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Mappings homework problem

  1. Oct 9, 2009 #1
    how do i show that [itex]f: [0,1) \rightarrow S^1; t \rightarrow (\cos{2 \pi t}, \sin{2 \pi t})[/itex] is a continuous bijection but not a homeomorphism.

    so i know i have to
    (i) show f is a continuous bijection
    (ii) show [itex]f^{-1}[/itex] is not continuous

    so for (i),
    surjectivity:
    if i take a point [itex](a,b) \in S^1[/itex] then f is surjective if [itex]\exists t \in [0,1)[/itex] such that [itex]\cos{2 \pi t}=a[/itex] and [itex]\sin{2 \pi t}=b[/itex]

    solving these for t i get [itex]t=\frac{1}{2 \pi} \arccos{a}[/itex] and [itex]t=\frac{1}{2 \pi} \arcsin{b}[/itex]

    how does this help prove surjectivity though? surely i am trying to find a unique value for [itex]t \in [0,1)[/itex]?

    then injectivity:
    say i take [itex]t_1 \neq t_2[/itex], i need to show [itex]f(t_1) \neq f(t_2)[/itex] but i keep getting confused...how do you show this?

    then for continuity:
    can i just say that since both cos and sin are continuous functions over the given domain then f will be continuous or do i have to do more work, an [itex]\epsilon - \delta[/itex] proof perhaps?

    thanks guys.
     
  2. jcsd
  3. Oct 9, 2009 #2

    Office_Shredder

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    Re: Mappings

    For surjectivity, you don't need a unique value of t (that's injectivity) but since t is between 0 and 1 anyway, you will get one.

    For injectivity work backwards. Prove that f(t1)=f(t2) implies t1=t2. Remember that if (a,b) = (c,d) then you get that a=c and b=d

    To show that f-1 is not continuous, you want to find an open set in your domain that maps to a non-open set in your codomain. Hint: The endpoints of your domain were chosen very specifically.
     
  4. Oct 9, 2009 #3
    Re: Mappings

    ok surely i do need a unique t for surjectivity because if

    [itex]t=\frac{1}{2 \pi} \arccos{a}[/itex] and [itex]t=\frac{1}{2 \pi} \arcsin{b}[/itex] gave different values of t then there would be no value of [itex]t \in [0,1)[/itex] that mapped to [itex](a,b) \in s^1[/itex]?

    is the following sufficient for injectivity:

    let [itex]f(t_1)=(a,b),f(t_2)=(c,d)[/itex]
    assume [itex]f(t_1)=f(t_2)[/itex] then
    [itex]a=c,b=d[/itex]
    then [itex]\cos{2 \pi t_1}=\cos{2 \pi t_2}[/itex] and [itex]\sin{2 \pi t_1}=\sin{2 \pi t_2}[/itex]
    [itex]\Rightarrow t_1=t_2[/itex]

    there is a hint for the continuity part : use the fact that open arcs are a basis for the topology on [itex]S^1[/itex].
    i guess i'm supposed to make use of that...

    so the domain is [0,1). i guess if i want to pick an open subset of that i could use something like [itex](t,1), t \neq 0[/itex]. is that what you meant?
     
    Last edited: Oct 9, 2009
  5. Oct 9, 2009 #4

    Office_Shredder

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    Re: Mappings

    For surjectivity: It's OK if more than one value of t maps to the same point (a,b) on your circle, you just need that there is at least one
     
  6. Oct 9, 2009 #5
    Re: Mappings

    no i meant that if say t1 mapped to a with the cos (2 pi t) and t2 mapped to b with the sin (2 pi t) then what value of t maps to (a,b)? none surely? then it wouldnt be surjective?
     
  7. Oct 10, 2009 #6
    Re: Mappings

    ok. so ive done the bijectivity. what do i do to show its continuous. can i just say sin and cos are continuous?
     
  8. Oct 10, 2009 #7

    Landau

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    Re: Mappings

    This is correct, but I hope you did not conclude t1=t2 from [itex]\cos{2 \pi t_1}=\cos{2 \pi t_2}[/itex] separately, and from [itex]\sin{2 \pi t_1}=\sin{2 \pi t_2}[/itex] separately. After all, [tex]\sin(2\pi t_1)=\sin(2\pi t_2)[/tex] implies [tex]t_1=t_2[/tex] or [tex]t_1=\pi-t_2[/tex].

    For continuity, yes: sine and cosine are continuous, from which it follows that f is continuous.
     
  9. Oct 10, 2009 #8
    Re: Mappings

    How can you prove anything without specifying the topology on the domain or codomain?
     
  10. Oct 10, 2009 #9
    Re: Mappings

    ok so id have to say that [itex]\sin{(2 \pi t_1)}=\sin{(2 \pi t_2)} \Rightarrow t_1=t_2[/itex] or [itex]t_1=\pi - t_2[/itex]
    and [itex]\cos{(2 \pi t_1)}=\cos{(2 \pi t_2)} \Rightarrow t_1=t_2[/itex] or [itex]t_1=\pi + t_2[/itex]
    and so [itex]t_1=t_2[/itex] is the solution that satisfies both simultaneously.
    how's that?

    what about the need to specify a topology on either [itex] [0,1)[/itex] or [itex]S^1[/itex]?

    ok. as for finding the open set in [0,1) that maps to a non open set in S^1.
    could i use something like (a,1)?
     
  11. Oct 10, 2009 #10

    Landau

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    Re: Mappings

    That's ok. You could also divide the equations to obtain [tex]\tan{(2 \pi t_1)}=\tan{(2 \pi t_2)}[/tex]. Since tan is bijective on our interval, this implies t1=t2.
     
  12. Oct 10, 2009 #11
    Re: Mappings

    ok thanks. can i use the interval (a,1) to show that the inverse isnt continuous or did u have a different set in mind?
     
  13. Oct 10, 2009 #12
    Re: Mappings

    could someone possibly go through the surjectivity again in more detail for the slower students amongst us (ie: myself??!)
     
  14. Oct 10, 2009 #13

    Office_Shredder

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    Re: Mappings

    I think you're safe assuming that the topologies are the metric topologies that you're familiar with.


    Wrong endpoint.... the image of that is going to be everything in the circle on the arc between (1,0) (corresponding to your endpoint 1) to whatever point a is mapped to, and won't include either. So you can see you still get an open set in S1

    The key here is that a set of the form [0,a) is open even though it 'looks' like it isn't; this is because there aren't any points in your interval to the left of 0 so small open balls around 0 are in fact contained in [0,a)

    Now consider what the image of [0,a) is
     
  15. Oct 10, 2009 #14
    Re: Mappings

    well the image of [0,a) is the open arc from (1,0) to whatever a is mapped to.
    but in this case it includes the point (1,0).
    why does this make it not open? is it do with open balls again? i think im getting confused with the defn of open in terms of balls and the defn of an open set.

    thanks.
     
  16. Oct 10, 2009 #15

    Office_Shredder

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    Re: Mappings

    Basically, an open set is one in which if a point is in the set, then points nearby are near the set

    An open ball around a point x is just all the points within some distance epsilon > 0, NOT including the boundary (points that are epsilon away).

    For the circle case, it has what's called an induced topology. Basically, a set in S1 is open if it's the intersection of S1 with an open set in R2. It happens to be that this reduces to something easy to work with... a set in S1 is open if given any point in the set, points nearby it on the arc are also in the set.

    So looking at the image of [0,a), (1,0) is in the image, but points below it (small negative y value) are not in the image, which means the image of that set is not open
     
  17. Oct 10, 2009 #16

    Office_Shredder

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    Re: Mappings

    Ok, so we want to find t between 0 and 1 so that given (a,b) a2+b2=1, then cos(2pit)=a, sin(2pit)=b. Taking arccosine and arcsine, we see that

    2pi*t = arccos(a), 2pi*t=arcsin(b).

    The way that I would solve this is that by the definition of cosine and sine, given a point on the unit circle there MUST be some value x so that (cos(x),sin(x)) is that point, for 0<=x<2pi. So pick t=x/2pi
     
  18. Oct 10, 2009 #17

    Landau

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    Re: Mappings

    To be precise (and to avoid confusion), I would say that [0,a) is open in the domain [0,1).
    You mean in :smile:
     
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