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I am studying about differentiable manifolds. My professor give me an example show that graph of a continuous function is a submanifold, but image of its is not in general.

$$f: \mathbb{R} \longrightarrow \mathbb{R}^2 $$ $$f(x)=(x^2,x^3)$$

$gr(f)=\{ (x,f(x))| x \in \mathbb{R}\} \subset \mathbb{R}^3$

$Im(f)=\{ (x,y)| x=y^{2/3} \}$

I do not know how to prove a subset is a submanifold or not. Could you give me the technique to prove it?

I proved that $Im(f)$ is a topological manifold, but how can I show that it does not have differentiable structure?

user69833
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  • In textbook, there is a property that $gr(f)$ is a submanifold from Implicit Function Theorem, but I would like to find more technique to show a set is a manifold, which is more convenient. – user69833 Sep 29 '14 at 13:55

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Projection onto the $y$-axis gives a chart that generates a smooth atlas. It's not a question of whether it has a smooth structure (all topological 1-manifolds can be given smooth structure). This deals with a function being a smooth submanifold.

I'm not sure your professor intended for a rigorous proof, though, since this seems like a classic case of "Oh! Look! There's a cusp in the graph!" (see $(0,0)$). Can you take it from there?

Rawlrus Rumbleroar
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