For a vector valued function $f:\mathbb R^n\rightarrow \mathbb R^n$ does existence of inverse implies one –one. In any linear transformation it does. But I don’t know for vector valued function. Also by Jacobean we can check for inverse, can we use the same for one-one?
existence of inverse implies one –one
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3You can prove more generarily that if $f,g$ are functions and $g \circ f$ is one to one, then $f$ is one to one. – 2012-12-21
2 Answers
You could prove this directly if you wanted, but it is implied by the word inverse. A function with a well defined inverse is always a bijection. If you wanted to prove it, take two elements of the range $f(x),f(y)$ such that $f(x)=f(y)$. By taking the inverse of both sides you see that $x=y$ and thus it is one to one (and by "taking the inverse of both sides" I mean evaluating the inverse at each of these points, and by the inverse being well defined, they remain equal).
The Jacobian will only tell you if locally there is an inverse, and so that means locally the function would be $1-1$. For example, $f(x)=x^2$ is locally $1-1$ on $[0,\infty)$.
This result even holds for just `set theoretic' functions. Try proving it yourself!
The jacobian matrix is the matrix of the derivative of $f$ if it exists, hence the jacobian says something about the `global' invers of the derivative, not about the global invers of $f$. However, the invers function theorem can be used to prove that $f$ is locally invertible.
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0Is linear algebra tag needed here for this problem? – 2012-12-21
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0@BabakSorouh, I'm not sure. The second question is about linear algebra, but my guess is that he is confused about which properties require linearity and which do not. – 2012-12-21
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0I think you are wrong about the Jacobian - it is defined for more than linear transformations. – 2012-12-21
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0@ThomasAndrews, if he just wants to check whether the function itself is invertible? I'm not saying anything about the derivative here. – 2012-12-21
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0That said, the "inverse function theorem," which uses the Jacobian, is only about local inverses, not about whether there is a global inverse. – 2012-12-21
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0@ThomasAndrews, I'll phrase it differently. – 2012-12-21
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0consider f(x,y)=(x^2,y^2), this function is not one-one, but its jacobian is non zero, so it has inverse. please explain how? – 2012-12-21
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0@ThomasAndrews then how to check for global inverse? – 2012-12-21
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0@AlkaGoyal, the jacobian being non zero, says something about the invertability of the derivative if it exists, not about $f$. – 2012-12-21
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0@ThomasAndrews, is it ok now? I ment something different than I wrote. – 2012-12-21
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0@Alka Goyal Jacobian for $f\colon\;\; (x,\,y)\mapsto(x^2,\, y^2)$ is $Jf(x,\,y)=\begin{pmatrix}2x &0 \\ 0&2y \end{pmatrix}$ and has (maximal) $\operatorname {rank}{Jf}=2$ for $x\ne{0} ;\; y\ne{0};$ thus $f$ is *locally* one-to-one for that $(x,\,y).$ – 2012-12-21
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0but (x,y) and (-x,-y) has same image, then how is this function one-one? – 2012-12-21
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0@AlkaGoyal, it is locally one-to one, not globally – 2012-12-21
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1@Alka Goyal $f$ is one-to-one only *locally*, i.e. in the some (maybe sufficiently small) neighbourhood of every point $(x,\,y)$ such that $x\ne{0} \wedge y\ne{0}$ – 2012-12-21