# Equivalence relation, product and quotient spaces

I have a problem with the following:

"Define a relation $\sim$ on $R^2$ by $(u,v) \sim (x,y)$ if and only if both $u-x$ and $v-y$ are integers. Show that for each point $(x,y) \in R^2$ there exists at least one point $(u,v) \in [0,1] \times [0,1]$ such that $(u,v) \sim (x,y)$ and deduce that $R^2/\sim$ is compact. ($\sim$ is an equivalence relation.)"

My guess is that it has something to do with homeomorphism and torus but could not move on. I would appreciate any help.

• Have you done the first part of the hint and shown that each point of the plane is equivalent to some point in the unit square? – Brian M. Scott Apr 11 '15 at 14:25
• No, I haven't. Not sure how to approach it. – Jim Apr 11 '15 at 14:27
• Think about the factional part of a number. – Brian M. Scott Apr 11 '15 at 14:27
• That’s fractional, of course! – Brian M. Scott Apr 11 '15 at 14:34

HINT: To show the first part of the hint, think about fractional parts of $x$ and $y$. Once you’ve done that, you know that $R^2/\!\sim$ is homeomorphic to $[0,1]^2/\!\sim$. Which points of $[0,1]^2$ are related? Can you see why $[0,1]^2/\!\sim$ is a torus?
• Hm, I don't find it intuitive at all that each point of $R^2$ is equivalent to some point of the unit square. Related points are $0$ and $1$ I guess. Is it torus because when I take points $(0,0)$, $(1,0)$ and try to glue them with $(0,1)$ and $(1,1)$ I will have to bend it to get the cylinder and then bend again it's open ends which would give me torus? – Jim Apr 11 '15 at 15:10
• @Jim: For the first part, given $\langle x,y\rangle\in\Bbb R^2$, consider the point $$\big\langle x-\lfloor x\rfloor,y-\lfloor y\rfloor\big\rangle\;.$$ There are more pairs of related points in the unit square. For instance, $\langle 0,y\rangle\sim\langle 1,y\rangle$ for each $y\in[0,1]$. Thus, the identifications force you to roll the square into a cylinder and then join the ends of the cylinder. – Brian M. Scott Apr 11 '15 at 15:14
• Ah, of course. For the compactness part, is it enough to say that $R^2/\!\sim$ is compact, since it is homeomorphic to $[0,1]^2 /\!\sim$ which is compact because it is closed and bounded? And do I really need to construct homeomorphism from the square to the torus or all of of the above is enough? – Jim Apr 11 '15 at 15:33
• @Jim: Once you’ve shown that it’s homeomorphic to $[0,1]^2/\!\sim$, you can definitely conclude that it’s compact, since it’s the continuous image of the compact set $[0,1]^2$. As for the rest, it depends on how fussy the person marking the answer is. It’s not actually terribly hard to write down an actual homeomorphism with, say, the set obtained by revolving the circle $(x-2)^2+y^2=1$ about the $y$-axis, but it is finicky and tedious. – Brian M. Scott Apr 11 '15 at 15:37
• Ok, I will think about that, probably not necessary then. But I need to construct the homeomorphism between $$R^2/\!\sim$$ and square or I just say as above? – Jim Apr 11 '15 at 15:42