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Consider a continuously differentiable function $f:\mathbb{R}^n\rightarrow\mathbb{R}^n$. Fix $y\in\mathbb{R}^n$. Assume $\{u,v\}=f^{-1}(\{y\})$. If Jacobian is positive at both $u$ and $v$, then can we draw out a contradiction?

Moonshine
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No. (Provided I understand your question , which could use more detail.) Take $\mathbb{C}$ and $f(z) = z^2$, and $y = 1$. (Identify with $\mathbb{R}^2$ appropriately.)

Elle Najt
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