Why is the cusp not a submanifold?

[quasar987]

My book says that the cusp y=x^2/3 is not an embedded submanifold of R². Why is that?

 

[jostpuur]

According to my notes the embedding is defined like this:

Let M and N be differentiable manifolds, and $$f:M\to N$$ a smooth ($$C^{\infty}$$) mapping. If for all $$p\in M$$ the tangent space mapping $$f_{*p}:T_p(M)\to T_{f(p)}(N)$$ is injective, and $$f:M\to f(M)$$ is a homeomorphism when $$f(M)$$ has the induced topology from N, then $$f$$ is an embedding of M in N.

If we set the natural differentiable structures on $$\mathbb{R}$$ and $$\mathbb{R}^2$$, then a mapping

$$f:\mathbb{R}\to\mathbb{R}^2,\quad\quad f(x)=(x, (x^2)^{1/3})$$

is not an embedding, because it is not smooth at origo.

 

[quasar987]

True, but an embedded submanifold is by definition (or characterisation) the image of a smooth embedding. Couldn't there be a pair (M, f) other than M=R and f given in your post such that f(M) = the cusp, and such that f is an smooth embedding?

 

[jostpuur]

So is the real question this: We give $$\mathbb{R}^2$$ the natural differentiable structure, define $$f:\mathbb{R}\to\mathbb{R}^2$$ like in my post, and ask that does this mapping somehow induce a differentiable structure on $$\mathbb{R}$$ so that $$f$$ becomes smooth?

Or perhaps the question is, that why cannot this $$f$$ induce differentiable structure in such way, assuming that the book's claim is right?

 

[quasar987]

The real question is this:
Consider R² with the natural differentiable structure, and C the subset of R² defined by the equation y=x^2/3. Is there a smooth manifold M and an embedding f:M-->R² whose image is C.

 

[mathwonk]

every line through the origin of that set has intersection number ≥ 2 with the set. for a manifold, the generic intersection number will be one.

 

[jostpuur]

Isn't the content of the posts #4 and #5 the same?

 

[quasar987]

every line through the origin of that set has intersection number ≥ 2 with the set. for a manifold, the generic intersection number will be one.

Well, the cusp is a smooth manifold, with smooth atlas consisting of the unique chart "projection onto the x coordinate".

Isn't the content of the posts #4 and #5 the same?

Well, it seems to me that the question in post #5 is more general than any of the 2 questions of post #4. Or perhaps they are equivalent?