Let $\mathcal{H}_1$ and $\mathcal{H}_2$ be two Hilbert spaces such that $\dim\mathcal{H}_2<\infty$, and let $F:\mathcal{H }_1\rightarrow\mathcal{H}_2$ be a Fréchet differentiable surjective map. Consider $x_1,x_2\in\mathcal{H}_1$ and suppose that $F(x_1)=F(x_2)$.
Question/Conjecture: Do we have $\mathrm{im}\left(\mathrm{d}F(x_1)\right)=\mathrm{im}\left(\mathrm{d}F(x_2)\right)$ ?
Partial answer: If $x_1$ and $x_2$ are regular, that is the differentials $\mathrm{d}F(x_1)$ and $\mathrm{d}F(x_2)$ are surjective, then the conjecture is obviously true. But I don't know what can happen if $x_1$ and $x_2$ are both singular (the differentials are not surjective), I don't even know if it is possible to have $x_1$ regular and $x_2$ singular?
While redacting the question, I realised that the situation $x_1$ regular and $x_2$ singular can occur. Thus, the conjecture is false. Indeed, take $F:\mathbb{R}\rightarrow\mathbb{R}$, and $x_1,x_2$ such that $F'(x_1)\neq0$ and $F'(x_2)=0$. Then, $\mathrm{im}(\mathrm{d}F(x_1))=\mathbb{R}$ and $\mathrm{im}(\mathrm{d}F(x_2))=0$.