[proofplan]
The only obstruction to invertibility of the semi-discrete mass map is the gauge symmetry obtained by adding the same constant to all weights. We remove this symmetry by restricting the weights to the zero-sum hyperplane $H$, and we view the masses in the affine hyperplane $A$. The assumed Jacobian formula identifies the derivative at $w$ with a connected weighted graph Laplacian, whose kernel consists exactly of constant vectors; hence its restriction to $H$ is an isomorphism. The finite-dimensional inverse function theorem then gives local invertibility, and the positivity condition follows by restricting the target neighbourhood to the open simplex.
[/proofplan]
[step:Record that the mass map takes values in the affine mass hyperplane]
For each $z\in\mathbb R^N$, the sets $P_1(z),\dots,P_N(z)$ cover $\Omega$. If $i\ne j$, then $P_i(z)\cap P_j(z)$ is contained in the affine hyperplane
\begin{align*}
\{x\in\mathbb R^n:2x\cdot(y_j-y_i)=|y_j|^2-|y_i|^2+z_j-z_i\},
\end{align*}
unless the hyperplane equation is empty. Since $y_i\ne y_j$, this affine hyperplane has $\mathcal L^n$-measure zero. Because $\mu_0$ is absolutely continuous with respect to $\mathcal L^n\!\restriction_\Omega$, every pairwise overlap has $\mu_0$-measure zero. Therefore finite additivity gives
\begin{align*}
\sum_{i=1}^N \mu_0(P_i(z))=\mu_0(\Omega)=1.
\end{align*}
Thus $\mathcal M(z)\in A$ for every $z\in\mathbb R^N$, and in particular $m=\mathcal M(w)\in A$.
[guided]
We first verify that the target space for the local inverse is really the affine hyperplane $A$, not all of $\mathbb R^N$. Fix $z=(z_1,\dots,z_N)\in\mathbb R^N$. By definition, every point $x\in\Omega$ has at least one index $i$ at which the finite list
\begin{align*}
|x-y_1|^2-z_1,\dots,|x-y_N|^2-z_N
\end{align*}
attains its minimum. Hence $x\in P_i(z)$ for at least one $i$, so the cells cover $\Omega$.
The only possible problem is double-counting on cell boundaries. If $x\in P_i(z)\cap P_j(z)$ with $i\ne j$, then the two defining quadratic functions agree at $x$, so
\begin{align*}
|x-y_i|^2-z_i=|x-y_j|^2-z_j.
\end{align*}
Expanding the squares and cancelling $|x|^2$ gives
\begin{align*}
2x\cdot(y_j-y_i)=|y_j|^2-|y_i|^2+z_j-z_i.
\end{align*}
Because the sites are distinct, $y_j-y_i\ne 0$, so this is the equation of an affine hyperplane in $\mathbb R^n$. Such a hyperplane has zero $\mathcal L^n$-measure. Since $\mu_0$ is defined by integration of the density $\rho_0$ against $\mathcal L^n\!\restriction_\Omega$, it is absolutely continuous with respect to $\mathcal L^n\!\restriction_\Omega$, and therefore each overlap $P_i(z)\cap P_j(z)$ has zero $\mu_0$-measure.
The cells are finitely many measurable polyhedral pieces of $\Omega$, they cover $\Omega$, and their pairwise overlaps have zero $\mu_0$-measure. Finite additivity therefore yields
\begin{align*}
\sum_{i=1}^N \mu_0(P_i(z))=\mu_0(\Omega)=1.
\end{align*}
Thus $\mathcal M(z)\in A$ for every $z\in\mathbb R^N$. This is why, after differentiating, the derivative of the restricted mass map lands in the tangent space of $A$, namely the zero-sum hyperplane $H$.
[/guided]
[/step]
[step:Remove the constant-weight gauge by restricting to $H$]
Let $\mathbf 1\in\mathbb R^N$ denote the vector
\begin{align*}
\mathbf 1=(1,\dots,1).
\end{align*}
For every $z\in\mathbb R^N$ and every $t\in\mathbb R$, the equality
\begin{align*}
P_i(z+t\mathbf 1)=P_i(z)
\end{align*}
holds for every $i\in\{1,\dots,N\}$, because subtracting the same constant $t$ from every expression $|x-y_i|^2-z_i$ does not change the set of minimizers. Hence
\begin{align*}
\mathcal M(z+t\mathbf 1)=\mathcal M(z).
\end{align*}
The affine hyperplane $w+H$ equals $H$ because $w\in H$. We therefore consider the restricted map
\begin{align*}
F:W\cap H&\to A
\end{align*}
\begin{align*}
z&\mapsto \mathcal M(z).
\end{align*}
Since $\mathcal M|_W$ is $C^1$ and $H,A$ are finite-dimensional affine submanifolds of $\mathbb R^N$, the map $F$ is $C^1$. Its derivative at $w$ is
\begin{align*}
DF_w:H&\to H
\end{align*}
\begin{align*}
h&\mapsto J\mathcal M_w h,
\end{align*}
where the target is $H=T_mA$.
[/step]
[step:Show that the restricted Jacobian is an isomorphism]
Let $L:\mathbb R^N\to\mathbb R^N$ be the linear map represented by the matrix $J\mathcal M_w$. By hypothesis, $L$ is the weighted graph Laplacian associated to the connected graph $G_w$ and the symmetric weights $a_{ij}$. Thus, for every $\xi=(\xi_1,\dots,\xi_N)\in\mathbb R^N$,
\begin{align*}
\xi\cdot L\xi=\sum_{1\le i<j\le N}a_{ij}(\xi_i-\xi_j)^2.
\end{align*}
Every summand on the right-hand side is nonnegative. If $\xi\in\ker L$, then
\begin{align*}
0=\xi\cdot L\xi=\sum_{1\le i<j\le N}a_{ij}(\xi_i-\xi_j)^2,
\end{align*}
so $\xi_i=\xi_j$ for every edge $\{i,j\}$ of $G_w$. Since $G_w$ is connected, $\xi_1=\cdots=\xi_N$. Hence
\begin{align*}
\ker L=\operatorname{span}\{\mathbf 1\}.
\end{align*}
If $\xi\in H\cap\ker L$, then $\xi=\alpha\mathbf 1$ for some $\alpha\in\mathbb R$ and
\begin{align*}
0=\sum_{i=1}^N \xi_i=N\alpha,
\end{align*}
so $\alpha=0$. Thus $L|_H$ is injective. Since $H$ is a finite-dimensional real vector space and $L(H)\subset H$, injectivity of $L|_H:H\to H$ implies bijectivity. Consequently
\begin{align*}
DF_w=L|_H:H\to H
\end{align*}
is a linear isomorphism.
[guided]
The main point is to prove that the only infinitesimal direction invisible to the mass map is the constant-weight direction, and that this direction has already been removed by imposing the zero-sum condition.
Let $L:\mathbb R^N\to\mathbb R^N$ denote the linear map whose matrix is $J\mathcal M_w$. By the hypothesis on the Jacobian, the entries of $L$ have the weighted graph Laplacian form
\begin{align*}
L_{ii}=\sum_{j\ne i}a_{ij}
\end{align*}
and
\begin{align*}
L_{ij}=-a_{ij}
\end{align*}
for $i\ne j$, where $a_{ij}=a_{ji}\ge 0$ and $a_{ij}>0$ exactly when $\{i,j\}$ is an edge of $G_w$.
We compute the quadratic form. For $\xi=(\xi_1,\dots,\xi_N)\in\mathbb R^N$, the diagonal part contributes
\begin{align*}
\sum_{i=1}^N \xi_i^2\sum_{j\ne i}a_{ij},
\end{align*}
and the off-diagonal part contributes
\begin{align*}
-\sum_{\substack{1\le i, j\le N, i\ne j}}a_{ij}\xi_i\xi_j.
\end{align*}
Using the symmetry $a_{ij}=a_{ji}$ and grouping the terms indexed by each unordered pair $\{i,j\}$ gives
\begin{align*}
\xi\cdot L\xi=\sum_{1\le i<j\le N}a_{ij}(\xi_i-\xi_j)^2.
\end{align*}
This identity is the weighted graph Laplacian energy identity. It shows at once that $L$ is positive semidefinite.
Now suppose $\xi\in\ker L$. Then $L\xi=0$, so $\xi\cdot L\xi=0$. By the energy identity,
\begin{align*}
0=\sum_{1\le i<j\le N}a_{ij}(\xi_i-\xi_j)^2.
\end{align*}
Each summand is nonnegative, so every summand with $a_{ij}>0$ must vanish. Therefore $\xi_i=\xi_j$ whenever $\{i,j\}$ is an edge of $G_w$. Since $G_w$ is connected, any two vertices can be joined by an edge path. Repeating the equality along such a path gives $\xi_i=\xi_j$ for all $i,j\in\{1,\dots,N\}$. Hence $\xi$ is a constant vector, meaning
\begin{align*}
\xi\in\operatorname{span}\{\mathbf 1\}.
\end{align*}
Conversely, $L\mathbf 1=0$ because every row sum of a graph Laplacian is zero. Thus
\begin{align*}
\ker L=\operatorname{span}\{\mathbf 1\}.
\end{align*}
Finally restrict to $H$. If $\xi\in H\cap\ker L$, then $\xi=\alpha\mathbf 1$ for some $\alpha\in\mathbb R$, and the condition $\xi\in H$ gives
\begin{align*}
0=\sum_{i=1}^N \xi_i=\sum_{i=1}^N \alpha=N\alpha.
\end{align*}
Because $N\ge 2$, in particular $N\ne 0$, we get $\alpha=0$. Therefore the restricted map $L|_H:H\to H$ is injective. The space $H$ has finite dimension $N-1$, so an injective linear map from $H$ to itself is also surjective. Hence $DF_w=L|_H$ is a linear isomorphism.
[/guided]
[/step]
[step:Apply the inverse function theorem on the affine hyperplanes]
The finite-dimensional inverse function theorem applies to the $C^1$ map
\begin{align*}
F:W\cap H\to A
\end{align*}
at the point $w$, because
\begin{align*}
DF_w:H\to H
\end{align*}
is a linear isomorphism. Therefore there exist neighbourhoods $U\subset W\cap H$ of $w$ and $V\subset A$ of $m=F(w)$ such that
\begin{align*}
F|_U:U\to V
\end{align*}
is bijective and has a $C^1$ inverse
\begin{align*}
(F|_U)^{-1}:V\to U.
\end{align*}
Since $F=\mathcal M|_{W\cap H}$, this says that
\begin{align*}
\mathcal M|_U:U\to V
\end{align*}
is bijective and has a continuous local inverse.
[/step]
[step:Restrict the local inverse to nearby positive mass vectors]
The set $\Delta_N^\circ$ is open in the relative topology of $A$, because it is the intersection of $A$ with the open set
\begin{align*}
\{q\in\mathbb R^N:q_i>0\text{ for every }i\in\{1,\dots,N\}\}.
\end{align*}
For every $m'\in V\cap\Delta_N^\circ$, the bijectivity of
\begin{align*}
\mathcal M|_U:U\to V
\end{align*}
gives a unique $w'\in U$ such that
\begin{align*}
\mathcal M(w')=m'.
\end{align*}
Because $U\subset H$, this weight is normalized:
\begin{align*}
\sum_{i=1}^N w'_i=0.
\end{align*}
This proves the stated local stability and uniqueness of normalized semi-discrete weights.
[/step]
