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

  1. Mar 20, 2014 #1
    I am reading Martin Crossley's book, Essential Topology.

    Example 5.43 on page 74 reads as follows:


    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:



    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)


    [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?


    Attached Files:

  2. jcsd
  3. Mar 21, 2014 #2
    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:

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