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Let $f:I\to \mathbb R^2$, defined by $f(t)=(x(t),y(t))$ a differentiable path. I would like to prove the following implication

$$\text{If $x'(t)\neq 0$ for every $t\in I$, then $f(I)$ is a graph of a differentiable function $\xi:J\to \mathbb R$}$$

I'm trying to use the immersion theorem which is the following:

Theorem: Let $U\subset \mathbb R^m$ be an open set. If $f:U\to \mathbb R^{m+n}$ is an immersion, then there is a diffeomorphism $h$ such that $h\circ f=i$, where $i$ is the canonical inclusion.

We know that $f'(x):\mathbb R\to\mathbb R^2$ is injective for every $x$, since $f'(x)$ is non-zero for every $x\in U$ (due to the fact $x'(t)\neq 0$). Then we know there are open sets $U\in \mathbb R^2$ and $W\subset \mathbb R\times \mathbb R$ and a diffeomorphism $h:U\to W$ such that $h\circ f(x)=(x,0)$.

I'm almost there, I need help to finish this proof.

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    So what part are you stuck on? Now that you have $h \circ f(x) = (x,0)$, are you just trying to identify what the function $\xi$ is and what the neighborhood $J$ is? Why don't $h \circ f$ and $U$ work? (I'm sure there is an obvious reason -- I am just being dumb and not thinking clearly.) – 2017-01-12
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    @William yes, we have to have $f(I)=\{(x,\xi(x))\}$, then what is $\xi$? – 2017-01-12
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    @user75086: As stated, the immersion theorem doesn't give you enough control to obtain the desired conclusion. (If you knew $h$ had the form $h(x, y) = (x, y - \phi(x))$ for some function $\phi$, you'd be in business.) Do you have an objection to the direct strategy of showing $x:I \to J$ is invertible, and $\xi = y \circ x^{-1}$? – 2017-01-12
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    @AndrewD.Hwang What is this function $x:I\to J$? thank you for your comment – 2017-01-12
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    The function $x$ is some antiderivative of $x'$, and $J = x(I)$ is its image. – 2017-01-13
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    Apparently my previous comment was left in haste; $x$ is the _first component of your curve_, which happens to be a _particular_ antiderivative of $x'$, but that's relevant only because $x' \neq 0$ implies $x$ is invertible and $x^{-1}$ is differentiable. – 2017-01-13
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    @AndrewD.Hwang Thank you – 2017-01-13

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