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Choose an actual pole $p_i$ of $f$; one exists by nonconstancy and compactness. In a local coordinate $z$ centered there, its simple pole has expansion $f(z)=a_{-1}/z+O(1)$ with $a_{-1}\ne 0$. HenceThe holomorphic inverse function theorem makes $1/f$ a local coordinate. Here the source notation $f^{-1}$ denotes the reciprocal, rather than an inverse map of the meromorphic function.

The chosen pole is among the fixed points, so $\phi$ maps a sufficiently small neighborhood of it into the coordinate neighborhood. If $f=f\circ \phi$, then $w=w\circ \phi$ there. Injectivity of this coordinate forces $\phi (x)=x$ throughout a nonempty open set. The stated identity principle then gives $\phi$ equal to the identity everywhere.

Choose centred charts so the coordinate expression $F$ satisfies $F(0)=0$. On a component where the map is nonconstant, the identity theorem makes $F$ nonconstant. Its Taylor expansion factors as $F(z)=z^{n}h(z)$, with $n\ge 1$ and $h(0)\ne 0$. In a sufficiently small disc, choose a holomorphic logarithm of $h$, and hence a holomorphic $n$th root $q$ with $q^n=h$. The function $zq(z)$ has nonzero derivative at zero. The holomorphic inverse function theorem, which says that a holomorphic function with nonzero derivative has a holomorphic local inverse, makes it a valid new source coordinate. In that coordinate,If the map is bijective, local injectivity forces $n=1$: for $n>1$, a small nonzero value has $n$ distinct preimages in this local disc. The inverse is therefore holomorphic in a neighbourhood of every point, and the local inverses agree, proving the map is biholomorphic. The nonconstant hypothesis must apply to the component containing $p$ if disconnected Riemann surfaces are allowed; otherwise a constant component is an exception to the proposed local form.

Choose coordinates $z$ at $p$ and $w$ at $f(p)$ vanishing at those points. The local expansion is $w\circ f\circ z^{-1}(\zeta )=a{\zeta }^v+O({\zeta }^{v+1})$, $a\ne 0$. Define the branching order $v_f(p)=v$. A change of either coordinate has nonzero linear coefficient, so it preserves the leading exponent, proving well-definedness.

Ramification means $v>1$, equivalently vanishing of the derivative in local coordinates. These zeros are isolated unless the local derivative is identically zero, which by analytic continuation would make $f$ constant. An infinite set of ramification points on a compact surface would have an accumulation point, contradicting isolation. Thus that set is finite.

A nonconstant holomorphic map of compact connected Riemann surfaces is onto, since its image is open by the local form and closed by compactness. It has a finite degree $d\ge 1$. The Riemann-Hurwitz formula isIf both genera equal $g>1$, positivity of $2g-2$ forces $d=1$ and every ramification term zero. The map is bijective and has nonzero derivative everywhere. The holomorphic inverse function theorem says that a holomorphic function with nonzero derivative has a local holomorphic inverse; these inverses patch, so $f$ is biholomorphic. For $g=1$, multiplication by two on $\mathbb{C}/\Lambda$ is a degree-four holomorphic map of an elliptic curve to itself and is not biholomorphic.

A fixed singularity of a complex differential equation has a location determined by the coefficients, independently of the initial value. A movable singularity of a complex differential equation belongs to an individual continued solution and changes location when the initial value changes. For example, $w^{\prime }=w^2$ has the movable pole $z=c$ in $w=-1/(z-c)$. A fixed exceptional location need not be singular for every solution.

The precise first-order form of the Painlevé determinateness theorem needed here is as follows. Let $w^{\prime }=A(z,w)$ be rational in $w$, with coefficients meromorphic in $z$ on the chosen independent-variable surface. Algebraic coefficients can be made single valued on their own surface. Exclude the fixed coefficient singularities and exceptional fibres where a reduced numerator and denominator vanish together, working also in the dependent coordinate $v=1/w$. If a solution is continued along an arc $\gamma :[0,1)\to U$ with endpoint $z_{\ast }={\lim }_{t\to 1}\gamma (t)$ outside that exceptional set, then $w(\gamma (t))$ has a definite finite or infinite limit. Any singularity there is a pole or an algebraic branch point; there is no movable essential or logarithmic singularity. The standard rectifiable-arc formulation is included in this statement.

Here is a proof. First suppose a sequence along the arc has $(z_n,w_n)\to (z_{\ast },w_{\ast })$ at a point where the vector field is holomorphic. Choose a smaller closed bidisc about this point. Uniform bounds on the vector field and its dependent-variable derivative give, by the integral contraction mapping proof, one radius $r>0$ on which every solution starting sufficiently close to $(z_{\ast },w_{\ast })$ exists and remains bounded. For large $n$, the solution through $(z_n,w_n)$ is therefore defined on a fixed disc about $z_{\ast }$. The entire sufficiently late tail of $\gamma$ lies in that disc. Uniqueness identifies the original continued solution with this local solution throughout that tail, so the solution extends through $z_{\ast }$ and has a limit. The same argument applies to an ordinary point in the reciprocal coordinate $v=1/w$.

If the limit did not exist, consider the cluster set $K$ in the dependent-variable Riemann sphere:Each closed tail is compact and connected, and these sets are nested; hence $K$ is nonempty, compact and connected. The preceding paragraph excludes every ordinary point from $K$. At $z_{\ast }$ the rational vector field has only finitely many nonordinary dependent values, in both coordinate charts: an entire fibre of singular points was excluded among the fixed exceptions. Thus $K$ is a connected subset of a finite set, so it is a singleton. Compactness then forces convergence to its unique point, contradicting the assumed lack of a limit. This proves determinateness.

It remains to determine the local type rather than merely the limit. At a finite limiting value $w_{\ast }$ which is a pole of the vector field, write the reduced equation as $w^{\prime }=P/Q$. Since $z_{\ast }$ is not exceptional, $P(z_{\ast },w_{\ast })\ne 0$, and the inverse equationis holomorphic near $(z_{\ast },w_{\ast })$. The holomorphic dependence of ordinary differential equations on parameters makes its local solution curves jointly holomorphic in $w$ and their initial $z$ values. In particular, the value at the transverse line $w=w_{\ast }$ is a holomorphic first integral labelling these curves. The original graph has constant label; taking its endpoint limit shows that it lies on the inverse solution through $(z_{\ast },w_{\ast })$. That inverse solution cannot be the vertical curve $z\equiv z_{\ast }$: this would require $Q(z_{\ast },w)$ to vanish identically near $w_{\ast }$, an excluded fixed singular fibre. Consequently its power series has a first nonzero term,Here $k\ge 2$ at a vector-field pole. Factor the right side as $(w-w_{\ast }{)}^k$ times a nonvanishing holomorphic function and take its local $k$th root. This gives a new dependent coordinate with nonzero derivative, and the holomorphic inverse function theorem expresses $w$ as a convergent power series in $(z-z_{\ast }{)}^{1/k}$. Thus the singularity is an algebraic branch point. At an infinite limit, repeat the argument for $v=1/w$. An ordinary zero of $v$ gives a pole of $w$, while a finite ramification of $v$ gives an algebraic pole branch. This proves the asserted local types and the theorem.

For the given equation, the finite-coordinate rational field isThe whole fibre $z=-1$ is singular in the displayed coefficient, and at $(z,w)=(0,0)$ the reduced numerator and denominator vanish together. Hence the finite fixed exceptional set is $\{-1,0\}$. Away from these two locations, the only possible finite dependent limits at a singularity are $w_{\ast }=\sigma z_{\ast }$, with $\sigma =\pm 1$.

At such a point the inverse equation isIt is holomorphic when $z_{\ast }\ne 0,-1$ and $w_{\ast }=\sigma z_{\ast }$. Its inverse solution satisfies $z_w(w_{\ast })=0$, and differentiation givesThereforeandThese are genuine square-root algebraic branch points. The inverse solutions can be based at arbitrary $z_{\ast }\notin \{0,-1\}$ and then continued to nearby ordinary initial points. Their distinct branch locations require different initial data, so these branch points are movable. The signs specify the two inverse branches near each of the two values $w_{\ast }=\pm z_{\ast }$.

There are no movable poles. Indeed the reciprocal equation isFor every finite $z\ne -1$ it is holomorphic at $v=0$, and $v\equiv 0$ is its unique solution with zero initial value. A nonzero reciprocal solution therefore cannot reach zero there. This rules out both an ordinary pole and an infinite algebraic branch at a movable finite location. Together with the theorem, it exhausts the finite movable singularities.

The fixed point zero must not be mistaken for a generic movable collision. It is ordinary for solutions with $w(0)\ne 0$. But solutions reaching $w=0$ can have a fixed square-root singularity. To construct them rather than just balance powers, set $z(w)=w^{2}h(w)$ in the inverse equation. The equation for $h$ isEquivalently,On a small disc this is a contraction mapping of the closed ball $\Vert h-1/2{\Vert }_{\infty }\le 1/4$: both its departure from $1/2$ and its Lipschitz constant are $O(r^2)$. Its unique holomorphic fixed point gives $z=w^2/2+O(w^4)$. Thusis a genuine branch at the fixed location zero. The zero solution itself extends through zero as a function, although the original displayed coefficient is undefined at $(0,0)$.

The point $-1$ can produce a fixed nonalgebraic singularity. For an explicit example take $z=-1+x$ real, $0<x<x_0<1$, and $w=iy$ with initial $y>2$. Put $q=y^2$ and $L=-\log x$. ThenAs $L$ increases, $q$ increases, its derivative is bounded above by $2$ and bounded below by a positive constant, and therefore the solution exists for all large $L$ and $q\to \infty$. The derivative then tends to $2$, so $q/L\to 2$. ConsequentlyThis is unbounded but grows more slowly than every negative power of $z+1$. It cannot be a meromorphic pole even after a finite ramified substitution, demonstrating an actual fixed singularity with logarithmic growth. The determinateness theorem does not exclude such behavior at fixed exceptions.

Finally, on the independent-variable Riemann sphere use $t=1/z$. For finite $w$ the transformed equation iswhich is ordinary at $t=0$. At simultaneous $z=\infty$, $w=\infty$, however, the reciprocal field iswith an indeterminate point at $(t,v)=(0,0)$. Thus infinity is an additional fixed exceptional location on the sphere, rather than an unavoidable singularity of solutions with finite limiting value. There are solutions escaping to infinity there: for real initial data $w(z_0)>z_0>0$, the difference $d=w-z$ stays positive, since its derivative becomes positive as $d\downarrow 0$. Moreover $d^{\prime }\le 0$ when $d\ge 1$, so $0<d\le \max (1,d(z_0))$. Such a solution continues for all positive $z$, with $w=z+O(1)\to \infty$.

In the finite plane, the fixed exceptional locations are $0,-1$; all other solution singularities are movable square-root branches at $w=\pm z$, and there are no movable poles. On the sphere, infinity is also a fixed exception at the infinite dependent value. Being fixed or movable describes the base location, not merely whether the coefficient contains the factors $w-z$ and $w+z$.

Near an isolated point $P\in A$, choose source and target coordinate charts, using continuity of $\alpha$ to keep a sufficiently small source neighbourhood in the target chart. The coordinate representation is holomorphic on the punctured neighbourhood and is bounded near $P$ by continuity. The Riemann removable singularity theorem extends it holomorphically across $P$, with the given continuous value. Repeating at all points of $A$ proves that $\alpha$ is analytic.

For a nonconstant holomorphic function $f$ on a connected Riemann surface, let $A$ be the set where its coordinate derivative vanishes. This condition is independent of the chart. The derivative cannot vanish identically on an open set, because the identity theorem would then make $f$ constant globally. Its zeros are isolated and closed locally, so $A$ is a closed discrete subset. At every other point the holomorphic inverse function theorem makes $f$ a local coordinate chart.

Now suppose $\alpha$ is a homeomorphism and $g=f\circ \alpha$ is analytic. Outside ${\alpha }^{-1}(A)$, use $f$ as a target coordinate to write locally $\alpha =f^{-1}\circ g$. This is analytic. The preimage ${\alpha }^{-1}(A)$ is discrete because $\alpha$ is a homeomorphism, and the first argument extends analyticity across it. An injective holomorphic map cannot have a zero derivative: a zero of order at least two would give a locally multiple-to-one map. Its local inverse is therefore holomorphic, proving that $\alpha$ is a conformal equivalence.

The plane $\mathbb{C}$ and the unit disk are homeomorphic by $z\mapsto z/(1+|z|)$, with inverse $w\mapsto w/(1-|w|)$. They are not conformally equivalent: a holomorphic equivalence from the plane to the disk would be a bounded nonconstant entire function, contradicting Liouville's theorem.

The unheaded definitions use the complex structure $J$ on the real tangent bundle. Compatibility means $g(Ju,Jv)=g(u,v)$. Its fundamental Hermitian form is $\omega (u,v)=g(Ju,v)$; the compatibility identity makes this real, alternating and of type $(1,1)$. The metric is a Kähler metric precisely when $d\omega =0$. In complex dimension one a real three-form is zero, so every compatible metric is a Kähler metric.

Now assume closedness and prove the Kähler normal holomorphic coordinates condition. Start with holomorphic coordinates $w$ centered at the point and make a complex-linear change so that the positive Hermitian coefficient matrix satisfies $h(0)=I$. Closedness of the $(1,1)$ form givesDefine $a_{i\bar{j},k}={\partial }_{w_k}{h}_{i\bar{j}}(0)$; it is symmetric in $i,k$. Choose new coordinates implicitly byThe derivative of this holomorphic map at zero is the identity, so the holomorphic inverse function theorem makes $z$ valid local coordinates. In the new coordinates,At zero, $h^{\prime }=I$, and differentiating givesHermitian symmetry makes all antiholomorphic first derivatives zero as well. Taylor's theorem for a smooth function now yieldsThus condition (a) implies condition (b), with the exact factor $i/2$ in the fundamental-form convention retained.

On ${\mathbb{C}}^{n+1}\setminus \{0\}$, let $\Omega =d{z}_0\wedge \cdots \wedge d{z}_n$ and let $E={\sum }_i{z}_i{\partial }_{z_i}$ be the Euler vector field. The numerator is its interior product of a differential form, soBoth numerator and denominator have homogeneous scaling degree $n+1$. Thus this meromorphic $n$-form is invariant under constant nonzero scalings, and ${\iota }_E\alpha =0$ makes it horizontal for the quotient to Complex projective space.

To check descent using the hint, take a local holomorphic section $Z$ of the quotient and another section $Z^{\prime }=\lambda Z$, where $\lambda$ is a nowhere-zero holomorphic function. In the evaluationevery term involving $d\lambda$ has two collinear arguments and vanishes. The remaining term is ${\lambda }^{n+1}{Z}^{\ast }({\iota }_E\Omega )$, while $f(\lambda Z)={\lambda }^{n+1}f(Z)$. Hence $Z^{\prime \ast }\alpha =Z^{\ast }\alpha$. The forms therefore glue to a meromorphic section of the canonical bundle of Complex projective space.

On the affine chart $z_j\ne 0$, use the section $z_j=1$ and list the remaining coordinates as $t_1,\dots ,t_n$ in increasing original-index order. If $F$ is the resulting polynomial, thenAt a point of $Y$, some derivative $F_{t_r}$ is nonzero. Otherwise all the homogeneous derivatives except possibly $f_{z_j}$ would vanish; the Euler homogeneous function theorem gives ${\sum }_i{z}_i{f}_{z_i}=(n+1)f=0$ there, forcing $f_{z_j}=0$ too. This contradicts the permitted smoothness criterion. The holomorphic inverse function theorem now makes $F$ a transverse coordinate, so the pole is simple. On a neighbourhood with $F_{t_r}\ne 0$, rearranging the wedge product of differential forms givesBy part (a), its Poincaré residue is consequentlyThe remaining $t$ coordinates form a holomorphic coordinate system on $Y$, and $F_{t_r}$ is nonzero in this neighbourhood. The displayed top form is therefore nowhere zero. These neighbourhoods cover $Y$ and part (a) guarantees agreement on their overlaps. The residue is a nowhere-vanishing global holomorphic section:This is the explicit residue trivialization for a degree n+1 projective hypersurface, consistent with the adjunction formula $K_Y\cong {\mathcal{O}}_Y((n+1)-n-1)$, but obtained here directly from the indicated form.

Fix $z_0\in \mathcal{U}$. We prove the Riemann mapping theorem in this case by maximizing a normalized derivative. Choose $a\notin \mathcal{U}$. Since $\mathcal{U}$ is a simply connected domain and $z-a$ never vanishes, it has a holomorphic square root $g$. This $g$ is injective, and $g(\mathcal{U})$ is disjoint from $-g(\mathcal{U})$: equality up to sign would first force the original points to coincide. Choose $b\in g(\mathcal{U})$ and $r>0$ with $B(b,r)\subset g(\mathcal{U})$. The ball $B(-b,r)$ is omitted, so $1/(g+b)$ is a bounded univalent function. Scaling and composing with an automorphism of the unit disk produces an injective holomorphic function $\varphi :\mathcal{U}\to \mathbb{D}$ with $\varphi (z_0)=0$ and ${\varphi }^{\prime }(z_0)>0$.

Let $\mathcal{F}$ be the family of all such normalized univalent functions. A Cauchy estimate in a small disk about $z_0$ bounds their derivatives, so $M={\sup }_{\varphi \in \mathcal{F}}{\varphi }^{\prime }(z_0)$ is finite and positive. Choose a maximizing sequence in this normal family. Montel theorem gives a subsequence converging uniformly on compact subsets to $\varphi$. Its derivative at $z_0$ is $M>0$. The maximum modulus principle puts its image in $\mathbb{D}$, and Hurwitz's theorem implies that a nonconstant limit of injective holomorphic functions is injective. Thus the maximum is attained.

If $a_1\in \mathbb{D}$ is omitted, then $a_1\ne 0$. Put $T_a(w)=(w-a)/(1-\overline{a}w)$. A holomorphic square root $v$ of $T_{a_1}\circ \varphi$ exists, is injective, and takes values in $\mathbb{D}$. Compose $v$ with $T_{v(z_0)}$ and a rotation to normalize it. Writing $s=|a_1|\in (0,1)$, the new derivative has magnitudea contradiction. Therefore $\varphi$ maps onto $\mathbb{D}$. The Cayley transform between the half-plane and disk now givesIts inverse is holomorphic by the holomorphic inverse function theorem, so this is a biholomorphism. The proper-subset hypothesis was used to choose $a\notin \mathcal{U}$; the whole complex plane cannot be mapped this way, by Liouville theorem.

Here “covering map” uses the course convention of a local homeomorphism; “regular” adds the evenly-covered product condition. The derivative of $p(z)=z^4$ is $4z^3$, which never vanishes on the upper half-plane $\mathbb{H}$. The holomorphic inverse function theorem therefore makes $p$ a local biholomorphism and hence a covering map in this sense.

It is not regular. The point $1$ has only the preimage $i$ in $\mathbb{H}$, whereas $e^{i\epsilon }$ for small positive $\epsilon$ has the two preimages with arguments $\epsilon /4$ and $\pi /2+\epsilon /4$. Thus the fibre cardinality is not locally constant near $1$, so no neighbourhood of $1$ has a preimage of the form $U\times \Delta$ for a fixed discrete set $\Delta$.

In local coordinates centred at $p$ and $f(p)$, a nonconstant rational map has the formThe point $p$ is a ramification point of a holomorphic map when its ramification index of a holomorphic map satisfies $e_p>1$. A branch value of a holomorphic map, also called a branch point in the target, is a value $f(p)$ of some ramification point.

Let $B$ be the finite set of branch values and put $R=f^{-1}(B)$. For every $w\notin B$, all points of $f^{-1}(w)$ have local degree one, so the holomorphic inverse function theorem supplies disjoint neighbourhoods on which $f$ is biholomorphic. Compactness of the fibre lets their target neighbourhoods be intersected to one evenly covered neighbourhood of $w$. Henceis an unramified covering map. Here the displayed restriction requires $R=f^{-1}(B)$; if “ramification point” is reserved only for points with $e_p>1$, then the source deletion must be written $f^{-1}(B)$.

The monodromy theorem says that analytic continuations along endpoint-fixed homotopic paths have the same terminal germ. Equivalently here, the path lifting theorem lifts a loop $\gamma$ based at $w\notin B$ from each point of $f^{-1}(w)$; taking the endpoint of each lift permutes that fibre. The permutation depends only on $[\gamma ]\in {\pi }_1({\mathbb{C}}_{\infty }\setminus B,w)$, giving the monodromy group of a covering.

For the stated function, make the Möbius change of coordinateThenThe critical points are $t=0,\pm 1/\sqrt{2},\infty$, with branch valuesA loop around $0$ interchanges the two roots born from $t=0$, so its branch cycle is a transposition. A loop around $-1/4$ simultaneously interchanges the two pairs that collide there, so its branch cycle is a product of two disjoint transpositions. With a suitable labelling these areThey generate a transitive group of order eight; their product is a four-cycle. Therefore the full monodromy group isthe dihedral group in its action on the four vertices of a square.

Sincethe critical points are exactlyThe holomorphic inverse function theorem shows that $f$ is conformal locally everywhere else.

For $z=x+iy$,Its imaginary part is positive throughout $S$. On the three boundary pieces,These intervals traverse the boundary of the upper half-plane. The inverse branchexists for $\Im \zeta >0$, proving that the image is preciselyThis is the hyperbolic-cosine half-strip map.

Suppose $F:\mathbb{D}\to \mathbb{C}$ were a holomorphic bijection. By the allowed nonvanishing derivative result, $F^{\prime }(z)\ne 0$ throughout $\mathbb{D}$. The holomorphic inverse function theorem gives a holomorphic inverse locally at every image point. Since $F$ is globally bijective, these local inverses agree with its global inverse $H:\mathbb{C}\to \mathbb{D}$.

Thus $H$ is an entire function with $|H|<1$. The Liouville theorem forces $H$ to be constant, contradicting its being an inverse. ThereforeThis is the analytic obstruction behind the fact that the complex plane and unit disc are homeomorphic but not conformally equivalent.