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Fix $z_0\notin \Gamma$ and put $\delta =\mathrm{dist}(z_0,\Gamma )>0$. If $|z-z_0|<\delta /2$, thenuniformly for $w\in \Gamma$. Multiplication by the bounded continuous function $f$ and integration therefore prove continuity of $F$ at $z_0$.

Take $\Gamma$ to be the positively oriented unit circle and $f\equiv 1$. By the Cauchy integral formula, $F(z)=1$ inside the circle and $F(z)=0$ outside it, so no continuous extension across $\Gamma$ exists.

For the final assertion, cover $K$ by finitely many sufficiently small closed axis-parallel squares whose slightly enlarged squares lie in $\Omega$, choosing the grid so that no boundary meets $K$. Apply the square Cauchy formula to $f$ on every selected square and add the results. Integrals over shared edges cancel with opposite orientations. The remaining finitely many oriented boundary polygons can be split into contours ${\Gamma }_j\subset \Omega \setminus K$, and their sum isfor every $z\in K$.

The Riemann-Roch theorem statesOn a genus-one curve $K\sim 0$. If $D$ has degree zero, then $D+P_0$ has degree one, and Riemann--Roch gives $\ell (D+P_0)=1$. Its unique nonzero section has an effective divisor of degree one, say $P$, soIf also $D\sim Q-P_0$, then $P\sim Q$; a nonconstant function with divisor $P-Q$ would define a degree-one map to ${\mathbb{P}}^1$, forcing the genus to be zero. Hence $P=Q$.

This identifies $X$ with ${\mathrm{Pic}}^0(X)$ by $P\mapsto [P-P_0]$ and defines $P+Q=R$ throughFor the stated cubic, the line through $A=(2,1)$ and $B=(-2,-1)$ is $y=x/2$. Its third intersection is $(1/4,1/8)$, so reflection in the $x$-axis gives

Because $P_0$ is an inflection point, a line section is linearly equivalent to $3P_0$. A point $P$ is an inflection point exactly when some line has intersection divisor $3P$, equivalentlyUnder the group-law identification this is precisely $3(P-P_0)=0$, so the inflection points are exactly the three-torsion points.

In local coordinates around $p\in R$ and $f(p)\in S$, a nonconstant analytic map has the formAfter shrinking the chart, $h$ has an analytic $m$th root, so a change of coordinate makes the map $z\mapsto z^m$. It maps small discs onto neighborhoods of the origin; hence every nonconstant analytic map of Riemann surfaces is open. If $R$ is compact and $S$ connected, $f(R)$ is compact and therefore closed, and it is also nonempty and open. Thus $f(R)=S$.

For $f:\mathbb{D}\to \mathbb{D}$ with $f(0)=0$, the removable-singularity theorem makes $g=f/z$ analytic. On $|z|=r$, $|g(z)|=|f(z)|/r\le 1/r$, and the maximum principle gives the same bound throughout ${\mathbb{D}}_r$. Applying the Schwarz lemma to an automorphism $h$ and to $h^{-1}$ shows that every automorphism fixing zero is

If $F:\mathbb{C}\to \mathbb{C}$ is an analytic isomorphism fixing zero, continuity of $F^{-1}$ implies that the preimage of every closed disc is compact. Hence $|F(z)|\to \infty$ as $|z|\to \infty$, and defining $F(\infty )=\infty$ gives an analytic isomorphism of the Riemann sphere.

Consequently an automorphism of $\mathbb{D}$ fixing zero does extend to $\mathbb{C}$. A general disc automorphism need not: for $a\ne 0$,has a finite pole and cannot extend to an entire isomorphism.