Proof
Let T_{m} be the set of tangent vectors at m (that is, the tangent space), and define
by assigning to a typical tangent vector its n local coordinates. Define an inverse
by the formula
G(v^{1}, v^{2}, . . . , v^{n})  = 
 
= 

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(v^{1}, v^{2}, . . . , v^{n}))  = 

, 
where F converts each of the summands to local coordinates. But we have seen that the local coordinates of /x^{i} are given by the Kronecker delta:
Substituting gives
j th coordinate of F(G(v))  =  F(G(v))^{j}  =  v^{i}_{i}^{j}  =  v^{j}  =  j th coordinate of v. 
In other words, F(G(v)) = v.
Conversely,
G(F(w))  =  w^{i}  x^{i} 
, 
where w^{i} 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  x^{i}  =  x^{i} 
Thus, the j th ambient coordinate of the vector G(F(w)) above is
G(F(w))_{j}  =  w^{i}  x^{i} 
=  w_{j}  , 
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.
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