A simply connected surface without a Green kernel is plane or sphere

The meaning of the hypothesis. By [F3] applied to the pole $p_0$, the envelope $g_X(\cdot ,p_0)$ is either $+\mathrm{\infty }$ everywhere on $X\setminus \{p_0\}$ or finite and strictly positive everywhere there; the hypothesis of the statement is the first alternative, so the second is excluded. Consequently, for every point $q\in X\setminus \{p_0\}$ and every real number $B$ there is a candidate $v\in {\mathcal{F}}_{p_0}$ with $v(q)>B$, because $+\mathrm{\infty }$ is the supremum of the values $v(q)$.

Three distinct points of the surface. By [F16] the space $X$ is uncountable, so we may choose distinct points $p_1,p_2\in X$ and a further point $r\in X\setminus \{p_1,p_2\}$; these choices are finite and explicit.

A claim: every bounded holomorphic function on $X$ is constant. We prove the claim. Let $h:X\to \mathbb{C}$ be holomorphic with $\mathrm{\mid }h\mathrm{\mid }\le M<+\mathrm{\infty }$. If $M=0$ then $h\equiv 0$ is constant, so assume $M>0$; set $c:=h(p_0)$ and $B(w):=(w-c)\mathrm{/}(2M)$. The map $B$ is affine and injective with $B(h(p_0))=0$, so $B\circ h$ is holomorphic on $X$ and is nonconstant whenever $h$ is nonconstant, and $\mathrm{\mid }B\circ h\mathrm{\mid }\le 1$ on $X$ because $\mathrm{\mid }h-c\mathrm{\mid }\le \mathrm{\mid }h\mathrm{\mid }+\mathrm{\mid }c\mathrm{\mid }\le 2M$. Suppose, toward a contradiction, that $h$ is nonconstant.

The zero set of $B\circ h$. Since $h$ is nonconstant and $X$ is connected, $B\circ h$ is nonconstant ([F1]) and hence not constant on any nonempty open subset and not identically zero ([F7]). Let $Z:=\{x\in X:B(h(x))=0\}$, the zero set of $B\circ h$. By [F7] every point of $Z$ is isolated in $Z$, so $Z$ is closed with empty interior, $X\setminus Z$ is dense in $X$ and $p_0\in Z$. Moreover, in a centred chart $z:U\to \mathbb{D}$ at $p_0$ [F2], the precedence of [F7] gives a holomorphic $u$ on $U$ with $u(p_0)\mathrm{\ne }0$ and an integer $m\ge 1$ such that $B\circ h=z^{m}u$ on $U$.

The function $u_{\epsilon }$ is subharmonic off $Z$. Fix $v\in {\mathcal{F}}_{p_0}$ and $\epsilon >0$ and set $u_{\epsilon }:=v+(1+\epsilon )\log \mathrm{\mid }B\circ h\mathrm{\mid }$ on $X\setminus Z$. The restriction of $v$ to $X\setminus Z$ is subharmonic [F2, F4]; the function $\log \mathrm{\mid }B\circ h\mathrm{\mid }$ is harmonic on $X\setminus Z$ by [F6]; a nonnegative multiple of a harmonic function is subharmonic and sums of subharmonic functions are subharmonic [F4]; hence $u_{\epsilon }$ is subharmonic on $X\setminus Z$.

The positive part is subharmonic on all of $X$. Put $\widehat{u}:=\max (u_{\epsilon },0)$ on $X\setminus Z$ and $\widehat{u}:=0$ on $Z$. At a point $z\in Z$ with $z\mathrm{\ne }{p}_0$, the function $v$ is upper semicontinuous and finite at $z$ [F2, F4], so $v\le C$ on some neighbourhood $N$ of $z$ which we may take so small that $N\cap Z=\{z\}$; since $B(h(z))=0$, the function $\log \mathrm{\mid }B\circ h\mathrm{\mid }$ tends to $-\mathrm{\infty }$ at $z$ [F6], so $u_{\epsilon }<0$ on a punctured neighbourhood of $z$ and $\widehat{u}=0$ there. At $p_0$ the unit pole condition gives $v\le -\log \mathrm{\mid }z\mathrm{\mid }+C_1$ near $p_0$ [F2] and the factorization of step 1.4 gives $\log \mathrm{\mid }B\circ h\mathrm{\mid }=m\log \mathrm{\mid }z\mathrm{\mid }+\log \mathrm{\mid }u\mathrm{\mid }$ with $u$ continuous and $u(p_0)\mathrm{\ne }0$, so $u_{\epsilon }\le ((1+\epsilon )m-1)\log \mathrm{\mid }z\mathrm{\mid }+C_3$ with $(1+\epsilon )m-1\ge \epsilon >0$, and again $u_{\epsilon }<0$ near $p_0$; hence $\widehat{u}=0$ near $p_0$. So $\widehat{u}$ is upper semicontinuous on $X$ [F4], it agrees on $X\setminus Z$ with the maximum of the subharmonic function $u_{\epsilon }$ and $0$, which is subharmonic [F4], and it is locally constant $0$ near every point of $Z$; the locality of subharmonicity [F5] therefore makes $\widehat{u}$ subharmonic on $X$.

The dipole with two pole discs. By step 1.2 the points $p_1,p_2$ are distinct, so the dipole supplier [F8] applies, with the Countable Choice ${\mathrm{A}\mathrm{C}}_{\omega }$ supplied by [A1]: there are disjoint coordinate discs $U_1\ni p_1$ and $U_2\ni p_2$ with compact closures, centred coordinates $z_1,z_2$, and a harmonic $G:X\setminus \{p_1,p_2\}\to \mathbb{R}$ such that $G+\log \mathrm{\mid }{z}_1\mathrm{\mid }$ is harmonic on $U_1$, $G-\log \mathrm{\mid }{z}_2\mathrm{\mid }$ is harmonic on $U_2$, and ${\sup }_{X\setminus (U_1\cup U_2)}\mathrm{\mid }G\mathrm{\mid }=:C<+\mathrm{\infty }$.

The maximum principle forces $u_{\epsilon }\le 0$. By step 2.1, $\widehat{u}$ is nonnegative and subharmonic on $X$, and it vanishes on a neighbourhood of $p_0$. Let $K$ be a compact support of $v$ from [F2]. If $X$ is compact, put $L:=X$; otherwise $\widehat{u}$ vanishes on $X\setminus K$, so put $L:=K$. If $\widehat{u}$ were positive somewhere, $M:={\sup }_L\widehat{u}$ would be positive and would bound $\widehat{u}$ on all of $X$. This maximum is finite and attained: the open sets $\{\widehat{u}<n\}$ cover the compact set $L$, giving an upper bound, and the closed nonempty superlevel sets $\{\widehat{u}\ge b\}$ for $b<M$ have the finite-intersection property, so compactness yields $x^{\ast }\in L$ with $\widehat{u}(x^{\ast })=M$. The strong maximum principle [F4] would then make $\widehat{u}\equiv M$ on connected $X$, contradicting its vanishing near $p_0$. Hence $\widehat{u}\equiv 0$, and $u_{\epsilon }\le 0$ on $X\setminus Z$.

A meromorphic function with a simple zero and a simple pole. On $U_1\setminus \{p_1\}$ the function $G$ differs from $-\log \mathrm{\mid }{z}_1\mathrm{\mid }$ by a harmonic function and on $U_2\setminus \{p_2\}$ it differs from $\log \mathrm{\mid }{z}_2\mathrm{\mid }$ by a harmonic function [F8]; that is, the exponents $m_1:=1$ and $m_2:=-1$ satisfy $G=-m_1\log \mathrm{\mid }{z}_1\mathrm{\mid }+h_1$ on $U_1\setminus \{p_1\}$ and $G=-m_2\log \mathrm{\mid }{z}_2\mathrm{\mid }+h_2$ on $U_2\setminus \{p_2\}$ with $h_1,h_2$ harmonic. The monodromy supplier [F9] applies with $Y:=X$, $P:=\{p_1,p_2\}$ and $u:=G$, because $X$ is simply connected: there is a meromorphic function $F:X\to \widehat{\mathbb{C}}$ with $\mathrm{\mid }F\mathrm{\mid }=e^{-G}$ on $X\setminus \{p_1,p_2\}$ and local forms $F=z_1{G}_1$ with $G_1(p_1)\mathrm{\ne }0$ on $U_1$ and $F=z_2^{-1}{G}_2$ with $G_2(p_2)\mathrm{\ne }0$ on $U_2$. Consequently $F$ has a simple zero at $p_1$, a simple pole at $p_2$, and no other zeros or poles.

The envelope must be finite: contradiction, so $h$ is constant. The set $Z$ has empty interior, so $X\setminus Z$ is dense in $X$ [F11], and $X\setminus \{p_0\}$ is a nonempty open subset of $X$ because $X$ has more than one point and is Hausdorff [F1, F16]; hence the dense set $X\setminus Z$ meets $X\setminus \{p_0\}$ and there is a point $q\in X\setminus (Z\cup \{p_0\})$. At that point $B(h(q))\mathrm{\ne }0$, and step 3.1 gives $v(q)\le -(1+\epsilon )\log \mathrm{\mid }B(h(q))\mathrm{\mid }$ for every $v\in {\mathcal{F}}_{p_0}$ and every $\epsilon >0$; taking the supremum over ${\mathcal{F}}_{p_0}$ [F2] and letting $\epsilon \downarrow 0$ yields $g_X(q,p_0)\le -\log \mathrm{\mid }B(h(q))\mathrm{\mid }<+\mathrm{\infty }$. This contradicts the hypothesis $g_X(q,p_0)=+\mathrm{\infty }$ [given]. Therefore the supposition of step 1.3 was false: every bounded holomorphic function $h:X\to \mathbb{C}$ is constant.

The auxiliary dipole at an arbitrary third point. Let $r\in X\setminus \{p_1,p_2\}$ be arbitrary (step 1.2). Applying [F8] and [F9] to the pair $(r,p_2)$ exactly as in steps 2.2 and 3.2, with the same exponents $1$ at the zero and $-1$ at the pole, produces discs $V_r\ni r$ and $V_2\ni p_2$ with compact closures, a harmonic $G_r$ on $X\setminus \{r,p_2\}$ with ${\sup }_{X\setminus (V_r\cup V_2)}\mathrm{\mid }{G}_r\mathrm{\mid }=:C^{\mathrm{\prime }}<+\mathrm{\infty }$ and the two unit log poles, and a meromorphic function $F_r:X\to \widehat{\mathbb{C}}$ with $\mathrm{\mid }{F}_r\mathrm{\mid }=e^{-G_r}$ on $X\setminus \{r,p_2\}$, a simple zero at $r$, a simple pole at $p_2$, and no other zeros or poles. Since $r\mathrm{\notin }\{p_1,p_2\}$ is neither the zero nor the pole of $F$, the value $F(r)$ is a nonzero complex number.

The quotient $H$ is holomorphic on $X$. Define $H$ on $X\setminus \{r,p_2\}$ by $H:=(F-F(r))\mathrm{/}{F}_r$; there both functions are holomorphic and $F_r\mathrm{\ne }0$ [F1, step 3.2, step 4.2]. We extend $H$ holomorphically across $r$ and $p_2$. Near $r$, in a centred chart $z_r$ at $r$, step 4.2 provides $F_r=z_{r}g$ with $g(r)\mathrm{\ne }0$, while $F-F(r)$ is holomorphic near $r$, vanishes at $r$, and is not identically zero there because $F$ is not constant on any nonempty open subset [F7, step 3.2]; by [F7] it therefore has finite order $k\ge 1$ at $r$, say $F-F(r)=z_r^{k}a$ with $a$ holomorphic, so $H=z_r^{k-1}a\mathrm{/}g$ is holomorphic at $r$. Near $p_2$, step 3.2 and step 4.2 give $F=z_2^{-1}{G}_2$ and $F_r=z_2^{-1}{G}_r^{\mathrm{\prime }}$ with $G_2(p_2)\mathrm{\ne }0\mathrm{\ne }{G}_r^{\mathrm{\prime }}(p_2)$, so $H=(G_2-z_{2}F(r))\mathrm{/}{G}_r^{\mathrm{\prime }}$ extends holomorphically to $p_2$ with value $G_2(p_2)\mathrm{/}{G}_r^{\mathrm{\prime }}(p_2)$. Hence $H$ is a holomorphic function $H:X\to \mathbb{C}$.

$H$ is bounded. Off $U_1\cup U_2$ we have $\mathrm{\mid }F\mathrm{\mid }=e^{-G}\le e^C$ and off $V_r\cup V_2$ we have $\mathrm{\mid }{F}_r\mathrm{\mid }=e^{-G_r}\ge e^{-C^{\mathrm{\prime }}}$ by steps 2.2 and 4.2, so on the open set $W:=X\setminus (U_1\cup U_2\cup V_r\cup V_2)$ the quotient satisfies $\mathrm{\mid }H\mathrm{\mid }\le (e^C+\mathrm{\mid }F(r)\mathrm{\mid })e^{C^{\mathrm{\prime }}}$. On each of the four compact sets ${\bar{U}}_1,{\bar{U}}_2,{\bar{V}}_r,{\bar{V}}_2$ the continuous function $\mathrm{\mid }H\mathrm{\mid }$ [F1, step 5.1] attains a maximum, a finite real number [F11]; let $B_0$ be the largest of these four numbers. Since $X=W\cup U_1\cup U_2\cup V_r\cup V_2\subseteq W\cup {\bar{U}}_1\cup {\bar{U}}_2\cup {\bar{V}}_r\cup {\bar{V}}_2$, the function $H$ is bounded on $X$, with $\mathrm{\mid }H\mathrm{\mid }\le \max \left\{\right(e^C+\mathrm{\mid }F(r)\mathrm{\mid })e^{C^{\mathrm{\prime }}},B_0\}$.

$H$ is a nonzero constant. The function $H$ of step 5.1 is holomorphic and bounded by step 6.1, so the claim proved in steps 1.3-4.1 gives that $H\equiv c$ is constant. Evaluating at $p_1$, using $F(p_1)=0$ (step 3.2) and that $F_r(p_1)$ is a nonzero complex number because $p_1\mathrm{\notin }\{r,p_2\}$ is neither the zero nor the pole of $F_r$ (step 4.2), gives $c=H(p_1)=(0-F(r))\mathrm{/}{F}_r(p_1)\mathrm{\ne }0$.

$F$ is injective. Fix $r\in X\setminus \{p_1,p_2\}$ and let $c\mathrm{\ne }0$ be the constant of step 7.1. Then $F-F(r)=c{F}_r$ on $X$, as both sides are holomorphic and agree on the nonempty open set $X\setminus \{r,p_2\}$ [F1, step 5.1, step 7.1]. If $x\in X\setminus \{r,p_2\}$ satisfies $F(x)=F(r)$, then $F_r(x)=(F(x)-F(r))\mathrm{/}c=0$, and the only zero of $F_r$ is $r$ (step 4.2), so $x=r$, a contradiction; hence $F^{-1}(F(r))=\{r\}$, since $F(p_2)=\mathrm{\infty }\mathrm{\ne }F(r)$ and $F(p_1)=0\mathrm{\ne }F(r)$. Since $r$ was an arbitrary point of $X\setminus \{p_1,p_2\}$, we have shown that $F^{-1}(F(r))=\{r\}$ for every such $r$. Now let $x,y\in X$ with $F(x)=F(y)=w$. If $w=0$ then $x=y=p_1$, the only zero of $F$, and if $w=\mathrm{\infty }$ then $x=y=p_2$, the only pole of $F$ (step 3.2). If $w\in \mathbb{C}\setminus \{0\}$ then $x,y\mathrm{\notin }\{p_1,p_2\}$, and the previous paragraph applied to $r:=x$ gives $y\in F^{-1}(F(x))=\{x\}$, so $y=x$. Therefore $F$ is injective.

$F$ is a local biholomorphism. Let $x\in X$. If $x\mathrm{\ne }{p}_2$, choose a chart $\psi :U\to \mathbb{C}$ at $x$ with $F(U)\subseteq \mathbb{C}$, shrinking $U$ if necessary; then $f:=F\circ {\psi }^{-1}$ is a holomorphic and injective function on the complex domain $\psi (U)$, because $F$ is holomorphic [F1] and injective (step 8.1), so by [F15] $f$ has nowhere-zero derivative and is biholomorphic onto its open image. If $x=p_2$, then by step 3.2 the function $z_2\mathrm{/}{G}_2=1\mathrm{/}F$ is holomorphic near $p_2$ with value $0$ at $p_2$ and is injective there, since $F$ is injective; as a chart expression on a disc it is an injective holomorphic function, so by [F15] it is biholomorphic onto its open image, and since $w\mapsto 1\mathrm{/}w$ is the chart ${\varphi }_{\mathrm{\infty }}$ of $\widehat{\mathbb{C}}$ at $\mathrm{\infty }$ [F10], the function $F=1\mathrm{/}(1\mathrm{/}F)$ is a biholomorphism from a neighbourhood of $p_2$ onto an open neighbourhood of $\mathrm{\infty }$. Hence $F$ is a local biholomorphism at every point of $X$.

$F$ is a biholomorphism onto its open image. By step 9.1 the map $F$ is open: for open $O\subseteq X$, each point of $F(O)$ has a neighbourhood on which $F$ is a biholomorphism onto an open set, so $F(O)$ is a union of open subsets of $\widehat{\mathbb{C}}$. Hence $F(X)$ is open in $\widehat{\mathbb{C}}$, and $F:X\to F(X)$ is a continuous bijection (step 8.1) that is an open map, therefore a homeomorphism, whose inverse is holomorphic because it is locally the inverse supplied by step 9.1. Thus $F$ is a biholomorphism from $X$ onto the open subset $F(X)$ of $\widehat{\mathbb{C}}$.

The image omits at most one point of the sphere. Suppose that $a,b\in \widehat{\mathbb{C}}\setminus F(X)$ are distinct. If $a\mathrm{\ne }\mathrm{\infty }$ put $T(z):=1\mathrm{/}(z-a)$, and if $a=\mathrm{\infty }$ let $T$ be the identity; in both cases $T$ is a Mobius transformation with $T(a)=\mathrm{\infty }$, hence a biholomorphism and a homeomorphism of $\widehat{\mathbb{C}}$ [F10]. Then $\mathrm{\Omega }:=T(F(X))$ is contained in $\widehat{\mathbb{C}}\setminus \{\mathrm{\infty }\}=\mathbb{C}$, and $T(b)\in \mathbb{C}\setminus \mathrm{\Omega }$ because $b\mathrm{\ne }a$ means $b$ is not the point sent to $\mathrm{\infty }$ and $b\mathrm{\notin }F(X)$; hence $\mathrm{\Omega }⊊\mathbb{C}$. As a continuous image of the connected space $F(X)$ the set $\mathrm{\Omega }$ is connected, and it is nonempty and open in $\mathbb{C}$ because $F(X)$ is open in $\widehat{\mathbb{C}}$ (step 10.1) and $T$ is a homeomorphism [F11]; so $\mathrm{\Omega }$ is a complex domain [F15]. Every based loop of $\mathrm{\Omega }$ is null: if $\gamma$ is a loop in $\mathrm{\Omega }$ based at $y_0$, then $T^{-1}\circ \gamma$ is a loop in $F(X)$ and $F^{-1}\circ T^{-1}\circ \gamma$ is a loop in $X$ because $F:X\to F(X)$ is a homeomorphism (step 10.1); this loop is null in $X$, the space $X$ being simply connected, so pushing forward along the continuous maps $F$ and $T$ and using the functoriality of [F12] makes $[\gamma ]$ the identity element. By [F13] the domain $\mathrm{\Omega }$ is homologically simply connected, and the Riemann mapping theorem [F14] applied at any $z_0\in \mathrm{\Omega }$ yields a biholomorphic map $g:\mathrm{\Omega }\to \mathbb{D}$. Then $g\circ T\circ F:X\to \mathbb{D}\subseteq \mathbb{C}$ is holomorphic [F1, F10], bounded, and injective, since $F,T,g$ are injective; because $X$ contains two distinct points [F16], an injective map on $X$ is not constant. This contradicts step 4.1, where the claim announced in step 1.3 was proved, so no two distinct points of $\widehat{\mathbb{C}}\setminus F(X)$ exist: the complement has at most one point.

Compact case: $X$ is the Riemann sphere. Suppose that $X$ is compact. Then $F(X)$ is compact, as a continuous image of $X$ [F11], and it is a nonempty open subset of $\widehat{\mathbb{C}}$ that is connected in the connected space $\widehat{\mathbb{C}}$ [F10, F11, step 10.1]. A compact subset of the Hausdorff space $\widehat{\mathbb{C}}$ is closed [F10, F11], so $F(X)$ is a clopen nonempty subset of the connected space $\widehat{\mathbb{C}}$, and therefore $F(X)=\widehat{\mathbb{C}}$ [F11]. Thus $F:X\to \widehat{\mathbb{C}}$ is a bijective holomorphic map with holomorphic inverse (step 10.1), that is, $X$ is biholomorphic to the Riemann sphere.

Noncompact case: $X$ is the complex plane. Suppose that $X$ is not compact. Then $F(X)$ is not compact, because $X$ and $F(X)$ are homeomorphic (step 10.1) and compactness is preserved by continuous maps in the inverse direction [F11]; in particular $F(X)\mathrm{\ne }\widehat{\mathbb{C}}$, since $\widehat{\mathbb{C}}$ is compact [F10]. By step 11.1 there is exactly one point $a\in \widehat{\mathbb{C}}\setminus F(X)$, that is, $F(X)=\widehat{\mathbb{C}}\setminus \{a\}$. Let $T_a$ be the Mobius transformation of [F10] with $T_a(a)=\mathrm{\infty }$; it restricts to a biholomorphism of $\widehat{\mathbb{C}}\setminus \{a\}$ onto $\mathbb{C}=\widehat{\mathbb{C}}\setminus \{\mathrm{\infty }\}$, so the composite $T_a\circ F:X\to \mathbb{C}$ is bijective, holomorphic and has holomorphic inverse, being a composite of the biholomorphisms $F$ and $T_a$ [F10, F15, step 10.1]. Hence $X$ is biholomorphic to the complex plane.

Conclusion and choice accounting. The two cases of steps 12.1 and 11.2 are exhaustive: either $X$ is compact or it is not. They prove the two assertions of the statement. The Axiom of Choice [A1] is used exactly through the Countable Choice consumed by the dipole supplier [F8] in steps 2.2 and 4.2 and through the Riemann mapping theorem [F14] in step 11.1; all other selections are finite and explicit (the points $p_1,p_2,r$ of step 1.2, a chart and its shrunk domain in step 9.1, and a basepoint of a loop in step 11.1).