For every $q\in L$, the line $pq$ meets the nonsingular conic in two points counted with multiplicity by the Bézout theorem. One is $p$; define $\alpha (q)$ as the second, using $\alpha (q)=p$ for the unique tangent line at $p$. Conversely, for $r\in C\setminus \{p\}$, intersect the line $pr$ with $L$, and send $p$ itself to the intersection of its tangent with $L$. Since $p\notin L$, each intersection with $L$ is unique. These constructions are inverse, so $\alpha :L\to C$ is a bijection preserving the stated collinearity. This is parametrization of a conic by projection.

For the particular conic take $p=[1:0:1]$ and $L=\{x=0\}$. If $q=[0:s:t]$, points on $pq$ have coordinates $[a:bs:a+bt]$. Substitution into the conic givesTaking the second intersection gives the explicit mapThe three coordinates do not vanish simultaneously. If $s+2t=0$, the formula gives $p$, as required for the tangent direction.

For rational points, put $r=s/t$ when $t\ne 0$. The full affine list istogether with the parameter $r=\infty$, which gives $(x,y)=(1,1)$. The denominator is $(r+1/2{)}^2+3/4>0$ for rational $r$. Conversely, every rational point other than $p$ determines a rational joining line and hence a rational point of $L$; the point $p$ is obtained at $r=-2$. Thus this lists all rational solutions.

For a degree-$d$ nonconstant holomorphic map from a compact connected Riemann surface of genus $g$ to the projective line, the Riemann-Hurwitz formula iswhere $e_p$ is the local degree. A meromorphic function defines a map to the sphere by giving its ordinary finite values away from poles and value infinity at each pole. The reciprocal function is holomorphic in a neighborhood of a pole, proving holomorphicity there. For use in Riemann-Hurwitz the function must be nonconstant; a nonzero constant still defines a holomorphic map, but of degree zero.

Let $X=\mathbb{C}/\Lambda$. The double pole of $\wp$ gives a degree-two map to the sphere. The torus has genus one, so total ramification is four. One simple ramification point is zero, above infinity. At the three nonzero half-period classes $h_i$, evenness and periodicity give ${\wp }^{\prime }(h_i)=0$. Each is a ramification point of local degree two, and these account for all remaining ramification. Put $e_i=\wp (h_i)$. They are distinct, since a degree-two fiber cannot contain two points each of local degree two.

At each $h_i$, $\wp -e_i$ has a double zero and ${\wp }^{\prime }$ has a simple zero. Consequently the elliptic quotienthas removable singularities there and no zeros or poles elsewhere except possibly zero. Near zero, $\wp =z^{-2}+O(z^2)$ and ${\wp }^{\prime }=-2z^{-3}+O(z)$, so $R\to 4$. Thus $R$ is holomorphic on the compact torus and constant, provingThe printed additional requirement that all three $e_i$ be nonzero is false. For the square lattice $\Lambda =\mathbb{Z}+i\mathbb{Z}$, invariance of the defining series under multiplication by $i$ gives $\wp (iz)=-\wp (z)$. At $h=(1+i)/2$, one has $ih=h-1$, so periodicity gives $\wp (h)=\wp (ih)=-\wp (h)$ and hence $\wp (h)=0$. This is one of the three branch values. Distinctness, which is all the cubic construction needs, remains valid.

DefineThe image lies on $Y^{2}Z=4(X-e_{1}Z)(X-e_{2}Z)(X-e_{3}Z)$. Distinct roots make its affine part smooth, and the point at infinity is smooth because the derivative of its equation with respect to $Z$ is nonzero there. Near zero the projective coordinates $X/Y=\wp /{\wp }^{\prime }\sim -z/2$ and $Z/Y=1/{\wp }^{\prime }\sim -z^3/2$ extend holomorphically. Every finite $X/Z$ value has two preimages under $\wp$ except at a branch value. Evenness pairs those preimages as $z,-z$, and oddness of ${\wp }^{\prime }$ distinguishes them unless they are the same half-period class. Therefore the two coordinates together give injectivity and reach both signs of $Y$, proving surjectivity onto the cubic. Thus $\Phi$ is a bijective holomorphic map to a nonsingular cubic, including when one branch value is zero.

An inflection point is a smooth point where the tangent line has intersection multiplicity at least three with the curve. For a nonsingular plane curve of degree $n\ge 3$, these are the intersections with its Hessian curve, whose degree is $3(n-2)$. The original curve is not a component of its Hessian: otherwise in each graph chart the tangent direction would have zero derivative identically, making the curve locally a line and hence a line globally. Bézout therefore gives the upper bound $3n(n-2)$, counting multiplicity. A nonsingular conic has no inflection points; lines are excluded from this finite-contact definition.

Take $F=y^{2}z-x^3+x{z}^2$. Its Hessian determinant isAt infinity the curve has only $O=[0:1:0]$, which is an inflection point: its tangent is $z=0$ and the restricted equation is $-x^3=0$. In the chart $z=1$, substitute $y^2=x^3-x$ into the Hessian equation to obtainThus the full answer isThe four complex values of $x$ are distinct and none is $0,1,-1$, so each gives two distinct nonzero values of $y$. Together with $O$ these are all nine inflection points, exhausting the Bézout bound.