let

be a regular surface with

. Then there exists a neighborhood V of p in S such that V is the graph of a differentiable function that is either z = f(x,y), y = g(x,z), or x = h(y, z).

he starts off by letting

be a parametrization of S in p and that x(u,v) = (x(u,v), y(u,v), z(u,v)), (u,v) belonging to U. he supposes that

. Then he composes the projection map

with the function so that we have

. Then the inverse function theorem applies so there exists neighborhoods V1 of q and V2 of

such that there is a diffeomorphic mapping from V2 to V2.

Then he says that it follows that

restricted to x(V1) = V is one to one and that there is a differentiable inverse

. I always seem to get confused when domains become restricted in differential geometry. i don't know how it follows that

is one to one and why that is even necessary in this proof.

the rest of his proof just says: observe that x is a homeomorphism so that V is a neighborhood of p in S. now if we compose

with

and we see that V is the graph of the differentiable function z = f(x,y). the remaining cases can be treated in the same way.

i'm confused about the middle part of the proof on how did he know that

is one to one and why did he even need that result in the first place. any help in clearing this up will be greatly appreciated.