Why f: R -> R^2 by f(t) = (t^2, t^3) is not a smooth manifold? Just learned the concept of smooth manifold. Really confused about it.

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- Apr 24th 2012, 05:15 PMljwrr1990Help on smooth manifold!
Why f: R -> R^2 by f(t) = (t^2, t^3) is not a smooth manifold? Just learned the concept of smooth manifold. Really confused about it.

- Apr 28th 2012, 11:32 PMRebesquesRe: Help on smooth manifold!
We cannot find a well defined tangent space at t=0.

- May 11th 2012, 12:13 PMmastermind2007Re: Help on smooth manifold!
It's because the inverse function is not differentiable at the point x = 0.

A smooth manifold M is a topological space where for every x in M there is a**diffeomorphism**$\displaystyle f: M \supseteq U \rightarrow V \subseteq \mathbb{R}^n \ with \ x \in U $ (which means that for every open subset of M there is a chart (=diffeomorphism) that maps U to an open subset of $\displaystyle \mathbb{R}^n $).