Torus T^2 homeomorphic to S^1 x S^1

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I am reading Martin Crossley's book, Essential Topology.

Example 5.43 on page 74 reads as follows:

attachment.php?attachmentid=67832&stc=1&d=1395294875.jpg


I am really struggling to get a good sense of why/how/wherefore Crossley came up with the maps f and g in EXAMPLE 5.43. How did he arrive at these maps?

Why/how does f map [itex]S^1 \times S^1[/itex] onto [itex]T^2[/itex] and how does one check/prove that this is in fact a valid mapping between these topological spaces.

Can anyone help in making the origins of these maps clear or perhaps just indicate the logic behind their design and construction? I am completely lacking a sense or intuition for this example at the moment ... ...

Definitions for [itex]T^2[/itex] and [itex]S^1[/itex] are as follows:

attachment.php?attachmentid=67833&stc=1&d=1395295113.jpg


attachment.php?attachmentid=67834&stc=1&d=1395295184.jpg


My ideas on how Crossley came up with f and g are totally bankrupt ... but to validate f (that is to check that it actually maps a point of [itex]S^1 \times S^1[/itex] onto [itex]T^2[/itex] - leaving out for the moment the concerns of showing that f is a continuous bijection ... ... I suppose one would take account of the fact that (x,y) and (x',y') are points of [itex]S^1[/itex] and so we have:

[itex]x^2 + y^2 = 1[/itex] ... ... ... ... (1)

and

[itex]x'^2 + y'^2 = 1[/itex] ... ... ... ... (2)

Then, keeping this in mind check that

[itex]((x' +2)x, (x' +2)y, y'[/itex] is actually a point on the equation for [itex]T^2[/itex], namely:

[itex]x^2 + y^2 + z^2 - 4 \sqrt{x^2 + y^2} = -3[/itex] ... ... ... (3)

So in (3) we must:

- replace x by (x' +2)x
- replace y by (x' +2)y
- replace z by y'

and then simplify and if necessary use (1) (2) to finally get -3.

Is that correct? Or am I just totally confused ?

Can someone please help?

Peter
 

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Look at the terms. What is ##(x'+2)x##? What is important about the number 2 here? (Hint: Look at your definition of the torus.)

I've always defined the torus as ##S^1\times S^1##. It's kind of "obvious" if you think about it. Just look at this picture:

Torus_cycles.png
 
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