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Write the rational function $A$ in reduced form. We first exclude a finite pole in its dependent variable at a generic point $(a,b)$. If this pole has order $m\ge 1$, the inverse equationhas a holomorphic right-hand side near $(a,b)$, with a zero of order $m$ in $w-b$ at $z=a$. Choosing a generic point avoids collisions of pole branches and zeros of the reduced numerator. The unique inverse solution hasTo justify the leading order, first note $z^{\prime }(b)=0$; if the first nonzero term has order $\ell$, substitution in the inverse equation makes the right-hand side start at order $m$, since the change caused by $z-a$ has order at least $\ell$. Thus $\ell -1=m$. Inverting produces an algebraic branch point of order $m+1$. Away from the fixed coefficient exceptions, $a$ can be chosen throughout an open set on the pole curve; inverse solutions based at these different $a$ give different initial values at a nearby ordinary point. These are movable singularities of a complex differential equation, contradicting the hypothesis. Hence the denominator has no genuine $w$-dependent factor: $A$ is a polynomial in $w$, with rational functions of $z$ as coefficients.

Let its degree in $w$ be $d$. Use the dependent coordinate $v=1/w$ on the Riemann sphere. Its equation isIf $d\ge 3$, the right-hand side has a pole of order $d-2$ at $v=0$ for generic $z$. The same inverse-equation argument gives a movable algebraic branch point of order $d-1$. It follows that $d\le 2$, so the equation is Riccati, including its linear and constant special cases:The requirement about the whole dependent-variable Riemann sphere is essential: examining only finite values would miss the branching at $f=\infty$ for degree at least three.

The linearization of a Riccati equation is particularly symmetric in the systemOn a simply connected domain where the coefficients are holomorphic, its solutions are holomorphic. Direct differentiation givesA nonzero initial vector never becomes the zero vector, by uniqueness for the linear differential equation, so the quotient is meromorphic. Choosing initial vector $(f(z_0),1)$ represents every finite initial value, and $(1,0)$ represents infinity. At fixed coefficient singularities, a single globally holomorphic vector solution need not exist; the assertion is local on ordinary domains, or on their appropriate continuation surface.

Finally, for any two finite solutions $u,v$ of the Riccati equation,Use this identity on the four differences inThe logarithmic derivative of this cross-ratio isThus $K$ is constant wherever the expression is initially defined, and the identity extends as an identity of meromorphic functions. Distinctness of $g,h,j$ is preserved at ordinary points by uniqueness, interpreted on the Riemann sphere. Solving the constant cross-ratio equation gives the Riccati cross-ratio superpositionFor $K=\infty$ this means $f=j$; $K=0$ gives $g$, and $K=1$ gives $h$. The constant is selected by the initial value of $f$, so the formula represents every solution, with its denominator zeros interpreted as poles.