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Let $\sim$ be the equivalence relation on $\mathbb{R}$ given by $a \sim b$ iff $a-b \in \mathbb{Z}$

Prove that the function $f: \mathbb{R}/{\sim} \to S^1$, $f([x]) = (\cos(2\pi x), \sin(2\pi x))$ is a homeomorphism.

$\mathbb{R}$ has standard topology and $S^1$ has subspace topology induced from $\mathbb{R^2}$

I just dont know how to show that the inverse is continuous.

Michael Hardy
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sarah
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  • Do you know what the topology of $\mathbb R$/~ is? – user99680 Dec 08 '13 at 02:28
  • yes, preimage of an open set under quotient map is open in $\mathbb{R}$ – sarah Dec 08 '13 at 02:30
  • Right. So you know every open set in $\mathbb R$ is a union of open intervals. Can you use this to figure out the general topology of the quotient? The idea is that if you know what an open set ( at least a basic open set ) is like, you can compute its inverse image , i.e., $(f^{-1})^{-1} (U)$ , where $U$ is open in the quotient. – user99680 Dec 08 '13 at 02:32
  • Think you can find the answer in these questions:

    http://math.stackexchange.com/questions/6357/quotient-topologies-and-equivalence-classes/6359#6359

    http://math.stackexchange.com/questions/6357/quotient-topologies-and-equivalence-classes/6362#6362

     – Agustí Roig Dec 08 '13 at 02:34
  • user99680, I'm not sure what you mean, could you explain more in detail? – sarah Dec 08 '13 at 03:07
  • agusti, thanks but its not what I'm looking for. I think user99680 has something but needs to explain more. – sarah Dec 08 '13 at 04:55

1 Answers1

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So what you have already shown is that

  • $f \colon {\mathbb R}/{\sim} \to S^1$ is continuous;
  • $f$ is bijective.

What is left is to show that $f^{-1}$ is continuous as well, or, equivalently in this case, that $f$ is open or closed.

You could start monkeying around with open subsets of ${\mathbb R}/{\sim}$ and trying to prove that their image is open. This is bound to involve lot of handwaving, not because it is particularly difficult, but because it is extremely tricky to write down properly. (Note, by the way, that typical $\epsilon$-$\delta$-computations involving the topology on ${\mathbb R}$ and ${\mathbb R}^2$ for the continuity of $f$ are hidden in the continuity of $\sin$ and $\cos$).

So, I'll resort to trickery. The one-shot solution to this particular problem is the following theorem.

Theorem. Every continuous map from a compact space to a Hausdorff space is closed.

This settles it: ${\mathbb R}/{\sim}$ is compact, because it can also be seen as the quotient $[0,1]/{\sim}$ and a quotient of a compact space is compact; and $S^1$ is Hausdorff. So $f$ is continuous, bijective, and closed, and therefore a homeomorphism.

Now maybe you don't like this, for instance because you haven't seen this theorem yet. So, here's another approach. I'll still try to avoid having to do $\epsilon$-$\delta$-computations; this time I'll hide them in the continuity of $\arccos \colon (-1,1) \to (0, \pi)$.

The map $(x,y) \mapsto \arccos(x)/(2\pi)$ is a continuous map from $\{(x,y) \in {\mathbb R}^2 \mid y > 0 \}$ to $(0, \frac{1}{2})$. Restricting to $\{(x,y) \in S^1 \mid y > 0\}$ and composing with the projection ${\mathbb R} \to {\mathbb R}/{\sim}$ gives a continuous map that is exactly the restriction of $f^{-1}$ to $\{(x,y) \in S^1 \mid y > 0\}$.

Similary, the map $(x,y) \mapsto \arccos(y)/(2\pi) + \frac{1}{4}$ is a continuous map from $\{(x,y) \in {\mathbb R}^2 \mid x < 0\}$ to $(\frac{1}{4},\frac{3}{4})$. Restricting this one to $\{(x,y) \in S^1 \mid x < 0\}$ and composing with the projection ${\mathbb R} \to {\mathbb R}/{\sim}$ gives a continuous map that is exactly the restriction of $f^{-1}$ to $\{(x,y) \in S^1 \mid x < 0\}$.

Now do this twice more to find that the restriction of $f^{-1}$ to each of those four open subsets of $S^1$ (with $y > 0$, $x < 0$, $y < 0$, $x > 0$) is continuous. Because those four open subsets cover $S^1$, $f^{-1}$ is continuous.

Magdiragdag
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