[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 $m=\mathcal M(w)\in A$.[/step]

[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: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.[/step]

[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: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]