Prove $f: S^1 \to S^1 \times S^1$ where $f(\theta) = (2\theta, 3\theta)$ is a homeomorphism 
Prove $f: S^1 \to S^1 \times S^1$ where $f(\theta) = (2\theta, 3\theta)$ is a homeomorphism onto its image.

We just need to prove continuity, one-to-one, and continuous inverse.
For one-to-one, I have to prove that if $(2\theta, 3\theta) = (2\Theta, 3\Theta)$, then $\theta = \Theta$. Given no other information, I don't know how to do this. I can't take the inverse of both sides, because one-to-one is exactly what I'm trying to prove.
For continuity and inverse continuity, I have to show that if a set is open in the image ($f(O) \subset S^1 \times S^1$), then the pre-image $O$ is open in $S^1$. And vice versa for inverse continuity. Firstly, I don't even know what open sets look like in $S^1 \times S^1$. I believe the open sets in $S^1$ are just the regular open sets in $\mathbb{R^2}$, but in the subspace topology of $S^1$. So if $(2\theta, 3\theta) \times (2\Theta, 3\Theta)$ is an open set, what can I do to show that $(\theta, \Theta)$ is an open set?

For the first part, if $(2\theta, 3\theta)=(2\Theta, 3\Theta)$, the $2\theta=2\Theta$ and $3\theta=3\Theta$.  Either of those imply that $\theta=\Theta$.
For the second part, imagine $S^1\times S^1$ as the unit square with the boundary appropriately identified.  Then the image of the map will be a set of diagonal lines across the inside of the square.  Pick a point on one of these lines in the interior of the square.  The inverse image of a small open circle around that point will be an open segment of the circle.  If the point is on the boundary of the square, then an open circle will look different, but have the same preimage.  
