${\ell }_p$ must cross $a_{qr}$, and returning arcs based at distinct punctures cannot be made disjoint.

Every orientation-preserving permutation of the three punctures is realizable, while a homeomorphism acting trivially on the punctures is isotopic to the identity. ThusIt permutes the labels in the displayed description. There are two vertex orbits, the three joining arcs and the three returning arcs, as summarized by the arc complex of the three-punctured sphere.

Suppose that two closed geodesics ${\gamma }_1$ and ${\gamma }_2$ were disjoint. Under the usual convention that a closed geodesic curve is simple, the Jordan curve theorem on the sphere shows that they bound an annulus $A\subset S$. Its Euler characteristic is zero. Give $A$ the induced orientation and apply the Gauss-Bonnet theorem:Both boundary components are geodesics, so their geodesic curvature vanishes. It follows thatThis is impossible because $K>0$ everywhere and $A$ has positive area. Hence the two closed geodesics must intersect, as summarized by the intersection of closed geodesics on a positively curved sphere.

Choose a spanning tree of the connected plane graph. A tree with $n$ vertices has $n-1$ edges and one face of a plane graph, so $n-(n-1)+1=2$. Add the remaining drawn edges one at a time. Each joins two existing vertices and closes a cycle in a graph; its interior lies in one current face of a plane graph, splitting that face of a plane graph into two by the Jordan curve theorem. Thus $m$ and $f$ increase together while $n$ stays fixed. The quantity $n-m+f$ remains two, proving Euler formula for a connected planar graph.

A Jordan curve in the two-sphere is carried to an equator by a homeomorphism of the sphere; equivalently, the closures of its two complementary regions are discs. This strengthens the separation assertion of the Jordan curve theorem and controls the boundary degrees of characteristic maps for cell decompositions of a surface.