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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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