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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.

Math101
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  • $S^1$ is the circle, yes? Not a ball or a sphere, right? – Brian Tung Mar 05 '21 at 21:39
  • @BrianTung Formally, $S^1={(x,y)\in\Bbb{R^2}|x^2+y^2=1}$ – Math101 Mar 05 '21 at 21:40
  • Yes, but you also called it the "unit ball sphere"; usually "sphere" refers to $S^2$, and "ball" typically refers to the sphere plus its interior. I wouldn't call $S^1$ either of those, myself. – Brian Tung Mar 05 '21 at 21:50
  • @BrianTung Edited it, thanks – Math101 Mar 05 '21 at 21:51
  • The lemma you proved will help you define a basis in the codomain of $f$. Then to prove continuity, you just check that the pre-images of these basis elements are open. – JLinsta Mar 05 '21 at 21:53
  • @JLinsta So basically I can define $B={(S^1\times S^1)\cap (a,b)\times (c,d)}$ as a basis, and look at $f^{-1}(S^1\times S^1)\cap f^{-1}((a,b)\times (c,d))$? – Math101 Mar 05 '21 at 22:02
  • The lemma will let you define $B = {[S^1 \cap (a,b)] \times [S^1 \cap (c,d)]}$ instead, which might be easier to work with. Without the lemma, you do not know that the topology generated by this basis is indeed the product topology on $S^1 \times S^1$. – JLinsta Mar 05 '21 at 22:09
  • @JLinsta Why is it easier to work with? I can't see it – Math101 Mar 05 '21 at 22:12
  • I just noticed a small mistake. $S^1$ is a subspace of $\mathbb{R}^2$, so $S^1 \times S^1$ is a subspace of $\mathbb{R}^4$. You should have $B = {(S^1 \times S^1) \cap [(a,b) \times (c,d) \times (e,f) \times (g,h)]}$. – JLinsta Mar 05 '21 at 22:12
  • @JLinsta Isn't it a subspace of $\Bbb{R^2}\times \Bbb{R^2}$ or are they formally the same? And does it change anything in the solution? – Math101 Mar 05 '21 at 22:14
  • They are the same, but $(a,b) \times (c,d)$ is a set in $\mathbb{R}^2$. So currently you have $S^1 \times S^1$ as a subset of $\mathbb{R}^2$. – JLinsta Mar 05 '21 at 22:15
  • @JLinsta So how could I write the basis in a way s.t I'm using the lemma? – Math101 Mar 05 '21 at 22:16
  • There is some miscommunication going on. I am thinking of $(a,b)$ as an open interval in $\mathbb{R}$. Are you thinking of it as a point in $\mathbb{R}^2$? – JLinsta Mar 05 '21 at 22:16
  • Let us continue this discussion in chat. – JLinsta Mar 05 '21 at 22:16

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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.

Henno Brandsma
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