# an isomorphism between the tangent space of a manifold to euclidean space

1) I was told in class many years ago that, the tangent space if the sphere $\mathbb{S}^2$at a point $p$, i.e.

$T_p\mathbb{S}^2$ is isomorphic to $\mathbb{R}^2$. Could anyone give me a proof of this? Why?

2) Is it true in general, that is, does that always exist an isomorphism between a general manifold of dimension $n$ to $\mathbb{R}^n$?

• The proof of this depends on your definition of $T_pM$. If you have a precise definition of $T_pM$, the proof is probably very easy for you to recreate in the most general setting (or look it up if its say $(\mathfrak{m}_p/\mathfrak{m}_p^2)^*$. If you don't have a rigorous definition for $T_pM$, then this will be a very hard question to answer. – PVAL-inactive Aug 23 '15 at 2:30
• For part 2: In general, for a smooth $n$-dimensional manifold, the tangent space at a point of the manifold will be a vector space isomorphic to $\mathbb R^n$. Proving this may be more or less difficult, depending on which of the many (mostly equivalent) definitions of manifold (and tangent space) you're using. – John Hughes Aug 23 '15 at 2:55
• What is hard to prove? $T_pM$ is an $n$-dimensional real vector space, no matter how you chose to define it (or more accurately, no matter how you chose to construct it). There is, up to isomorphism, only one $n$-dimensional real vector space. – Paul Sinclair Aug 23 '15 at 3:08
• Why $T_p M$ is an $n$-dimensional real vector space? – math101 Aug 23 '15 at 4:12