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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.

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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 $\Bbb R^2/\Bbb Z^2$. –  Brian M. Scott May 6 '12 at 19:35
    
Usually the quotient space is given a topology by setting the quotient map continuous. What other information are you given? –  Thomas E. May 6 '12 at 19:41
    
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. –  Bananas May 6 '12 at 19:57
    
@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 $$\mathbb{T}^2 = \Bigl\{ \{(x+m,y+n)\colon m,n\in\mathbb{Z}\}\Bigm| x,y\in\mathbb{R}\Bigr\}.$$ –  Arturo Magidin May 6 '12 at 20:15

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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.

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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! –  Bananas May 6 '12 at 20:46

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