This question is related to this one A question about the inverse theorem function in $\mathbb{R}^n$. In the response I received, I was told that the issue was related to the singularities, as I do not have a strong background in this area, I'll ask the question again stating exactly the situation.

Let $A\in\mathbb{R}^{n\times n}$ and $\eta\in\mathbb{R}^n$ such that the familly $(A^{n-1}\eta,\ldots,\eta)$ is a basis of $\mathbb{R}^n$. We define the mapping \begin{equation*} \begin{matrix} G_{\eta} \quad : \quad&\mathbb{R}^{n}&\rightarrow&\mathbb{R}^{n}\\ & (t_{1},\cdots,t_{n}) &\mapsto &\prod_{i=1}^{n}(A+t_{i}I)\eta \end{matrix}. \end{equation*}

$G_{\eta}$ is continuously differentiable and one can readily see that the jacobian determinant of $G_{\eta}$ is given by $$\mid\frac{\partial G_{\eta}(t_{1},\ldots,t_{n})}{\partial(t_{1},\ldots,t_{n})}|= \mathrm{det}(A^{n-1}\eta,\quad A^{n-2}\eta,\quad\ldots\quad ,\eta)V(t_{1},\cdots,t_{n}),$$ where $V(t_{1},\cdots,t_{n})=\displaystyle\prod_{1\leq i<j\leq n}(t_i-t_j)$.

Then for $(t_{1},\cdots,t_{n})\in\mathbb{R}^{n}$ such that $V(t_{1},\cdots,t_{n})\neq0$, we have from the Inverse theorem function the existence of a neighborhood V of $(t_{1},\cdots,t_{n})$ in $\mathbb{R}^{n}$ such that $G_{\eta}(V)$ is a neighborhood of $G_{\eta}(t_{1},\cdots,t_{n})$ in $\mathbb{R}^{n}$. I wonder if this reamain true even for $(t_{1},\cdots,t_{n})\in\mathbb{R}^{n}$ such that $V(t_{1},\cdots,t_{n})=0$.


Near any point where $V$ has a zero of order 1, i.e. a point where precisely two of the $t_i$ are equal, the map looks like a fold singularity, so doesn't have any such neighborhood. It looks like $(x,y,\dots) \mapsto (x^2,y,\dots)$. I think you can find everything you need in this nice introduction


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