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I've been trying to solve this exercise: If $f:\mathbb R^3\rightarrow \mathbb R^2$ is continuously differentiable, then $f$ is not injectve.

I was thinking in use the implicit function theorem, in the sense that if I can see that the matrix $$df(a)=\begin{bmatrix}\frac{df}{dx}&\frac{df}{dy} \end{bmatrix},$$ where $y=(y_1,y_2)$, has $\frac{df}{dy}$ non singular, it follows that exist a $g:\mathbb R\rightarrow \mathbb R$ continuously differentiable, and that $f(x,g(x))=c$ for all $x\in U\subset\mathbb R$, for some open set $U$, therefore $f$ is not injective.

The problem is that I do not know how to show the existence of such $a$.

Thanks in advance.

ett
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    See https://math.stackexchange.com/questions/159446/why-isnt-there-a-continuously-differentiable-injection-into-a-lower-dimensional? – Robert Z Jul 13 '17 at 14:05
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    There is no guarantee that the derivative of $f$ can be rank $2$ at some point, but if the rank cannot be $2$, then it should be "easier" to show that $f$ is not injective. For an informal argument for this, observe that a function having derivative of rank $0$ everywhere is constant and hence not injective. – edm Jul 13 '17 at 14:20
  • Do you have to stick to using implicit function theorem? It seems easier to prove using inverse function theorem (with the help of a hint given in a textbook). – edm Jul 13 '17 at 14:43

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