Proving two set are homeomorphic I want to deduce that the following sets: $\Sigma_1=\{(x,y,z,w)\in\Bbb{R^4}|x^2+y^2=1,z^2+w^2=1\}$ and $S^1\times S^1$ (where $S^1$ is simply the unit circle) are homeomorphic. The question hints that first I need to prove that:
"If $A$ is a subspace of $X$ and $B$ is a subspace of $Y$ then the product topology on $A\times B$ is the same as the topology $A\times B
$ inherits as a subspace of $X\times Y$".
My thought was simply to define $f(x,y,z,w)=((x,y),(z,w))$ which obviously is $1-1$ and onto. Now I can't see how what I proved helps me to prove continuity of the function I defined. Any help would be apperciated.
 A: Your map $f: \Sigma_1 \to S^1 \times S^1$ works fine. It's clearly a bijection and continuity follows from the universal property for products (see my post here where I also show the hint you were given): $p_1 \circ f = \pi_{1,2}\restriction_{\Sigma_1}$, where $p_1,p_2: S^1 \times S^1 \to S^1$ are the two projections on the codomain, and $\pi_{1,2}: \Bbb R^4 \to \Bbb R^2$ is just the "projection" $(x_1,x_2,x_3,x_4) \to (x_1,x_2)$ which is also continuous by standard facts (restrictions of continuous maps are continuous etc), and $p_2 \circ f = \pi_{3,4}\restriction_{\Sigma_1}$ (similar considerations). So $f$ is continuous. I won't bother with doing the same for the inverse, because that is already "automatically" continuous as $f$ is closed, going from the compact domain $\Sigma_1$ to a Hausdorff (even metric) $S^1 \times S^1$. That's all there is to it. I don't see how your hint would actually be needed here. It's just index juggling on products, continuous by the universal property.
