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Math Help - Invertible function bijective only?

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    Member Greengoblin's Avatar
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    Invertible function bijective only?

    Why must an invertble function f^{-1}:T\rightarrow S be bijective? I understand the need to be injective, since for any a,b\in S, f(a)\ne f(b), but why must it also be surjective to be a well defined function? Could someone please show with an example?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Greengoblin View Post
    Why must an invertble function f^{-1}:T\rightarrow S be bijective? I understand the need to be injective, since for any a,b\in S, f(a)\ne f(b), but why must it also be surjective to be a well defined function? Could someone please show with an example?
    lets say it wasn't surjective. what would that mean. it would not hit all the points in the domain of f, and thus, the function f would not be defined, since it doesn't have its domain, and so there would be no inverse function, since f cannot map back to f^{-1}
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    Member Greengoblin's Avatar
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    hi, thanks, when you say 'does not hit all the points in the domain' do you mean for example that there are some values belonging to the domain, for which the function does not have a value? So for example f:\mathbb{R}\rightarrow\mathbb{R} where f(x)=+\sqrt{x} is not actually a well defined function since for negative \{x\in\mathbb{R}:x<0\} (in the domain), the function doesn't make sense? Meaning we need to restrict the domain.

    So we would have a similar problem if we tried to get the inverse of a non-surjective function?

    I was focusing on a function not being well-defined when there is more than one image for any x in the domain, andforgot that a function is also not well-defined when there are some x in the domain that have no value in the range. So is this the problem?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Greengoblin View Post
    hi, thanks, when you say 'does not hit all the points in the domain' do you mean for example that there are some values belonging to the domain, for which the function does not have a value? So for example f:\mathbb{R}\rightarrow\mathbb{R} where f(x)=+\sqrt{x} is not actually a well defined function since for negative \{x\in\mathbb{R}:x<0\} (in the domain), the function doesn't make sense? Meaning we need to restrict the domain.
    no, the domain is already the nonnegative numbers for the square-root function.

    So we would have a similar problem if we tried to get the inverse of a non-surjective function?
    by not hitting the points, i mean there are some values that do not get mapped to. for instance, the function f : \mathbb{Z} \mapsto \mathbb{Z} defined by f(x) = 2x is not surjective, since there are elements in the codomain that are not mapped to, namely, the odd integers. it is not a matter of of the function being undefined there or anything, as it is fine to plug in odd integers into our function, but the outputs miss a lot of elements.

    I was focusing on a function not being well-defined when there is more than one image for any x in the domain, andforgot that a function is also not well-defined when there are some x in the domain that have no value in the range. So is this the problem?
    no, this has nothing to do with well-definedness (my own word ). if a function is not well defined, it simply means that you can have one input value that maps to two or more output values. here we are worried about not having input values in the first place, which is what would happen for the function f if its inverse were not surjective. we would miss some elements in the domain of f and hence, we cannot get a function from the domain of f to the domain of its inverse, which means it could not be the inverse in the first place
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  5. #5
    Member Greengoblin's Avatar
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    Ok thanks, I see what you're saying.
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