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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.