I doing a dynamic systems course and don't really know much topology yet, but the final work for the course requires me to prove for $\mathbb{T}^{2} = \{(x+m,y+n) : m,n \in \mathbb{Z}\} $ that the canonical projection $\pi: \mathbb{R}^{2} \rightarrow \mathbb{T}^{2}$ defined by $\pi (x,y) = \{(x+m,y+n) : m,n \in \mathbb{Z}\}$ is continuous.

I did some googling and found that in a product space all canonical projections are continuous, but I really have no idea on how to prove that since I'm not really familiar with this kind of thing at all. I couldn't really find anything else of use that I understood.

I'd greatly appreciate at least a hint on how to proceed, I've already mailed my professor but it'll take a while to get an answer and I'd like to get working on this as soon as possible.

  • 1
    $\begingroup$ Are you sure that you've given the correct definition of $\Bbb T^2$? I suspect that it's intended to be the quotient space $\Bbb R^2/\Bbb Z^2$. $\endgroup$ – Brian M. Scott May 6 '12 at 19:35
  • $\begingroup$ Usually the quotient space is given a topology by setting the quotient map continuous. What other information are you given? $\endgroup$ – T. Eskin May 6 '12 at 19:41
  • $\begingroup$ It defines $\mathbb{T}^{2}$ like the equivalence relation I said, and then goes on to say that it's equivalent to $\frac{\mathbb{R}^{2}}{\mathbb{Z}^{2}}$. I'm not really given other information. What I just realized is that if the definitions above make sense $\pi$ is continuous iff $\mathbb{T}^{2}$ is continuous, I'm not sure how that helps me though. $\endgroup$ – Bananas May 6 '12 at 19:57
  • $\begingroup$ @Collman: The problem is in your notation. The sets you describe are the equivalence classes you want, but that means that $\mathbb{T}^2$ should be the set of all such sets. That is, the correct way to write down what you meant is that $$\mathbb{T}^2 = \Bigl\{ \{(x+m,y+n)\colon m,n\in\mathbb{Z}\}\Bigm| x,y\in\mathbb{R}\Bigr\}.$$ $\endgroup$ – Arturo Magidin May 6 '12 at 20:15

The problem seems to be a complete triviality: it's not clear that you actually have anything to do.

You have an equivalence relation $\sim$ defined on $\Bbb R^2$ by $\langle x,y\rangle\sim\langle u,v\rangle$ iff $u-x,v-y\in\Bbb Z$. If $p=\langle x,y\rangle$, the $\sim$-equivalence class of $p$ is then $[p]=\{\langle x+m,y+n\rangle:m,n\in\Bbb Z\}$. $\Bbb T^2$ is defined to be the quotient space of $\Bbb R^2$ by the relation $\sim$. By definition the points of this quotient space are the $\sim$-equivalence classes: $\Bbb T^2=\{[p]:p\in\Bbb R^2\}$. By definition of quotient space $\Bbb T^2$ has the coarsest topology that makes the map $\pi:\Bbb R^2\to\Bbb T^2:p\mapsto[p]$ continuous, so there really isn't anything to prove.

It's possible, I suppose, that you're to use the following equivalent definition of the quotient topology instead:

A set $U\subseteq\Bbb T^2$ is open if and only if $\pi^{-1}[U]$ is open in $\Bbb R^2$.

But this still makes the problem a complete triviality, since it instantly implies that $\pi$ is continuous.

  • $\begingroup$ I guess that solves the exercise then, I'll look up more on the matter on my own to see if I can understand it a little better. Thanks! $\endgroup$ – Bananas May 6 '12 at 20:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.