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 Apr 12 awarded Popular Question Apr 12 awarded Popular Question Dec 28 awarded Necromancer Jun 28 awarded Yearling Feb 18 awarded Popular Question Dec 10 awarded Caucus Dec 5 awarded Tumbleweed Nov 29 accepted If $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ smooth, $g(x,y)= x^3 + y^3$ and $g \circ f \equiv 0$, then $\det Df \equiv 0$ Nov 29 comment If $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ smooth, $g(x,y)= x^3 + y^3$ and $g \circ f \equiv 0$, then $\det Df \equiv 0$ Yes, you are right. Then I think it should be like this: $D(g(f)=Dg(f)Df = \nabla g (f) Df = \begin{bmatrix} 3f_1^2 & 3f_2^2 \end{bmatrix} \begin{bmatrix} \nabla f_1 \\ \nabla f_2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \end{bmatrix}$ which implies that $(Df)^T \begin{bmatrix} 3f_1^2 \\ 3f_2^2 \end{bmatrix} = 0$. Then either $f_1=f_2=0$ or $(Df)^T$ is singular. Nov 29 revised If $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ smooth, $g(x,y)= x^3 + y^3$ and $g \circ f \equiv 0$, then $\det Df \equiv 0$ edited title Nov 29 asked If $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ smooth, $g(x,y)= x^3 + y^3$ and $g \circ f \equiv 0$, then $\det Df \equiv 0$ Nov 29 accepted If $f: U \rightarrow \mathbb{R}^n$ differentiable such that $|f(x)-f(y)| \geq c |x-y|$ for all $x,y \in U$, then $\det \mathbf{J}_f(x) \neq 0$ Nov 29 accepted $f: \mathbb{R}^2-\{0\} \rightarrow \mathbb{R}$ is continuously differentiable and $f(\alpha x) = \alpha^2f(x)$, then $x \cdot \nabla f(x) = 2 f(x)$ Nov 29 asked $f: \mathbb{R}^2-\{0\} \rightarrow \mathbb{R}$ is continuously differentiable and $f(\alpha x) = \alpha^2f(x)$, then $x \cdot \nabla f(x) = 2 f(x)$ Nov 29 asked If $f: U \rightarrow \mathbb{R}^n$ differentiable such that $|f(x)-f(y)| \geq c |x-y|$ for all $x,y \in U$, then $\det \mathbf{J}_f(x) \neq 0$ Nov 28 accepted If $||\mathbf{J}_f(x) - I_n|| < \frac{1}{2n}$ for all $x \in \mathbb{R^n}$, then $f$ is a diffeomorphism Nov 28 comment If $||\mathbf{J}_f(x) - I_n|| < \frac{1}{2n}$ for all $x \in \mathbb{R^n}$, then $f$ is a diffeomorphism @RobertLewis Yes, that is right. Nov 28 comment If $||\mathbf{J}_f(x) - I_n|| < \frac{1}{2n}$ for all $x \in \mathbb{R^n}$, then $f$ is a diffeomorphism @RobertLewis I cannot prove that it is a norm and I could not find a proof on the internet. Nov 28 asked If $a \frac{\partial}{\partial x} (f+g) = \sin(f-g)$ and $f= f_0 + af_1 + a^2 f_2 + a^3 f_3+…$, then finding $f_0, f_1, f_2$ and $f_3$ Nov 28 asked If $||\mathbf{J}_f(x) - I_n|| < \frac{1}{2n}$ for all $x \in \mathbb{R^n}$, then $f$ is a diffeomorphism