# Showing a function is bijective

In a lecture we were trying to show a torus $T^{2}_{a,b}$ is homeomorphic to the planar model $I/\thicksim = \left\{ (x,y)| 0 \le x \le 1, 0 \le y \le 1 \right\}$

I need to show,

$\overline{f}([(x,y)]) = (a+b\cos(u))\left(\cos(v)\mathbf{i}+\sin(v)\mathbf{j}\right)+b\sin(u)\mathbf{k}$

is a bijective function. Does it suffice for me to find an inverse function to show bijectivity?

Ok in the lecture to show something is homeomorphic, we needed to show several things but one thing that confused me is how to show bijectivity. Now I know what it means for something to be bijective, but as for showing this, that's another question.

As for recognising answers, how do I do this properly?

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You haven't accepted answers on most of your questions. Please take into account that this feature exists not only to acknowledge the efforts of those who answered, but also to mark the questions as satisfactorily dealt with so people don't unnecessarily go back to them. –  joriki Aug 16 '11 at 12:40
@kahen, I don't know whether you have ever taught topology. I have, and I'm not the least bit surprised to see students using words like "homeomorphic" whilst not understanding their meaning. Once, on a test, I described two spaces, and asked whether they were homeomorphic. One student answered, "the first one is, but the second one isn't." Go figure. –  Gerry Myerson Aug 16 '11 at 12:52
@Gerry: This reminds me of a story my friend likes to tell of someone in his class who asked why a quadrilateral necessarily has four corners. The teacher asked her to draw one that doesn't, she went up to the blackboard, thought for a while, said "ah, I see" and sat down again. I wish I knew what went through her mind... –  joriki Aug 16 '11 at 12:55
@Gerry: Isn't that the story that was not isomorphic, in contrast to mine that was, from the comments on that long MO thread? :-) –  Asaf Karagila Aug 16 '11 at 13:27
@Asaf, if I've changed/misrepresented/stolen something from an MO thread, I plead not guilty by reason of comedic license. A good joke justifies many sins. –  Gerry Myerson Aug 17 '11 at 0:52

Your notation disconcerts me a little bit. I think there is a mistake somewhere.

First, your description of $I/\!\!\sim$ seems to me to be incorrect: what is on the right-hand side is what is called usually $I^2 = [0,1]^2 = [0,1]\times [0,1]$; the unit square of the plane $\mathbb{R}^2$. When you write $\text{"something"}/\!\!\sim$, you usually mean that your are quotienting out by some equivalence relation. I guess that, in our case, this relation is the one generated by

$$(x,y) \sim (x',y') \qquad \Longleftrightarrow \qquad \begin{cases} x = x' & \text{if}\qquad y=0 \qquad\text{and}\qquad y'= 1 \\ y = y' & \text{if}\qquad x=0 \qquad\text{and}\qquad x' = 1\\ \end{cases}$$

Then, I guess that you originally had the usual parametrization (the way topologists write it at least) of the torus in $\mathbb{R}^3$, $f : [0,1]^2 \longrightarrow \mathbb{T}^2$, generated by rotating a circumference of radius $a$ along a circumference of radius $b$:

$$f(x,y) = ((a + b\cos (2\pi x))\cos (2\pi y) , (a + b\cos (2\pi x)) \sin (2\pi y) , b\sin (2\pi x ) )$$

(If these $2\pi$ bother you, you can delete them, but then you have to change your unit square $[0,1]^2$ for $[0,2\pi]^2$.)

And then, your professor should have said something like: "Since this parametrization is compatible with the former equivalence relation $\sim$, it passes to the quotient inducing a well-defined map"

$$\overline{f} : I^2/\!\!\sim \longrightarrow \mathbb{T}^2$$

which is

$$\overline{f}\overline{(x,y)} = f(x,y) =((a + b\cos (2\pi x))\cos (2\pi y) , (a + b\cos (2\pi x)) \sin (2\pi y) , b\sin (2\pi x ) ) \ .$$

Am I right?

Well, all this assumes that you previously knew that parametrization of the torus and, in this case, bijectivity is now obvious. But, just in case you haven't seen that $f$ before, look just at the two first components:

$$((a + b\cos (2\pi x))\cos (2\pi y) , (a + b\cos (2\pi x)) \sin (2\pi y) ) \ .$$

Keeping $x$ fixed for a moment, this is a parametrization of a circumference of radius $a + b\cos (2\pi x)$: when $y$ goes from $0$ to $1$ in the unit interval $[0,1]$, we go round the circumference of centre $(0,0)$ and that radius. Right?

Ok, so let's assume that, for $(x,y), (x',y') \in [0,1]^2$ we could have

$$(a + b\cos (2\pi x))\cos (2\pi y) = (a + b\cos (2\pi x'))\cos (2\pi y')$$

and

$$(a + b\cos (2\pi x)) \sin (2\pi y) = (a + b\cos (2\pi x')) \sin (2\pi y') \ .$$

Since these are the coordinates of points in circumferences of radii $a + b\cos (2\pi x)$ and $a + b\cos (2\pi x')$, if they are the same point, these radii must be equal. Hence

$$a + b\cos (2\pi x) = a + b\cos (2\pi x') \qquad \Longleftrightarrow \qquad \cos (2\pi x) = \cos (2\pi x') \ .$$

Now, look at the third coordinate:

$$b\sin (2\pi x) = b \sin (2\pi x' ) \qquad \Longleftrightarrow \qquad \sin (2\pi x) = \sin (2\pi x') \ .$$

So, we would have both:

$$\cos (2\pi x) = \cos (2\pi x') \qquad\text{and}\qquad \sin (2\pi x) = \sin (2\pi x')$$

and, since $x,x' \in [0,1]$, if $x\neq x'$, this is only possible if one of them is $0$ and the other one is $1$.

Then we would get that

$$\cos (2\pi y) = \cos (2\pi y') \qquad\text{and}\qquad \sin (2\pi y) = \sin (2\pi y')$$

and conclude that also, if $y\neq y'$, then $y=0$ and $y'=1$ or vice versa.

So, our map $f$ is "almost" injective, except for these points on the perimetre of the unit square $[0,1]^2$ we have just found. But these points are the ones we are identifying with the equivalence relation $\sim$. Thus $\overline{f}$ is injective.

As for surjectivity: every point on $\mathbb{T}^2$ can be located saying in which meridian and parallel it lies, right? Well, this is exactly what you are doing when you say your point on the torus has coordinates $(x,y)$: the first one tells you the parallel and the second one the meridian. Varying $(x,y) \in [0,1]^2$ you go through all meridians and parallels.

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