Cutoff extension across a puncture in complex dimension two

The singleton $\{0\}$ is compact, so [F3] gives $\chi =1$ near $0$ with support $K_{\chi }\subset B(0,1)$. By [F4], $K_{\chi }$ is closed; it is bounded because it lies in the unit ball, hence compact by [F5]. Since $f$ is holomorphic, [F14] makes it complex differentiable at each point; [F12] makes it smooth, and [F13] gives $\bar{\mathrm{\partial }}f=0$ (the dictionary [F5] identifies the library's coordinates $0,1$ with $z_1,z_2$ here). Since $\chi =1$ near $0$, $f_0$ is identically zero near the puncture and therefore smooth on $B(0,2)$. The bidegree formula [F6] makes $g=\bar{\mathrm{\partial }}{f}_0$ a smooth $(0,1)$ form. On $G$, the product rule [F19] gives $g=-f\bar{\mathrm{\partial }}\chi$. Outside $K_{\chi }$, $\chi$ vanishes on a neighborhood, so $g=0$ there; hence $\mathrm{supp}g\subseteq K_{\chi }\subset B(0,1)$. The support is closed by [F4] and bounded, so [F5] makes it compact. Thus $g$ is zero near $\mathrm{\partial }B(0,2)$, its zero extension is smooth, and [F7] gives $\bar{\mathrm{\partial }}g=0$ on all of ${\mathbb{C}}^2$.

The set $G$ is open: if $z\in G$ and $r=\mathrm{\parallel }z\mathrm{\parallel }$, choose $\delta =\frac{1}{2}\min (r,2-r)>0$. For $w\in B(z,\delta )$, [F16] gives $0<\mathrm{\parallel }w\mathrm{\parallel }<2$, and [F17] makes this ball an open neighborhood contained in $G$. The ball notation and norm topology are those of [F8]. To connect two points of $G$, move each radially to the unit sphere; these paths remain at positive norm below $2$ and are continuous by [F18]. Join their endpoints by a path in the unit sphere using [F9] and [F20]. Thus $G$ is path-connected, hence connected by [F10].

Let $E=\{z\in {\mathbb{C}}^2:\mathrm{\parallel }z\mathrm{\parallel }>1\}$. It lies in ${\mathbb{C}}^2\setminus \mathrm{supp}g$ by step 1.1 and is unbounded. For each $z\in E$, a radial path joins $z$ to $2z\mathrm{/}\mathrm{\parallel }z\mathrm{\parallel }$ while keeping the norm greater than $1$; [F18] ensures the path is continuous. The radius-$2$ sphere is path-connected by [F9] and [F20] after rescaling the unit sphere in ${\mathbb{R}}^4\cong {\mathbb{C}}^2$; the ball and sphere notation is that of [F8]. Thus $E$ is path-connected and connected by [F10]. It lies in a connected component by [F11]; that component is unbounded because it contains $E$, and therefore is the unique unbounded component named in [F1].

The full AC assumption [F2] permits applying [F1] to the smooth compactly supported closed $(0,1)$ form $g$ from step 1.1, giving $u\in C_c^{\mathrm{\infty }}({\mathbb{C}}^2)$ with $\bar{\mathrm{\partial }}u=g$ and $u=0$ on the unique unbounded component identified in step 2.1. Therefore $u=0$ on $E$. Since $\chi =0$ on $E$, the correction $h=u+\chi f$ vanishes on $A=\{1<\mathrm{\parallel }z\mathrm{\parallel }<2\}\subset G$. The annulus is nonempty because $(3\mathrm{/}2,0)\in A$. For any $z\in A$, set $ϵ=\frac{1}{2}\min (\mathrm{\parallel }z\mathrm{\parallel }-1,2-\mathrm{\parallel }z\mathrm{\parallel })>0$; [F16] gives $B(z,ϵ)\subset A$, and [F17] says this metric ball, with notation from [F8], is open. Thus $A$ is open.

On $G$, [F19] and step 1.1 give $\bar{\mathrm{\partial }}h=\bar{\mathrm{\partial }}u+\bar{\mathrm{\partial }}(\chi f)=g+f\bar{\mathrm{\partial }}\chi +\chi \bar{\mathrm{\partial }}f=0$. The function $h$ is smooth, so [F13]–[F14] make it holomorphic on $G$. By step 1.2, $G$ is connected and open; since $h=0$ on $A$, [F15] gives $h=0$ throughout $G$. Thus on $G$, $F=f_0-u=(1-\chi )f+\chi f=f$.

On $B(0,2)$, $F=f_0-u$ is smooth and $\bar{\mathrm{\partial }}F=\bar{\mathrm{\partial }}{f}_0-g=0$ by step 3.1. The Cauchy–Riemann criterion [F13], followed by the definition [F14], makes $F$ holomorphic on the whole ball. For $f\equiv 1$, step 1.1 gives $g=-\bar{\mathrm{\partial }}\chi$ and step 4.1 gives $u=-\chi$ on $G$, so the formula yields $F=1$ there. Every neighborhood of $0$ in the ball contains nonzero points of $G$, so continuity gives $F(0)=1$. If $f\equiv 0$, then $f_0=g=0$ and the unique solution in [F1] is $u=0$, since the zero function is a compactly supported solution; hence $F=0$. [F1, F13, F14, step 1.1, step 3.1, step 4.1, given, algebra] $\mathrm{\square }$