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Math Help - show the map is a bijection

  1. #1
    Junior Member
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    show the map is a bijection

    Let \phi : \mathbb{R}^2 \rightarrow \mathbb{R}^3, so:

     (u,v) \rightarrow (u^3,u+v^3,v)

    Show this is a homeomorphism:

    To show injectivity: We want to show that:
    \phi(u_1,v_1)=\phi(u_2,v_2) \rightarrow (u_1,v_1)= (u_2,v_2)

    So if we set:

    u_{1}^{3}&=&u_{2}^{3}

    u_1+v_1^3=u_2+v_2^3

    v_1=v_2

    Then from the third equation we get v_1=v_2, and can sub back in to find that u_1=u_2.


    The surjectivity is where I'm getting stuck. If we let any (x,y,z) be a point in \mathbb{R}^3

    Then we get v=z and subbing in to the equation for y means that we get u = y-z^3.

    So I guess I go on to say that given any (x,y,z) I can find a (u,v) such that \phi(u,v)=(x,y,z)

    However, I think this is wrong since what happens with the x value in this case? I mean we also get that u=x^{\frac{1}{3}}

    I'm lost here can someone help? I don't think what I did initially was enough?

    Thanks!
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  2. #2
    Member ModusPonens's Avatar
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    Re: show the map is a bijection

    I knew something was fishy with this. Check and see that the point (2,2,2), for example, doesn't belong to the image of the function.
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