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Math Help - Prove that Phi^-1:W->V is an isomorphism

  1. #1
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    Prove that Phi^-1:W->V is an isomorphism

    Suppose V and W are vector spaces over a field F.

    Now suppose \varphi :V\rightarrow W is an isomorphism. The inverse of this is defined as \varphi^{-1}:W\rightarrow V. Prove that \varphi^{-1}:W\rightarrow V is an isomorphism, by verifying that it is linear and bijective.

    So far, this is what I have available.

    \forall w\in W, \exists v\in V such that \varphi (v)=w. This is for the surjectivity of \varphi.
    Additionally, the aforementioned v is unique, which is for the injectivity of \varphi.
    As such, I can define \varphi^{-1}(w)=v.
    This implies that \varphi(\varphi^{-1}(w))=\varphi(v)=w and \varphi^{-1}(\varphi(v))=\varphi^{-1}(w)=v

    My guess is that I should do the same things for \varphi^{-1}, but just in reverse of \varphi.
    What I mean is:
    \forall v\in V, \exists w\in W such that \varphi^{-1}(w)=v (surjectivity of \varphi^{-1}).
    Since the aforementioned w will be unique, this implies that \varphi^{-1} is injective.
    Both of these together would imply that \varphi^{-1} is bijective.

    I haven't gotten to the linearity part yet, as I'd like to check to make sure I'm on the right track, or if I'm assuming too much.
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Runty View Post
    Suppose V and W are vector spaces over a field F.

    Now suppose \varphi :V\rightarrow W is an isomorphism. The inverse of this is defined as \varphi^{-1}:W\rightarrow V. Prove that \varphi^{-1}:W\rightarrow V is an isomorphism, by verifying that it is linear and bijective.

    So far, this is what I have available.

    \forall w\in W, \exists v\in V such that \varphi (v)=w. This is for the surjectivity of \varphi.
    Additionally, the aforementioned v is unique, which is for the injectivity of \varphi.
    As such, I can define \varphi^{-1}(w)=v.
    This implies that \varphi(\varphi^{-1}(w))=\varphi(v)=w and \varphi^{-1}(\varphi(v))=\varphi^{-1}(w)=v

    My guess is that I should do the same things for \varphi^{-1}, but just in reverse of \varphi.
    What I mean is:
    \forall v\in V, \exists w\in W such that \varphi^{-1}(w)=v (surjectivity of \varphi^{-1}).
    Since the aforementioned w will be unique, this implies that \varphi^{-1} is injective.
    Both of these together would imply that \varphi^{-1} is bijective.

    I haven't gotten to the linearity part yet, as I'd like to check to make sure I'm on the right track, or if I'm assuming too much.
    A function is a bijection if and only if it has an inverse.

    This map \varphi is just a function. We know it has an inverse (as a function), which is just \varphi. We also know that its inverse has an inverse,

    (\varphi^{-1})^{-1} = \varphi.

    Thus, \varphi^{-1} is a bijection because it has an inverse.

    Do you understand that?

    That said, you working did seem fine. It was just the long way round!

    If you have any problems showing linearity, ask. But do try it first...
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  3. #3
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    Quote Originally Posted by Swlabr View Post
    A function is a bijection if and only if it has an inverse.

    This map \varphi is just a function. We know it has an inverse (as a function), which is just \varphi. We also know that its inverse has an inverse,

    (\varphi^{-1})^{-1} = \varphi.

    Thus, \varphi^{-1} is a bijection because it has an inverse.

    Do you understand that?
    I'd forgotten that bit about bijections. Thanks for reminding me. Now just to get to the linearity part.
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