February 10, 2013

http://coursework.tylerlogic.com/courses/math501/homework03

The set S is the cylinder {(x,y,z) ∈ ℝ

With this definition, f^{-1}(1) is exactly S since S is the set of all points of ℝ which have x^{2} + y^{2} = 1. Since the differential
df_{(x0,y0,z0)} is the left multiplication by the matrix (2x_{0},2y_{0},0), then the only point critical point of the differential is
(0,0,0). Putting the previous two ideas together then gives us that 1 is a regular point of the mapping f since f^{-1}(1) = S
doesn’t contain the origin of ℝ^{3}. Hence, because f is smooth, then do Carmo’s Proposition 2 of Section 2-2 informs us that
S is regular.

Denote the open disc, {(x,y,z) ∈ ℝ

Closed Disc
The closed disc D is not a regular surface because if it were, then by the definition of a regular surface there would exist
a neighborhood V ⊂ ℝ^{3} of (0,1) ∈D such that D∩V is homeomorphic to some open set of ℝ^{2}, say U. However, the point
(0,1) is in D ∩ V but is not an interior point, which contradicts the fact that U, being open in ℝ^{2}, contains only interior
points.

Open Disc
Setting U ⊂ ℝ^{2} to the open set {(x,y) ∈ ℝ^{2} | x^{2} + y^{2} < 1} and defining f : U → ℝ by f(u,v) = 0, then D is simply the
graph {(u,v,f(u,v)) | (u,v) ∈ U} of ℝ^{3}. So D is a regular surface by do Carmo’s first proposition of chapter
two.

Let S be the set {(x,y,z) ∈ ℝ

Regular Surface
Define f : ℝ^{3} → ℝ by f(x,y,z) = x^{2} -y^{2} -z. With this definition, f is smooth and df_{p} for p = (x_{0},y_{0},z_{0}) is the linear
map corresponding to the matrix (2x_{0},-2y_{0},-1). Since the third parameter of the matrix is -1, then df_{p} is
surjective. Since p was arbitrary, we know the differential to be surjective at all points of ℝ^{3}, and thus all
points of ℝ are regular values of the map f. Hence by do Carmo’s second proposition of section 2-2, f^{-1}(0) is
a regular surface, but f^{-1}(0) is all points that satify 0 = x^{2} - y^{2} - z, which is exactly the set S. So S is
regular.

Parametrizations
For the following, note that S can equivalently be given by the equation x^{2} - y^{2} - z = 0.

The parametrization defined by

covers S due to the following.

The parametrization defined by

covers S due to the following.

Coverings of the parametrizations ???

For the π

Letting d = u^{2} + v^{2} for convience we have the following.

The function f(x) = x

and this is undefined at x = 0.