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

• I
• cianfa72

#### cianfa72

TL;DR Summary
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).

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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.

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##.

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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.

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