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If $f\in \mathcal{C}^1(\mathbb{R}^2)$. Prove that there exists a continuous one-one function $g:[0,1]\to \mathbb{R}^2$ such that $f\circ g:[0,1]\to \mathbb{R}^2 $ is constant.

Would anybody give me some hint to do this problem?

I am pi
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  • I don't see how it is possible. For instance $f(x,y)=x$ then $f\circ g$ constant means $g(t)=(x_0,y(t))$ but then $x_0$ is certainly not one-to-one with $\mathbb R$. – zwim Dec 14 '17 at 03:45
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    There should exist contour lines on the map of the surface of $f$, some of which are locally homeomorphic to $[0,1]$. – kimchi lover Dec 14 '17 at 03:47
  • @zwim: Look again. The function $g(t) = (x_0, t)$ is one-to-one. – Nate Eldredge Dec 14 '17 at 04:06
  • @NateEldredge it is injective but one-to-one only on its image. when you say one-to-one from [0,1] to R^2 it is implied R^2 is the image. Anyway, OP wording is imprecise. – zwim Dec 14 '17 at 04:08
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    @zwim: In all the books I have read, "one-to-one" is a synonym of "injective", and there is nothing imprecise about it. You might argue the literal meaning suggests it should mean "bijective" but it just doesn't. – Nate Eldredge Dec 14 '17 at 04:10

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