Prove that a standard torus is diffeomorphic to $ \mathbb S^1\times \mathbb S^1$

Question

I was asked to prove that a standard torus(which means we don't consider those pathological cases where it intersects with itself, e.g horn torus) is diffeomorphic to $ \mathbb S^1\times \mathbb S^1$.

I was thinking if we could prove it this way: Since every point on the torus can be uniquely defined with a pair of angles $(\theta_1, \theta_2)$. Then we construct a diffeomorphism $\phi(\theta_1, \theta_2)=(\tilde{\theta}_1 ,\tilde{\theta}_2)$ which maps every point on the torus to every point on $\mathbb S^1 \times \mathbb S^1$. Since the map is definitely bijective and smooth with a smooth inverse. We're basically done...

THERE MUST BE SOMETHING WRONG I THINK.

Thanks a lot for everyone's help!

Answer 1

For completeness: $$F(\theta,\phi) = (a\cos\theta,a\sin\theta,0)+(0,b\cos\phi,b\sin\phi)$$ is a bijection between $S^1\times S^1$ and the torus as a subset of $\mathbb R^3$. Things to check:

1. $F$ is smooth
2. $F$ is a bijection
3. The inverse of $F$ is smooth.

Part 3 is harder than 1 and 2. One approach is to show that the the derivative matrix $DF$ has rank 2, which allows for the inverse function theorem to be used.

The fact that the standard torus is indeed a submanifold of $\mathbb R^3$ can be shown by considering its non-parametric equation $(\sqrt{x^2+y^2}-a)^2+z^2=b^2$ and using the fact that the gradient of the left-hand side is nonzero on the surface.