text:
\section{Introduction}
There has been a recent flurry of activity in
probability~\cite{MoOdOl:10,Mossel:10} and computer science~\cite{KKMO:04,
KindlerOdonnell:12,Raghavendra:08}
around a certain paper of Borell~\cite{Borell:85} on inequalities
satisfied by the Ornstein-Uhlenbeck process.
In his paper, Borell proved a theorem, which is somewhat complicated to
state, showing that certain quantities only decrease under
Ehrhard symmetrization~\cite{Ehrhard:84}. He then derived two simpler
corollaries, about hitting times for the Ornstein-Uhlenbeck process,
from this general result.

We recall that the Ornstein-Uhlenbeck process on $\mathbb{R}^n$ is the Gaussian process
$\{X_t: t \in \mathbb{R}\}$ with mean zero and covariance $\mathbb{E} X_s X_t^T = e^{|t-s|} I_n$.
This is a Markov process, as may be seen by the construction
$X_t = e^{-t} B_{e^{2t}}$ for a Brownian motion $B_t$, and the stationary
measure of $X_t$ is the standard Gaussian measure, $\gamma_n$.
For a set $A \subset \mathbb{R}^n$, we denote its exit time under $X_t$ by
$e_A = \inf \{t \ge 0: X_t \not \in A\}$.
Although they were originally written in terms of hitting times instead of
exit times, Borell's two corollaries of his general inequality may be
written as follows, in which \emph{half-space} means a set of the form
$\{x \in \mathbb{R}^n : x\cdot a \le b\}$, and half-spaces are \emph{parallel} if they
have the same normal vector.
\begin{theorem}[Borell]\label{thm:exit-time}
If $B \subset \mathbb{R}^n$ is a half-space with $\gamma_n(B) = \gamma_n(A)$
then $e_B$ stochastically dominates $e_A$; i.e., for every $t \ge 0$,
\[
 \mathbb{P}(e_A \ge t) \le \mathbb{P}(e_B \ge t).
\]
\end{theorem}

\begin{theorem}[Borell]\label{thm:occupation-time}
If $B_1$ and $B_2$ are parallel half-spaces with
$\gamma_n(B_i) = \gamma_n(A_i)$ then
\[
 \mathbb{E} \int_0^{t \land e_{A_1}} 1_{A_2}(X_s)\ ds
 \le \mathbb{E} \int_0^{t \land e_{B_1}} 1_{B_2}(X_s)\ ds.
\]
\end{theorem}

There is a third corollary of Borell's general inequality that did not appear
in his original paper~\cite{Borell:85}, but has nevertheless become widely
applied in theoretical computer science, particularly in the study
of hardness of approximation.

\begin{theorem}[Borell]\label{thm:noise-sense}
If $B_1$ and $B_2$ are parallel half-spaces with
$\gamma_n(B_i) = \gamma_n(A_i)$ then for any $t > 0$,
\[
 \mathbb{P}(X_0 \in A_1, X_t \in A_2) \le \mathbb{P}(X_0 \in B_1, X_t \in B_2).
\]
\end{theorem}

In the special case $A_1 = A_2$, this inequality is sometimes interpreted
as showing that half-spaces are the most ``noise stable'' sets. Here,
we think of $X_t$ as being a noisy version of $X_0$, and so a set $A$ is
noise stable if the event $\{X_0 \in A\}$ tends to agree with the
event $\{X_t \in A\}$. Theorem~\ref{thm:noise-sense} implies that this
overlap is maximized, over all sets with a fixed Gaussian measure, by
half-spaces. Using an invariance principle, Mossel et al.~\cite{MoOdOl:10}
deduced from Theorem~\ref{thm:noise-sense} a similar inequality on
the discrete cube (although the statement on the cube is necessarily more
complicated, because the direction of a half-space's normal vector becomes
important); that work then laid the foundation for many applications in
theoretical computer science (for a few examples, see~\cite{KKMO:04,KindlerOdonnell:12,Raghavendra:08}).

Note that in Theorem~\ref{thm:noise-sense}, the joint distribution of
$(X_0, X_t)$ has mean zero and covariance
$\big(\begin{smallmatrix} 1 & \rho \\ \rho & 1\end{smallmatrix}\big) \otimes I_n$,
where $\rho = e^{-t}$.
Our main result is a multivariate generalization of Theorem~\ref{thm:noise-sense},
which allows for more than two Gaussian vectors and a more general
covariance structure than that endowed by the
Ornstein-Uhlenbeck process. Our generalization is also strong enough to recover
Theorems~\ref{thm:exit-time} and~\ref{thm:occupation-time}; we are thankful
to Michel Ledoux for pointing this out.

 \begin{theorem}\label{thm:main}
  Let $M = (m_{ij})$
  be a $k \times k$ positive semidefinite matrix with $m_{ij} \ge 0$,
  and let $X = (X_1, \dots, X_k)$ be a $kn$-dimensional Gaussian vector
  with covariance $M$.
  For any measurable $A_1, \dots, A_k \subset \mathbb{R}^n$,
  \begin{equation}\label{eq:main}
   \mathbb{P}(X_i \in A_i \text{ for all } i)
   \le \mathbb{P}(X_i \in B_i \text{ for all } i)
  \end{equation}
  whenever $B_i$ is a collection of parallel half-spaces with
  $\gamma_n(B_i) = \gamma_n(A_i)$.

  Moreover, if equality is attained in~\eqref{eq:main} then there exists
  a collection $(B_1, \dots, B_k)$ of parallel half-spaces such that
  $A_i = B_i$ up to sets of measure zero.
 \end{theorem}

 By setting $k = 2$,
 Theorem~\ref{thm:main} recovers Theorem~\ref{thm:noise-sense}.
 We should remark that a generalization along these lines,
 but nevertheless different from Theorem~\ref{thm:main}, has already
 been discovered: Isaksson and Mossel~\cite{IsakssonMossel:12} showed that
 the inequality~\eqref{eq:main} also holds
 under the hypothesis that the off-diagonal elements of $M^{-1}$
 are non-positive. In other words, our hypothesis is that every pair
 $X_i, X_j$ is positively correlated, while~\cite{IsakssonMossel:12} assumed
 that every pair $X_i, X_j$ is conditionally positively correlated given
 all the other variables. Neither of these conditions is strictly stronger than
 the other, and in fact both would suffice for recovering
 Theorems~\ref{thm:exit-time} and~\ref{thm:occupation-time}.
 However,~\cite{IsakssonMossel:12} did not characterize the equality cases
 of~\eqref{eq:main} under their hypothesis.

 Next, we will show how Theorem~\ref{thm:main} may be used to recover
 Theorem~\ref{thm:exit-time}. This reduction is quite similar to one by
 Burchard and Schmuckenschlager~\cite{BurchardSchmuckenschlager:01}, who
 were studying exit times of Brownian motion on manifolds. (In that case,
 the study of exit times has a fairly long history;
 see~\cite{BurchardSchmuckenschlager:01} for references.) As we will see, though,
 our approach to Theorem~\ref{thm:main} is quite different to that
 of Burchard and Schmuckenschlager, who studied two-point symmetrizations.

 Let $X_t$ be the Ornstein-Uhlenbeck process,
 and consider the finite dimensional marginal
 $(X_{t_1}, \dots, X_{t_k})$ for a sequence of times $t_1 < \cdots < t_k$.
 This is a mean-zero Gaussian vector with covariance $M \otimes I_n$, where
 $m_{ij} = e^{-|t_i - t_j|}$. Clearly, then, $M$ satisfies the hypothesis
 of Theorem~\ref{thm:main} and in particular, we have
 \begin{equation}\label{eq:discrete-hitting-time}
  \mathbb{P}(X_{t_i} \in A \text{ for all } i) \le \mathbb{P}(X_{t_i} \in B \text{ for all } i)
 \end{equation}
 when $B$ is a half-space with $\gamma_n(A) = \gamma_n(B)$.
 This is essentially a discrete version of Theorem~\ref{thm:exit-time},
 since the event $\{X_{t_i} \in A \text{ for all } i\}$ is a discretization
 of $\{e_A \ge t_k\}$.

 To complete the proof, we need to show that one can
 take limits.
 Setting $t_i = it/k$ in~\eqref{eq:discrete-hitting-time}, we have
 \begin{align*}
 \mathbb{P}(e_A \ge \tau)
 &= \mathbb{P}(X_t \in A \text{ for all } 0 < t < \tau) \\
 &\le \mathbb{P}(X_{i\tau/k} \in A \text{ for all } i = 1, \dots, k) \\
 &\le \mathbb{P}(X_{i\tau/k} \in B \text{ for all } i = 1, \dots, k),
 \end{align*}
 where the last inequality follows from~\eqref{eq:discrete-hitting-time}.
 Next, we send $k \to \infty$. Recall that $X_t$ is uniformly continuous
 with probability 1. Hence for any $\epsilon > 0$, we make take $k$ large enough
 so that $|s - t| \le 1/k$ implies $|X_s - X_t| \le \epsilon$; for this $k$,
 \[
 \mathbb{P}(X_{i\tau/k} \in B \text{ for all } i = 1, \dots, k) \le
 \mathbb{P}(X_t \in B_\epsilon \text{ for all } 0 < t < \tau)
 = \mathbb{P}(e_{B_{\epsilon}} \ge \tau),
 \]
 where $B_\epsilon$ is the $\epsilon$-enlargement of $B$:
 $B_\epsilon = \{x \in R^n: d(x, B) \le \epsilon\}$. We have shown,
 therefore, that for any $\epsilon > 0$,
 $\mathbb{P}(e_A \ge \tau) \le \mathbb{P}(e_{B_\epsilon} \ge \tau)$. It only remains to
 show, then, that $\mathbb{P}(e_{B_\epsilon} \ge \tau)$ converges to
 $\mathbb{P}(e_B \ge \tau)$ as $\epsilon \to 0$.

 Consider instead the equivalent statement that $\mathbb{P}(e_{B_\epsilon} < \tau)$ converges
 to $\mathbb{P}(e_B < \tau)$.
 Since $B$ is closed and $X_t$ has continuous paths,
 $e_B < \tau$ implies that there is some $\epsilon > 0$ with
 $e_{B_\epsilon} < \tau$. That is, the function $1_{\{e_{B_\epsilon} < \tau\}}$
 converges pointwise (and upwards) to $1_{\{e_B < \tau\}}$ as $\epsilon \to 0$.
 By the monotone convergence theorem, it follows that $\mathbb{P}(e_{B_\epsilon} < \tau)$
 converges to $\mathbb{P}(e_B < \tau)$ as $\epsilon \to 0$. Hence
 \[
  \mathbb{P}(e_A \ge \tau) \le \lim_{\epsilon \to 0} \mathbb{P}(e_{B_\epsilon} \ge \tau)
  = \mathbb{P}(e_B \ge \tau)
 \]
 and so we have recovered Theorem~\ref{thm:exit-time}.

 We have mentioned already that it is also possible to recover
 Theorem~\ref{thm:exit-time}
 from the result in~\cite{IsakssonMossel:12}. Indeed, the matrix $M$ with
 entries $m_{ij} = e^{-|t_i - t_j|}$ does satisfy the hypothesis
 in~\cite{IsakssonMossel:12} (namely that the off-diagonal entries of its
 inverse are non-positive), although this is certainly less obvious
 then the fact that $M$ satisfies the conditions of Theorem~\ref{thm:main}.

 Let us also indicate how Theorem~\ref{thm:occupation-time} is recovered.
 We want to show that
 \begin{equation}\label{eq:occupation}
  \mathbb{E} \int_0^{t\land e_{A_1}} 1_{A_2}(X_s)\ ds
 \end{equation}
 is only increased when $A$ is replaced by $B$ (recall that $B_1$ and $B_2$ are
 parallel half-spaces satisfying $\gamma_n(B_i) = \gamma_n(A_i)$.
 We may suppose that $A_2 \subset A_1$, since if not then~\eqref{eq:occupation}
 may be trivially made larger by moving some of $A_2$'s mass inside $A_1$;
 if this is impossible because $\gamma_n(A_2) > \gamma_n(A_1)$
 then~\eqref{eq:occupation} is trivially bounded by $t \land e_{A_1}$,
 which, by Theorem~\ref{thm:exit-time}, is stochastically dominated
 by $t \land e_{B_1}$, and this in turn is equal to the right hand side
 of~\eqref{eq:occupation} with $B$ replacing $A$.

 Now that we have reduced to the case $A_2 \subset A_1$, we may
 discretize~\eqref{eq:occupation} as
 \[
  \frac{1}{k} \sum_{i=1}^k \mathbb{P}(X_{t_j} \in A_1 \text{ for all } j < i \text{ and } X_{t_i} \in A_2).
 \]
 By~\eqref{eq:discrete-hitting-time}, this is only increased when $A$ is
 replaced by $B$. To recover Theorem~\ref{thm:occupation-time} from here,
 it suffices to take a limit in much the same manner as before; we omit
 the details.

 \section{The Ornstein-Uhlenbeck semigroup}

 We will prove Theorem~\ref{thm:main} by differentiating a particular
 functional under the Ornstein-Uhlenbeck semigroup.
 This proof method has a long history, beginning with Varopoulos'
 work~\cite{Varopoulos:85} connecting the heat semigroup with Sobolev
 inequalties. More recently, and more apropos of this work, Bakry
 and Ledoux~\cite{BakryLedoux:96} proved the Gaussian isoperimetric inequality
 by differentiating Bobkov's functional~\cite{Bobkov:97} under the
 Ornstein-Uhlenbeck semigroup. We will follow quite a similar approach here,
 using a generalization of a functional that was introduced by Mossel
 and the author~\cite{MosselNeeman:12b} to prove Theorem~\ref{thm:main} in the case
 $k=2$.

 We define the Ornstein-Uhlenbeck semigroup $\{P_t: t \ge 0\}$, which
 acts on functions $f: \mathbb{R}^n \to \mathbb{R}^n$, by
 \[
  (P_t f)(x) = \int_{\mathbb{R}^n} f(e^{-t} x + \sqrt{1 - e^{-2t}}y)\ d\gamma_n(y),
 \]
 where $\gamma_n$ is the standard Gaussian measure on $\mathbb{R}^n$.
 Equivalently, $(P_t f)(X_0) = \mathbb{E} (f(X_t) \mid X_0)$, where $\{X_t: t \ge 0\}$
 is the Ornstein-Uhlenbeck process from the previous section. From either
 definition, one can easily see that $P_0$ is the identity operator, while
 $P_t f \to \mathbb{E} f$ as $t \to \infty$.

 One remarkable property of the Ornstein-Uhlenbeck semigroup is that
 it has very nice formulas for its commutation with smooth functions.
 In particular, for any $F = (f_1, \dots, f_k): \mathbb{R}^n \to \mathbb{R}^k$
 and for any smooth $\Psi: \mathbb{R}^k \to \mathbb{R}$,
 there is the formula (see, eg., ~\cite{Ledoux:00})
 \begin{equation}\label{eq:diffusion}
  \diff{}{s} P_s \Psi(P_{t-s} F)
  = \sum_{i,j = 1}^k \partial^2_{ij} \Psi(P_{t-s} F) \inr{\nabla P_{t-s} f_i}
  {\nabla P_{t-s} f_j}.
 \end{equation}

 We begin with an observation that comes, essentially, from
 applying~\eqref{eq:diffusion} to the function that is $\Psi$ composed
 with a linear operator. In the following, $\odot$ denotes the
 Hadamard (elementwise) product between two matrices and $H_\Psi$ denotes
 the Hessian matrix of $\Psi$.
 \begin{proposition}\label{prop:linear-diffusion}
  Let $M$ be a $k \times k$ positive-definite matrix with $m_{ii} = 1$ and let
  $X = (X_1, \dots, X_k)$ be a $kn$-dimensional Gaussian vector with
  mean zero and covariance $M \otimes I_n$. If $\Psi: [0, 1]^k \to \mathbb{R}$ satisfies
  $M \odot H_\Psi \le 0$ then for all measurable
  $F = (f_1, \dots, f_k): \mathbb{R}^n \to [0, 1]^k$,
  \[
   \mathbb{E} \Psi(F(X)) \le \Psi(\mathbb{E} F).
  \]
 \end{proposition}

 We remark that the assumption $m_{ii} = 1$ in
 Proposition~\ref{prop:linear-diffusion} is not necessary, but it makes our
 notation simpler since otherwise we need to consider Ornstein-Uhlenbeck semigroups
 with different stationary measures.

 Before proving Proposition~\ref{prop:linear-diffusion}, we introduce
 some notation that will be useful in what follows:
 for any $f: \mathbb{R}^n \to \mathbb{R}$
and any $n \times m$ matrix $M$, denote the function $f \circ M : \mathbb{R}^m \to \mathbb{R}$
by $f^M$.

\begin{proof}
Let $Q = (q_{ij})$ be the positive definite square root of $M \otimes I_n$,
and for $i = 1, \dots, k$, let $Q_i$ be the $n \times kn$ matrix
consisting of rows $(i-1)n + 1$ through $in$ of $Q$.
Let $Z$ be a standard Gaussian vector in $\mathbb{R}^{kn}$, and
note that
$Q Z = (Q_1 Z, \dots, Q_k Z)$ is a $kn$-dimensional Gaussian vector with
mean 0 and covariance $M \otimes I_n$ (i.e., $QZ$ has the same distribution as $X$).
We consider the quantity
\[
 F(s, t, z) = \left(P_s \Psi(P_{t-s} f_1^{Q_1}, \dots, P_{t-s} f_k^{Q_k})\right)(z)
\]
for $s, t \in [0, \infty)$ and $z \in \mathbb{R}^{kn}$.
First, let us check how $P_t$ commutes with linear transformations.
Since $Q \otimes I_n$ is the square root of $M \otimes I_n$,
we have $Q_i^T Q_i = I_n$ and so
\begin{align}
 (P_t f_i^{Q_i})(x)
 &= \int_{\mathbb{R}^n} f_i(e^{-t} Q_i x + \sqrt{1-e^{-2t}} Q_i y)\: d\gamma_{kn}(y) \notag \\
 &= \int_{\mathbb{R}^n} f_i(e^{-t} Q_i x + \sqrt{(1-e^{-2t})} y)\: d\gamma_n(y) \notag \\
 &= (P_t f_i)^{Q_i}(x).
 \label{eq:Pt-linear}
\end{align}
Of course, the gradient commutes with linear transformations as
$\nabla f^A = A^T (\nabla f)^A$.
Combining this with~\eqref{eq:Pt-linear},
\[
\nabla P_{t-s} f_i^{Q_i} = \nabla (P_{t-s} f_i)^{Q_i}
= Q_i^T (\nabla P_{t-s} f_i)^{Q_i}.
\]
In particular,
\begin{align}
\inr{\nabla P_{t-s} f_i^{Q_i}}{\nabla P_{t-s} f_j^{Q_j}}
&= \inr{Q_i^T (\nabla P_{t-s} f_i)^{Q_i}}
{Q_j^T (\nabla P_{t-s} f_j)^{Q_i}} \notag \\
&= m_{ij} \inr{(\nabla P_{t-s} f_i)^{Q_i}}{(\nabla P_{t-s} f_j)^{Q_i}}.
\label{eq:grad-linear}
\end{align}
For brevity, let $v_i = (\nabla P_{t-s} f_i)^{Q_i}$.
Then, by~\eqref{eq:diffusion} and~\eqref{eq:grad-linear},
\begin{align}
 \pdiff{F(s,t,\cdot)}{s}
 &= P_s \sum_{i,j=1}^k
 \inr{\nabla P_{t-s} f_i^{Q_i}}{\nabla P_{t-s} f_j^{Q_j}}
 \partial^2_{ij} \Psi(P_{t-s} F^Q) \notag \\
 &= P_s \sum_{i,j=1}^k
 m_{ij} \inr{v_i}{v_j}
 \partial^2_{ij} \Psi(P_{t-s} F^Q).
\label{eq:exchangeable-derivative}
\end{align}
Note that if $v^T = (v_1^T \dots v_k^T)$ then the last line may be rewritten as
\begin{equation}\label{eq:exchangeable-derivative-2}
P_s  \left(v^T ((M \odot H_\Psi) \otimes I_n) v\right).
\end{equation}
In particular, if $M \odot H_\Psi \le 0$ then
 $\pdiff{F(s,t,z)}{s} \le 0$ for every $s, t$ and $z$. Hence,
 $\lim_{t \to \infty} F(t,t,Z) \le \lim_{t\to\infty} F(0,t,Z)$. But since
 $(Q_1 Z, \dots, Q_k Z)$ has the same distribution as $(X^{(1)}, \dots, X^{(k)})$,
 $\mathbb{E} F(t,t,Z)$ converges to $\mathbb{E} \Psi(F(X))$
 and $\mathbb{E} F(0,t,Z)$ converges to $\Psi(\mathbb{E} F(X))$.
 \end{proof}

 With hardly any extra effort, the proof of Proposition~\ref{prop:linear-diffusion}
 also allows us to characterize its equality cases. Indeed, if
 $\mathbb{E} \Psi(F) = \Psi(\mathbb{E} F)$ then we must have $\pdiff{F(s, t, z)}{s} = 0$ for
 every $s$, $t$, and $z$. Going back to~\eqref{eq:exchangeable-derivative}
 and~\eqref{eq:exchangeable-derivative-2}, we see that
 $P_s \left(v^T ((M \odot H_\Psi) \otimes I_n) v\right)$ must be
 identically zero, and hence
 $((M \odot H_\Psi) \otimes I_n) v = 0$. In other words, we have
 the following corollary:
 \begin{corollary}\label{cor:linear-diffusion}
  Under the hypothesis of Proposition~\ref{prop:linear-diffusion}, if
  $\mathbb{E} \Psi(F) = \Psi(\mathbb{E} F)$ then for every $t > 0$,
  \[
   \big((M \odot H_\Psi(P_t F^Q)) \otimes I_n\big)
   \begin{pmatrix}
    (\nabla P_t f_1)^{Q_1}\\
    \vdots \\
    (\nabla P_t f_k)^{Q_k}
   \end{pmatrix}
   = 0 \text{ pointwise.}
  \]
 \end{corollary}

 \section{Proof of Theorem~\ref{thm:main}}

 Before proving Theorem~\ref{thm:main}, note that by translating $X$,
 $A_i$, and $B_i$, it suffices to consider the
 case in which $X$ has mean zero.
 Moreover, by scaling each $X_i$, $A_i$, and $B_i$, we may assume
 that $m_{ii} = 1$ for each $i$ (here and in the previous sentence
 we are using the fact that a collection of parallel half-spaces
 remains one under translation and scaling).
 We may also assume that $M$ is strictly positive definite, since if not then
 the distribution of $X$ is supported on a subspace, onto which we may project.

 Consider the function
 \begin{equation}\label{eq:def-J}
  J(x_1, \dots, x_k; M)
  = \mathbb{P}\left(X_i \le \Phi^{-1}(x_i) \text{ for all } i\right).
 \end{equation}
 Note that $\mathbb{P}(X_i \le \Phi^{-1}(x_i)) = x_i$; in particular, the collection
 $(B_1, \dots, B_k)$ with $B_i = \{y \in \mathbb{R}^n: y_1 \le x_i\}$ is a set of
 parallel half-spaces with $\mathbb{P}(X_i \in B_i) = x_i$. Since every such set of
 parallel half-spaces may be obtained by applying a fixed rotation to each
 $B_i$, Theorem~\ref{thm:main} is equivalent to the statement
 \[
  \mathbb{P}(X_i \in A_i \text{ for all } i) \le J(\gamma_n(A_1), \dots, \gamma_n(A_k); M).
 \]
 Next, note that if $x_1, \dots, x_k \in \{0, 1\}$ then
 $J(x_1, \dots, x_k)$ is 1 if all the $x_i$ are 1, and 0 otherwise. In
 particular,
 \[
  \mathbb{P}(X_i \in A_i \text{ for all } i) = \mathbb{E} J(1_{A_1}(X_1), \dots, 1_{A_k}(X_k); M)
 \]
 and hence~\eqref{eq:main} is equivalent to the statement
 \begin{equation}\label{eq:functional}
  \mathbb{E} J(f_1(X_1), \dots, f_k(X_k); M) \le J(\mathbb{E} f_1, \dots, \mathbb{E} f_k; M)
 \end{equation}
 in the special case $f_i = 1_{A_i}$. In fact, we will prove~\eqref{eq:functional}
 for general measurable functions $f_i: \mathbb{R}^n \to [0, 1]$.

 Unsurprisingly, the proof of~\eqref{eq:functional} goes through
 Proposition~\ref{prop:linear-diffusion}. The main task left, therefore,
 is to compute the Hessian of $J$ and check that it satisfies the hypothesis
 of Proposition~\ref{prop:linear-diffusion}.

 \begin{proposition}\label{prop:hess-J}
  If $m_{ii} = 1$ and $m_{ij} \ge 0$
  then $J$ satisfies $M \odot H_J \le 0$.

 \end{proposition}

 For a vector $v$, let $v_{\hat i}$ denote $v$ without the $i$th entry,
 and for a square matrix $M$, let $M_{\hat i}$ denote $M$ without the $i$th row
 and column.
 For a square matrix $M$, let $\overline{M_i}$ denote the Schur complement of
 $M_{\hat i}$ in $M$. In other words, $\overline{M_i}$ is defined by
 \[\overline{M_i}^{-1} = (M^{-1})_{\hat i}.\]
 A well-known formula for conditional distributions
 of Gaussian vectors~\cite{Eaton:83} states if $X$ has mean zero and covariance
 $M$ satisfying $m_{ii} = 1$,
 then conditioned on $X_i = x_i$, $X_{\hat i}$ has mean $x_i M_{i \hat i}$
 (where $M_i$ denotes the $i$th row of $M$, so $M_{i\hat i}$ is the $i$th
 row of $M$ with its $i$th element removed) and covariance
 $\overline{M_i}$.

 To compute the first derivatives of $J$, let
 \[
  K(x_1, \dots, x_k; M) = \mathbb{P}\left(X_i \le x_i \text{ for all } i\right),
 \]
 and note that
 $J(x_1, \dots, x_k) = K(\Phi^{-1}(x_1), \dots, \Phi^{-1}(x_k))$.
  Now, for any $i$ we may write $K$ as
 \[
  K(x; M) = \int_0^{x_i} \phi(y)
  \mathbb{P}(X_j \le x_j \text{ for all } j \ne i \mid X_i = y)\ dy,
 \]
 from which we see that
 \[
  \partial_i K(x; M)
  = \phi(x_i) \mathbb{P}(X_j \le x_j \text{ for all } j \ne i \mid X_i = x_i).
 \]
 Now, given that $X_i = x_i$, $X_{\hat i}$ has mean $M_{i\hat i} x_i$
 and covariance $\overline{M_i}$; hence,
 \[
  \mathbb{P}(X_j \le x_j \text{ for all } j \ne i \mid X_i = x_i)
  = K(x_{\hat i} - M_{i\hat i} x_i; \overline{M_i}),
 \]
 and so we have the formula
 \[
  \partial_i K(x; M) = \phi(x_i) K(x_{\hat i} - M_{i\hat i} x_i; \overline{M_i})
 \]
 (bear in mind that this formula is only valid under the assumption $m_{ii} = 1$;
 if not then $m_{ii}$ makes an appearance in the formula also).
 Applying the chain rule and the identity
 $\diff{}{x} \Phi^{-1}(x) = 1/\phi(\Phi^{-1}(x))$, we have
 \begin{equation}\label{eq:diff_J}
  \partial_i J(x; M)
  = K(\Phi^{-1}(x_{\hat i}) - M_{i\hat i} \Phi^{-1}(x_i); \overline{M_i})
 \end{equation}
 (where by $\Phi^{-1}(x_{\hat i})$, we mean the vector
 obtained by applying $\Phi^{-1}$ to $x_{\hat i}$ element-wise.)
 Now let $I(x) = \phi(\Phi^{-1}(x))$ and define, for $j \ne i$,
 \[
  J_{ij}(x; M) = I(x_i) \partial_j K(\Phi^{-1}(x_{\hat i}) - M_{i\hat i} \Phi^{-1}(x_i); \overline{M_i});
 \]
 by the chain rule applied to~\eqref{eq:diff_J}, we have
 \begin{align}
  \partial_i \partial_j J(x; M)
  &= \frac{1}{I(x_j)}
  \partial_j K(\Phi^{-1}(x_{\hat i}) - M_{i\hat i} \Phi^{-1}(x_i); \overline{M_i}) \notag \\
  &= \frac{1}{I(x_i) I(x_j)} J_{ij}(x; M).
  \label{eq:2diff-J-mixed}
 \end{align}
 It is worth mentioning that this last equation shows that in fact
 $J_{ij} = J_{ji}$. This is not obvious from the definition of $J_{ij}$, although
 it may also be checked by the tedious process of calculating the derivative
 in that definition.

 To compute the repeated second derivatives of $J$, we use~\eqref{eq:diff_J}
 and the chain rule to write
 \begin{align}
  \partial_i^2 J(x; M)
  &= -\sum_{j \ne i} \frac{m_{ij}}{I(x_i)} \partial_j K(\Phi^{-1}(x_{\hat i}) - M_{i\hat i} \Phi^{-1}(x_i); \overline{M_i}) \notag \\
  &= -\frac{1}{I^2(x_i)} \sum_{j \ne i} m_{ij} J_{ij}(x; M).
  \label{eq:2diff-J-diag}
 \end{align}
 Now let $\mathcal{I}(x)$ be the $k \times k$ matrix with $1/I(x_i)$ as the $i$th
 diagonal entry.
 Note that by~\eqref{eq:2diff-J-mixed}, the $ij$ entry of
 $M \odot H_J$ is given by $\frac{m_{ij}}{I(x_i) I(x_j)} J_{ij}$,
 while the $ii$ entry of $M \odot H_J$ is just given
 by~\eqref{eq:2diff-J-diag} (since $m_{ii} = 1$).
 Hence we may write
 \begin{equation}\label{eq:hess-J}
  M \odot H_J(x; M) = \mathcal{I}(x) A(x) \mathcal{I}(x),
 \end{equation}
 where $a_{ij} = m_{ij} J_{ij}$ and $a_{ii} = - \sum_{j \ne i} a_{ij}$.

 \begin{lemma}\label{lem:neg-def}
  If $A$ is a matrix such that $a_{ij} \ge 0$ for $i \ne j$ and
  $a_{ii} = -\sum_{j \ne i} a_{ij}$ then $A \le 0$.
 \end{lemma}

 \begin{proof}
  In fact, the proof follows immediately from some well-known facts in linear
  algebra, such as the fact that $-A$ is diagonally dominant. However,
  we may also give a simple proof by noting that the quadratic form of $A$ is
  nothing but
  \[
   v^T A v = - \sum_{i<j} a_{ij} (v_i - v_j)^2 \le 0.
  \]
  This formula also leads us to the observation that if all of the $a_{ij}$
  are strictly positive (which is the case for the matrix $A$
  in~\eqref{eq:hess-J}) then the kernel of $A$ is the span of the all-ones
  vector. This observation is irrelevant to Lemma~\ref{lem:neg-def}, but
  we will use it later.
 \end{proof}

 Combining Lemma~\ref{lem:neg-def} with~\eqref{eq:hess-J}, we have proved
 Proposition~\ref{prop:hess-J}, and hence proved~\eqref{eq:main}.

 To complete the proof of Theorem~\ref{thm:main}, it remains to characterize
 the equality cases; for this, we will use Corollary~\ref{cor:linear-diffusion}:
 if $\mathbb{E} J(F) = J(\mathbb{E} F)$ then for every $t > 0$,
  \begin{align}
   0 &= \big((M \odot H_\Psi(P_t F^Q)) \otimes I_n\big)
   \begin{pmatrix} (\nabla P_t f_1)^{Q_1}\\ \vdots \\ (\nabla P_t f_k)^{Q_k} \end{pmatrix} \notag \\
   &= \big((\mathcal{I}(P_t F^Q) A(P_t F^Q) \mathcal{I}(P_t F^Q)) \otimes I_n\big)
   \begin{pmatrix} (\nabla P_t f_1)^{Q_1}\\ \vdots \\ (\nabla P_t f_k)^{Q_k} \end{pmatrix} \notag \\
   &= (\mathcal{I}(P_t F^Q)\otimes I_n) ( A(P_t F^Q)\otimes I_n)( \mathcal{I}(P_t F^Q)) \otimes I_n)
   \begin{pmatrix} (\nabla P_t f_1)^{Q_1}\\ \vdots \\ (\nabla P_t f_k)^{Q_k} \end{pmatrix},
   \label{eq:equality-1}
  \end{align}
  where the second equality follows from~\eqref{eq:hess-J}. Since
  $\mathcal{I}$ is always non-singular, we may drop the first instance of it
  from~\eqref{eq:equality-1}. Defining $w_i = \Phi^{-1} \circ f_i$, we have
  $\nabla w_i = \nabla f_i / I(f_i)$. By multiplying out the last two terms
  of~\ref{eq:equality-1}, we have
  \[
   0 = ( A(P_t F^Q)\otimes I_n)
   \begin{pmatrix} \frac{(\nabla P_t f_1)^{Q_1}}{I(P_t f_1)^{Q_1}} \\ \vdots \\ \frac{(\nabla P_t f_k)^{Q_k}}{I(P_t f_k)^{Q_k}} \end{pmatrix}
   = ( A(P_t F^Q)\otimes I_n)
    \begin{pmatrix} (\nabla w_1)^{Q_1} \\ \vdots \\ (\nabla w_k)^{Q_k} \end{pmatrix}
  \]
  Now, recall from the proof of Lemma~\ref{lem:neg-def}
  that the kernel of $A$ is $(1 \dots 1)^T$. It follows then that if
  \[
   (A \otimes I_n)
    \begin{pmatrix} (\nabla w_1)^{Q_1} \\ \vdots \\ (\nabla w_k)^{Q_k} \end{pmatrix}
    = 0
  \]
  then the $(\nabla w_i)^{Q_i}$ are all equal. Since this holds pointwise, and
  since the distribution of $Z$ is fully supported on $\mathbb{R}^{kn}$, we see
  that $\nabla w_i$ must all be almost surely equal to the same constant, $a$ say.
  Since each $w_i$ is a smooth function, we have
  $w_i(x) = a \cdot x + b_i$ for some $b_i$. Recalling the definition of $w_i$,
  we have $(P_t f_i)(x) = \Phi(a \cdot x + b_i)$. Carlen and
  Kerce~\cite{CarlenKerce:01} observed
  (and this observation was subsequently used in~\cite{MosselNeeman:12a} and~\cite{MosselNeeman:12b})
  that under this condition, and if $f_i = 1_{A_i}$, then $A_i$
  is a half-space and $a$ is normal to it. Since we have the same $a$ for every
  $A_i$, it follows that $A_1, \dots, A_k$ is a family of parallel half-spaces,
  which completes the proof of Theorem~\ref{thm:main}.

  In order to be more self-contained, let us sketch a proof
  (from~\cite{MosselNeeman:12b}) of why $(P_t 1_A)(x) = \Phi(a \cdot x + b_i)$ implies
  that $A$ is a half-space. First, one checks that if $A$ is a half-space
  and $\nu$ its outward unit normal, then
  \[
    (P_t 1_A)(x) = \Phi(k_t \nu \cdot x + b)
  \]
  for some $b \in \mathbb{R}$, where $k_t = (e^{2t} - 1)^{-1/2}$. Moreover, as
  $A$ ranges over all half-spaces with normal $\nu$ then $b$ ranges over $\mathbb{R}$.
  After checking that $P_t$ is one-to-one, this implies that if
  $(P_t 1_A)(x) = \Phi(a \cdot x + b)$ with $|a| = k_t$, then $A$ is a half-space
  with normal $a$. It remains to see what happens when $|a| \ne k_t$. First of
  all, Bakry and Ledoux showed that $|\nabla (\Phi^{-1} \circ P_t f)| \le k_t$
  for any $f: \mathbb{R}^n \to [0, 1]$; hence $|a| \le k_t$. But if $|a| < k_t$ then
  there is some $s > 0$ with $|a| = k_{t + s}$. It follows from the previous
  argument, then, that there is a half-space $B$ with
  $P_{t+s} 1_B = \Phi(a \cdot x + b) = P_t 1_A$. We then have
  $P_s 1_B = 1_A$, which is a contradiction since $P_s 1_B$ is always a smooth
  function.

 \bibliographystyle{plain}


text:
\section{Introduction}

Let $\alpha:(\C,S) \to(\C^{2},0)$ be a map germ of finite $\A_e$-codimension. D. Mond shows in \cite{M2} that the image Milnor number of $\alpha$, $\mu_I(\alpha)$, is determined by the number of branches $r(X,0)$ of the curve and the number $\delta(X,0)$ of nodes appearing in a stable perturbation of $\alpha$:
$$
\mu_I(\alpha)=\delta(X,0)-r(X,0)+1.
$$

D. Mond also proves in \cite{M2} a relation between the image Milnor number and the $\A_{e}$-codimension.

\begin{teo}[\cite{M2}]\label{Mond2}
Let $\alpha: (\C,S) \to (\C^{2},0)$ be a map germ of finite $\A_{e}$-codimension. Then,
$$\A_{e}\mbox{-}\codim(\alpha) \leq \mu_{I}(\alpha),
$$
with equality if $\alpha$ is weighted homogeneous.
\end{teo}

Inspired by the previous inequality, the first and second authors consider in \cite{DJ} map germs $f:(X,0)\to(\C^2,0)$, where $(X,0)$ is a plane curve. They define the image Milnor number of $f$ and  obtain the similar inequality.

\begin{teo}[\cite{DJ}] Let $(X,0)$ be a plane curve and let $f:(X,0)\to(\C^2,0)$ be a finite map germ of degree 1 onto its image $(Y,0)$.
Then, $$\A_e\text{-}\codim(X,f) \leq \mu_{I}(f),$$
with equality if and only if $(Y,0)$ is weighted homogeneous.
\end{teo}

Furthermore, following the proof of this result, it is possible obtain an equality,
$$\A_{e}\mbox{-}\codim(X,f) + \mu(Y,0) - \tau(Y,0) = \mu_{I}(f).$$

In this work, we consider map germs $f:(X,0)\to(\C^2,0)$, where $(X,0)$ is an isolated complete intersection singularity (ICIS) of dimension one, then we can consider the $\A_e$-codimension, $\A_e$-$\codim(X,f)$, in the sense of \cite{MM}.
On the other hand, we define the image Milnor number in this case as the number of vanishing cycles in the image of a stable perturbation $f_s:X_s\to B_{\epsilon}$, where $B_{\epsilon}$ is a ball of radius $\epsilon$ centered at the origin in $\C^2$. Here, stable means that $X_s$ is smooth and $f_s$ is stable in the usual sense.
We show that $\A_e$-$\codim(X,f)= \mu_I(f)$, if $(X,0)$ is irreducible and both $(X,0)$ and $f$ are weighted homogeneous (with the same weights).

Furthermore, if we consider $(X,0)$ parametrized by a map $\alpha:(\C,0)\to(\C^n,0)$, we have also the following equalities.
$$
\mu_{I}(f)=\delta(X,0)+\mu_{I}(f\circ\alpha)
$$
and
$$
\A_e\mbox{-}\codim(f)=\A_e\mbox{-}\codim(f\circ\alpha)-\frac{1}{n-1}\A_e\mbox{-}\codim(\alpha)
$$

\section{Maps on singular varieties}

In this section, we give the basic definitions we will use in the work.
First, we introduce some notations. We denote $\mathcal{O}_{n}$ the local ring of the analytic functions germs $f:(\C^n,0) \to \C$, $(X,0)\subset (\C^n,0)$ is a germ of analytic variety (possibly with singularities),
$I(X,0)$ is the ideal of $\mathcal O_n$ of functions vanishing on $(X,0)$,  $\O_{X,0}=\O_n/I(X,0)$ is the local ring of $(X,0)$,
$\Theta_n$ is the $\mathcal O_n$-module of the vector fields in $(\C^n,0)$ and $\Theta_{X,0}$ is the $\mathcal{O}_{X,0}$-module of vector fields tangents on $(X,0)$.

We refer to \cite{MM} for the general definition of $\A_e$-codimension for analytic map germs $f:(X,0)\to(\C^p,0)$. The $\A_e$-codimension is equal to
$$
\A_{e}\mbox{-}\codim(f)=\dim_\C\frac{\Theta(f)}{tf(\Theta_{X,0})+ \omega f(\Theta_{p})},
$$
where $\Theta(f)$ is the $\mathcal{O}_{X,0}$-module of vector fields along $f$, that is, holomorphic germs $\xi:(X,0)\to T\C^p$ such that $\pi\circ\xi=f$ (where $\pi:T\C^p\to\C^p$ is the canonical projection). The map
$tf:\Theta_{X,0}\to\Theta(f)$ is the morphism of $\mathcal{O}_{X,0}$-modules given by $tf(\xi)=df\circ \xi$
and $\omega f:\Theta_p\to\Theta(f)$  is the morphism of $\mathcal{O}_{p}$-modules given by $\omega f(\eta)=\eta\circ f$ (where $\Theta(f)$ is considered as an $\mathcal{O}_{p}$-module via $f^*:\O_p\to\O_{X,0}$). We can denote $T\A_{e} f=tf(\Theta_{X,0})+ \omega f(\Theta_{p})$.

We say that $f$ is \emph{$\mathscr A$-finite} if this codimension is finite. We say that $f$ has \emph{finite singularity type} if
$$
\dim_\C\frac{\Theta(f)}{tf(\Theta_{X,0})+(f^*m_p)\Theta_{p}}<\infty.
$$

In the case that $(X,0)$ is an ICIS and $f$ has finite singularity type, we have the following important result due to Mond and Montaldi \cite{MM}:

\begin{thm}\label{estabilizacao} Let $(X,0)$ be an ICIS and assume $f:(X,0)\to(\C^p,0)$ has finite singularity type. The minimal number of parameters in a versal unfolding of $f$ is equal to the number
$$
\mathscr A_e\text{-}\codim(X,f):=\mathscr A_e\text{-}\codim(f)+\tau(X,0),
$$
where $\tau(X,0)$ is the Tjurina number of $(X,0)$, that is, the minimal number of parameters in a versal deformation of $(X,0)$.
\end{thm}

As a corollary of this theorem, we have the following consequences. The second one is a generalization of the Mather-Gaffney criterion (see \cite{wall}).

\begin{cor}\label{infinitesimal} Let $(X,0)$ be an ICIS and assume $f:(X,0)\to(\C^p,0)$ has finite singularity type.
\begin{enumerate}
\item $f$ is stable if and only if $X$ is smooth and $f$ is stable in the usual sense.
\item $f$ is $\mathscr A$-finite if and only if $f$ has isolated instability (i.e., there is a representative $f:X\to B_\epsilon$ such that for any $y\in B_\epsilon\setminus\{0\}$, the multigerm of $f$ at $f^{-1}(y)\cap S$ is stable).
\end{enumerate}
\end{cor}

Another consequence of the theorem is the existence of stabilizations, at least in the range of Mather's nice dimensions. Given an analytic map germ $f:(X,0)\to(\C^p,0)$, a \emph{stabilization} is a 1-parameter unfolding $F:(\mathcal X,0)\to (\C\times\C^p)$ with the property that for all $s\ne0$ small enough, the map $f_s:X_s\to B_\epsilon$ is stable, where $B_\epsilon$ is a ball of radius $\epsilon$ in $\C^p$.
By Theorem \ref{estabilizacao}, if $f:(X,0)\to(\C^p,0)$ is $\mathscr A$-finite, a stabilization of $f$ exists if $(r,p)$ are nice dimensions in the Mather's sense ($r=\dim(X,0)$).

We will consider $f:(X,0)\to(\C^2,0)$, where $(X,0)\subset(\C^n,0)$ is a curve ICIS.
Observe that by Corollary \ref{infinitesimal}, $f$ is stable if and only if $X$ is smooth and $f$ is an immersion with only transverse double points, called nodes.
As a consequence, $f$ is $\mathscr A$-finite if and only if $f$ is finite and has degree one onto its image.

\section{Weighted homogeneous maps and varieties}

We introduce the definition of weighted homogeneous for variety germs and map germs.
We fix positive integer numbers $w_1,\cdots,w_n$ such that $\gdc(w_1,\cdots,w_n)=1$.
Given $h \in \mathcal{O}_n$, we say that $h$ is weighted homogeneous of type $(w_1,\cdots,w_n;d)$ if it satisfies
$$h(t^{w_1}x_1,\cdots,t^{w_n}x_n)=t^d h(x_1,\cdots,x_n), \quad \forall x \in \C^n, \quad \forall t \in \C.$$
In this case, we call $(w_1,\cdots,w_n)$ the weights of $h$ and $d$ the weighted degree of $h$.
Let $(X,0)\subset(\C^n,0)$ be a variety germ, which is the zero set of an ideal $I \subset \mathcal{O}_n$, we say that $(X,0)$ is weighted homogeneous of type $(w_1,\cdots,w_n;d_1,\cdots,d_m)$
if $I$ can be generated by weighted homogeneous map germs $\phi_1,\cdots,\phi_m$ where each $\phi_i$ is weighted homogeneous of type $(w_1,\cdots,w_n,d_i)$, $i=1,\cdots,n$.
Finally, let $f:(X,0)\to (\C^p,0)$ be a map germ, with $(X,0)\subset (\C^n,0)$ a weighted homogeneous variety germ of type $(w_1,\cdots,w_n;d_1,\cdots,d_m)$.
We say that $f$ is weighted homogeneous if $f$  is the restriction of a map germ $\tilde{f}=(f_1,\cdots,f_p):(\C^n,0)\to(\C^p,0)$ where each $f_j \in \mathcal{O}_n$ is weighted homogeneous
and we say that $f$ is {\it consistent} with $(X,0)$ if each $f_j$ is weighted homogeneous with the weights $(w_1,\cdots,w_n)$, we observe that the weighted degree may be different.

It follows from \cite{NOT} that if $(X,0)$ is a curve ICIS weighted homogeneous of type $(w_1,\cdots,w_n;d_1,\cdots,d_{n-1})$, then the Milnor number satisfies
$$\mu(X,0)=\frac{d_1\cdots d_{n-1}\left(d_1 + \cdots+ d_{n-1} - w_1 - \cdots - w_n\right)}{w_1 \cdots w_n}+1.$$
Moreover, if $(X,0)$ is irreducible, then $\frac{d_1\cdots d_{n-1}}{w_1 \cdots w_n}=1$, thus the above equality is given by
$$\mu(X,0)=d_1 + \cdots + d_{n-1} - w_1 - \cdots - w_n + 1.$$

Wahl \cite{wahl} shows how to compute the generators of $\Theta_{X,0}$ when $(X,0)$ is a weighted homogeneous ICIS.
In particular, if we suppose $(X,0)=h^{-1}(0)$ a curve, where $h=(h_1, \cdots, h_{n-1})$, then $\Theta_{X,0}$  is generated by vector fields $h_i \frac{\partial}{\partial x_j}$ with $i=1,\cdots,n-1$ and $j=1, \cdots, n$, the Euler field $\epsilon= (w_1x_1,\cdots,w_nx_n)$ and $\mathcal H = (\xi_1,\cdots,\xi_n),$
where
\begin{align*}
\mathcal H &= \left| \begin{array}{ccc}
\frac{\partial}{\partial x_1}&\cdots & \frac{\partial}{\partial x_n}  \\
\frac{\partial h_1}{\partial x_1}&\cdots & \frac{\partial h_{1}}{\partial x_n}  \\
\vdots                                         & \ddots & \vdots \\
\frac{\partial h_{n-1}}{\partial x_1}&\cdots & \frac{\partial h_{n-1}}{\partial x_n}\\
\end{array} \right| \\
& = (-1)^{1+1} \left| \begin{array}{cccc}
\frac{\partial \widehat{h}_1}{\partial x_1} &\frac{\partial h_1}{\partial x_2}&\cdots & \frac{\partial h_{1}}{\partial x_n}  \\
\vdots                   & \vdots                      & \ddots & \vdots \\
\frac{\partial h_{n-1}}{\partial x_1} & \frac{\partial h_{n-1}}{\partial x_2}&\cdots & \frac{\partial h_{n-1}}{\partial x_n}\\
\end{array} \right| \frac{\partial}{\partial x_1} + \cdots \\
& \quad +(-1)^{n+1} \left| \begin{array}{cccc}
\frac{\partial h_1}{\partial x_1}&\cdots & \frac{\partial h_{1}}{\partial x_{n-1}}& \frac{\partial \widehat{h}_{1}}{\partial x_{n}}  \\
\vdots                                         & \ddots & \vdots & \vdots\\
\frac{\partial h_{n-1}}{\partial x_1}&\cdots & \frac{\partial h_{n-1}}{\partial x_{n-1}}& \frac{\partial h_{n-1}}{\partial x_{n}}\\
\end{array} \right| \frac{\partial}{\partial x_{n}}\\
&=\xi_1 \frac{\partial}{\partial x_1} + \cdots + \xi_n \frac{\partial}{\partial x_n},
\end{align*}
we use \ $\widehat{}$ \ to indicate that we should exclude this column.

Let $J_1,J_2\in\O_{X,0}$ be defined as
\begin{align*}
J_i &= \left| \begin{array}{ccc}
\frac{\partial h_1}{\partial x_1} &\cdots & \frac{\partial h_{1}}{\partial x_n}\\
\vdots                                         & \ddots & \vdots                                                 \\
\frac{\partial h_{n-1}}{\partial x_1}   &\cdots & \frac{\partial h_{n-1}}{\partial x_n} \\
\frac{\partial f_i}{\partial x_1}   & \cdots &  \frac{\partial f_i}{\partial x_n} \\ \end{array} \right|  =(-1)^{n-1}\left(\frac{\partial f_i}{\partial x_1} \xi_1+  \cdots+
\frac{\partial f_i}{\partial x_n}\xi_n\right),\\
\end{align*}
with $i=1,2$.

We have the matrix
$$df= \left( \begin{array}{ccc}
\frac{\partial f_1}{\partial {x}_1}&\cdots & \frac{\partial f_1}{\partial {x}_{n}}  \\
\frac{\partial f_2}{\partial x_1}&\cdots & \frac{\partial f_2}{\partial x_n}  \\
\end{array} \right).
$$

Thus, $tf(\Theta_{X,0})$ is generated by vector fields
$$\left(\frac{\partial f_1}{\partial {x}_1} h_1, \frac{\partial f_2}{\partial {x}_1}h_1\right),\cdots,\left(\frac{\partial f_1}{\partial {x}_n}h_1,\frac{\partial f_2}{\partial {x}_n}h_1\right),\cdots,$$
$$\left(\frac{\partial f_1}{\partial {x}_1}h_{n-1},\frac{\partial f_2}{\partial {x}_1}h_{n-1}\right),\cdots,\left(\frac{\partial f_1}{\partial {x}_n}h_{n-1},\frac{\partial f_2}{\partial {x}_n}h_{n-1}\right),$$
\begin{align*}
df \circ \epsilon & = \left(w_1x_1\frac{\partial f_1}{\partial {x}_1}+\cdots+w_nx_n\frac{\partial f_1}{\partial {x}_{n}},w_1x_1\frac{\partial f_2}{\partial {x}_1}+\cdots+w_nx_n\frac{\partial f_2}{\partial {x}_{n}}\right)\\
&=(l_1 \beta_ 1 f_1,l_2 \beta_2 f_2)\\
\end{align*}
and
\begin{align*}
df\circ\mathcal H& = \left(\xi_1 \frac{\partial f_1}{\partial x_1}+ \cdots+  \xi_n \frac{\partial f_1}{\partial x_{n}} , \xi_1 \frac{\partial f_2}{\partial x_1}+ \cdots+  \xi_n \frac{\partial f_2}{\partial x_{n}}\right)\\
&= (-1)^{n+1} \left(J_1,J_2\right).
\end{align*}

\section{Maps between curves}

Let $(X,0)\subset(\C^n,0)$ be an irreducible curve ICIS, weighted homogeneous of type $(w_1,\cdots,w_n;d_1,\cdots,d_{n-1})$.
Apart from the case that $(X,0)$ is smooth (which is a trivial case), it follows that $(X,0)$ can be parametrized by a monomial map
$\alpha:(\C,0)\to(\C^n,0)$ of the form
$$ \alpha(t)=(\alpha_1 t^{w_1},\cdots,\alpha_n t^{w_n}),\quad (\alpha_1,\cdots,\alpha_n)\in\C^n \setminus\{0\}.$$

The local ring of $(X,0)$ is $\O_{X,0}=\C\{t^{w_1},\cdots,t^{w_n}\}$, which is a subring of $\O_1$. We denote by $\Gamma_X$ the associated semigroup. Since $(X,0)$ is irreducible, $\Gamma_X$ has a conductor $c$, which satisfies that $c-1 \notin \Gamma_X$ and if $n\in \N$, with $n \geq c$, then $n\in \Gamma_X$. We have that $\Gamma_X$ is symmetric, because $(X,0)$ is an ICIS, thus $\#\N\setminus \Gamma_X=c/2$.

The main invariant of $(X,0)$ is the delta invariant, which is in this case
$$\delta_X=\dim_{\C}\frac{\C\left \{ t \right \}}{\C\left \{ t^{w_1},\cdots,t^{w_n} \right \}}=\#\N\setminus \Gamma_X.$$
Since $(X,0)$ is irreducible, the Milnor number $\mu(X,0)$, which to simplify the notation we will denote by $\mu_X$, is $\mu_X=2\delta_X=c$, by Milnor's formula.
For more details, see for instance \cite{HeHer}.

Let $f:(X,0)\to(\C^2,0)$ be a finite map germ of degree 1 onto its image $(Y,0)$ and $f$ is consistent with $(X,0)$. We denote by $l_1,l_2$ the weighted degrees of $f_1,f_2$, respectively.
Then, $(Y,0)$ is also a weighted homogeneous irreducible plane curve with weights $(l_1, l_2)$ and it admits a parametrization of the form
$$f(\alpha(t))=(\beta_1 t^{l_1},\beta_2 t^{l_2}),\quad \beta_1,\beta_2\in\C\setminus\{0\}.$$
We have $\O_{Y,0}=\C\{t^{l_1},t^{l_2}\}\subset \O_{X,0}$,  $\Gamma_Y=\langle l_1,l_2\rangle\subset \Gamma_X$ and $\delta_Y=(l_1-1)(l_2-1)/2$. Finally, we assume that $(Y,0)$ has defining equation $g(u,v)=0$, where $g\in\C\{u,v\}$ is weighted homogeneous.
Thus, $g(u,v)=\beta_{2}^{l_1}u^{l_2}-\beta_{1}^{l_2}v^{l_1}$ is weighted homogeneous  of type $(l_1, l_2;l_1l_2)$.

We consider $f=i\circ \bar{f}$, where $\bar{f}:(X,0)\to(Y,0)$ is the restriction of $f$ and $i:(Y,0)\to(\C^2,0)$ is the inclusion map. We have,
$$
J_{g}\mathcal{O}_{X,0} = \left \{ a \left(\frac{\partial g}{\partial u}\circ f \right)+ b \left(\frac{\partial g}{\partial v}\circ f \right) ; \  a, b \in \mathcal{O}_{X,0} \right \},
$$
and

$
J_{g}\mathcal{O}_{Y,0} = \left \{\left(a \circ \bar{f} \right) \left( \left(\frac{\partial g}{\partial u} \circ i \right) \circ  \bar{f} \right) + \left( b \circ \bar{f} \right) \left( \left( \frac{\partial g}{\partial v}  \circ i \right) \circ \bar{f} \right) ; \  a, b \in \mathcal{O}_{Y,0} \right \},
$
where $J_g$ be the Jacobian ideal of $g$, thus $J_{g}\mathcal{O}_{Y,0}  \subset J_{g}\mathcal{O}_{X,0}$.

We consider the evaluation map
$$ev: \frac{\Theta(f)}{tf(\Theta_{X,0})+ wf(\Theta_{2})} \longrightarrow \frac{J_{g}\mathcal{O}_{X,0}}{J_{g}\mathcal{O}_{Y,0}}$$
given by $ev([\xi]) = [\xi(g)]$.

\begin{lem}\label{sobrejetora}
The evaluation map is well-defined and surjective, hence
$$ \A_{e}\mbox{-}\codim(f) = \mbox{\rm dim}_{\C} \ker(ev) + \mbox{\rm dim}_{\C} \frac{J_{g}\mathcal{O}_{X,0}}{J_{g}\mathcal{O}_{Y,0}}.$$
\end{lem}
\begin{proof}
Let us see that $ev$ is well-defined.
If $\xi \in tf(\Theta_{X,0})$, then $\xi = tf(\eta)$, for some $\eta \in \Theta_{X,0}$, hence $\eta(l)=0$ in $\mathcal{O}_{X,0}$, for all $l\in I(X,0)$.
Since $g \circ \tilde f = \lambda_1 h_1 + \cdots \lambda_{n-1} h_{n-1}$, with $\lambda_i \in \mathcal{O}_{n}$, $i=1,\cdots,n$, we have
$$
\xi(g) =tf(\eta)(g)=(df\circ \eta)(g) = \eta(g \circ \tilde{f})= \eta(\lambda_1 h_1 + \cdots \lambda_{n-1} h_{n-1})=0.$$

If $\xi \in \omega f(\Theta_{2})$, then $\xi = \zeta \circ f$, for some $\zeta \in \Theta_{2}$. Write  $\zeta = \tilde{a} \frac{\partial}{\partial u} + \tilde{b} \frac{\partial}{\partial v}$, with $\tilde a,\tilde b\in\O_2$. Thus,
$$\xi=\zeta \circ f =  \left( \tilde{a} \circ f \right) \frac{\partial}{\partial u} + \left(\tilde{b} \circ f \right) \frac{\partial}{\partial v}.
$$

Then,
\begin{align*}
\xi(g)& =\left( \tilde{a} \circ f \right) \frac{\partial g}{\partial u} \circ f + \left(\tilde{b} \circ f \right) \frac{\partial g}{\partial v} \circ f \\
& = \left( \left(\tilde{a} \circ i \right) \circ \bar{f} \right) \left( \left(\frac{\partial g}{\partial u} \circ i \right) \circ  \bar{f} \right) + \left( \left(\tilde{b} \circ i \right) \circ \bar{f} \right) \left( \left( \frac{\partial g}{\partial v}  \circ i \right) \circ \bar{f} \right),
\end{align*}
with $\tilde{a} \circ i ,\tilde{b} \circ i \in \mathcal{O}_{Y,0}$. Hence, $\xi(g) \in J_{g}\mathcal{O}_{Y,0}$.

If $\xi= a \frac{\partial}{\partial u} + b \frac{\partial}{\partial v}$, with $a$, $b \in \mathcal{O}_{X,0}$, then
$\xi(g) = a \frac{\partial g}{\partial u} \circ f + b \frac{\partial g}{\partial v} \circ f$, hence $ev$ is surjective.
\end{proof}

The evaluation map is not injective in general, so we need to know its kernel in order to compute the $\A_e$-codimension of $f$.
To this we need the following result.

\begin{lem} \label{lemmalambda} With the above notation, we have
$$
\frac{\partial g}{\partial u} \circ f  = C \lambda_f J_2  \quad \mbox{and} \quad \frac{\partial g}{\partial v} \circ f = - C  \lambda_f  J_1,
$$
when $\lambda_f=t^{\mu_Y-\mu_X}\in \O_{X,0}$.
\end{lem}
\begin{proof}
By definitions, we obtain
$$
\frac{\partial g}{\partial u} \circ f  =  l_2 \beta_{2}^{l_1} \beta_1^{l_2-1}t^{l_1(l_2-1)} \quad \mbox{and}  \quad \frac{\partial g}{\partial v} \circ f  =  -l_1 \beta_{1}^{l_2} \beta_2^{l_1-1} t^{l_2(l_1-1)}.
$$

We also have
\begin{align*}
J_j&=\left| \begin{array}{ccc}
t^{d_1-w_1}\frac{\partial h_1}{\partial x_1}(\alpha_1,\cdots,\alpha_n) &\cdots & t^{d_1-w_n}\frac{\partial h_{1}}{\partial x_n}(\alpha_1,\cdots,\alpha_n)\\
\vdots                                         & \ddots & \vdots                                                 \\
t^{d_{n-1}-w_1}\frac{\partial h_{n-1}}{\partial x_1} (\alpha_1,\cdots,\alpha_n)  &\cdots & t^{d_{n-1}-w_n}\frac{\partial h_{n-1}}{\partial x_n}(\alpha_1,\cdots,\alpha_n) \\
t^{l_j-w_1}\frac{\partial f_j}{\partial x_1} (\alpha_1,\cdots,\alpha_n)  & \cdots &  t^{l_j-w_n}\frac{\partial f_j}{\partial x_n}(\alpha_1,\cdots,\alpha_n) \\ \end{array} \right|.
\end{align*}
We denote $\frac{\partial h_i}{\partial x_k}(\alpha_1,\cdots,\alpha_n)=\gamma_{i,k}$ with $i=1,\cdots,n-1$, $k=1,\cdots,n$ and $\frac{\partial f_j}{\partial x_k} (\alpha_1,\cdots,\alpha_n)=\zeta_{j,k}$ with $j=1,2$ and $k=1,\cdots,n$.

Furthermore, since $h_1, \cdots, h_{n-1}, f_1, f_2$ are weighted homogeneous with weights $w_1, \cdots, w_n$, they satisfies:
$$d_ih_i=w_1 x_1\frac{\partial h_i}{\partial x_1}+\cdots+w_n x_n\frac{\partial h_i}{\partial x_n},$$
$$l_jf_j=w_1 x_1\frac{\partial f_j}{\partial x_1}+\cdots+w_n x_n\frac{\partial f_j}{\partial x_n},$$
with $i=1,\cdots,n-1$ and $j=1,2$. Thus,
$$w_1 \alpha_1t^{w_1}t^{d_i-w_1}\gamma_{i,1}+\cdots+w_n \alpha_n t^{w_n} t^{d_i-w_n}\gamma_{i,n}=0,$$
$$w_1 \alpha_1t^{w_1}t^{l_j-w_1}\zeta_{j,1}+\cdots+w_n \alpha_n t^{w_n} t^{l_j-w_n}\zeta_{j,n}=l_j\beta_j t^{l_j},$$
with $i=1,\cdots,n-1$ and $j=1,2$.

Rewriting in matrix form
$$\left( \begin{array}{ccc}
t^{d_1-w_1}\gamma_{1,1} &\cdots & t^{d_1-w_n}\gamma_{1,n}\\
\vdots                                         & \ddots & \vdots                                                 \\
t^{d_{n-1}-w_1}\gamma_{n-1,1} &\cdots & t^{d_{n-1}-w_n}\gamma_{n-1,n}\\
t^{l_j-w_1}\zeta_{j,1} & \cdots &  t^{l_j-w_n}\zeta_{j,n} \\ \end{array} \right)\left( \begin{array}{c}
 w_1 \alpha_1t^{w_1}\\
\vdots                  \\
w_n \alpha_n t^{w_n} \\ \end{array} \right)= \left( \begin{array}{c}
 0\\
\vdots                  \\
0\\
l_j\beta_j t^{l_j} \\
 \end{array} \right).
$$

By Cramer's Rule,
$$w_1 \alpha_1 t^{w_1} J_j= \left| \begin{array}{cccc}
 0 & t^{d_1-w_2}\gamma_{1,2} &\cdots & t^{d_1-w_n}\gamma_{1,n}\\
\vdots   & \vdots                                      & \ddots & \vdots                                                 \\
0& t^{d_{n-1}-w_2}\gamma_{n-1,2} &\cdots & t^{d_{n-1}-w_n}\gamma_{n-1,n}\\
l_j\beta_j t^{l_j}& t^{l_j-w_2}\zeta_{j,2} & \cdots &  t^{l_j-w_n}\zeta_{j,n} \\ \end{array} \right|.
$$

Thus,
\begin{align*}
J_j&=(-1)^{1+n} \frac{l_j\beta_j}{w_1\alpha_1} t^{l_j-w_1}\left| \begin{array}{ccc}
  t^{d_1-w_2}\gamma_{1,2} &\cdots & t^{d_1-w_n}\gamma_{1,n}\\
\vdots                                      & \ddots & \vdots                                                 \\
t^{d_{n-1}-w_2}\gamma_{n-1,2} &\cdots & t^{d_{n-1}-w_n}\gamma_{n-1,n}\\  \end{array} \right| \\
&= l_j\beta_j t^{l_j - w_1 + d_1+ \cdots + d_{n-1}- w_2 - \cdots - w_n} A, \\
\end{align*}
with $A\neq 0$.

We obtain,
$$
J_1 = l_1 \beta_1 A  t^{l_1+\mu_X-1} \quad \mbox{and} \quad J_2 = l_2 \beta_2 A t^{l_2+\mu_X-1}
$$

Therefore,
\begin{align*}
\frac{\partial g}{\partial u} \circ f  &=  l_2 \beta_{2}^{l_1} \beta_1^{l_2-1}t^{l_1(l_2-1)}= \frac{ \beta_2^{l_1-1}\beta_1^{l_2-1}}{A} t^{\mu_Y - \mu_X} l_2 \beta_2 A  t^{l_2+\mu_X-1}\\
&= C \lambda_f J_2
\end{align*}
and
\begin{align*}
\frac{\partial g}{\partial v} \circ f & =  - l_1 \beta_{1}^{l_2} \beta_2^{l_1-1}t^{l_2(l_1-1)}=- \frac{ \beta_1^{l_2-1}\beta_2^{l_1-1}}{A} t^{\mu_Y - \mu_X} l_1 \beta_1 A  t^{l_1+\mu_X-1}\\
&= -C \lambda_f J_1.
\end{align*}
\end{proof}

\begin{lem} \label{nucleo}
We consider the map
$$\widetilde{ev}: \frac{\Theta(f)}{tf(\Theta_{X,0})} \longrightarrow J_g \O_{X,0},$$
given by $\widetilde{ev}([\xi])=\xi(g)$. Then, $\ker(ev) \cong \ker(\widetilde{ev})$.
\end{lem}
\begin{proof}
We observe that the map $\widetilde{ev}$ is well-defined and it is surjective.

We define the morphism $\ker(ev)  \to \ker(\widetilde{ev})$ given by $\left[\xi\right] \mapsto [\xi]$, where the classes have to be considered in the respective quotients.

We claim that if $[\xi] \in \ker(ev)$, there exists representative $\xi$ such that
$$ \xi(g) = a \frac{\partial g}{\partial u} \circ f + b \frac{\partial g}{\partial v} \circ f = 0.$$
In fact, if $[\xi] \in \ker(ev)$, we choose a representative $\xi = a \frac{\partial}{\partial u} + b \frac{\partial}{\partial v}$, $a$, $b \in \mathcal{O}_{X,0}$, thus $\xi(g) = a \frac{\partial g}{\partial u} \circ f + b \frac{\partial g}{\partial v} \circ f$.
On the other hand, since $[\xi] \in \ker(ev)$, we have $\xi(g) \in J_{g}\mathcal{O}_{Y,0}$, so,
$$\xi(g) = \left( \tilde{a} \circ i \circ \bar{f} \right) \left( \left(\frac{\partial g}{\partial u} \circ i \right) \circ  \bar{f} \right) + \left( \tilde{b} \circ i \circ \bar{f} \right) \left( \left( \frac{\partial g}{\partial v}  \circ i \right) \circ \bar{f} \right),$$
with $\tilde{a} \circ i, \tilde{b} \circ i  \in \mathcal{O}_{Y,0}$. Therefore,
$$\left(  a - \tilde{a} \circ i \circ \bar{f} \right) \left( \frac{\partial g}{\partial u}  \circ  f \right) + \left(b - \tilde{b}  \circ i \circ \bar{f} \right)  \left( \frac{\partial g}{\partial v}   \circ f \right) = 0.$$
We define $\bar{a} = a - \tilde{a} \circ i \circ \bar{f}$, $\bar{b} = b - \tilde{b} \circ i \circ \bar{f}$ and $\bar{\xi} = \bar{a} \frac{\partial}{\partial u} + \bar{b} \frac{\partial}{\partial v}$, thus
\begin{align*}
\xi - \bar{\xi}& =  a \frac{\partial}{\partial u} + b \frac{\partial}{\partial v} -  \left( \bar{a} \frac{\partial}{\partial u} + \bar{b} \frac{\partial}{\partial v} \right) =
\left( \tilde{a} \circ i \circ \bar{f}\right) \frac{\partial}{\partial u} + \left(\tilde{b}  \circ i \circ \bar{f}\right) \frac{\partial }{\partial v}\\
&=  \left(\tilde{a}  \circ f\right) \frac{\partial}{\partial u} +\left( \tilde{b}   \circ f \right) \frac{\partial }{\partial v} =
 \tilde{\zeta} \circ f,
\end{align*}
where $\tilde{\zeta}=\tilde{a} \frac{\partial}{\partial u} + \tilde{b} \frac{\partial }{\partial v}$, with $\tilde{a}, \tilde{b} \in \mathcal{O}_{2}$.
Thus, $\xi - \bar{\xi} \in wf(\Theta_{2})$, i.e., $[\xi]=[\bar{\xi}]$ with $\bar{\xi}(g) = 0$.  This prove the claim.

Now, we will show that the morphism is well-defined. In fact, let $[\xi] = [\tilde{\xi}] \in \ker(ev)$, i.e., $\xi + T\A_{e} f = \tilde{\xi}+T\A_{e} f $, then $\xi - \tilde{\xi} \in T\A_{e} f$, with $\xi(g)=\tilde{\xi}(g)=0$.
We will show that $\xi - \tilde{\xi} \in tf(\Theta_X)$.

Since  $\xi - \tilde{\xi} \in T\A_{e} f$, then  $\xi - \tilde{\xi} = df \circ \eta + \rho \circ f$, $\eta \in \Theta_X$, $\rho \in \Theta_2$, with
$0 = \xi(g) - \tilde{\xi}(g) = df \circ \eta(g) + \rho \circ f(g)$, moreover $df \circ \eta(g) = 0$, thus $\rho \circ f(g)=0$.

We have $\rho =(a,b)$, with $a, b \in \mathcal{O}_{2}$, then, $\rho \circ f = (a \circ f,b \circ f)$ such that
$ (a \circ f) \frac{\partial g}{\partial u} \circ f + (b \circ f) \frac{\partial g}{\partial v}\circ f =0$. Thus,
$$\left( a \frac{\partial g}{\partial u} + b \frac{\partial g}{\partial v}\right)\circ f =0, \  \mbox{in}  \  \mathcal{O}_{Y}$$
i.e.,
$$ a \frac{\partial g}{\partial u} + b \frac{\partial g}{\partial v} = \kappa g, \ \kappa \in \mathcal{O}_2.$$
Since $g$ is weighted homogeneous,
$$ l_1 l_2 g = l_1 u \frac{\partial g}{\partial u}+l_2 v \frac{\partial g}{\partial v},$$
so,
$$l_1 l_2 \left(a \frac{\partial g}{\partial u} + b \frac{\partial g}{\partial v}\right) = l_1 l_2 \kappa  g = \kappa \left(l_1 u \frac{\partial g}{\partial u}+l_2 v \frac{\partial g}{\partial v}\right),$$

$$(l_1 l_2 a - \kappa l_1 u) \frac{\partial g}{\partial u} + (l_1 l_2 b - \kappa l_2 v) \frac{\partial g}{\partial v} = 0,$$
then
$$ l_1 l_2 a - \kappa l_1 u = - \frac{\partial g}{\partial v} \quad \mbox{and} \quad l_1 l_2 b - \kappa l_2 v = \frac{\partial g}{\partial u}.$$

Therefore,
$$(l_1 l_2 (a \circ f) - (\kappa \circ f)  l_1 \beta_1 f_1) = - \frac{\partial g}{\partial v} \circ f \quad \mbox{and} \quad  (l_1 l_2 (b\circ f) - (\kappa \circ f)  l_2 \beta_2 f_2)= \frac{\partial g}{\partial u} \circ f$$

Thus,
\begin{align*}
l_1 l_2 (\rho \circ f) &= (l_1 l_2 (a \circ f) ,l_1 l_2 (b \circ f))\\
&= (\kappa \circ f )(l_1 \beta_1 f_1,l_2 \beta_2 f_2) + \left(- \frac{\partial g}{\partial v} \circ f, \frac{\partial g}{\partial u} \circ f\right).
\end{align*}

We have
$$(\kappa \circ f) ( l_1 \beta_1 f_1, l_2 \beta_2 f_2) \in tf(\Theta_X),$$
we will show that
$$\left(- \frac{\partial g}{\partial v} \circ f,\frac{\partial g}{\partial u} \circ f\right) \in tf(\Theta_X).$$

Since $ \frac{\partial g}{\partial u} \circ f  = C \lambda_f J_2$ and $\frac{\partial g}{\partial v} \circ f = - C  \lambda_f  J_1$, then
$$\left(- \frac{\partial g}{\partial v} \circ f,\frac{\partial g}{\partial u} \circ f\right) = C  \lambda_f  (J_1,J_2) \in tf(\Theta_X).$$

Therefore, $\rho \circ f \in tf(\Theta_X)$ and, consequently,  $$\xi - \tilde{\xi} = df \circ \eta + \rho \circ f \in tf(\Theta_X).$$

Since $t f(\Theta_X) \subseteq t f (\Theta_X) + \omega f(\Theta_2)$, the map is injective and it is surjective.
\end{proof}

By using the coordinates in source and target, we can identify $\Theta(f)$ with $\O_{X,0}^2=\O_{X,0}\oplus\O_{X,0}$. The module $\Theta_{X,0}$ can be seen as a submodule of $\O_{X,0}^n$. With these identifications, the morphism $tf:\Theta_{X,0}\to\Theta(f)$ is the restriction of the map $tf:\O_{X,0}^n\to\O_{X,0}^2$, whose matrix in the canonical basis is the Jacobian matrix of $f$.

The following result describes the elements of $\ker(\widetilde{ev})$.

\begin{lem} \label{caracterizacao} Let $k = l_2-l_1$. Then,
$$
\ker(\widetilde{ev}) \cong \frac{ \C\left \{ (l_1 \beta_1 t^{r},l_2 \beta_2 t^{r + k}):\ (t^r, t^{r + k}) \in \mathcal{O}^2_{X,0}, r \in \mathbb{N}  \right \} }{ \left \langle (l_1 \beta_1 t^{l_1},l_2 \beta_2 t^{l_1+k}),(l_1 \beta_1 A t^{l_1+\mu_X-1},l_2 \beta_2 A t^{l_1+\mu_X-1+k})  \right \rangle }.
$$
\end{lem}
\begin{proof}
For each $(a,b)\in \O_{X,0}^2$, we have:
$$\widetilde{ev}([(a,b)])=a \frac{\partial g}{\partial u}\circ f+b \frac{\partial g}{\partial v}\circ f=a l_2 \beta_2^{l_1}\beta_1^{l_2-1} t^{l_1 l_2 - l_1} - b l_1 \beta_1^{l_2}\beta_2^{l_1 -1} t^{l_1 l_2 - l_2}.
$$
Then, $[(a,b)]\in \ker(\widetilde{ev})$ if and only if $(a,b)=c(l_1 \beta_1 t^{s_1},l_2 \beta_2 t^{s_2} )$ such that $l_1 l_2-l_1+s_1=l_1 l_2 - l_2 +s_2$, for some $c\in\O_{X,0}$. It follows that $\ker(\widetilde{ev})$ is generated by monomial pairs of the form
$(l_1 \beta_1 t^r,l_2 \beta_2 t^{r+k})$, for some $r\in\N$ such that $(t^r,t^{r+k})\in\O^2_{X,0}$.

We remember that $tf(\Theta_X)$ is generated on $\mathcal{O}_{X,0}$ by vectors fields
$$\left(\frac{\partial f_1}{\partial {x}_1} h_1, \frac{\partial f_2}{\partial {x}_1}h_1\right),\cdots,\left(\frac{\partial f_1}{\partial {x}_n}h_1,\frac{\partial f_2}{\partial {x}_n}h_1\right),\cdots,$$
$$\left(\frac{\partial f_1}{\partial {x}_1}h_{n-1},\frac{\partial f_2}{\partial {x}_1}h_{n-1}\right),\cdots,\left(\frac{\partial f_1}{\partial {x}_n}h_{n-1},\frac{\partial f_2}{\partial {x}_n}h_{n-1}\right),$$
$$ df \circ \epsilon = (l_1 \beta_1 f_1,l_2 \beta_2 f_2) \quad \mbox{and} \quad df\circ\mathcal H= (-1)^{n+1}( J_1, J_2).$$

Since $\O_{X,0}=\C\{t^{w_1},\cdots,t^{w_n}\}$ and $\O_{Y,0}=\C\{t^{l_1},t^{l_2}\}\subset \O_{X,0}$, we have
$$(l_1 \beta_1 f_1,l_2 \beta_2 f_2)=(l_1 \beta_1 t^{l_1},l_2 \beta_2 t^{l_1 +k}).$$

We remember that
$$
J_1 = l_1 \beta_1 A  t^{l_1+\mu_X-1} \quad \mbox{and} \quad J_2 = l_2 \beta_2 A t^{l_2+\mu_X-1}
$$

Thus,
$$df\circ\mathcal H = (-1)^{n+1}(  l_1 \beta_1 A  t^{l_1+\mu_X-1}, l_2 \beta_2 A t^{l_1+\mu_X-1+k}).$$
\end{proof}

\begin{teo}\label{dimensao}
We denote by $s$ the number of elements of the form $(t^{r},t^{r+k})$ in $\mathcal{O}^2_{X,0}$ such that $0\leq r<\mu_X$.
Then,
$$
\mbox{dim}_{\C} \ker(ev)= l_1 - \delta_X + s -1
$$
and
$$
\mbox{dim}_{\C} \frac{\mathcal{O}_{X,0}}{\langle J_1,J_2 \rangle} =l_{1} + \delta_X + s - 1
$$
\end{teo}
\begin{proof}
By Lemma \ref{nucleo} and  Lemma \ref{caracterizacao},
$$
\ker(ev) \cong \frac{ \C\left \{ (l_1 \beta_1 t^{r},l_2 \beta_2 t^{r + k}):\ (t^r, t^{r + k}) \in \mathcal{O}^2_{X,0}, r \in \mathbb{N}  \right \} }{ \left \langle (l_1 \beta_1 t^{l_1},l_2 \beta_2 t^{l_1+k}),(l_1 \beta_1 A t^{l_1+\mu_X-1},l_2 \beta_2 A t^{l_1+\mu_X-1+k})  \right \rangle }.
$$

We consider the set $\Gamma= \left\{ l_1+ i: \ i\in\Gamma_X\right\}\cup\left\{l_1+\mu_X-1\right\}$ which is a subset of  $\Gamma_X$.
We denoted by $\Gamma_X \oplus \Gamma_X$ the associated semigroup to $\O^{2}_{X,0}$ e we take the sets:
$$
\Gamma_{k}=\left \{ (r,r+k)\in \Gamma_ X \oplus \Gamma_X \right\} \quad \mbox{and} \quad \Gamma_{\Theta}=\left\{ (l,l+k):\ l \in \Gamma \right\} \subset \Gamma_{k}
$$

We can identify the elements of $\ker(ev)$ with the elements of $\Gamma_{k} \setminus \Gamma_{\Theta}$.
Therefore, we compute the number of elements in $\Gamma_{k} \setminus \Gamma_{\Theta}$.

We observe that $\Gamma_{k}$ is equal to
\begin{align*}
\underbrace{\left\{(r,r+k):\ 0 \leq r < \mu_X \right\}}_{s \ \text{elements}} \cup \underbrace{\left\{(\mu_X,\mu_X+k),(\mu_X+1,\mu_X+1+k), \cdots \right\}}_{\text{all elements}}
\end{align*}
and $\Gamma_{\Theta}$ is
\begin{align*}
&\overbrace{\left\{ (l_1+p,l_1+p+k):\ 0 \leq p < \mu_X,\ p\in\Gamma_X \right\}}^{\delta_X \ \mbox{elements}}\\
 \cup & \underbrace{\left\{ (l_{1}+ \mu_X - 1,l_1+ \mu_X - 1 +k), (l_1 + \mu_X,l_1+\mu_X+k),\cdots\right\}}_{\text{all elements}}.
\end{align*}

Then, we describe $\Gamma_{k} \setminus \Gamma_{\Theta}$ as
\begin{align*}
\overbrace{\left\{(r,r+k):\ 0 \leq r < \mu_X \right\}}^{s \ \text{elements}}\cup\overbrace{\left\{(\mu_X,\mu_X+k),\cdots, (l_1+\mu_X-2,l_1+\mu_X-2+k)\right\}}^{l_1-1 \ \text{elements}} \\
\end{align*}
\begin{align*}
\setminus \underbrace{\left\{ (l_1+p,l_1+p+k):\ 0 \leq p < \mu_X,\  p\in\Gamma_X \right\}}_{\delta_X \ \mbox{elements}}
\end{align*}

Therefore, $$\mbox{dim}_{\C} \ker(ev) = s +  l_1 - 1 - \delta_X = l_1 - \delta_X + s - 1.$$

Now, we compute the dimension of $\O_X / \langle J_1,J_2 \rangle$.
We consider the sets
$$
\Gamma_{J_1}= \left\{ l_1+\mu_X-1 + i: \ i\in\Gamma_X\right\}
$$
and
$$
\Gamma_{J_2}= \left\{ l_1+\mu_X-1 + k+ j: \ j \in\Gamma_X\right\}
$$
which are subset of  $\Gamma_X$.
We can identify the elements of $\O_X / \langle J_1,J_2\rangle$ with the elements of $\Gamma_{X} \setminus (\Gamma_{J_1} \cup \Gamma_{J_2})$.
Therefore, we compute the number of elements in $\Gamma_{X} \setminus (\Gamma_{J_1} \cup \Gamma_{J_2})$.

We observe that there exists common elements in $\Gamma_{J_1}$ and $\Gamma_{J_2}$, we need to know the elements.
We have that all the elements after $l_1 + \mu_X -1 + k+ \mu_X$ are common elements in $\Gamma_{J_1}$ and $\Gamma_{J_2}$.
Then, we consider the subsets
$$
A=\left \{ l_1 + \mu_X - 1 + i :\ 0\leq i \leq \mu_X + k, i \in \Gamma_X \right \}
$$
and
$$
B=\left \{ l_1 + \mu_X -1 + k + j :\ 0\leq j < \mu_X , j \in \Gamma_X \right \}.
$$
Given $r$ and $r+k$ in $\Gamma_X$ such that $0 \leq r < \mu_X$, the elements common in $A$ and $B$ are the elements of form:
$$l_1 + \mu_X -1 +(r+k) = l_1 + \mu_X - 1 + k + r.$$
Since $s$ is the number of elements $r \in \Gamma_X$, $0 \leq r < \mu_X$, such that $r+k \in \Gamma_X$, then $A\cap B$ contained $s$ common elements.
We have that $A$ contained $\delta_X +  k + 1$ elements and $B$ contained $\delta_X$ elements, thus $A\cup B$ contained $\mu_X + k + 1 - s$ elements.

Thus, $\Gamma_{X} \setminus (\Gamma_{J_1} \cup \Gamma_{J_2})$ is equivalent to
\begin{align*}
\overbrace{\left\{j \in \Gamma_X; 0\leq j < \mu_X\right\}}^{\delta_X \ \mbox{elements}}\cup\overbrace{\left\{\mu_X,\cdots,l_1+\mu_X + \mu_X + k - 1 \right\}}^{l_1 + \mu_X + k  \ \mbox{elements}}
\end{align*}
\begin{align*}
\setminus \underbrace{\left \{ l_1 + \mu_X - 1 + i :\ 0\leq i \leq \mu_X  \right \}\cup \left \{ l_1 + \mu_X -1+ k + j :\ 0\leq j < \mu_X \right \}}_{\mu_X + k + 1 - s  \ \mbox{elements}}.
\end{align*}
Therefore,
$$\mbox{dim}_{\C} \frac{\mathcal{O}_{X,0}}{\langle J_1,J_2 \rangle} = \delta_X + l_1 + \mu_X + k - (\mu_X+ k + 1 - s) = l_1 + \delta_X + s -1.$$
\end{proof}

\begin{cor}\label{corollarydimensions} Let $(X,0)\subset(\C^n,0)$ be an irreducible curve ICIS, weighted homogeneous and let $f:(X,0)\to(\C^2,0)$ be a finite map germ of degree 1 onto its image $(Y,0)$ and $f$ is consistent with $(X,0)$. Then,
$$ \A_{e}\mbox{-}\codim(f) =  \dim_{\C} \frac{\mathcal{O}_{X,0}}{\langle J_1,J_2 \rangle} -\mu_X + \dim_{\C} \frac{J_{g}\mathcal{O}_{X,0}}{J_{g}\mathcal{O}_{Y,0}}.$$
\end{cor}
\begin{proof} By Lemma \ref{sobrejetora},
$$ \A_{e}\mbox{-}\codim(f) =  \dim_{\C} \ker(ev) + \dim_{\C} \frac{J_{g}\mathcal{O}_{X,0}}{J_{g}\mathcal{O}_{Y,0}}.$$

By Theorem \ref{dimensao},
$$\dim_{\C} \frac{\mathcal{O}_{X,0}}{\langle J_1,J_2 \rangle} - \dim_{\C} \ker(ev)= l_{1} + \delta_X + s - 1- \left( l_1 - \delta_X + s -1 \right)  = \mu_X.$$

Therefore,
$$ \A_{e}\mbox{-}\codim(f) = \dim_{\C} \frac{\mathcal{O}_{X,0}}{\langle J_1,J_2 \rangle} -\mu_X  + \dim_{\C} \frac{J_{g}\mathcal{O}_{X,0}}{J_{g}\mathcal{O}_{Y,0}}.$$
\end{proof}

\section{The image Milnor number}

The first and second authors in \cite{DJ}, define the image Milnor number $\mu_I(f)$ of $f:(X,0)\to(\C^2,0)$, where $(X,0)$ is a plane curve and $f$ is $\A$-finite.
In this work, naturally, we generalize the concept of the image Milnor number to the case that $(X,0)$ is a space curve ICIS.
We consider $(X,0)\subset(\C^n,0)$ a curve ICIS and $f:(X,0)\to(\C^2,0)$ be a finite map germ of degree 1 onto its image $(Y,0)$.
We suppose $f$ has finite singularity type, it follows from Theorem \ref{estabilizacao} that $f$ always admits a stabilisation $F: (\mathcal{X},0) \to (\C \times \C^2,0)$, given by $F(s,x)=(s,f_s(x))$. This means that $f_s:X_s\to B_\epsilon$ is stable, for all $s\ne0$ small enough, where $B_\epsilon$ is a small enough ball centered at the origin in $\C^2$.  Thus the image $Y_s=f_s(X_s)$ has the homotopy type of a wedge of 1-spheres.

\begin{defi} The \emph{image Milnor number} $\mu_I(f)$ is
the number of 1-spheres in $Y_s$.
\end{defi}

Analogous to case where $(X,0)$ is a plane curve \cite{DJ}, $\mu_I(f)$ is well-defined, that is, it is independent of the stabilisation and of the representative. The image Milnor number satisfies the equality.
$$\mu_{I}(f) = \frac{\mu(Y,0) + \mu(X,0)}{2}.$$

For the next results, we use the definition of the delta invariant of $f$, $\delta(f)$, which was introduced in \cite{NT} for maps  of degree 1 between curves.

\begin{defi}[\cite{NT}] Let $f:(X,0)\to(Y,0)$ be a holomorphic map of degree 1 between curves $(X,0)$ and $(Y,0)$. The \emph{delta invariant} of $f$ is
$$
\delta(f)=\dim_\C\frac{\O_{X,0}}{f^* \O_{Y,0}}.
$$
\end{defi}

This number also satisfies $\delta(Y,0)=\delta(X,0)+\delta(f)$ when $(X,0)$ and $(Y,0)$ are irreducible curves.
Therefore, when $(X,0)$ and $(Y,0)$ are irreducible curves  $\mu(Y,0)-\mu(X,0)=2\delta(f)$ (see \cite{NT}), thus
$$
\mu_I(f)=\mu(Y,0) - \delta(f)=\mu(X,0)+\delta(f).
$$

Now we consider again, $(X,0)\subset(\C^n,0)$ an irreducible curve ICIS and weighted homogeneous of type $(w_1,\cdots,w_n;d_1,\cdots,d_{n-1})$, and $f:(X,0)\to(\C^2,0)$ a finite map germ of degree 1 onto its image $(Y,0)$, such that $f$ is consistent with $(X,0)$.

We take $\mathcal{C}$ the ideal generated by $\lambda_f$ in $\O_{X,0}$. Then we have the following result.

\begin{prop}\label{delta}
Let $(X,0)\subset(\C^n,0)$ be an irreducible curve ICIS and weighted homogeneous of type $(w_1,\cdots,w_n;d_1,\cdots,d_{n-1})$ and let $f:(X,0)\to(\C^2,0)$ be a finite map germ of degree 1 onto its image $(Y,0)$ and consistent with $(X,0)$.
Then,
$$
\dim_{\C}\frac{\O_{X,0}}{\mathcal{C}} = 2\delta(f) \quad \mbox{and} \quad \dim_{\C}\frac{\O_{Y,0}}{\mathcal{C}} = \delta(f).
$$
Furthermore, we have that $\frac{\mathcal{O}_{X,0}}{\langle J_1,J_2 \rangle}$ is isomorphic to $\frac{\mathcal{C}}{J_g \mathcal{O}_{X,0}}$.
\end{prop}
\begin{proof}
By Lemma \ref{lemmalambda},
$$
\lambda_f=t^{\mu_Y-\mu_X}=t^{2\delta(f)}.
$$
Therefore, the weighted degree of $\lambda_f$ with the weights $w_1,\cdots,w_n$ is $2\delta(f)$.
Denoted by $(h_1,\cdots,h_{n-1}):(\C^n,0)\to (\C^{n-1},0)$ the map that generates $(X,0)$,  by Bezout theorem, we obtain
\begin{align*}
\dim_{\C}\frac{\O_{X,0}}{\mathcal{C}} &= \dim_{\C}\frac{\O_{X,0}}{\langle \lambda_f \rangle} = \dim_{\C}\frac{\O_{2}}{\langle h_1,\cdots,h_{n-1}, \lambda_f \rangle}\\
&=\frac{d_1\cdots d_{n-1}2\delta(f)}{w_1 \cdots w_n}=2\delta(f).
\end{align*}

Therefore,
$$
\dim_{\C}\frac{\O_{X,0}}{\mathcal{C}} = 2\delta(f).
$$

To obtain the second equality, we first show that $\mathcal{C} \subset \O_{Y,0}$. We recall that $\mathcal{C}$ is the ideal generated by $\lambda_f=t^{2\delta(f)}$ in $\O_{X,0}$.
Then, we need to show that $\sigma t^{2\delta(f)} \in \O_{Y,0}$, for all $\sigma \in \O_{X,0}$.
To see this, we consider the numerical semigroups $\Gamma_X$ and $\Gamma_Y$ associated the curves $(X,0)$ and $(Y,0)$ respectively. We recall that $\Gamma_X$ and $\Gamma_Y$ are symmetric.

We show that if $2\delta(f) \in \Gamma_Y$, then $t^{2\delta(f)} \in \O_{Y,0}$.
In fact,  since $\Gamma_Y$ is symmetric we have $2\delta(f) \in \Gamma_Y$ or $\mu_Y-1-2\delta(f) \in \Gamma_Y$.
We suppose $\mu_Y-1-2\delta(f) \in \Gamma_Y$, then
$$
\mu_Y-1-2\delta(f)= \mu_Y - 1 - \mu_Y + \mu_X = \mu_X-1 \in \Gamma_Y \subset \Gamma_X.
$$
But $\mu_X-1 \notin \Gamma_X$, because $\mu_X$ is the conductor of $\Gamma_X$.
Therefore, $2\delta(f) \in \Gamma_Y$, thus $t^{2\delta(f)} \in \O_{Y,0}$.

Now, let $\sigma=t^{a} \in \O_{X,0}$ with $a \in \Gamma_X$, then $2\delta(f)+a \in \Gamma_Y$.
In fact, we have $2\delta(f)+a \in \Gamma_Y$ or $\mu_Y-1-(2\delta(f)+a) \in \Gamma_Y$.
If $\mu_Y-1-(2\delta(f)+a) \in \Gamma_Y$, then
$$
\mu_Y-1-\mu_Y+\mu_X- a= \mu_X-1-a \in \Gamma_Y \subset \Gamma_X.
$$
Since $\Gamma_X$ is symmetric, we have $a \notin \Gamma_X$, but $\sigma=t^{a} \in \O_{X,0}$.
Thus, $\mu_Y-1-(2\delta(f)+a) \notin \Gamma_Y$ and  consequently $2\delta(f)+a \in \Gamma_Y$.
Therefore, $\mathcal{C} \subset \O_{Y,0}$.
We obtain,
$$
\dim_{\C}\frac{\O_{Y,0}}{\mathcal{C}}=\dim_{\C}\frac{\O_{X,0}}{\mathcal{C}} - \dim_\C\frac{\O_{X,0}}{f^* \O_{Y,0}} = 2\delta(f) -\delta(f) = \delta(f).
$$

To complete the proof, we observe that multiplication by $\lambda_f$ gives an isomorphism $\phi:\O_{X,0}\to\mathcal C$. Moreover, $\phi(\langle J_1,J_2 \rangle)=J_g \mathcal{O}_{X,0}$ and hence, it induces an isomorphism
$$\frac{\mathcal{O}_{X,0}}{\langle J_1,J_2 \rangle} \longrightarrow \frac{\mathcal{C}}{J_g \mathcal{O}_{X,0}}.$$
\end{proof}

\begin{teo}\label{teoprincipal}
Let $(X,0)\subset(\C^n,0)$ be an irreducible weighted homogeneous curve ICIS and let $f:(X,0)\to(\C^2,0)$ be a finite map germ of degree 1 onto its image $(Y,0)$ and consistent with $(X,0)$.
Then,
$$\A_{e}\mbox{-}\codim(X,f) = \mu_{I}(f).$$
Moreover, let $\alpha:(\C,0)\to(X,0)$ be a parametrisation of $(X,0)$. Then,
$$
\mu_{I}(f)=\delta(X,0)+\mu_{I}(f\circ\alpha)
$$
and
$$
\A_e\mbox{-}\codim(f)=\A_e\mbox{-}\codim(f\circ\alpha)-\frac{1}{n-1}\A_e\mbox{-}\codim(\alpha)
$$
\end{teo}
\begin{proof}
By Corollary \ref{corollarydimensions} and by Proposition \ref{delta}, we obtain
\begin{align*}
\A_{e}\mbox{-}\codim(f) &= \dim_{\C} \frac{\mathcal{O}_{X,0}}{\langle J_1,J_2 \rangle}  -  \mu_X + \dim_{\C} \frac{J_{g}\mathcal{O}_{X,0}}{J_{g}\mathcal{O}_{Y,0}}\\
&= \dim \frac{\mathcal{C}}{J_g \mathcal{O}_{X,0}} - \mu_X + \dim_{\C} \frac{J_{g}\mathcal{O}_{X,0}}{J_{g}\mathcal{O}_{Y,0}}.
\end{align*}

We have the exact sequence,
$$ 0 \longrightarrow \frac{J_{g}\mathcal{O}_{X,0}}{J_{g}\mathcal{O}_{Y,0}} \longrightarrow \frac{\mathcal{C}}{J_g \mathcal{O}_{Y,0}}\longrightarrow \frac{\mathcal{C}}{J_g \mathcal{O}_{X,0}} \longrightarrow 0,$$
thus, by the exact sequence and by Proposition \ref{delta},
\begin{align*}
\A_{e}\mbox{-}\codim(f) &=\dim_{\C} \frac{\mathcal{C}}{J_g \mathcal{O}_{Y,0}} - \mu_X\\
&=\dim_{\C} \frac{\O_{Y,0}}{J_g \mathcal{O}_{Y,0}}  - \dim_{\C} \frac{\O_{Y,0}}{\mathcal{C}} - \mu_X \\
&= \mu_Y - \delta(f) - \mu_X = \delta(f).
\end{align*}

Since $(X,0)$ is weighted homogeneous, we have $\mu_X=\tau(X,0)$, hence
$$\A_{e}\mbox{-}\codim(X,f) = \A_{e}\mbox{-}\codim(f) + \mu_X =  \delta(f) + \mu_X = \mu_I(f).$$

To prove the second part, we observe that $f\circ\alpha:(\C,0)\to(Y,0)$ is a parametrisation of $(Y,0)\subset(\C^2,0)$, then by \cite{M2},
$$
\A_e\mbox{-}\codim(f\circ\alpha)=\mu_{I}(f\circ\alpha)=\delta(Y,0).
$$
By \cite{NT}, $\delta(Y,0)=\delta(X,0)+\delta(f)$, and by definition of image Milnor number, we obtain
\begin{align*}
\mu_I(f)&=\mu(X,0)+\delta(f)=\mu(X,0)+\delta(Y,0)-\delta(X,0)\\
&=\delta(X,0)+\delta(Y,0)=\delta(X,0)+\mu_I(f\circ\alpha).
\end{align*}

We also have,
\begin{align*}
\A_e\mbox{-}\codim(f)&= \delta(f)= \delta(Y,0) - \delta(X,0) \\
&=\A_e\mbox{-}\codim(f\circ\alpha)-\delta(X,0).
\end{align*}
It follows from \cite{HHR} that  $\A_e\mbox{-}\codim(\alpha)=(n-1) \delta(X,0)$, thus
\begin{align*}
\A_e\mbox{-}\codim(f)&= \A_e\mbox{-}\codim(f\circ\alpha)-\frac{1}{n-1}\A_e\mbox{-}\codim(\alpha).
\end{align*}
\end{proof}

We observe that, in particular, if $(X,0)$ is a plane curve, by \cite{M2}, $\mu_I(\alpha)=\delta(X,0)$ and we obtain
$$
\mu_{I}(f)=\mu_{I}(\alpha)+\mu_{I}(f\circ\alpha).
$$

\begin{exam}\label{example1}
We consider $(X,0)$ a weighted homogeneous  curve defined by $h(x,y) = x^{2}-y^{3}$  and $f:(X,0) \to (\C^{2},0)$ a map defined by $f(x,y)=(x,y^2)$.
Then, $(Y,0)$ is weighted homogeneous plane curve defined by $g(u,v) = u^{4}-v^{3}$.

We have, $\mu(X,0) =2$, $\mu(Y,0)=6$, then
$$\mu_{I}(f)=(\mu(Y,0) + \mu(X,0))/2 = (6 + 2)/2 = 4.$$
By \cite{MM},
$$ \A_{e}\mbox{-}\codim(X,f)=\dim_{\C} \frac{\Theta(\gamma)}{t\gamma(\Theta_{2})+\gamma^{*}(\Derlog D(G))},
$$
where $G:\C^{3} \to \C^4$ defined by $G(x,y,s)=(x,y^2,x^2-y^3+sy,s)$ is a stability of map $(h,\bar{f}): \C^2 \to \C^3$, given by $(h,\bar{f})(x,y)=(x^2-y^3,x,y^2)$.
We observe that, $D(G)=Im(G)$, because $G$ is no surjective, then $\Derlog D(G)=\Derlog \mbox{Im}(G)$.
Hence, to compute $ \A_{e}\mbox{-}\codim(X,f)$, we find $H: \C^4 \to \C$ such that $\mbox{Im}(G)=H^{-1}(0)$. We do this using the Singular software,
$$H(z,u,v,w)= z^2-v^3-2u^2z+2v^2w-vw^2+u^4.$$
We use the map $H$  to compute $\Derlog \mbox{Im}(G)$ and then we can calculate
$$\dim_{\C} \frac{\Theta(\gamma)}{t\gamma(\Theta_{2})+\gamma^{*}(\Derlog \mbox{Im}(G))}.$$
Therefore,
$$ \A_{e}\mbox{-}\codim(X,f)=\dim_{\C} \frac{\Theta(\gamma)}{t\gamma(\Theta_{2})+\gamma^{*}(\Derlog \mbox{Im}(G))} = 4 = \mu_{I}(f). $$
and
$$\A_{e}\mbox{-}\codim(f)= \A_{e}\mbox{-}\codim(X,f)-\mu(X,0)=4-2=2=\delta(f).$$

Now, we consider $\alpha:(\C,0)\to(X,0)$, given by $\alpha(t)=(t^3,t^2)$, the parametrisation of $(X,0)$. Then,
$\A_{e}\mbox{-}\codim(\alpha)=\mu_I(\alpha)=\delta(X,0)=1$ and $\A_{e}\mbox{-}\codim(f\circ\alpha)=\mu_I(f\circ\alpha)=\delta(Y,0)=3$.
Therefore,
$$
\mu_I(f)=\mu_I(\alpha)+\mu_I(f\circ\alpha).
$$
and
$$
\A_{e}\mbox{-}\codim(f)= \A_{e}\mbox{-}\codim(f\circ\alpha)- \A_{e}\mbox{-}\codim(\alpha).
$$
\end{exam}

\begin{exam}
We consider $(X,0)\subset (\C^3,0)$ a weighted homogeneous curve defined by $h(x,y,z) = (x^{3}-y^{2},xy-z)$ with parametrization $\alpha:(\C,0)\to (\C^3,0)$ defined by $\alpha(t)=(t^2,t^3,t^5)$  and $f:(X,0) \to (\C^{2},0)$ a map defined by $f(x,y,z)=(x,y^2z)$, thus $f$ is consistent with $(X,0)$. Then, $(Y,0)$ is a weighted homogeneous plane curve defined by $g(u,v) = u^{11}-v^{2}$ with parametrization $\beta:(\C,0)\to (\C^2,0)$ defined by $\beta(t)=(t^2,t^{11})$.

We use the Singular software to calculate the Milnor number, we have $\mu(X,0)=2$ and $\mu(Y,0)=10$.
Thus, $\mu_{I}(f)=(\mu(Y,0) + \mu(X,0)) / 2 = (10 + 2) / 2 = 6$.

By \cite{MM},
$$ \A_{e}\mbox{-}\codim(X,f)=\dim_{\C} \frac{\Theta(\gamma)}{t\gamma(\Theta_{2})+\gamma^{*}(\Derlog D(G))},
$$
where $G:\C^{4} \to \C^5$ defined by $G(x,y,z,s)=(x^3-y^2,xy-z,x,y^2z+sy,s)$ is a stability of map $(h,\bar{f}): \C^3 \to \C^4$, given by $(h,\bar{f})(x,y,z)=(x^3-y^2,xy-z,x,y^2z)$.
As in example \ref{example1},  we use the Singular software, thus,
$$ \A_{e}\mbox{-}\codim(X,f)=\dim_{\C} \frac{\Theta(\gamma)}{t\gamma(\Theta_{2})+\gamma^{*}(\Derlog \mbox{Im}(G))} = 6=\mu_{I}(f).$$
\end{exam}

\begin{exam}
We consider $(X,0)\subset (\C^3,0)$ a weighted homogeneous curve defined by $h(x,y,z) = (x^{5}-y^{2},xy-z)$ with parametrization $\alpha:(\C,0)\to (\C^3,0)$ defined by $\alpha(t)=(t^2,t^5,t^7)$  and $f:(X,0) \to (\C^{2},0)$ a map defined by $f(x,y,z)=(x^2,y+z)$. We observe that $f$ is not consistent with $(X,0)$. Then, $(Y,0)$ is a plane curve defined by $g(u,v) = u^{7}-2u^6 +4u^3 v^2 + u^5-v^{4}$ with parametrization $\beta:(\C,0)\to (\C^2,0)$ defined by $\beta(t)=(t^4,t^5 + t^7)$.

We use the Singular software to calculate the Milnor number and the Tjurina number, we have $\mu(X,0)=4$, $\tau(X,0)=4$, $\mu(Y,0)=12$ and $\tau(Y,0)=11$.
Thus, $(Y,0)$ is not weighted homogeneous. We also have, $\mu_{I}(f)=(\mu(Y,0) + \mu(X,0)) / 2 = (12 + 4) / 2 = 8$.

By \cite{MM},
$$ \A_{e}\mbox{-}\codim(X,f)=\dim_{\C} \frac{\Theta(\gamma)}{t\gamma(\Theta_{2})+\gamma^{*}(\Derlog D(G))},
$$
where $G:\C^{4} \to \C^5$ defined by $G(x,y,z,s)=(x^5-y^2+sx,xy-z,x^2,y+z,s)$ is a stability of map $(h,\bar{f}): \C^3 \to \C^4$, given by $(h,\bar{f})(x,y,z)=(x^5-y^2,xy-z,x^2,y+z)$.
As in example \ref{example1},  we use the Singular software, thus,
$$ \A_{e}\mbox{-}\codim(X,f)=\dim_{\C} \frac{\Theta(\gamma)}{t\gamma(\Theta_{2})+\gamma^{*}(\Derlog \mbox{Im}(G))} = 7.$$
Therefore,
$$ \A_{e}\mbox{-}\codim(X,f)<\mu_{I}(f).$$
We observe that the hypotheses $f$ consistent with of $(X,0)$ is necessary to obtain an equality.
\end{exam}

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