Clarification about submanifold definition in ##\mathbb R^2##

[cianfa72]

TL;DR

Clarification about submanifold definition and diffeomorphisms involved in $\mathbb R^2$

Hi,
a clarification about the following: consider a smooth curve $γ:\mathbb R→\mathbb R^2$. It is a injective smooth map from $\mathbb R$ to $\mathbb R^2$. The image of $\gamma$ (call it $\Gamma$) is itself a smooth manifold with dimension 1 and a regular/embedded submanifold of $\mathbb R^2$.

Since it is a regular submanifold there is a global chart $(\phi, \mathbb R^2)$ such that $\Gamma$ is represented as $(x,0),x∈\mathbb R$. Restricting such a chart to $\Gamma$ we get the manifold structure on it w.r.t. the inclusion map $i:\Gamma ↪\mathbb R^2$ is an embedding. So far so good.

My point is: is the above map $γ:\mathbb R→\mathbb R^2$ a diffeomorphism onto its image ? I believe the answer is positive.
$\Gamma$ indeed is diffeomorphic to $\mathbb R$ by definition of chart. Now the composition $g = \phi \circ \gamma$ is a continuous injective map $g: \mathbb R \rightarrow \mathbb R$. By virtue of Invariance of domain theorem $g$ is an homeomorphism hence the map $\gamma$ is actually a differentiable homeomorphism onto its image (i.e. a diffeomorphism onto the image).

 

[Office_Shredder]

I haven't done differential geometry in a while, so sorry if I say something wrong here. I think since $g$ isn't the identity map, you haven't proven that the inverse of $\gamma$ is differentiable.

Consider the map $\gamma: x\to (x^3,0)$. Then the inverse is easy to compute as $(z,0)\to z^{1/3}$. This is not differentiable. There are obviously choices of $\phi:\Gamma\to \mathbb{R}$ for which $\phi \circ \gamma$ is smooth (e.g. $\phi(z,0)=z$), but this doesn't mean $\gamma$ is a diffeomorphism.

 

[cianfa72]

I haven't done differential geometry in a while, so sorry if I say something wrong here. I think since $g$ isn't the identity map, you haven't proven that the inverse of $\gamma$ is differentiable.

Ah yes, that makes sense. Indeed from Invariance of domain theorem it follows that $g$ is a smooth homeomorphism $g: \mathbb R \rightarrow \mathbb R$ (i.e. it is a differentiable/smooth homeomorphism however we cannot claim the inverse map $g^{-1}$ is a differentiable/smooth map too).

So, from a general point of view, even though the image $\Gamma$ of a smooth curve $\gamma$ is a regular/embedded submanifold of $\mathbb R^2$, the map $\gamma$ itself may or may not be a diffeomorphism onto its image $\Gamma$.

 

[Office_Shredder]

I think $\gamma$ is locally a diffeomorphism with its image whenever its derivative is an injective linear map.

 

[cianfa72]

I think $\gamma$ is locally a diffeomorphism with its image whenever its derivative is an injective linear map.

Yes, that is the definition of smooth immersion: if $\gamma$ is an immersion from $\mathbb R$ in $\mathbb R^2$ then it is a local diffeomorphism.