I was just wondering why we don't define the tangent space of a smooth manifold at a point $p$ to be $\{p\} \times \mathbb{R}^{n}$, rather than using derivations, germs, or equivalence classes of curves? I just started learning the material, so maybe I should just keep on reading...
tangent space clarification
5
$\begingroup$
general-topology
manifolds
-
1Even for embedded manifolds in some $\mathbb{R}^m$, if we did, then any smooth manifolds would be parallelizable, i.e. the tangent bundle $TM = M \times \mathbb{R}^n$, which is false. You can take $M=S^2.$ See http://en.wikipedia.org/wiki/Hairy_ball_theorem – 2012-01-26