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
0
$\begingroup$
real-analysis
-
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