Proving the parametrization of a Torus imbedded in R3 is a Quotient map

[Smtih]

Homework Statement

Let b > a > 0. Consider the map F : [0, 1] X [0, 1] -> R3
defined by
F(s, t) = ((b+a cos(2PIt)) cos(2PIs), (b+a cos(2PIt)) sin(2PIs), a sin(2PIt)).
This is the parametrization of a Torus.
Show F is a quotient map onto it's image.

Homework Equations

Proving that any subset U of the image of F is open if and only if F−1(U) is open in[0,1]X[0,1]

The Attempt at a Solution

The first part of the definition is obvious. The map has to be surjective if it's a map onto it's image. This second part means reasonably little to me and a nudge in the right direction would be great.

 

[mighty2000]

To prove that F is a quotient map, we need to show that for any subset U of the image of F, U is open if and only if F^-1(U) is open in [0, 1] x [0, 1].

First, let's assume that U is open in the image of F. This means that for any point x in U, there exists an open neighborhood N around x such that N is contained in U. Since F is surjective, for every point y in N, there exists a point (s, t) in [0, 1] x [0, 1] such that F(s, t) = y. This means that F^-1(N) is an open neighborhood of (s, t) in [0, 1] x [0, 1]. Since F is continuous, F^-1(N) is mapped to an open neighborhood of y in the image of F, which is contained in U. Therefore, F^-1(U) is open in [0, 1] x [0, 1].

Conversely, assume that F^-1(U) is open in [0, 1] x [0, 1]. This means that for any point (s, t) in F^-1(U), there exists an open neighborhood N' around (s, t) such that N' is contained in F^-1(U). Since F is continuous, F(N') is an open neighborhood of F(s, t) in the image of F. Since F(s, t) is in U, F(N') is contained in U. Therefore, U is open in the image of F.

Hence, we have shown that for any subset U of the image of F, U is open if and only if F^-1(U) is open in [0, 1] x [0, 1]. Therefore, F is a quotient map onto its image, which is the parametrization of a Torus.