Proof

Let Tm be the set of tangent vectors at m (that is, the tangent space), and define

F: Tm En

by assigning to a typical tangent vector its n local coordinates. Define an inverse

G: En Tm

by the formula

G(v1, v2, . . . , vn)=  v1 x1 + v2 x2 +  ...  + vn xn
= vi xi (using the Einstein summation convention).

First of all, F and G are both linear functions, by construction (proof of that fact is left as n exercise!). More important, we can verify that F and G are inverses as follows:

F(G(v))= F(G(v1, v2, . . . , vn)) =
F
vi xi
,

where F converts each of the summands to local coordinates. But we have seen that the local coordinates of /xi are given by the Kronecker delta:

j th coordinate of /xi = ij.

Substituting gives

 j th coordinate of F(G(v)) = F(G(v))j = viij = vj = j th coordinate of v.

In other words, F(G(v)) = v.

Conversely,

 G(F(w)) = wi xi ,

where wi are the local coordinates of the vector w. Is this the same vector as w? Well, let us look at its ambient coordinates; since if two vectors have the same ambient coordinates, they are certainly the same vector! But we already know that

 j th Ambient coordinate of xi = yjxi

Thus, the j th ambient coordinate of the vector G(F(w)) above is

 G(F(w))j = wi yjxi = wj ,

where the last step is the conversion formula for ambient coordinates. In other words, G(F(w)) = w, and we are done.

That is why we use local coordinates; there is no need to specify a path every time we want a tangent vector! All we need to do is specify its local coordinates.