The Holomorphic Inverse Function Theorem and Weierstrass Preparation — Examples

The slice $f(0,z_2)=-z_2$ has a simple zero at $0$, so $f$ is regular in $z_2$ of order $1$.

The identity $f=(-1)(z_2-z_1^2)$ already has the required form: the factor $-1$ is a unit and $z_2-z_1^2$ is monic in $z_2$ with lower coefficient $-z_1^2$ vanishing at the origin. Thus [L1] produces exactly this preparation.

On the slice $z_1=0$ one has $f(0,z_2)=0$ for every $z_2$. So the last-variable restriction vanishes identically rather than to a finite exact order.

Step 1.1 contradicts the requirement in [L1]. Therefore the germ is not regular in $z_2$ at the origin.

Direct substitution gives $\left(z_1{z}_2\right)\circ T=z_1{z}_2+z_2^2\mathrm{.}$

Setting $z_1=0$ in step 1.1 yields the slice $z_2^2$, which has exact order $2$ at the origin. So [L1] makes the transformed germ regular in $z_2$ of order $2$, exhibiting the coordinate-change mechanism promised by [L2].

The identity $z_1=0\cdot (z_2^2-z_1)+z_1$ is immediate.

The remainder $z_1$ has degree $0$ in $z_2$, hence degree $<2$. Therefore [L1] identifies this displayed identity as the unique Weierstrass division of $z_1$ by $z_2^2-z_1$.

One has $f(0,1)=0$ and $\frac{\mathrm{\partial }f}{\mathrm{\partial }{z}_2}(0,1)=2\mathrm{\ne }0.$

Therefore [L1] yields neighbourhoods of $0$ and $1$ and a unique holomorphic function $\phi$ such that $f(z_1,\phi (z_1))=0$ for all nearby $z_1$. So near $(0,1)$ the zero set is the holomorphic graph $z_2=\phi (z_1)$.

The complex Jacobian of $F$ is $J_{\mathbb{C}}F(z_1,z_2)=\left(\begin{array}{cc}{e}^{z_1}& 0\\ 0& 1\end{array}\right),$ so $\det J_{\mathbb{C}}F(z_1,z_2)=e^{z_1}\mathrm{\ne }0$ for every $(z_1,z_2)$.

By [L2], $e^{2\pi i}=e^{i\pi }{e}^{i\pi }=(-1)(-1)=1$, so $F(z_1+2\pi i,z_2)=(e^{z_1+2\pi i},z_2)=(e^{z_1},z_2)=F(z_1,z_2)\mathrm{.}$ Thus distinct points have the same image, and $F$ is not injective despite step 1.1.

Fact [L1] provides a holomorphic map whose complex Jacobian determinant is never zero.

The same fact [L1] also shows that this map identifies $(z_1,z_2)$ and $(z_1+2\pi i,z_2)$. So the claimed global injectivity conclusion fails.

Fact [L1] says that an invertible complex Jacobian gives a local biholomorphism near each point, not a global inverse on the whole domain.

Fact [L2] realizes exactly that gap: the map $(z_1,z_2)\mapsto (e^{z_1},z_2)$ satisfies the local hypothesis everywhere, yet it is not injective and therefore has no global inverse. So the claim is false.

Fact [L1] gives a specific nonzero germ that fails the claimed property in the original coordinates.

Fact [L2] shows that the coordinate-change lemma is doing real work rather than decorating an already-true statement. Therefore the claim is false.

The factorization $z_2^2=1\cdot z_2^2$ is a genuine preparation by [L1].

If neither the unit condition nor the degree condition is retained, then also $z_2^2=z_2\cdot z_2$ is allowed, and [L2] says the second factor is a Weierstrass polynomial of degree $1$. This differs from the genuine degree-$2$ preparation in step 1.1, so the relaxed factorization is not unique.

Fact [L1] states the exact negation of the claimed phenomenon.

Therefore such an isolated zero cannot occur, and the claim is false.