Curve Selection Conjecture I have the following conjecture which I cannot seem to settle either way:
Let $f:[0,1]\to\mathbb R^2$ be a differentiable function such that $f(0)=(0,0)$.
Then there exists a continuous function $g:[0,1]\to\mathbb R^2$ such that:
1) $g(0)=(0,0)$
2) $g([0,1])\cap f([0,1])=\{(0,0)\}$.
3) $g$ is not a constant function. (Thanks to user68061 for pointing this out.)
Basically what I am trying to prove is that if we have a differentiable curve in $\mathbb R^2$ which passes through origin then we can find a continuous curve in $\mathbb R^2$ which intersects the given curve only at origin.
Does anybody know if this is an already known result or if there exists a counterexample?
Thanks in advance for your help.
 A: First define $F_0: [0,1] \rightarrow \mathbb{R}^2$ so that $F_0(0)=(1,0)$ and for the first half it goes around a unit circle, and then for the second half it goes along the line from $(1,0)$ to $(1/2,0)$ making sure to have the derivative vanish exponentially fast at the beginning, middle, and end to be safe.
Next define $F_1: [0,1] \rightarrow \mathbb{R}^2$ by $F_1(x)=\frac{1}{2^{n-1}}F_0(2^n(x-\frac{2^{n-1}-1}{2^{n-1}}))$ on $[\frac{2^{n-1}-1}{2^{n-1}},\frac{2^{n}-1}{2^{n}}]$ and $F(1)=(0,0)$
I'm not entirely sure I got the indices right there, but the point is taking $f(x)=F_1(1-x)$ we should have the image be a line segment from (0,0) to (1,0) along with circles of radius $\frac{1}{2^n}$.
Then the image of $g$ can't contain any points other than 0 without crossing $f$ by the Jordan curve theorem.
A: A counterexample is
$$
f(t)=t^2\Bigl(\sin\Bigl(\frac1t\Bigr),\cos\Bigl(\frac1t\Bigr)\Bigr),\quad t\ne0,\qquad f(0)=(0,0).
$$
Edit
This is a counterexample if $g'(0)\ne(0,0)$. See Dejan Govc's comment.
