tangent space clarification

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...

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Among other things, that definition doesn't allow you to correctly describe the relationship between the tangent spaces of two nearby points (see en.wikipedia.org/wiki/…). I also have no idea how you would give a coordinate-free definition of the derivative of a smooth map at a point this way. – Qiaochu Yuan Jan 26 '12 at 23:06
Even 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 en.wikipedia.org/wiki/Hairy_ball_theorem – Ehsan M. Kermani Jan 26 '12 at 23:09